Introductory Chemistry - 1st Canadian Edition

Introductory Chemistry - 1st Canadian Edition

David W. Ball and Jessie A. Key

BCcampus

Victoria, B.C.

Contents

1

Acknowledgments

Jessie A. Key and David W. Ball

I would like to acknowledge the team at BCcampus for all their hard work on this project. Project managers Amanda Coolidge and Clint Lalonde, and the entire editorial team were instrumental to the success of this work. As well, I would like to thank my colleagues at Vancouver Island University for their support.

Jessie A. Key

September 2014

The decision to write a new textbook from scratch is not one to be taken lightly. The author becomes a saint to some and a sinner to others—and the feedback from the “others” is felt more acutely than the feedback from the “some”! Ultimately, the decision to write a new book comes from the deep feeling that an author can make a positive contribution to the field, and that it is ultimately time well invested.

It also helps that there are people supporting the author both personally and professionally. The first person to thank must be Jennifer Welchans of Flat World Knowledge. I have known Jen for years; indeed, she was instrumental in getting me to write my first academic book, a math review book that is still available through another publisher. We reconnected recently, and I learned that she was working for a new publisher with some interesting publishing ideas. With her urging, the editorial director and I got together, first by phone and then in person, to discuss this project. With all the enthusiasm and ideas that Flat World Knowledge brought to the table, it was difficult not to sign on and write this book. So thanks, Jen—again. Hopefully this won’t be the last book we do together.

Thanks also to Michael Boezi, editorial director at Flat World Knowledge, for his enthusiastic support. Jenn Yee, project manager at Flat World Knowledge, did a great job of managing the project and all of its pieces—manuscript, answers to exercises, art, reviews, revisions, and all the other things required to put a project like this together. Vanessa Gennarelli did a great job of filling in when necessary (although Jenn should know better than to take a vacation during a project). Kudos to the technology team at Flat World Knowledge, who had the ultimate job of getting this book out: Brian Brennan, David Link, Christopher Loncar, Jessica Carey, Jon Gottfried, Jon Williams, Katie Damo, Keith Avery, Mike Shnaydman, Po Ki Chui, and Ryan Lowe. I would also like to thank the production team at Scribe Inc., including Stacy Claxton, Chrissy Chimi, Melissa Tarrao, and Kevin McDermott. This book would not exist without any of these people.

Thanks to Mary Grodek and Bill Reiter of Cleveland State University’s Marketing Department for assistance in obtaining a needed photograph.

A project like this benefits from the expertise of external reviewers. I would like to thank the following people for their very thoughtful evaluation of the manuscript at several stages:

Thanks especially to ANSR Source, who performed accuracy checks on various parts of the text. Should any inaccuracies remain, they are the responsibility of the author. I hope that readers will let me know if they find any; one of the beauties of the Flat World process is the ability to update the textbook quickly, so that it will be an even better book tomorrow.

I am looking forward to seeing how the Flat World Knowledge model works with this book, and I thank all the adopters and users in advance for their help in making it a better text.

David W. Ball

February 2011

 

 

2

Dedication

David W. Ball and Jessie A. Key

David W. Ball:

For Gail, with thanks for her support of this book and all the other projects in my life.

I scorn to change my state with kings.

William Shakespeare, Sonnet 29

 

 

Jessie A. Key:

For all of the amazing chemistry educators that I have had the privilege of learning from.

“If I have seen further, it is from standing on the shoulders of giants.”

Isaac Newton, 1676

3

About the book

Introductory Chemistry- 1st Canadian Edition was adapted by Jessie A. Key from David Ball’s textbook Introductory Chemistry with support from BCcampus Open Education. For information about what was changed in this adaptation, refer to the Copyright statement at the bottom of the home page.

BCcampus Open Education began in 2012 as the B.C. Open Textbook Project with the goal of making post-secondary education in British Columbia more accessible by reducing student costs through the use of openly licenced textbooks and other OER. BCcampus supports the post-secondary institutions of British Columbia as they adapt and evolve their teaching and learning practices to enable powerful learning opportunities for the students of B.C. BCcampus Open Education is funded by the British Columbia Ministry of Advanced Education, Skills & Training, and the Hewlett Foundation.

Open textbooks are open educational resources (OER) created and shared in ways so that more people have access to them. This is a different model than traditionally copyrighted materials. OER are defined as teaching, learning, and research resources that reside in the public domain or have been released under an intellectual property license that permits their free use and re-purposing by others."Open Educational Resources," Hewlett Foundation, https://hewlett.org/strategy/open-educational-resources/ (accessed September 27, 2018). Our open textbooks are openly licensed using a Creative Commons licence, and are offered in various e-book formats free of charge, or as printed books that are available at cost. For more information about this project, see the BCcampus Open Education website. If you are an instructor who is using this book for a course, please fill out our Adoption of an Open Textbook form.

I

Chapter 1. What is Chemistry

1

Introduction

If you are reading these words, you are likely starting a chemistry course. Get ready for a fantastic journey through a world of wonder, delight, and knowledge. One of the themes of this book is “chemistry is everywhere,” and indeed it is; you would not be alive if it weren’t for chemistry because your body is a big chemical machine. If you don’t believe it, don’t worry. Every chapter in this book contains examples that will show you how chemistry is, in fact, everywhere. So enjoy the ride—and enjoy chemistry.

What is chemistry? Simply put, chemistry is the study of the interactions of matter with other matter and with energy. This seems straightforward enough. However, the definition of chemistry includes a wide range of topics that must be understood to gain a mastery of the topic or even take additional courses in chemistry. In this book, we will lay the foundations of chemistry in a topic-by-topic fashion to provide you with the background you need to successfully understand chemistry.

2

Some Basic Definitions

Learning Objectives

  1. Learn the basic terms used to describe matter.

 The definition of chemistry—the study of the interactions of matter with other matter and with energy—uses some terms that should also be defined. We start the study of chemistry by defining some basic terms.

Matter. A book is matter, a computer is matter, food is matter, and dirt in the ground is matter. Sometimes matter may be difficult to identify. For example, air is matter, but because it is so thin compared to other matter (e.g., a book, a computer, food, and dirt), we sometimes forget that air has mass and takes up space. Things that are not matter include thoughts, ideas, emotions, and hopes.

Example 1

Which of the following is matter and not matter?

  1. a hot dog
  2. love
  3. a tree

Solution

  1. A hot dog has mass and takes up space, so it is matter.
  2. Love is an emotion, and emotions are not matter.
  3. A tree has mass and takes up space, so it is matter.

Test Yourself

Which of the following is matter and not matter?

  1. the moon
  2. an idea for a new invention

Answer

  1. The moon is matter.
  2. The invention itself may be matter, but the idea for it is not.

Figure 1.1 The Phases of Matter

Chemistry recognizes three fundamental phases of matter: solid (left), liquid (middle), and gas (right). “Ice cubes” by Darren Hester is licensed under the Creative Commons Attribution-Share Alike 2.5 Generic; “Glass of Water” by Greg Riegler is licensed under the Creative Commons Attribution 2.0 Generic; Tea Time by Vélocia is licensed under the Creative Commons Attribution-NonCommercial- NoDerivs 2.0 Generic.
Chemistry recognizes three fundamental phases of matter: solid (left), liquid (middle), and gas (right).
“Ice cubes” by Darren Hester is licensed under the Creative Commons Attribution-Share Alike 2.5 Generic; “Glass of Water” by Greg Riegler is licensed under the Creative Commons Attribution 2.0 Generic; Tea Time by Vélocia is licensed under the Creative Commons Attribution-NonCommercial- NoDerivs 2.0 Generic.

To understand matter and how it changes, we need to be able to describe matter. There are two basic ways to describe matter: physical properties and chemical properties. Physical properties are characteristics that describe matter as it exists. Some of many physical characteristics of matter are shape, color, size, and temperature. An important physical property is the phase (or state) of matter. The three fundamental phases of matter are solid, liquid, and gas (see Figure 1.1 “The Phases of Matter”).

Figure 1.2 Chemical Properties

The fact that this match burns is a chemical property of the match.

Chemical properties are characteristics of matter that describe how matter changes form in the presence of other matter. Does a sample of matter burn? Burning is a chemical property. Does it behave violently when put in water? This reaction is a chemical property as well (Figure 1.2 “Chemical Properties”). In the following chapters, we will see how descriptions of physical and chemical properties are important aspects of chemistry.

Figure 1.3 Physical Changes

The solid ice melts into liquid water—a physical change.

If matter always stayed the same, chemistry would be rather boring. Fortunately, a major part of chemistry involves change. A physical change occurs when a sample of matter changes one or more of its physical properties. For example, a solid may melt (Figure 1.3 “Physical Changes”), or alcohol in a thermometer may change volume as the temperature changes. A physical change does not affect the chemical composition of matter.

A chemical change is the process of demonstrating a chemical property, such as the burning match in Figure 1.2 “Chemical Properties”. As the matter in the match burns, its chemical composition changes, and new forms of matter with new physical properties are created. Note that chemical changes are frequently accompanied by physical changes, as the new matter will likely have different physical properties from the original matter.

Example 2

Describe each process as a physical change or a chemical change.

  1. Water in the air turns into snow.
  2. A person’s hair is cut.
  3. Bread dough becomes fresh bread in an oven.

Solution

  1. Because the water is going from a gas phase to a solid phase, this is a physical change.
  2. Your long hair is being shortened. This is a physical change.
  3. Because of the oven’s temperature, chemical changes are occurring in the bread dough to make fresh bread. These are chemical changes. (In fact, a lot of cooking involves chemical changes.)

Test Yourself

Identify each process as a physical change or a chemical change.

  1. A fire is raging in a fireplace.
  2. Water is warmed to make a cup of coffee.

Answers

  1. chemical change
  2. physical change

A sample of matter that has the same physical and chemical properties throughout is called a substance. Sometimes the phrase pure substance is used, but the word pure isn’t needed. The definition of the term substance is an example of how chemistry has a specific definition for a word that is used in everyday language with a different, vaguer definition. Here, we will use the term substance with its strict chemical definition.

Chemistry recognizes two different types of substances: elements and compounds. An element is the simplest type of chemical substance; it cannot be broken down into simpler chemical substances by ordinary chemical means. There are about 115 elements known to science, of which 80 are stable. (The other elements are radioactive, a condition we will consider in Chapter 15 “Nuclear Chemistry”.) Each element has its own unique set of physical and chemical properties. Examples of elements include iron, carbon, and gold.

A compound is a combination of more than one element. The physical and chemical properties of a compound are different from the physical and chemical properties of its constituent elements; that is, it behaves as a completely different substance. There are over 50 million compounds known, and more are being discovered daily. Examples of compounds include water, penicillin, and sodium chloride (the chemical name for common table salt).

Elements and compounds are not the only ways in which matter can be present. We frequently encounter objects that are physical combinations of more than one element or compound. Physical combinations of more than one substance are called mixtures. There are two types of mixtures. A heterogeneous mixture is a mixture composed of two or more substances. It is easy to tell, sometimes by the naked eye, that more than one substance is present. A homogeneous mixture is a combination of two or more substances that is so intimately mixed that the mixture behaves as a single substance. Another word for a homogeneous mixture is solution. Thus, a combination of salt and steel wool is a heterogeneous mixture because it is easy to see which particles of the matter are salt crystals and which are steel wool. On the other hand, if you take salt crystals and dissolve them in water, it is very difficult to tell that you have more than one substance present just by looking—even if you use a powerful microscope. The salt dissolved in water is a homogeneous mixture, or a solution (Figure 1.4 “Types of Mixtures”).

Figure 1.4 Types of Mixtures

On the left, the combination of two substances is a heterogeneous mixture because the particles of the two components look different. On the right, the salt crystals have dissolved in the water so finely that you cannot tell that salt is present. The homogeneous mixture appears like a single substance. “flour and cocoa mixture” by Jessica and Lon Binder is licensed under Creative Commons Attribution-NonCommercial-NoDerivs 2.0 Generic; “a glass of water” by Bryan is licensed under a Creative Commons Attribution-NoDerivs 2.0 Generic.
On the left, the combination of two substances is a heterogeneous mixture because the particles of the two components look different. On the right, the salt crystals have dissolved in the water so finely that you cannot tell that salt is present. The homogeneous mixture appears like a single substance. “flour and cocoa mixture” by Jessica and Lon Binder is licensed under Creative Commons Attribution-NonCommercial-NoDerivs 2.0 Generic; “a glass of water” by Bryan is licensed under a Creative Commons Attribution-NoDerivs 2.0 Generic.

Example 3

Identify the following combinations as heterogeneous mixtures or homogenous mixtures.

  1. soda water (Carbon dioxide is dissolved in water.)
  2. a mixture of iron metal filings and sulfur powder (Both iron and sulfur are elements.)

Solution

  1. Because carbon dioxide is dissolved in water, we can infer from the behaviour of salt crystals dissolved in water that carbon dioxide dissolved in water is (also) a homogeneous mixture.
  2. Assuming that the iron and sulfur are simply mixed together, it should be easy to see what is iron and what is sulfur, so this is a heterogeneous mixture.

Test Yourself

Are the following combinations homogeneous mixtures or heterogeneous mixtures?

  1. the human body
  2. an amalgam, a combination of some other metals dissolved in a small amount of mercury

Answers

  1. heterogeneous mixture
  2. homogeneous mixture

There are other descriptors that we can use to describe matter, especially elements. We can usually divide elements into metals and nonmetals, and each set shares certain (but not always all) properties. A metal is an element that is solid at room temperature (although mercury is a well-known exception), is shiny and silvery, conducts electricity and heat well, can be pounded into thin sheets (a property called malleability), and can be drawn into thin wires (a property called ductility). A nonmetal is an element that is brittle when solid, does not conduct electricity or heat very well, and cannot be made into thin sheets or wires (Figure 1.5 “Semimetals”). Nonmetals also exist in a variety of phases and colors at room temperature. Some elements have properties of both metals and nonmetals and are called semimetals (or metalloids). We will see later how these descriptions can be assigned rather easily to various elements.

Figure 1.5 Semimetals

On the left is some elemental mercury, the only metal that exists as a liquid at room temperature. It has all the other expected properties of a metal. On the right, elemental sulfur is a yellow nonmetal that usually is found as a powder. “Pouring liquid mercury bionerd” is licensed under the Creative CommonsAttribution 3.0 Unported; “Sulphur-vulcano” by Heidi Soosalu is licensed under the Creative Commons Attribution-Share Alike 3.0 Unported.
On the left is some elemental mercury, the only metal that exists as a liquid at room temperature. It has all the other expected properties of a metal. On the right, elemental sulfur is a yellow nonmetal that usually is found as a powder.
“Pouring liquid mercury bionerd” is licensed under the Creative CommonsAttribution 3.0 Unported; “Sulphur-vulcano” by Heidi Soosalu is licensed under the Creative Commons Attribution-Share Alike 3.0 Unported.

Figure 1.6 “Describing Matter” is a flowchart of the relationships among the different ways of describing matter.

Figure 1.6 Describing Matter

Describing Matter

This flowchart shows how matter can be described.

Chemistry Is Everywhere: In the Morning

Most people have a morning ritual, a process that they go through every morning to get ready for the day. Chemistry appears in many of these activities.

  • If you take a shower or bath in the morning, you probably use soap, shampoo, or both. These items contain chemicals that interact with the oil and dirt on your body and hair to remove them and wash them away. Many of these products also contain chemicals that make you smell good; they are called fragrances.
  • When you brush your teeth in the morning, you usually use toothpaste, a form of soap, to clean your teeth. Toothpastes typically contain tiny, hard particles called abrasives that physically scrub your teeth. Many toothpastes also contain fluoride, a substance that chemically interacts with the surface of the teeth to help prevent cavities.
  • Perhaps you take vitamins, supplements, or medicines every morning. Vitamins and other supplements contain chemicals your body needs in small amounts to function properly. Medicines are chemicals that help combat diseases and promote health.
  • Perhaps you make some fried eggs for breakfast. Frying eggs involves heating them enough so that a chemical reaction occurs to cook the eggs.
  • After you eat, the food in your stomach is chemically reacted so that the body (mostly the intestines) can absorb food, water, and other nutrients.
  • If you drive or take the bus to school or work, you are using a vehicle that probably burns gasoline, a material that burns fairly easily and provides energy to power the vehicle. Recall that burning is a chemical change.

These are just a few examples of how chemistry impacts your everyday life. And we haven’t even made it to lunch yet!

Figure 1.7 Chemistry in Real Life

Examples of chemistry can be found everywhere—such as in personal hygiene products, food, and motor vehicles. “Soaps and Shampoos” by Takashi Ota is licensed under Creative Commons Attribution 2.0 Generic; “English Breakfast” is licensed under the Creative Commons Attribution-Share Alike 3.0 Unported; “Langley, Trans-Canada Highway” by James is licensed under the Creative Commons Attribution- Share Alike 3.0 Unported.
Examples of chemistry can be found everywhere—such as in personal hygiene products, food, and motor vehicles.
“Soaps and Shampoos” by Takashi Ota is licensed under Creative Commons Attribution 2.0 Generic; “English Breakfast” is licensed under the Creative Commons Attribution-Share Alike 3.0 Unported; “Langley, Trans-Canada Highway” by James is licensed under the Creative Commons Attribution- Share Alike 3.0 Unported.

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Video source: The chemical world by keyj (https://viuvideos.viu.ca/media/The+Chemical+World/0_ixlxmwe8)

Key Takeaways

  • Chemistry is the study of matter and its interactions with other matter and energy.
  • Matter is anything that has mass and takes up space.
  • Matter can be described in terms of physical properties and chemical properties.
  • Physical properties and chemical properties of matter can change.
  • Matter is composed of elements and compounds.
  • Combinations of different substances are called mixtures.
  • Elements can be described as metals, nonmetals, and semimetals.

Exercises

  1. Identify each as either matter or not matter.

a)  a book

b)  hate

c)  light

d)  a car

e)  a fried egg

2.  Give an example of matter in each phase: solid, liquid, or gas.

3. Does each statement represent a physical property or a chemical property?

a)  Sulfur is yellow.

b)  Steel wool burns when ignited by a flame.

c)  A gallon of milk weighs over eight pounds.

4.  Does each statement represent a physical property or a chemical property?

a)  A pile of leaves slowly rots in the backyard.

b)  In the presence of oxygen, hydrogen can interact to make water.

c)  Gold can be stretched into very thin wires.

5.  Does each statement represent a physical change or a chemical change?

a)  Water boils and becomes steam.

b)  Food is converted into usable form by the digestive system.

c)  The alcohol in many thermometers freezes at about −40 degrees Fahrenheit.

6.  Does each statement represent a physical change or a chemical change?

a) Graphite, a form of elemental carbon, can be turned into diamond, another form of carbon, at very high temperatures and pressures.

b)  The house across the street has been painted a new color.

c)  The elements sodium and chlorine come together to make a new substance called sodium chloride.

7.  Distinguish between an element and a compound. About how many of each are known?

8.  What is the difference between a homogeneous mixture and a heterogeneous mixture?

9.  Identify each as a heterogeneous mixture or a homogeneous mixture.

a)  Salt is mixed with pepper.

b)  Sugar is dissolved in water.

c)  Pasta is cooked in boiling water.

10.  Identify each as a heterogeneous mixture or a homogeneous mixture.

a)  air

b)  dirt

c)  a television set

11.  In Exercise 9, which choices are also solutions?

12.  In Exercise 10, which choices are also solutions?

13.  Why is iron considered a metal?

14.  Why is oxygen considered a nonmetal?

15.  Distinguish between a metal and a nonmetal.

16.  What properties do semimetals have?

17.  Elemental carbon is a black, dull-looking solid that conducts heat and electricity well. It is very brittle and cannot be made into thin sheets or long wires. Of these properties, how does carbon behave as a metal? How does carbon behave as a nonmetal?

18.  Pure silicon is shiny and silvery but does not conduct electricity or heat well. Of these properties, how does silicon behave as a metal? How does silicon behave as a nonmetal?

Answers

1.

a)  matter

b)  not matter

c)  not matter

d)  matter

3.

a)  physical property

b)  chemical property

c)  physical property

5.

a)  physical change

b)  chemical change

c) physical change

7. An element is a fundamental chemical part of a substance; there are about 115 known elements. A compound is a combination of elements that acts as a different substance; there are over 50 million known substances.

9.

a)  heterogeneous

b)  homogeneous

c)  heterogeneous

11. Choice b is a solution.

13. Iron is a metal because it is solid, is shiny, and conducts electricity and heat well.

15. Metals are typically shiny, conduct electricity and heat well, and are malleable and ductile; nonmetals are a variety of colors and phases, are brittle in the solid phase, and do not conduct heat or electricity well.

17. Carbon behaves as a metal because it conducts heat and electricity well. It is a nonmetal because it is black and brittle and cannot be made into sheets or wires.

3

Chemistry as a Science

Learning Objectives

  1. Learn what science is and how it works.

Chemistry is a branch of science. Although science itself is difficult to define exactly, the following definition can serve as starting point. Science is the process of knowing about the natural universe through observation and experiment. Science is not the only process of knowing (e.g., the ancient Greeks simply sat and thought), but it has evolved over more than 350 years into the best process that humanity has devised to date to learn about the universe around us.

The process of science is usually stated as the scientific method, which is rather naïvely described as follows: (1) state a hypothesis, (2) test the hypothesis, and (3) refine the hypothesis. Actually, however, the process is not that simple. (For example, I don’t go into my lab every day and exclaim, “I am going to state a hypothesis today and spend the day testing it!”) The process is not that simple because science and scientists have a body of knowledge that has already been identified as coming from the highest level of understanding, and most scientists build from that body of knowledge.

An educated guess about how the natural universe works is called a hypothesis. A scientist who is familiar with how part of the natural universe works—say, a chemist—is interested in furthering that knowledge. That person makes a reasonable guess—a hypothesis—that is designed to see if the universe works in a new way as well. Here’s an example of a hypothesis: “if I mix one part of hydrogen with one part of oxygen, I can make a substance that contains both elements.”

Most good hypotheses are grounded in previously understood knowledge and represent a testable extension of that knowledge. The scientist then devises ways to test if that guess is or is not correct. That is, the scientist plans experiments. Experiments are tests of the natural universe to see if a guess (hypothesis) is correct. An experiment to test our previous hypothesis would be to actually mix hydrogen and oxygen and see what happens. Most experiments include observations of small, well-defined parts of the natural universe designed to see results of the experiments.

Why do we have to do experiments? Why do we have to test? Because the natural universe is not always so obvious, experiments are necessary. For example, it is fairly obvious that if you drop an object from a height, it will fall. Several hundred years ago (coincidentally, near the inception of modern science), the concept of gravity explained that test. However, is it obvious that the entire natural universe is composed of only about 115 fundamental chemical building blocks called elements? This wouldn’t seem true if you looked at the world around you and saw all the different forms matter can take. In fact, the concept of the element is only about 200 years old, and the last naturally occurring element was identified about 80 years ago. It took decades of tests and millions of experiments to establish what the elements actually are. These are just two examples; a myriad of such examples exists in chemistry and science in general.

When enough evidence has been collected to establish a general principle of how the natural universe works, the evidence is summarized in a theory. A theory is a general statement that explains a large number of observations. “All matter is composed of atoms” is a general statement, a theory, that explains many observations in chemistry. A theory is a very powerful statement in science. There are many statements referred to as “the theory of _______” or the “______ theory” in science (where the blanks represent a word or concept). When written in this way, theories indicate that science has an overwhelming amount of evidence of its correctness. We will see several theories in the course of this text.

A specific statement that is thought to be never violated by the entire natural universe is called a law. A scientific law is the highest understanding of the natural universe that science has and is thought to be inviolate. For example, the fact that all matter attracts all other matter—the law of gravitation—is one such law. Note that the terms theory and law used in science have slightly different meanings from those in common usage; theory is often used to mean hypothesis (“I have a theory…”), whereas a law is an arbitrary limitation that can be broken but with potential consequences (such as speed limits). Here again, science uses these terms differently, and it is important to apply their proper definitions when you use these words in science.

There is an additional phrase in our definition of science: “the natural universe.” Science is concerned only with the natural universe. What is the natural universe? It’s anything that occurs around us, well, naturally. Stars; planets; the appearance of life on earth; and how animals, plants, and other matter function are all part of the natural universe. Science is concerned with that—and only that.

Of course, there are other things that concern us. For example, is the English language part of science? Most of us can easily answer no; English is not science. English is certainly worth knowing (at least for people in predominantly English-speaking countries), but why isn’t it science? English, or any human language, isn’t science because ultimately it is contrived; it is made up. Think of it: the word spelled b-l-u-e represents a certain color, and we all agree what color that is. But what if we used the word h-a-r-d-n-r-f to describe that color? (See Figure 1.9 “English Is Not Science”.) That would be fine—as long as everyone agreed. Anyone who has learned a second language must initially wonder why a certain word is used to describe a certain concept; ultimately, the speakers of that language agreed that a particular word would represent a particular concept. It was contrived.

Figure 1.9 English Is Not Science

Blue Square

 

How would you describe this color? Blue or hardnrf? Either way, you’re not doing science.

That doesn’t mean language isn’t worth knowing. It is very important in society. But it’s not science. Science deals only with what occurs naturally.

Example 4

Which of the following fields would be considered science?

  1. geology, the study of the earth
  2. ethics, the study of morality
  3. political science, the study of governance
  4. biology, the study of living organisms

Solution

  1. Because the earth is a natural object, the study of it is indeed considered part of science.
  2. Ethics is a branch of philosophy that deals with right and wrong. Although these are useful concepts, they are not science.
  3. There are many forms of government, but all are created by humans. Despite the fact that the word science appears in its name, political science is not true science.
  4. Living organisms are part of the natural universe, so the study of them is part of science.

Test Yourself

Which is part of science, and which is not?

  1. dynamics, the study of systems that change over time
  2. aesthetics, the concept of beauty

Answers

  1. science
  2. not science

The field of science has gotten so big that it is common to separate it into more specific fields. First, there is mathematics, the language of science. All scientific fields use mathematics to express themselves—some more than others. Physics and astronomy are scientific fields concerned with the fundamental interactions between matter and energy. Chemistry, as defined previously, is the study of the interactions of matter with other matter and with energy. Biology is the study of living organisms, while geology is the study of the earth. Other sciences can be named as well. Understand that these fields are not always completely separate; the boundaries between scientific fields are not always readily apparent. Therefore, a scientist may be labelled a biochemist if he or she studies the chemistry of biological organisms.

Finally, understand that science can be either qualitative or quantitative. Qualitative implies a description of the quality of an object. For example, physical properties are generally qualitative descriptions: sulfur is yellow, your math book is heavy, or that statue is pretty. A quantitative description represents the specific amount of something; it means knowing how much of something is present, usually by counting or measuring it. As such, some quantitative descriptions would include 25 students in a class, 650 pages in a book, or a velocity of 66 miles per hour. Quantitative expressions are very important in science; they are also very important in chemistry.

Example 5

Identify each statement as either a qualitative description or a quantitative description.

  1. Gold metal is yellow.
  2. A ream of paper has 500 sheets in it.
  3. The weather outside is snowy.
  4. The temperature outside is 24 degrees Fahrenheit.

Solution

  1. Because we are describing a physical property of gold, this statement is qualitative.
  2. This statement mentions a specific amount, so it is quantitative.
  3. The word snowy is a description of how the day is; therefore, it is a qualitative statement.
  4. In this case, the weather is described with a specific quantity—the temperature. Therefore, it is quantitative.

Test Yourself

Are these qualitative or quantitative statements?

  1. Roses are red, and violets are blue.
  2. Four score and seven years ago….

Answers

  1. qualitative
  2. quantitative

Food and Drink App: Carbonated Beverages

Some of the simple chemical principles discussed in this chapter can be illustrated with carbonated beverages: sodas, beer, and sparkling wines. Each product is produced in a different way, but they all have one thing in common. They are solutions of carbon dioxide dissolved in water.

Carbon dioxide is a compound composed of carbon and oxygen. Under normal conditions, it is a gas. If you cool it down enough, it becomes a solid known as dry ice. Carbon dioxide is an important compound in the cycle of life on earth.

Even though it is a gas, carbon dioxide can dissolve in water, just like sugar or salt can dissolve in water. When that occurs, we have a homogeneous mixture, or a solution, of carbon dioxide in water. However, very little carbon dioxide can dissolve in water. If the atmosphere were pure carbon dioxide, the solution would be only about 0.07% carbon dioxide. In reality, the air is only about 0.03% carbon dioxide, so the amount of carbon dioxide in water is reduced proportionally.

However, when soda and beer are made, manufacturers do two important things: they use pure carbon dioxide gas, and they use it at very high pressures. With higher pressures, more carbon dioxide can dissolve in the water. When the soda or beer container is sealed, the high pressure of carbon dioxide gas remains inside the package. (Of course, there are more ingredients in soda and beer besides carbon dioxide and water.)

When you open a container of soda or beer, you hear a distinctive hiss as the excess carbon dioxide gas escapes. But something else happens as well. The carbon dioxide in the solution comes out of solution as a bunch of tiny bubbles. These bubbles impart a pleasing sensation in the mouth, so much so that the soda industry sold over 225 billion servings of soda in the United States alone in 2009.

Some sparkling wines are made in the same way—by forcing carbon dioxide into regular wine. Some sparkling wines (including champagne) are made by sealing a bottle of wine with some yeast in it. The yeast ferments, a process by which the yeast converts sugars into energy and excess carbon dioxide. The carbon dioxide produced by the yeast dissolves in the wine. Then, when the champagne bottle is opened, the increased pressure of carbon dioxide is released, and the drink bubbles just like an expensive glass of soda.

Soda, beer, and sparkling wine take advantage of the properties of a solution of carbon dioxide in water. “cola with ice” by Simon Cousins is licensed under the Creative Commons Attribution 2.0 Generic license. “Beer” by Will Vanlue is licensed under Cre- ative Commons the Attribution-NonCommercial-ShareAlike 2.0 Generic. “Sparkling Water” by T. Faltings is licensed under the Creative Commons Attribution-NonCommercial-ShareAlike 2.0 Generic.
Soda, beer, and sparkling wine take advantage of the properties of a solution of carbon dioxide in water.
“cola with ice” by Simon Cousins is licensed under the Creative Commons Attribution 2.0 Generic license. “Beer” by Will Vanlue is licensed under Cre- ative Commons the Attribution-NonCommercial-ShareAlike 2.0 Generic. “Sparkling Water” by T. Faltings is licensed under the Creative Commons Attribution-NonCommercial-ShareAlike 2.0 Generic.

Key Takeaways

  • Science is a process of knowing about the natural universe through observation and experiment.
  • Scientists go through a rigorous process to determine new knowledge about the universe; this process is generally referred to as the scientific method.
  • Science is broken down into various fields, of which chemistry is one.
  • Science, including chemistry, is both qualitative and quantitative.

 

Exercises

  1. Describe the scientific method.

  2. What is the scientific definition of a hypothesis? Why is the phrase a hypothesis is just a guess an inadequate definition?

  3. Why do scientists need to perform experiments?

  4. What is the scientific definition of a theory? How is this word misused in general conversation?

  5. What is the scientific definition of a law? How does it differ from the everyday definition of a law?

  6. Name an example of a field that is not considered a science.

  7. Which of the following fields are studies of the natural universe?

a)  biophysics (a mix of biology and physics)

b)  art

c)  business

8.  Which of the following fields are studies of the natural universe?

a)  accounting

b)  geochemistry (a mix of geology and chemistry)

c)  astronomy (the study of stars and planets [but not the earth])

 

9.  Which of these statements are qualitative descriptions?

a)  The Titanic was the largest passenger ship build at that time.

b)  The population of the United States is about 306,000,000 people.

c)  The peak of Mount Everest is 29,035 feet above sea level.

 

10.  Which of these statements are qualitative descriptions?

a)  A regular movie ticket in Cleveland costs $6.00.

b)  The weather in the Democratic Republic of the Congo is the wettest in all of Africa.

c)  The deepest part of the Pacific Ocean is the Mariana Trench.

 

11.  Of the statements in Exercise 9, which are quantitative?

 

12.  Of the statements in Exercise 10, which are quantitative?

Answers

1. Simply stated, the scientific method includes three steps: (1) stating a hypothesis, (2) testing the hypothesis, and (3) refining the hypothesis.
3. Scientists perform experiments to test their hypotheses because sometimes the nature of natural universe is not obvious.
5. A scientific law is a specific statement that is thought to be never violated by the entire natural universe. Everyday laws are arbitrary limits that society puts on its members.
7. (a) yes (b) no (c) no
9. (a) qualitative (b) not qualitative (c) not qualitative
11. Statements b and c are quantitative.

II

Chapter 2. Measurements

4

Introduction to Measurements

Data suggest that a male child will weigh 50% of his adult weight at about 11 years of age. However, he will reach 50% of his adult height at only 2 years of age. It is obvious, then, that people eventually stop growing up but continue to grow out. Data also suggest that the average human height has been increasing over time. In industrialized countries, the average height of people increased 5.5 inches from 1810 to 1984. Most scientists attribute this simple, basic measurement of the human body to better health and nutrition.

Stature Percentile

In 1983, an Air Canada airplane had to make an emergency landing because it unexpectedly ran out of fuel; ground personnel had filled the fuel tanks with a certain number of pounds of fuel, not kilograms of fuel. In 1999, the Mars Climate Orbiter spacecraft was lost attempting to orbit Mars because the thrusters were programmed in terms of English units, even though the engineers built the spacecraft using metric units. In 1993, a nurse mistakenly administered 23 units of morphine to a patient rather than the “2–3” units prescribed. (The patient ultimately survived.) These incidents occurred because people weren’t paying attention to quantities.

Chemistry, like all sciences, is quantitative. It deals with quantities, things that have amounts and units. Dealing with quantities is very important in chemistry, as is relating quantities to each other. In this chapter, we will discuss how we deal with numbers and units, including how they are combined and manipulated.

5

Expressing Numbers

Learning Objective

1. Learn to express numbers properly.

 

Quantities have two parts: the number and the unit. The number tells “how many.” It is important to be able to express numbers properly so that the quantities can be communicated properly.

Standard notation is the straightforward expression of a number. Numbers such as 17, 101.5, and 0.00446 are expressed in standard notation. For relatively small numbers, standard notation is fine. However, for very large numbers, such as 306,000,000, or for very small numbers, such as 0.000000419, standard notation can be cumbersome because of the number of zeros needed to place nonzero numbers in the proper position.

Scientific notation is an expression of a number using powers of 10. Powers of 10 are used to express numbers that have many zeros:

100 = 1
101 = 10
102 = 100 = 10 × 10
103 = 1,000 = 10 × 10 × 10
104 = 10,000 = 10 × 10 × 10 × 10

and so forth. The raised number to the right of the 10 indicating the number of factors of 10 in the original number is the exponent. (Scientific notation is sometimes called exponential notation.) The exponent’s value is equal to the number of zeros in the number expressed in standard notation.

Small numbers can also be expressed in scientific notation but with negative exponents:

10−1 = 0.1 = 1/10
10−2 = 0.01 = 1/100
10−3 = 0.001 = 1/1,000
10−4 = 0.0001 = 1/10,000

and so forth. Again, the value of the exponent is equal to the number of zeros in the denominator of the associated fraction. A negative exponent implies a decimal number less than one.

A number is expressed in scientific notation by writing the first nonzero digit, then a decimal point, and then the rest of the digits. The part of a number in scientific notation that is multiplied by a power of 10 is called the coefficient. Then determine the power of 10 needed to make that number into the original number and multiply the written number by the proper power of 10. For example, to write 79,345 in scientific notation,

79,345 = 7.9345 × 10,000 = 7.9345 × 104

Thus, the number in scientific notation is 7.9345 × 104. For small numbers, the same process is used, but the exponent for the power of 10 is negative:

0.000411 = 4.11 × 1/10,000 = 4.11 × 10−4

Typically, the extra zero digits at the end or the beginning of a number are not included.

Example 1

Express these numbers in scientific notation.

  1. 306,000
  2. 0.00884
  3. 2,760,000
  4. 0.000000559

Solution

  1. The number 306,000 is 3.06 times 100,000, or 3.06 times 105. In scientific notation, the number is 3.06 × 105.
  2. The number 0.00884 is 8.84 times 1/1,000, which is 8.84 times 10−3. In scientific notation, the number is 8.84 × 10−3.
  3. The number 2,760,000 is 2.76 times 1,000,000, which is the same as 2.76 times 106. In scientific notation, the number is written as 2.76 × 106. Note that we omit the zeros at the end of the original number.
  4. The number 0.000000559 is 5.59 times 1/10,000,000, which is 5.59 times 10−7. In scientific notation, the number is written as 5.59 × 10−7.

Test Yourself

Express these numbers in scientific notation.

  1. 23,070
  2. 0.0009706

Answers

  1. 2.307 × 104
  2. 9.706 × 10−4

Another way to determine the power of 10 in scientific notation is to count the number of places you need to move the decimal point to get a numerical value between 1 and 10. The number of places equals the power of 10. This number is positive if you move the decimal point to the right and negative if you move the decimal point to the left.

Many quantities in chemistry are expressed in scientific notation. When performing calculations, you may have to enter a number in scientific notation into a calculator. Be sure you know how to correctly enter a number in scientific notation into your calculator. Different models of calculators require different actions for properly entering scientific notation. If in doubt, consult your instructor immediately.

Key Takeaways

  • Standard notation expresses a number normally.
  • Scientific notation expresses a number as a coefficient times a power of 10.
  • The power of 10 is positive for numbers greater than 1 and negative for numbers between 0 and 1.
This calculator shows only the coefficient and the power of 10 to represent the number in scientific notation. Thus, the number being displayed is 3.84951 × 1018, or 3,849,510,000,000,000,000. Source: “Casio”Asim Bijarani is licensed under Creative Commons Attribution 2.0 Generic
This calculator shows only the coefficient and the power of 10 to represent the number in scientific notation. Thus, the number being displayed is 3.84951 × 1018, or 3,849,510,000,000,000,000.
Source: “Casio”Asim Bijarani is licensed under Creative Commons Attribution 2.0 Generic

Exercises

  1. Express these numbers in scientific notation.

a)  56.9

b)  563,100

c)  0.0804

d)  0.00000667

 

2.  Express these numbers in scientific notation.

a)  −890,000

b)  602,000,000,000

c)  0.0000004099

d)  0.000000000000011

 

3.  Express these numbers in scientific notation.

a)  0.00656

b)  65,600

c)  4,567,000

d)  0.000005507

 

4.  Express these numbers in scientific notation.

a)  65

b)  −321.09

c)  0.000077099

d)  0.000000000218

 

5.  Express these numbers in standard notation.

a)  1.381 × 105

b)  5.22 × 10−7

c)  9.998 × 104

 

6.  Express these numbers in standard notation.

a)  7.11 × 10−2

b)  9.18 × 102

c)  3.09 × 10−10

 

7.  Express these numbers in standard notation.

a)  8.09 × 100

b)  3.088 × 10−5

c)  −4.239 × 102

 

8.  Express these numbers in standard notation.

a)  2.87 × 10−8

b)  1.78 × 1011

c)  1.381 × 10−23

 

9. These numbers are not written in proper scientific notation. Rewrite them so that they are in proper scientific notation.

a)  72.44 × 103

b)  9,943 × 10−5

c)  588,399 × 102

 

10.  These numbers are not written in proper scientific notation. Rewrite them so that they are in proper scientific notation.

a)  0.000077 × 10−7

b)  0.000111 × 108

c)  602,000 × 1018

 

11.  These numbers are not written in proper scientific notation. Rewrite them so that they are in proper scientific notation.

a)  345.1 × 102

b)  0.234 × 10−3

c)  1,800 × 10−2

 

12.  These numbers are not written in proper scientific notation. Rewrite them so that they are in proper scientific notation.

a)  8,099 × 10−8

b)  34.5 × 100

c)  0.000332 × 104

 

13.  Write these numbers in scientific notation by counting the number of places the decimal point is moved.

a)  123,456.78

b)  98,490

c)  0.000000445

 

14.  Write these numbers in scientific notation by counting the number of places the decimal point is moved.

a)  0.000552

b)  1,987

c)  0.00000000887

 

15.  Use your calculator to evaluate these expressions. Express the final answer in proper scientific notation.

a)  456 × (7.4 × 108) = ?

b)  (3.02 × 105) ÷ (9.04 × 1015) = ?

c)  0.0044 × 0.000833 = ?

 

16.  Use your calculator to evaluate these expressions. Express the final answer in proper scientific notation.

a)  98,000 × 23,000 = ?

b)  98,000 ÷ 23,000 = ?

c)  (4.6 × 10−5) × (2.09 × 103) = ?

 

17.  Use your calculator to evaluate these expressions. Express the final answer in proper scientific notation.

a)  45 × 132 ÷ 882 = ?

b) [(6.37 × 104) × (8.44 × 10−4)] ÷ (3.2209 × 1015) = ?

 

18.  Use your calculator to evaluate these expressions. Express the final answer in proper scientific notation.

a)  (9.09 × 108) ÷ [(6.33 × 109) × (4.066 × 10−7)] = ?

b)  9,345 × 34.866 ÷ 0.00665 = ?

Answers

1.
a)  5.69 × 101

b)  5.631 × 105

c)  8.04 × 10−2

d)  6.67 × 10−6

3.
a)  6.56 × 10−3

b)  6.56 × 104

c)  4.567 × 106

d)  5.507 × 10−6

5.
a) 138,100

b)  0.000000522

c)  99,980

7.
a)  8.09
b) 0.00003088

c)  −423.9

9.
a)  7.244 × 104
b)   9.943 × 10−2
c)  5.88399 × 107
11.
a)  3.451 × 104

b)  2.34 × 10−4

c)  1.8 × 101

13.
a)  1.2345678 × 105

b)  9.849 × 104
c)  4.45 × 10−7

15.
a) 3.3744 × 1011
b) 3.3407 × 10−11
c) 3.665 × 10−6

17.
a)  6.7346 × 100

b) 1.6691 × 10−14

 

6

Significant Figures

Learning Objectives

  1. Apply the concept of significant figures to limit a measurement to the proper number of digits.
  2. Recognize the number of significant figures in a given quantity.
  3. Limit mathematical results to the proper number of significant figures.

If you use a calculator to evaluate the expression 337/217, you will get the following:

337217=1.55299539171…

and so on for many more digits. Although this answer is correct, it is somewhat presumptuous. You start with two values that each have three digits, and the answer has twelve digits? That does not make much sense from a strict numerical point of view.

Consider using a ruler to measure the width of an object, as shown in Figure 2.6 “Expressing Width”. The object is definitely more than 1 cm long, so we know that the first digit in our measurement is 1. We see by counting the tick marks on the ruler that the object is at least three ticks after the 1. If each tick represents 0.1 cm, then we know the object is at least 1.3 cm wide. But our ruler does not have any more ticks between the 0.3 and the 0.4 marks, so we can’t know exactly how much the next decimal place is. But with a practiced eye we can estimate it. Let us estimate it as about six-tenths of the way between the third and fourth tick marks, which estimates our hundredths place as 6, so we identify a measurement of 1.36 cm for the width of the object.

Figure 2.6 Expressing Width

image

What is the proper way to express the width of this object?

Does it make any sense to try to report a thousandths place for the measurement? No, it doesn’t; we are not exactly sure of the hundredths place (after all, it was an estimate only), so it would be fruitless to estimate a thousandths place. Our best measurement, then, stops at the hundredths place, and we report 1.36 cm as proper measurement.

This concept of reporting the proper number of digits in a measurement or a calculation is called significant figures. Significant figures (sometimes called significant digits) represent the limits of what values of a measurement or a calculation we are sure of. The convention for a measurement is that the quantity reported should be all known values and the first estimated value. The conventions for calculations are discussed as follows.

Example 3

Use each diagram to report a measurement to the proper number of significant figures.

  1. image
  2. Ruler

Solution

  1. The arrow is between 4.0 and 5.0, so the measurement is at least 4.0. The arrow is between the third and fourth small tick marks, so it’s at least 0.3. We will have to estimate the last place. It looks like about one-third of the way across the space, so let us estimate the hundredths place as 3. Combining the digits, we have a measurement of 4.33 psi (psi stands for “pounds per square inch” and is a unit of pressure, like air in a tire). We say that the measurement is reported to three significant figures.
  2. The rectangle is at least 1.0 cm wide but certainly not 2.0 cm wide, so the first significant digit is 1. The rectangle’s width is past the second tick mark but not the third; if each tick mark represents 0.1, then the rectangle is at least 0.2 in the next significant digit. We have to estimate the next place because there are no markings to guide us. It appears to be about halfway between 0.2 and 0.3, so we will estimate the next place to be a 5. Thus, the measured width of the rectangle is 1.25 cm. Again, the measurement is reported to three significant figures.

Test Yourself

What would be the reported width of this rectangle?

Rectangle

Answer

0.63 cm

In many cases, you will be given a measurement. How can you tell by looking what digits are significant? For example, the reported population of the United States is 306,000,000. Does that mean that it is exactly three hundred six million or is some estimation occurring?

The following conventions dictate which numbers in a reported measurement are significant and which are not significant:

  1. Any nonzero digit is significant.
  2. Any zeros between nonzero digits (i.e., embedded zeros) are significant.
  3. Zeros at the end of a number without a decimal point (i.e., trailing zeros) are not significant; they serve only to put the significant digits in the correct positions. However, zeros at the end of any number with a decimal point are significant.
  4. Zeros at the beginning of a decimal number (i.e., leading zeros) are not significant; again, they serve only to put the significant digits in the correct positions.

So, by these rules, the population figure of the United States has only three significant figures: the 3, the 6, and the zero between them. The remaining six zeros simply put the 306 in the millions position.

Example 4

Give the number of significant figures in each measurement.

  1. 36.7 m
  2. 0.006606 s
  3. 2,002 kg
  4. 306,490,000 people

Solution

  1. By rule 1, all nonzero digits are significant, so this measurement has three significant figures.
  2. By rule 4, the first three zeros are not significant, but by rule 2 the zero between the sixes is; therefore, this number has four significant figures.
  3. By rule 2, the two zeros between the twos are significant, so this measurement has four significant figures.
  4. The four trailing zeros in the number are not significant, but the other five numbers are, so this number has five significant figures.

Test Yourself

Give the number of significant figures in each measurement.

  1. 0.000601 m
  2. 65.080 kg

Answers

  1. three significant figures
  2. five significant figures

How are significant figures handled in calculations? It depends on what type of calculation is being performed. If the calculation is an addition or a subtraction, the rule is as follows: limit the reported answer to the rightmost column that all numbers have significant figures in common. For example, if you were to add 1.2 and 4.71, we note that the first number stops its significant figures in the tenths column, while the second number stops its significant figures in the hundredths column. We therefore limit our answer to the tenths column.

Sig Figs 1

We drop the last digit—the 1—because it is not significant to the final answer.

The dropping of positions in sums and differences brings up the topic of rounding. Although there are several conventions, in this text we will adopt the following rule: the final answer should be rounded up if the first dropped digit is 5 or greater and rounded down if the first dropped digit is less than 5.

Sig Figs 2

Example 5

Express the final answer to the proper number of significant figures.

  1. 101.2 + 18.702 = ?
  2. 202.88 − 1.013 = ?

Solution

  1. If we use a calculator to add these two numbers, we would get 119.902. However, most calculators do not understand significant figures, and we need to limit the final answer to the tenths place. Thus, we drop the 02 and report a final answer of 119.9 (rounding down).
  2. A calculator would answer 201.867. However, we have to limit our final answer to the hundredths place. Because the first number being dropped is 7, which is greater than 7, we round up and report a final answer of 201.87.

Test Yourself

Express the answer for 3.445 + 90.83 − 72.4 to the proper number of significant figures.

Answer

21.9

If the operations being performed are multiplication or division, the rule is as follows: limit the answer to the number of significant figures that the data value with the least number of significant figures has. So if we are dividing 23 by 448, which have two and three significant figures each, we should limit the final reported answer to two significant figures (the lesser of two and three significant figures):

23448=0.051339286…=0.051

The same rounding rules apply in multiplication and division as they do in addition and subtraction.

Example 6

Express the final answer to the proper number of significant figures.

  1. 76.4 × 180.4 = ?
  2. 934.9 ÷ 0.00455 = ?

Solution

  1. The first number has three significant figures, while the second number has four significant figures. Therefore, we limit our final answer to three significant figures: 76.4 × 180.4 = 13,782.56 = 13,800.
  2. The first number has four significant figures, while the second number has three significant figures. Therefore we limit our final answer to three significant figures: 934.9 ÷ 0.00455 = 205,472.5275… = 205,000.

Test Yourself

Express the final answer to the proper number of significant figures.

  1. 22.4 × 8.314 = ?
  2. 1.381 ÷ 6.02 = ?

Answers

  1. 186
  2. 0.229

As you have probably realized by now, the biggest issue in determining the number of significant figures in a value is the zero. Is the zero significant or not? One way to unambiguously determine whether a zero is significant or not is to write a number in scientific notation. Scientific notation will include zeros in the coefficient of the number only if they are significant. Thus, the number 8.666 × 106 has four significant figures. However, the number 8.6660 × 106 has five significant figures. That last zero is significant; if it were not, it would not be written in the coefficient. So when in doubt about expressing the number of significant figures in a quantity, use scientific notation and include the number of zeros that are truly significant.

qrcode.23437479

Video source: Significant figures by keyj (https://viuvideos.viu.ca/media/Significant+Figures/0_t8xwe4s9)

Key Takeaways

  • Significant figures in a quantity indicate the number of known values plus one place that is estimated.
  • There are rules for which numbers in a quantity are significant and which are not significant.
  • In calculations involving addition and subtraction, limit significant figures based on the rightmost place that all values have in common.
  • In calculations involving multiplication and division, limit significant figures to the least number of significant figures in all the data values.

Exercises

  1. Express each measurement to the correct number of significant figures.

    a) image

    b) Ruler-2

  2. Express each measurement to the correct number of significant figures.

    a) image

    b)   Ruler-3

  3. How many significant figures do these numbers have?

a)  23

b)  23.0

c)  0.00023

d)  0.0002302

4.  How many significant figures do these numbers have?

a)  5.44 × 108

b)  1.008 × 10−5

c)  43.09

d)  0.0000001381

5.  How many significant figures do these numbers have?

a)  765,890

b)  765,890.0

c)  1.2000 × 105

d)  0.0005060

6)  How many significant figures do these numbers have?

a)  0.009

b)  0.0000009

c)  65,444

d)  65,040

7.  Compute and express each answer with the proper number of significant figures, rounding as necessary.

a)  56.0 + 3.44 = ?

b)  0.00665 + 1.004 = ?

c)  45.99 − 32.8 = ?

d)  45.99 − 32.8 + 75.02 = ?

8.  Compute and express each answer with the proper number of significant figures, rounding as necessary.

a)  1.005 + 17.88 = ?

b)  56,700 − 324 = ?

c)  405,007 − 123.3 = ?

d)  55.5 + 66.66 − 77.777 = ?

9.  Compute and express each answer with the proper number of significant figures, rounding as necessary.

a)  56.7 × 66.99 = ?

b)  1.000 ÷ 77 = ?

c)  1.000 ÷ 77.0 = ?

d)  6.022 × 1.89 = ?

10.  Compute and express each answer with the proper number of significant figures, rounding as necessary.

a)  0.000440 × 17.22 = ?

b)  203,000 ÷ 0.044 = ?

c)  67 × 85.0 × 0.0028 = ?

d)  999,999 ÷ 3,310 = ?

11.  Write the number 87,449 in scientific notation with four significant figures.

12.  Write the number 0.000066600 in scientific notation with five significant figures.

13.  Write the number 306,000,000 in scientific notation to the proper number of significant figures.

14.  Write the number 0.0000558 in scientific notation with two significant figures.

15.  Perform each calculation and limit each answer to three significant figures.

a)  67,883 × 0.004321 = ?

b)  (9.67 × 103) × 0.0055087 = ?

16.  Perform each calculation and limit each answer to four significant figures.

a)  18,900 × 76.33 ÷ 0.00336 = ?

b)  0.77604 ÷ 76,003 × 8.888 = ?

Answers

1.

a)  375 psi

b)  1.30 cm

3.

a)  two

b)  three

c)  two

d)  four

5.

a)  five

b)  seven

c)  five

d)  four

7.

a)  59.4

b)  1.011

c)  13.2

d)  88.2

9.

a)  3.80 × 103

b)  0.013

c)  0.0130

d)  11.4

11.

a)  8.745 × 104

b)  6.6600 × 10−5

13.

a)  293

b)  53.3

7

Converting Units

Learning Objective

1. Convert from one unit to another unit of the same type.

In Section 2.2 “Expressing Units”, we showed some examples of how to replace initial units with other units of the same type to get a numerical value that is easier to comprehend. In this section, we will formalize the process.

Consider a simple example: how many feet are there in 4 yards? Most people will almost automatically answer that there are 12 feet in 4 yards. How did you make this determination? Well, if there are 3 feet in 1 yard and there are 4 yards, then there are 4 × 3 = 12 feet in 4 yards.

This is correct, of course, but it is informal. Let us formalize it in a way that can be applied more generally. We know that 1 yard (yd) equals 3 feet (ft):

1 yd = 3 ft

In math, this expression is called an equality. The rules of algebra say that you can change (i.e., multiply or divide or add or subtract) the equality (as long as you don’t divide by zero) and the new expression will still be an equality. For example, if we divide both sides by 2, we get

1/2 yd = 3/2 feet

We see that one-half of a yard equals 3/2, or one and a half, feet—something we also know to be true, so the above equation is still an equality. Going back to the original equality, suppose we divide both sides of the equation by 1 yard (number and unit):

1/1 yd = 3 ft/ 1 yd

The expression is still an equality, by the rules of algebra. The left fraction equals 1. It has the same quantity in the numerator and the denominator, so it must equal 1. The quantities in the numerator and denominator cancel, both the number and the unit:

1/1 yd = 3 ft / 1 yd (cancelled units crossed out)

When everything cancels in a fraction, the fraction reduces to 1:

1 = 3 ft/1 yd

We have an expression, 3 ft1 yd, that equals 1. This is a strange way to write 1, but it makes sense: 3 ft equal 1 yd, so the quantities in the numerator and denominator are the same quantity, just expressed with different units. The expression 3 ft1 yd is called a conversion factor, and it is used to formally change the unit of a quantity into another unit. (The process of converting units in such a formal fashion is sometimes called dimensional analysis or the factor label method.)

To see how this happens, let us start with the original quantity:

4 yd

Now let us multiply this quantity by 1. When you multiply anything by 1, you don’t change the value of the quantity. Rather than multiplying by just 1, let us write 1 as 3 ft1 yd:

4 yd x (3ft/1yd)

The 4 yd term can be thought of as 4 yd/1; that is, it can be thought of as a fraction with 1 in the denominator. We are essentially multiplying fractions. If the same thing appears in the numerator and denominator of a fraction, they cancel. In this case, what cancels is the unit yard:

4 yd x (3 ft/ 1 yd) showing units cancel

That is all that we can cancel. Now, multiply and divide all the numbers to get the final answer:(4 x 3 ft)/1 = 12 ft/1 = 12 ft

Again, we get an answer of 12 ft, just as we did originally. But in this case, we used a more formal procedure that is applicable to a variety of problems.

How many millimeters are in 14.66 m? To answer this, we need to construct a conversion factor between millimeters and meters and apply it correctly to the original quantity. We start with the definition of a millimeter, which is

1 mm = 1/1,000 m

The 1/1,000 is what the prefix milli- means. Most people are more comfortable working without fractions, so we will rewrite this equation by bringing the 1,000 into the numerator of the other side of the equation:

1,000 mm = 1 m

Now we construct a conversion factor by dividing one quantity into both sides. But now a question arises: which quantity do we divide by? It turns out that we have two choices, and the two choices will give us different conversion factors, both of which equal 1:

conversion facts equaling 1 m / 1000 mm

conversion factor 1m / 1000 mm

Which conversion factor do we use? The answer is based on what unit you want to get rid of in your initial quantity. The original unit of our quantity is meters, which we want to convert to millimeters. Because the original unit is assumed to be in the numerator, to get rid of it, we want the meter unit in the denominator; then they will cancel. Therefore, we will use the second conversion factor. Canceling units and performing the mathematics, we get

14.66 m x (1000 mm/1 m) = 14660 mm

Note how m cancels, leaving mm, which is the unit of interest.

The ability to construct and apply proper conversion factors is a very powerful mathematical technique in chemistry. You need to master this technique if you are going to be successful in this and future courses.

Example 7

  1. Convert 35.9 kL to liters.
  2. Convert 555 nm to meters.

Solution

  1. We will use the fact that 1 kL = 1,000 L. Of the two conversion factors that can be defined, the one that will work is 1,000 L/1 kL. Applying this conversion factor, we get

    35.9 kL x (1000 L/1 kL) = 35900 L

  2. We will use the fact that 1 nm = 1/1,000,000,000 m, which we will rewrite as 1,000,000,000 nm = 1 m, or 109 nm = 1 m. Of the two possible conversion factors, the appropriate one has the nm unit in the denominator: 1 m/109 nm. Applying this conversion factor, we get

    555 nm x (1 m/ 10^9 nm) = 5.55 x 10^-7 m

    In the final step, we expressed the answer in scientific notation.

Test Yourself

  1. Convert 67.08 μL to liters.
  2. Convert 56.8 m to kilometers.

Answers

  1. 6.708 × 10−5 L
  2. 5.68 × 10−2 km

What if we have a derived unit that is the product of more than one unit, such as m2? Suppose we want to convert square meters to square centimeters? The key is to remember that m2 means m × m, which means we have two meter units in our derived unit. That means we have to include two conversion factors, one for each unit. For example, to convert 17.6 m2 to square centimeters, we perform the conversion as follows:

17.6 m^2 = 17.6 (mxm) x (100cm/1m) x (100cm/1m)=176000cm^2

Example 8

How many cubic centimeters are in 0.883 m3?

Solution

With an exponent of 3, we have three length units, so by extension we need to use three conversion factors between meters and centimeters. Thus, we have

0.883m^3 x (100cm/1m) x (100cm/1m) x (100cm/1m) = 883000 cm^3

You should demonstrate to yourself that the three meter units do indeed cancel.

Test Yourself

How many cubic millimeters are present in 0.0923 m3?

Answer

9.23 × 107 mm3

Suppose the unit you want to convert is in the denominator of a derived unit; what then? Then, in the conversion factor, the unit you want to remove must be in the numerator. This will cancel with the original unit in the denominator and introduce a new unit in the denominator. The following example illustrates this situation.

Example 9

Convert 88.4 m/min to meters/second.

Solution

We want to change the unit in the denominator from minutes to seconds. Because there are 60 seconds in 1 minute (60 s = 1 min), we construct a conversion factor so that the unit we want to remove, minutes, is in the numerator: 1 min/60 s. Apply and perform the math:

88.4m/m x 1min/60s = 1.47 m/s

Notice how the 88.4 automatically goes in the numerator. That’s because any number can be thought of as being in the numerator of a fraction divided by 1.

Test Yourself

Convert 0.203 m/min to meters/second.

Answer

0.00338 m/s or 3.38 × 10−3 m/s

Figure 2.8 How Fast Is Fast?

A common garden snail moves at a rate of about 0.2 m/min, which is about 0.003 m/s, which is 3 mm/s! Source: “Grapevine snail”by Jürgen Schoneris licensed under the Creative Commons Attribution-Share Alike 3.0 Unported license.
A common garden snail moves at a rate of about 0.2 m/min, which is about 0.003 m/s, which is 3 mm/s!
Source: “Grapevine snail”by Jürgen Schoneris licensed under the Creative Commons Attribution-Share Alike 3.0 Unported license.

Sometimes there will be a need to convert from one unit with one numerical prefix to another unit with a different numerical prefix. How do we handle those conversions? Well, you could memorize the conversion factors that interrelate all numerical prefixes. Or you can go the easier route: first convert the quantity to the base unit, the unit with no numerical prefix, using the definition of the original prefix. Then convert the quantity in the base unit to the desired unit using the definition of the second prefix. You can do the conversion in two separate steps or as one long algebraic step. For example, to convert 2.77 kg to milligrams:

2.77 kg x 1000 g/1kg = 2770 g (convert to the base unit of grams)
2770 g x 1000 mg/1g = 2770000 mg = 2.77x10^6 mg (convert to the desired unit)

Alternatively, it can be done in a single multistep process:2.77kg x 1000g/1kg x 1000 mg/1g = 2770000 mg = 2.77 x 10^6 mg

You get the same answer either way.

Example 10

How many nanoseconds are in 368.09 μs?

Solution

You can either do this as a one-step conversion from microseconds to nanoseconds or convert to the base unit first and then to the final desired unit. We will use the second method here, showing the two steps in a single line. Using the definitions of the prefixes micro- and nano-,

368.09 us x 1s/10^6us x 10^9ns /1s = 368090 ns = 3.608 x 10^5 ns

Test Yourself

How many milliliters are in 607.8 kL?

Answer

6.078 × 108 mL

When considering the significant figures of a final numerical answer in a conversion, there is one important case where a number does not impact the number of significant figures in a final answer—the so-called exact number. An exact number is a number from a defined relationship, not a measured one. For example, the prefix kilo- means 1,000—exactly 1,000, no more or no less. Thus, in constructing the conversion factor

1000 g/1 kg

neither the 1,000 nor the 1 enter into our consideration of significant figures. The numbers in the numerator and denominator are defined exactly by what the prefix kilo- means. Another way of thinking about it is that these numbers can be thought of as having an infinite number of significant figures, such as

1000.0000000....g/1.000000000... kg

The other numbers in the calculation will determine the number of significant figures in the final answer.

Example 11

A rectangular plot in a garden has the dimensions 36.7 cm by 128.8 cm. What is the area of the garden plot in square meters? Express your answer in the proper number of significant figures.

Solution

Area is defined as the product of the two dimensions, which we then have to convert to square meters and express our final answer to the correct number of significant figures, which in this case will be three.

36.7 cm x 128.8 cm x 1 m/100cm x 1 m/100 cm = 0.472696 m^2 = 0.473 m^2

The 1 and 100 in the conversion factors do not affect the determination of significant figures because they are exact numbers, defined by the centi- prefix.

Test Yourself

What is the volume of a block in cubic meters whose dimensions are 2.1 cm × 34.0 cm × 118 cm?

Answer

0.0084 m3

Chemistry Is Everywhere: The Gimli Glider

On July 23, 1983, an Air Canada Boeing 767 jet had to glide to an emergency landing at Gimli Industrial Park Airport in Gimli, Manitoba, because it unexpectedly ran out of fuel during flight. There was no loss of life in the course of the emergency landing, only some minor injuries associated in part with the evacuation of the craft after landing. For the remainder of its operational life (the plane was retired in 2008), the aircraft was nicknamed “the Gimli Glider.”

The 767 took off from Montreal on its way to Ottawa, ultimately heading for Edmonton, Canada. About halfway through the flight, all the engines on the plane began to shut down because of a lack of fuel. When the final engine cut off, all electricity (which was generated by the engines) was lost; the plane became, essentially, a powerless glider. Captain Robert Pearson was an experienced glider pilot, although he had never flown a glider the size of a 767. First Officer Maurice Quintal quickly determined that the aircraft would not be able make it to Winnipeg, the next large airport. He suggested his old Royal Air Force base at Gimli Station, one of whose runways was still being used as a community airport. Between the efforts of the pilots and the flight crew, they managed to get the airplane safely on the ground (although with buckled landing gear) and all passengers off safely.

What happened? At the time, Canada was transitioning from the older English system to the metric system. The Boeing 767s were the first aircraft whose gauges were calibrated in the metric system of units (liters and kilograms) rather than the English system of units (gallons and pounds). Thus, when the fuel gauge read 22,300, the gauge meant kilograms, but the ground crew mistakenly fueled the plane with 22,300 pounds of fuel. This ended up being just less than half of the fuel needed to make the trip, causing the engines to quit about halfway to Ottawa. Quick thinking and extraordinary skill saved the lives of 61 passengers and 8 crew members—an incident that would not have occurred if people were watching their units.

qrcode.23437561

Video source: Unit conversion by keyj (https://viuvideos.viu.ca/media/Unit+Conversion/0_o671v9j6)

Key Takeaways

  • Units can be converted to other units using the proper conversion factors.
  • Conversion factors are constructed from equalities that relate two different units.
  • Conversions can be a single step or multistep.
  • Unit conversion is a powerful mathematical technique in chemistry that must be mastered.
  • Exact numbers do not affect the determination of significant figures.

Exercises

  1. Write the two conversion factors that exist between the two given units.

a)  milliliters and liters

b)  microseconds and seconds

c)  kilometers and meters

2.  Write the two conversion factors that exist between the two given units.

a)  kilograms and grams

b)  milliseconds and seconds

c)  centimeters and meters

3.  Perform the following conversions.

a)  5.4 km to meters

b)  0.665 m to millimeters

c)  0.665 m to kilometers

4.  Perform the following conversions.

a)  90.6 mL to liters

b)  0.00066 ML to liters

c)  750 L to kiloliters

5.  Perform the following conversions.

a)  17.8 μg to grams

b)  7.22 × 102 kg to grams

c)  0.00118 g to nanograms

6.  Perform the following conversions.

a)  833 ns to seconds

b)  5.809 s to milliseconds

c)  2.77 × 106 s to megaseconds

7.  Perform the following conversions.

a)  9.44 m2 to square centimeters

b)  3.44 × 108 mm3 to cubic meters

8.  Perform the following conversions.

a)  0.00444 cm3 to cubic meters

b)  8.11 × 102 m2 to square nanometers

9.  Why would it be inappropriate to convert square centimeters to cubic meters?

10.  Why would it be inappropriate to convert from cubic meters to cubic seconds?

11.  Perform the following conversions.

a)  45.0 m/min to meters/second

b)  0.000444 m/s to micrometers/second

c)  60.0 km/h to kilometers/second

12.  Perform the following conversions.

a)  3.4 × 102 cm/s to centimeters/minute

b)  26.6 mm/s to millimeters/hour

c)  13.7 kg/L to kilograms/milliliters

13.  Perform the following conversions.

a)  0.674 kL to milliliters

b)  2.81 × 1012 mm to kilometers

c)  94.5 kg to milligrams

14.  Perform the following conversions.

a)  6.79 × 10−6 kg to micrograms

b)  1.22 mL to kiloliters

c)  9.508 × 10−9 ks to milliseconds

15.  Perform the following conversions.

a)  6.77 × 1014 ms to kiloseconds

b)  34,550,000 cm to kilometers

16.  Perform the following conversions.

a)  4.701 × 1015 mL to kiloliters

b)  8.022 × 10−11 ks to microseconds

17.  Perform the following conversions. Note that you will have to convert units in both the numerator and the denominator.

a)  88 ft/s to miles/hour (Hint: use 5,280 ft = 1 mi.)

b)  0.00667 km/h to meters/second

18.  Perform the following conversions. Note that you will have to convert units in both the numerator and the denominator.

a)  3.88 × 102 mm/s to kilometers/hour

b)  1.004 kg/L to grams/milliliter

19.  What is the area in square millimeters of a rectangle whose sides are 2.44 cm × 6.077 cm? Express the answer to the proper number of significant figures.

20.  What is the volume in cubic centimeters of a cube with sides of 0.774 m? Express the answer to the proper number of significant figures.

21.  The formula for the area of a triangle is 1/2 × base × height. What is the area of a triangle in square centimeters if its base is 1.007 m and its height is 0.665 m? Express the answer to the proper number of significant figures.

22.  The formula for the area of a triangle is 1/2 × base × height. What is the area of a triangle in square meters if its base is 166 mm and its height is 930.0 mm? Express the answer to the proper number of significant figures.

Answers

1.

a)  1,000 mL/1 L and 1 L/1,000 mL

b)  1,000,000 μs/1 s and 1 s/1,000,000 μs

c)  1,000 m/1 km and 1 km1,000 m

3.

a)  5,400 m

b)  665 mm

c)  6.65 × 10−4 km

5.

a)  1.78 × 10−5 g

b)  7.22 × 105 g

c) 1.18 × 106 ng

7.

a)  94,400 cm2

b)  0.344 m3

9.

One is a unit of area, and the other is a unit of volume.

11.

a)  0.75 m/s

b)  444 µm/s

c)  1.666 × 10−2 km/s

13.

a)  674,000 mL

b)  2.81 × 106 km

c)  9.45 × 107 mg

15.

a)  6.77 × 108 ks

b)  345.5 km

17.

a)  6.0 × 101 mi/h

b)  0.00185 m/s

19.

1.48 × 103 mm2

21.

3.35 × 103 cm2

8

Other Units: Temperature and Density

Learning Objectives

  1. Learn about the various temperature scales that are commonly used in chemistry.
  2. Define density and use it as a conversion factor.

There are other units in chemistry that are important, and we will cover others in the course of the entire book. One of the fundamental quantities in science is temperature. Temperature is a measure of the average amount of energy of motion, or kinetic energy, a system contains. Temperatures are expressed using scales that use units called degrees, and there are several temperature scales in use. In the United States, the commonly used temperature scale is the Fahrenheit scale (symbolized by °F and spoken as “degrees Fahrenheit”). On this scale, the freezing point of liquid water (the temperature at which liquid water turns to solid ice) is 32 °F, and the boiling point of water (the temperature at which liquid water turns to steam) is 212 °F.

Science also uses other scales to express temperature. The Celsius scale (symbolized by °C and spoken as “degrees Celsius”) is a temperature scale where 0 °C is the freezing point of water and 100 °C is the boiling point of water; the scale is divided into 100 divisions between these two landmarks and extended higher and lower. By comparing the Fahrenheit and Celsius scales, a conversion between the two scales can be determined:

oC = (oF-32) x 5/9

oF = (oC x 9/5) + 32

Using these formulas, we can convert from one temperature scale to another. The number 32 in the formulas is exact and does not count in significant figure determination.

Example 12

  1. What is 98.6 °F in degrees Celsius?
  2. What is 25.0 °C in degrees Fahrenheit?

Solution

  1. Using the first formula from above, we have

    °C = (98.6 – 32) × 5/9 = 66.6 × 5/9 = 37.0 °C

  2. Using the second formula from above, we have

    °F = (25.0 × 9/5) + 3/2 = 45.0 + 32 = 77.0 °F

Test Yourself

  1. Convert 0 °F to degrees Celsius.
  2. Convert 212 °C to degrees Fahrenheit.

Answers

  1. −17.8 °C
  2. 414 °F

The fundamental unit of temperature (another fundamental unit of science, bringing us to four) in SI is the kelvin (K). The Kelvin temperature scale (note that the name of the scale capitalizes the word Kelvin, but the unit itself is lowercase) uses degrees that are the same size as the Celsius degree, but the numerical scale is shifted up by 273.15 units. That is, the conversion between the Kelvin and Celsius scales is as follows:

K = °C + 273.15
°C = K − 273.15

For most purposes, it is acceptable to use 273 instead of 273.15. Note that the Kelvin scale does not use the word degrees; a temperature of 295 K is spoken of as “two hundred ninety-five kelvins” and not “two hundred ninety-five degrees Kelvin.”

The reason that the Kelvin scale is defined this way is because there exists a minimum possible temperature called absolute zero. The Kelvin temperature scale is set so that 0 K is absolute zero, and temperature is counted upward from there. Normal room temperature is about 295 K, as seen in the following example.

Example 13

If normal room temperature is 72.0 °F, what is room temperature in degrees Celsius and kelvins?

Solution

First, we use the formula to determine the temperature in degrees Celsius:

°C = (72.0 – 32) × 5/9 = 40.0 × 5/9 = 22.2 °C

Then we use the appropriate formula above to determine the temperature in the Kelvin scale:

K = 22.2 °C + 273.15 = 295.4 K

So, room temperature is about 295 K.

Test Yourself

What is 98.6 °F on the Kelvin scale?

Answer

310.2 K

Figure 2.9 “Fahrenheit, Celsius, and Kelvin Temperatures” compares the three temperature scales. Note that science uses the Celsius and Kelvin scales almost exclusively; virtually no practicing chemist expresses laboratory-measured temperatures with the Fahrenheit scale. (In fact, the United States is one of the few countries in the world that still uses the Fahrenheit scale on a daily basis. The other two countries are Liberia and Myanmar [formerly Burma].

Figure 2.9 Fahrenheit, Celsius, and Kelvin Temperatures

Temperatures

 

A comparison of the three temperature scales.

Density is a physical property that is defined as a substance’s mass divided by its volume:

density = mass/volume  or d = m/V

Density is usually a measured property of a substance, so its numerical value affects the significant figures in a calculation. Notice that density is defined in terms of two dissimilar units, mass and volume. That means that density overall has derived units, just like velocity. Common units for density include g/mL, g/cm3, g/L, kg/L, and even kg/m3. Densities for some common substances are listed in Table 2.2 “Densities of Some Common Substances”.

Table 2.2 Densities of Some Common Substances

Substance Density (g/mL or g/cm3)
water 1.0
gold 19.3
mercury 13.6
air 0.0012
cork 0.22–0.26
aluminum 2.7
iron 7.87

Because of how it is defined, density can act as a conversion factor for switching between units of mass and volume. For example, suppose you have a sample of aluminum that has a volume of 7.88 cm3. How can you determine what mass of aluminum you have without measuring it? You can use the volume to calculate it. If you multiply the given volume by the known density (from Table 2.2 “Densities of Some Common Substances”), the volume units will cancel and leave you with mass units, telling you the mass of the sample:

7.88 cm3 × 2.7 g/cm= 21 g of aluminum

where we have limited our answer to two significant figures.

Example 14

What is the mass of 44.6 mL of mercury?

Solution

Use the density from Table 2.2 “Densities of Some Common Substances” as a conversion factor to go from volume to mass:

44.6 mL × 13.6 g/mL = 607 g

The mass of the mercury is 607 g.

Test Yourself

What is the mass of 25.0 cm3 of iron?

Answer

197 g

Density can also be used as a conversion factor to convert mass to volume—but care must be taken. We have already demonstrated that the number that goes with density normally goes in the numerator when density is written as a fraction. Take the density of gold, for example:

d = 19.3 g/1 mL

Although this was not previously pointed out, it can be assumed that there is a 1 in the denominator:

That is, the density value tells us that we have 19.3 grams for every 1 milliliter of volume, and the 1 is an exact number. When we want to use density to convert from mass to volume, the numerator and denominator of density need to be switched—that is, we must take the reciprocal of the density. In so doing, we move not only the units but also the numbers:

1/d = 1mL/19.3g

This reciprocal density is still a useful conversion factor, but now the mass unit will cancel and the volume unit will be introduced. Thus, if we want to know the volume of 45.9 g of gold, we would set up the conversion as follows:

45.9 g x 1mL/19.3g = 2.38 mL

Note how the mass units cancel, leaving the volume unit, which is what we’re looking for.

Example 15

A cork stopper from a bottle of wine has a mass of 3.78 g. If the density of cork is 0.22 g/cm3, what is the volume of the cork?

Solution

To use density as a conversion factor, we need to take the reciprocal so that the mass unit of density is in the denominator. Taking the reciprocal, we find

1/d = 1cm^3/0.22g

We can use this expression as the conversion factor. So

3.78 g x 1 cm^3/0.22g = 17.2 cm^3

Test Yourself

What is the volume of 3.78 g of gold?

Answer

0.196 cm3

Care must be used with density as a conversion factor. Make sure the mass units are the same, or the volume units are the same, before using density to convert to a different unit. Often, the unit of the given quantity must be first converted to the appropriate unit before applying density as a conversion factor.

Food and Drink App: Cooking Temperatures

Because degrees Fahrenheit is the common temperature scale in the United States, kitchen appliances, such as ovens, are calibrated in that scale. A cool oven may be only 150°F, while a cake may be baked at 350°F and a chicken roasted at 400°F. The broil setting on many ovens is 500°F, which is typically the highest temperature setting on a household oven.

People who live at high altitudes, typically 2,000 ft above sea level or higher, are sometimes urged to use slightly different cooking instructions on some products, such as cakes and bread, because water boils at a lower temperature the higher in altitude you go, meaning that foods cook slower. For example, in Cleveland water typically boils at 212°F (100°C), but in Denver, the Mile-High City, water boils at about 200°F (93.3°C), which can significantly lengthen cooking times. Good cooks need to be aware of this.

At the other end is pressure cooking. A pressure cooker is a closed vessel that allows steam to build up additional pressure, which increases the temperature at which water boils. A good pressure cooker can get to temperatures as high as 252°F (122°C); at these temperatures, food cooks much faster than it normally would. Great care must be used with pressure cookers because of the high pressure and high temperature. (When a pressure cooker is used to sterilize medical instruments, it is called an autoclave.)

Other countries use the Celsius scale for everyday purposes. Therefore, oven dials in their kitchens are marked in degrees Celsius. It can be confusing for US cooks to use ovens abroad—a 425°F oven in the United States is equivalent to a 220°C oven in other countries. These days, many oven thermometers are marked with both temperature scales.

Key Takeaways

  • Chemistry uses the Celsius and Kelvin scales to express temperatures.
  • A temperature on the Kelvin scale is the Celsius temperature plus 273.15.
  • The minimum possible temperature is absolute zero and is assigned 0 K on the Kelvin scale.
  • Density relates a substance’s mass and volume.
  • Density can be used to calculate volume from a given mass or mass from a given volume.

Exercises

  1. Perform the following conversions.

a)  255°F to degrees Celsius

b)  −255°F to degrees Celsius

c)  50.0°C to degrees Fahrenheit

d)  −50.0°C to degrees Fahrenheit

2.  Perform the following conversions.

a)  1,065°C to degrees Fahrenheit

b)  −222°C to degrees Fahrenheit

c)  400.0°F to degrees Celsius

d)  200.0°F to degrees Celsius

 

3.  Perform the following conversions.

a)  100.0°C to kelvins

b)  −100.0°C to kelvins

c)  100 K to degrees Celsius

d)  300 K to degrees Celsius

 

4.  Perform the following conversions.

a)  1,000.0 K to degrees Celsius

b)  50.0 K to degrees Celsius

c)  37.0°C to kelvins

d)  −37.0°C to kelvins

 

5.  Convert 0 K to degrees Celsius. What is the significance of the temperature in degrees Celsius?

 

6.  Convert 0 K to degrees Fahrenheit. What is the significance of the temperature in degrees Fahrenheit?

 

7.  The hottest temperature ever recorded on the surface of the earth was 136°F in Libya in 1922. What is the temperature in degrees Celsius and in kelvins?

 

8.  The coldest temperature ever recorded on the surface of the earth was −128.6°F in Vostok, Antarctica, in 1983. What is the temperature in degrees Celsius and in kelvins?

 

9.  Give at least three possible units for density.

 

10.  What are the units when density is inverted? Give three examples.

 

11.  A sample of iron has a volume of 48.2 cm3. What is its mass?

 

12.  A sample of air has a volume of 1,015 mL. What is its mass?

 

13.  The volume of hydrogen used by the Hindenburg, the German airship that exploded in New Jersey in 1937, was 2.000 × 108 L. If hydrogen gas has a density of 0.0899 g/L, what mass of hydrogen was used by the airship?

 

14.  The volume of an Olympic-sized swimming pool is 2.50 × 109 cm3. If the pool is filled with alcohol (d = 0.789 g/cm3), what mass of alcohol is in the pool?

 

15.  A typical engagement ring has 0.77 cm3 of gold. What mass of gold is present?

 

16.  A typical mercury thermometer has 0.039 mL of mercury in it. What mass of mercury is in the thermometer?

 

17.  What is the volume of 100.0 g of lead if lead has a density of 11.34 g/cm3?

 

18.  What is the volume of 255.0 g of uranium if uranium has a density of 19.05 g/cm3?

 

19.  What is the volume in liters of 222 g of neon if neon has a density of 0.900 g/L?

 

20.  What is the volume in liters of 20.5 g of sulfur hexafluoride if sulfur hexafluoride has a density of 6.164 g/L?

 

21.  Which has the greater volume, 100.0 g of iron (d = 7.87 g/cm3) or 75.0 g of gold (d = 19.3 g/cm3)?

 

22.  Which has the greater volume, 100.0 g of hydrogen gas (d = 0.0000899 g/cm3) or 25.0 g of argon gas (d = 0.00178 g/cm3)?

Answers

1.

a)  124°C

b)  −159°C

c)  122°F

d)  −58°F

3.

a)  373 K

b)  173 K

c)  −173°C

d)  27°C

5.

−273°C. This is the lowest possible temperature in degrees Celsius.

7.

57.8°C; 331 K

9.

g/mL, g/L, and kg/L (answers will vary)

11.

379 g

13.

1.80 × 107 g

15.

15 g

17.

8.818 cm3

19.

247 L

21.

The 100.0 g of iron has the greater volume.

9

Expressing Units

Learning Objectives

  1. Learn the units that go with various quantities.
  2. Express units using their abbreviations.
  3. Make new units by combining numerical prefixes with units.

A number indicates “how much,” but the unit indicates “of what.” The “of what” is important when communicating a quantity. For example, if you were to ask a friend how close you are to Lake Erie and your friend says “six,” then your friend isn’t giving you complete information. Six what? Six miles? Six inches? Six city blocks? The actual distance to the lake depends on what units you use.

Chemistry, like most sciences, uses the International System of Units, or SI for short. (The letters SI stand for the French “le Système International d’unités.”) SI specifies certain units for various types of quantities, based on seven fundamental units for various quantities. We will use most of the fundamental units in chemistry. Initially, we will deal with three fundamental units. The meter (m) is the SI unit of length. It is a little longer than a yard (see Figure 2.3 “The Meter”). The SI unit of mass is the kilogram (kg), which is about 2.2 pounds (lb). The SI unit of time is the second (s).

Figure 2.3 The Meter

image

The SI standard unit of length, the meter, is a little longer than a yard.

To express a quantity, you need to combine a number with a unit. If you have a length that is 2.4 m, then you express that length as simply 2.4 m. A time of 15,000 s can be expressed as 1.5 × 104 s in scientific notation.

Sometimes, a given unit is not an appropriate size to easily express a quantity. For example, the width of a human hair is very small, and it doesn’t make much sense to express it in meters. SI also defines a series of numerical prefixes that refer to multiples or fractions of a fundamental unit to make a unit more conveniently sized for a specific quantity. Table 2.1 “Multiplicative Prefixes for SI Units” lists the prefixes, their abbreviations, and their multiplicative factors. Some of the prefixes, such as kilo-, mega-, and giga-, represent more than one of the fundamental unit, while other prefixes, such as centi-, milli-, and micro-, represent fractions of the original unit. Note, too, that once again we are using powers of 10. Each prefix is a multiple of or fraction of a power of 10.

Table 2.1 Multiplicative Prefixes for SI Units

Prefix Abbreviation Multiplicative Amount
giga- G 1,000,000,000 ×
mega- M 1,000,000 ×
kilo- k 1,000 ×
deci- d 1/10 ×
centi- c 1/100 ×
milli- m 1/1,000 ×
micro- μ* 1/1,000,000 ×
nano- n 1/1,000,000,000 ×
pico- p 1/1,000,000,000,000 ×
* The letter μ is the Greek letter lowercase equivalent to an m and is called “mu” (pronounced “myoo”).

To use the fractions to generate new units, simply combine the prefix with the unit itself; the abbreviation for the new unit is the combination of the abbreviation for the prefix and the abbreviation of the unit. For example, the kilometer (km) is 1,000 × meter, or 1,000 m. Thus, 5 kilometers (5 km) is equal to 5,000 m. Similarly, a millisecond (ms) is 1/1,000 × second, or one-thousandth of a second. Thus, 25 ms is 25 thousandths of a second. You will need to become proficient in combining prefixes and units. (You may recognize that one of our fundamental units, the kilogram, automatically has a prefix-unit combination, the kilogram. The word kilogram means 1,000 g.)

In addition to the fundamental units, SI also allows for derived units based on a fundamental unit or units. There are many derived units used in science. For example, the derived unit for area comes from the idea that area is defined as width times height. Because both width and height are lengths, they both have the fundamental unit of meter, so the unit of area is meter × meter, or meter2 (m2). This is sometimes spoken as “square meters.” A unit with a prefix can also be used to derive a unit for area, so we can also have cm2, mm2, or km2 as acceptable units for area.

Volume is defined as length times width times height, so it has units of meter × meter × meter or meter3 (m3), sometimes spoken as “cubic meters.” The cubic meter is a rather large unit, however, so another unit is defined that is somewhat more manageable: the liter (L). A liter is 1/1,000th of a cubic meter and is a little more than 1 quart in volume (see Figure 2.4 “The Liter”). Prefixes can also be used with the liter unit, so we can speak of milliliters (1/1,000th of a liter; mL) and kiloliters (1,000 L; kL).

Figure 2.4 The Liter

image

The SI unit of volume, the liter, is slightly larger than 1 quart.

Another definition of a liter is one-tenth of a meter cubed. Because one-tenth of a meter is 10 cm, then a liter is equal to 1,000 cm3 (Figure 2.5 “The Size of 1 Liter”). Because 1 L equals 1,000 mL, we conclude that 1 mL equals 1 cm3; thus, these units are interchangeable.

Figure 2.5 The Size of 1 Liter

Size of a Liter

 

One liter equals 1,000 cm3, so 1 cm3 is the same as 1 mL.

Units not only are multiplied together but also can be divided. For example, if you are traveling at one meter for every second of time elapsed, your velocity is 1 meter per second, or 1 m/s. The word per implies division, so velocity is determined by dividing a distance quantity by a time quantity. Other units for velocity include kilometers per hour (km/h) or even micrometers per nanosecond (μm/ns). Later, we will see other derived units that can be expressed as fractions.

Example 2

  1. A human hair has a diameter of about 6.0 × 10−5 m. Suggest an appropriate unit for this measurement and write the diameter of a human hair in terms of that unit.
  2. What is the velocity of a car if it goes 25 m in 5.0 s?

Solution

  1. The scientific notation 10−5 is close to 10−6, which defines the micro- prefix. Let us use micrometers as the unit for hair diameter. The number 6.0 × 10−5 can be written as 60 × 10−6, and a micrometer is 10−6 m, so the diameter of a human hair is about 60 μm.
  2. If velocity is defined as a distance quantity divided by a time quantity, then velocity is 25 meters/5.0 seconds. Dividing the numbers gives us 25/5.0 = 5.0, and dividing the units gives us meters/second, or m/s. The velocity is 5.0 m/s.

Test Yourself

  1. Express the volume of an Olympic-sized swimming pool, 2,500,000 L, in more appropriate units.
  2. A common garden snail moves about 6.1 m in 30 min. What is its velocity in meters per minute (m/min)?

Answers

  1. 2.5 ML
  2. 0.203 m/min

Key Takeaways

  • Numbers tell “how much,” and units tell “of what.”
  • Chemistry uses a set of fundamental units and derived units from SI units.
  • Chemistry uses a set of prefixes that represent multiples or fractions of units.
  • Units can be multiplied and divided to generate new units for quantities.

 

Exercises

  1. Identify the unit in each quantity.

a)  2 boxes of crayons

b)  3.5 grams of gold

 

2.  Identify the unit in each quantity.

a)  32 oz of cheddar cheese

b)  0.045 cm3 of water

 

3.  Identify the unit in each quantity.

a)  9.58 s (the current world record in the 100 m dash)

b)  6.14 m (the current world record in the pole vault)

 

4.  Identify the unit in each quantity.

a)  2 dozen eggs

b)  2.4 km/s (the escape velocity of the moon, which is the velocity you need at the surface to escape the moon’s gravity)

 

5.  Indicate what multiplier each prefix represents.

a)  k

b)  m

c)  M

 

6.  Indicate what multiplier each prefix represents.

a)  c

b)  G

c)  μ

 

7.  Give the prefix that represents each multiplier.

a)  1/1,000th ×

b)  1,000 ×

c)  1,000,000,000 ×

 

8.  Give the prefix that represents each multiplier.

a)  1/1,000,000,000th ×

b)  1/100th ×

c)  1,000,000 ×

 

9. Complete the following table with the missing information.

 

Unit Abbreviation
kilosecond
mL
Mg
centimeter

 

10.Complete the following table with the missing information.

Unit Abbreviation
kilometer per second
second
cm3
μL
nanosecond

11.  Express each quantity in a more appropriate unit. There may be more than one acceptable answer.

a)  3.44 × 10−6 s

b)  3,500 L

c)  0.045 m

 

12.  Express each quantity in a more appropriate unit. There may be more than one acceptable answer.

a)  0.000066 m/s (Hint: you need consider only the unit in the numerator.)

b)  4.66 × 106 s

c)  7,654 L

 

13.  Express each quantity in a more appropriate unit. There may be more than one acceptable answer.

a)  43,600 mL

b)  0.0000044 m

c)  1,438 ms

 

14.  Express each quantity in a more appropriate unit. There may be more than one acceptable answer.

a)  0.000000345 m3

b)  47,000,000 mm3

c)  0.00665 L

 

15.  Multiplicative prefixes are used for other units as well, such as computer memory. The basic unit of computer memory is the byte (b). What is the unit for one million bytes?

 

16.  You may have heard the terms microscale or nanoscale to represent the sizes of small objects. What units of length do you think are useful at these scales? What fractions of the fundamental unit of length are these units?

 

17.  Acceleration is defined as a change in velocity per time. Propose a unit for acceleration in terms of the fundamental SI units.

 

18.  Density is defined as the mass of an object divided by its volume. Propose a unit of density in terms of the fundamental SI units.

Answers

1.

a)  boxes of crayons

b)  grams of gold

3.

a)  seconds

b)  meters

5.

a)  1,000 ×

b)  1/1,000 ×

c)  1,000,000 ×

7.

a)  milli-

b)  kilo-

c)  giga-

9.

Unit Abbreviation
kilosecond ks
milliliter mL
megagram Mg
centimeter cm

11.

a)  3.44 μs

b)  3.5 kL

c)  4.5 cm

13.

a)  43.6 L

b)  4.4 µm

c)  1.438 s

15. megabytes (Mb)

17. meters/second2

10

End-of-Chapter Material

Additional Exercises

  1. Evaluate 0.00000000552 × 0.0000000006188 and express the answer in scientific notation. You may have to rewrite the original numbers in scientific notation first.
  2. Evaluate 333,999,500,000 ÷ 0.00000000003396 and express the answer in scientific notation. You may need to rewrite the original numbers in scientific notation first.

  3. Express the number 6.022 × 1023 in standard notation.

  4. Express the number 6.626 × 10−34 in standard notation.

  5. When powers of 10 are multiplied together, the powers are added together. For example, 102 × 103 = 102+3 = 105. With this in mind, can you evaluate (4.506 × 104) × (1.003 × 102) without entering scientific notation into your calculator?

  6. When powers of 10 are divided into each other, the bottom exponent is subtracted from the top exponent. For example, 105/103 = 105−3 = 102. With this in mind, can you evaluate (8.552 × 106) ÷ (3.129 × 103) without entering scientific notation into your calculator?

  7. Consider the quantity two dozen eggs. Is the number in this quantity “two” or “two dozen”? Justify your choice.

  8. Consider the quantity two dozen eggs. Is the unit in this quantity “eggs” or “dozen eggs”? Justify your choice.

  9. Fill in the blank: 1 km = ______________ μm.

  10. Fill in the blank: 1 Ms = ______________ ns.

  11. Fill in the blank: 1 cL = ______________ ML.

  12. Fill in the blank: 1 mg = ______________ kg.

  13. Express 67.3 km/h in meters/second.

  14. Express 0.00444 m/s in kilometers/hour.

  15. Using the idea that 1.602 km = 1.000 mi, convert a speed of 60.0 mi/h into kilometers/hour.

  16. Using the idea that 1.602 km = 1.000 mi, convert a speed of 60.0 km/h into miles/hour.

  17. Convert 52.09 km/h into meters/second.

  18. Convert 2.155 m/s into kilometers/hour.

  19. Use the formulas for converting degrees Fahrenheit into degrees Celsius to determine the relative size of the Fahrenheit degree over the Celsius degree.

  20. Use the formulas for converting degrees Celsius into kelvins to determine the relative size of the Celsius degree over kelvins.

  21. What is the mass of 12.67 L of mercury?

  22. What is the mass of 0.663 m3 of air?

  23. What is the volume of 2.884 kg of gold?

  24. What is the volume of 40.99 kg of cork? Assume a density of 0.22 g/cm3.

 

Answers

1.

3.42 × 10−183.

602,200,000,000,000,000,000,000

5.

4.520 × 1067.

The quantity is two; dozen is the unit.

9.

1,000,000,000

11.

1/100,000,000

13.

18.7 m/s

15.

96.1 km/h

17.

14.47 m/s

19.

One Fahrenheit degree is nine-fifths the size of a Celsius degree.

21.

1.72 × 105 g

23.

149 mL

III

Chapter 3. Atoms, Molecules, and Ions

11

Introduction to Atoms, Molecules, and Ions

Although not an SI unit, the angstrom (Å) is a useful unit of length. It is one ten-billionth of a meter, or 10−10 m. Why is it a useful unit? The ultimate particles that compose all matter are about 10−10 m in size, or about 1 Å. This makes the angstrom a natural—though not approved—unit for describing these particles.

The angstrom unit is named after Anders Jonas Ångström, a nineteenth-century Swedish physicist. Ångström’s research dealt with light being emitted by glowing objects, including the sun. Ångström studied the brightness of the different colors of light that the sun emitted and was able to deduce that the sun is composed of the same kinds of matter that are present on the earth. By extension, we now know that all matter throughout the universe is similar to the matter that exists on our own planet.

The Sun

Anders Jonas Ångstrom, a Swedish physicist, studied the light coming from the sun. His contributions to science were sufficient to have a tiny unit of length named after him, the angstrom, which is one ten-billionth of a meter.

The basic building block of all matter is the atom. Curiously, the idea of atoms was first proposed in the fifth century BCE, when the Greek philosophers Leucippus and Democritus proposed their existence in a surprisingly modern fashion. However, their ideas never took hold among their contemporaries, and it wasn’t until the early 1800s that evidence amassed to make scientists reconsider the idea. Today, the concept of the atom is central to the study of matter.

12

Acids

Learning Objectives

  1. Define acid.
  2. Name a simple acid.

There is one other group of compounds that is important to us—acids—and these compounds have interesting chemical properties. Initially, we will define an acid as an ionic compound of the H+ cation dissolved in water. (We will expand on this definition in Chapter 12 “Acids and Bases”.) To indicate that something is dissolved in water, we will use the phase label (aq) next to a chemical formula (where aq stands for “aqueous,” a word that describes something dissolved in water). If the formula does not have this label, then the compound is treated as a molecular compound rather than an acid.

Acids have their own nomenclature system. If an acid is composed of only hydrogen and one other element, the name is hydro- + the stem of the other element + -ic acid. For example, the compound HCl(aq) is hydrochloric acid, while H2S(aq) is hydrosulfuric acid. (If these acids were not dissolved in water, the compounds would be called hydrogen chloride and hydrogen sulfide, respectively. Both of these substances are well known as molecular compounds; when dissolved in water, however, they are treated as acids.)

If a compound is composed of hydrogen ions and a polyatomic anion, then the name of the acid is derived from the stem of the polyatomic ion’s name. Typically, if the anion name ends in -ate, the name of the acid is the stem of the anion name plus -ic acid; if the related anion’s name ends in -ite, the name of the corresponding acid is the stem of the anion name plus -ous acid. Table 3.9 “Names and Formulas of Acids” lists the formulas and names of a variety of acids that you should be familiar with. You should recognize most of the anions in the formulas of the acids.

Table 3.9 Names and Formulas of Acids

Formula Name
HC2H3O2 acetic acid
HClO3 chloric acid
HCl hydrochloric acid
HBr hydrobromic acid
HI hydriodic acid
HF hydrofluoric acid
HNO3 nitric acid
H2C2O4 oxalic acid
HClO4 perchloric acid
H3PO4 phosphoric acid
H2SO4 sulfuric acid
H2SO3 sulfurous acid
Note: The “aq” label is omitted for clarity.

Example 10

Name each acid without consulting Table 3.9 “Names and Formulas of Acids”.

  1. HBr
  2. H2SO4

Solution

  1. As a binary acid, the acid’s name is hydro- + stem name + -ic acid. Because this acid contains a bromine atom, the name is hydrobromic acid.
  2. Because this acid is derived from the sulfate ion, the name of the acid is the stem of the anion name + -ic acid. The name of this acid is sulfuric acid.

Test Yourself

Name each acid.

  1. HF
  2. HNO2

Answers

  1. hydrofluoric acid
  2. nitrous acid

All acids have some similar properties. For example, acids have a sour taste; in fact, the sour taste of some of our foods, such as citrus fruits and vinegar, is caused by the presence of acids in food. Many acids react with some metallic elements to form metal ions and elemental hydrogen. Acids make certain plant pigments change colors; indeed, the ripening of some fruits and vegetables is caused by the formation or destruction of excess acid in the plant. In Chapter 12 “Acids and Bases”, we will explore the chemical behaviour of acids.

Acids are very prevalent in the world around us. We have already mentioned that citrus fruits contain acid; among other compounds, they contain citric acid, H3C6H5O7(aq). Oxalic acid, H2C2O4(aq), is found in spinach and other green leafy vegetables. Hydrochloric acid not only is found in the stomach (stomach acid) but also can be bought in hardware stores as a cleaner for concrete and masonry. Phosphoric acid is an ingredient in some soft drinks.

Key Takeaways

  • An acid is a compound of the H+ ion dissolved in water.
  • Acids have their own naming system.
  • Acids have certain chemical properties that distinguish them from other compounds.

 

Exercises

  1. Give the formula for each acid.

a)  perchloric acid

b)  hydriodic acid

 

2.  Give the formula for each acid.

a)  hydrosulfuric acid

b)  phosphorous acid

 

3.  Name each acid.

a)  HF(aq)

b)  HNO3(aq)

c)  H2C2O4(aq)

 

4.  Name each acid.

a)  H2SO4(aq)

b)  H3PO4(aq)

c)  HCl(aq)

 

5.  Name an acid found in food.

 

6  Name some properties that acids have in common.

Answers

1.

a)  HClO4(aq)

b)  HI(aq)

3.

a)  hydrofluoric acid

b)  nitric acid

c)  oxalic acid

5.

oxalic acid (answers will vary)

 

13

Ions and Ionic Compounds

Learning Objectives

  1. Know how ions form.
  2. Learn the characteristic charges that ions have.
  3. Construct a proper formula for an ionic compound.
  4. Generate a proper name for an ionic compound.

So far, we have discussed elements and compounds that are electrically neutral. They have the same number of electrons as protons, so the negative charges of the electrons is balanced by the positive charges of the protons. However, this is not always the case. Electrons can move from one atom to another; when they do, species with overall electric charges are formed. Such species are called ions. Species with overall positive charges are termed cations, while species with overall negative charges are called anions. Remember that ions are formed only when electrons move from one atom to another; a proton never moves from one atom to another. Compounds formed from positive and negative ions are called ionic compounds.

Individual atoms can gain or lose electrons. When they do, they become monatomic ions. When atoms gain or lose electrons, they usually gain or lose a characteristic number of electrons and so take on a characteristic overall charge. Table 3.6 “Monatomic Ions of Various Charges” lists some common ions in terms of how many electrons they lose (making cations) or gain (making anions). There are several things to notice about the ions in Table 3.6 “Monatomic Ions of Various Charges”. First, each element that forms cations is a metal, except for one (hydrogen), while each element that forms anions is a nonmetal. This is actually one of the chemical properties of metals and nonmetals: metals tend to form cations, while nonmetals tend to form anions. Second, most atoms form ions of a single characteristic charge. When sodium atoms form ions, they always form a 1+ charge, never a 2+ or 3+ or even 1− charge. Thus, if you commit the information in Table 3.6 “Monatomic Ions of Various Charges” to memory, you will always know what charges most atoms form. (In Chapter 9 “Chemical Bonds”, we will discuss why atoms form the charges they do.)

Table 3.6 Monatomic Ions of Various Charges

Ions formed by losing a single electron H+
Na+
K+
Rb+
Ag+
Au+
Ions formed by losing two electrons Mg2+
Ca2+
Sr2+
Fe2+
Co2+
Ni2+
Cu2+
Zn2+
Sn2+
Hg2+
Pb2+
Ions formed by losing three electrons Sc3+
Fe3+
Co3+
Ni3+
Au3+
Al3+
Cr3+
Ions formed by losing four electrons Ti4+
Sn4+
Pb4+
Ions formed by gaining a single electron F
Cl
Br
I
Ions formed by gaining two electrons O2−
S2−
Se2−
Ions formed by gaining three electrons N3−
P3−

Third, there are some exceptions to the previous point. A few elements, all metals, can form more than one possible charge. For example, iron atoms can form 2+ cations or 3+ cations. Cobalt is another element that can form more than one possible charged ion (2+ and 3+), while lead can form 2+ or 4+ cations. Unfortunately, there is little understanding which two charges a metal atom may take, so it is best to just memorize the possible charges a particular element can have.

Note the convention for indicating an ion. The magnitude of the charge is listed as a right superscript next to the symbol of the element. If the charge is a single positive or negative one, the number 1 is not written; if the magnitude of the charge is greater than 1, then the number is written before the + or − sign. An element symbol without a charge written next to it is assumed to be the uncharged atom.

Naming an ion is straightforward. For a cation, simply use the name of the element and add the word ion (or if you want to be more specific, add cation) after the element’s name. So Na+ is the sodium ion; Ca2+ is the calcium ion. If the element has more than one possible charge, the value of the charge comes after the element name and before the word ion. Thus, Fe2+ is the iron two ion, while Fe3+ is the iron three ion. In print, we use roman numerals in parentheses to represent the charge on the ion, so these two iron ions would be represented as the iron(II) cation and the iron(III) cation, respectively.

For a monatomic anion, use the stem of the element name and append the suffix -ide to it, and then add ion. This is similar to how we named molecular compounds. Thus, Cl is the chloride ion, and N3− is the nitride ion.

Example 6

Name each species.

  1. O2−
  2. Co
  3. Co2+

Solution

  1. This species has a 2− charge on it, so it is an anion. Anions are named using the stem of the element name with the suffix -ide added. This is the oxide anion.
  2. Because this species has no charge, it is an atom in its elemental form. This is cobalt.
  3. In this case, there is a 2+ charge on the atom, so it is a cation. We note from Table 3.6 “Monatomic Ions of Various Charges” that cobalt cations can have two possible charges, so the name of the ion must specify which charge the ion has. This is the cobalt(II) cation.

Test Yourself

Name each species.

  1. P3−
  2. Sr2+

Answers

  1. the phosphide anion
  2. the strontium cation

Chemical formulas for ionic compounds are called ionic formulas. A proper ionic formula has a cation and an anion in it; an ionic compound is never formed between two cations only or two anions only. The key to writing proper ionic formulas is simple: the total positive charge must balance the total negative charge. Because the charges on the ions are characteristic, sometimes we have to have more than one of a cation or an anion to balance the overall positive and negative charges. It is conventional to use the lowest ratio of ions that are needed to balance the charges.

For example, consider the ionic compound between Na+ and Cl. Each ion has a single charge, one positive and one negative, so we need only one ion of each to balance the overall charge. When writing the ionic formula, we follow two additional conventions: (1) write the formula for the cation first and the formula for the anion next, but (2) do not write the charges on the ions. Thus, for the compound between Na+ and Cl, we have the ionic formula NaCl (Figure 3.5 “NaCl = Table Salt”). The formula Na2Cl2 also has balanced charges, but the convention is to use the lowest ratio of ions, which would be one of each. (Remember from our conventions for writing formulas that we don’t write a 1 subscript if there is only one atom of a particular element present.) For the ionic compound between magnesium cations (Mg2+) and oxide anions (O2−), again we need only one of each ion to balance the charges. By convention, the formula is MgO.

Figure 3.5 NaCl = Table Salt

 

The ionic compound NaCl is very common. Source: “Kosher Salt” by stlbites.com is licensed under the Creative Commons Attribution-NoDerivs 2.0 Generic.
The ionic compound NaCl is very common.
Source: “Kosher Salt” by stlbites.com is licensed under the Creative Commons Attribution-NoDerivs 2.0 Generic.

For the ionic compound between Mg2+ ions and Cl ions, we now consider the fact that the charges have different magnitudes, 2+ on the magnesium ion and 1− on the chloride ion. To balance the charges with the lowest number of ions possible, we need to have two chloride ions to balance the charge on the one magnesium ion. Rather than write the formula MgClCl, we combine the two chloride ions and write it with a 2 subscript: MgCl2.

What is the formula MgCl2 telling us? There are two chloride ions in the formula. Although chlorine as an element is a diatomic molecule, Cl2, elemental chlorine is not part of this ionic compound. The chlorine is in the form of a negatively charged ion, not the neutral element. The 2 subscript is in the ionic formula because we need two Cl ions to balance the charge on one Mg2+ ion.

Example 7

Write the proper ionic formula for each of the two given ions.

  1. Ca2+ and Cl
  2. Al3+ and F
  3. Al3+ and O2−

Solution

  1. We need two Cl ions to balance the charge on one Ca2+ ion, so the proper ionic formula is CaCl2.
  2. We need three F ions to balance the charge on the Al3+ ion, so the proper ionic formula is AlF3.
  3. With Al3+ and O2−, note that neither charge is a perfect multiple of the other. This means we have to go to a least common multiple, which in this case will be six. To get a total of 6+, we need two Al3+ ions; to get 6−, we need three O2− ions. Hence the proper ionic formula is Al2O3.

Test Yourself

Write the proper ionic formulas for each of the two given ions.

  1. Fe2+ and S2−
  2. Fe3+ and S2−

Answers

  1. FeS
  2. Fe2S3

Naming ionic compounds is simple: combine the name of the cation and the name of the anion, in both cases omitting the word ion. Do not use numerical prefixes if there is more than one ion necessary to balance the charges. NaCl is sodium chloride, a combination of the name of the cation (sodium) and the anion (chloride). MgO is magnesium oxide. MgCl2 is magnesium chloride—not magnesium dichloride.

In naming ionic compounds whose cations can have more than one possible charge, we must also include the charge, in parentheses and in roman numerals, as part of the name. Hence FeS is iron(II) sulfide, while Fe2S3 is iron(III) sulfide. Again, no numerical prefixes appear in the name. The number of ions in the formula is dictated by the need to balance the positive and negative charges.

Example 8

Name each ionic compound.

  1. CaCl2
  2. AlF3
  3. Co2O3

Solution

  1. Using the names of the ions, this ionic compound is named calcium chloride. It is not calcium(II) chloride because calcium forms only one cation when it forms an ion, and it has a characteristic charge of 2+.
  2. The name of this ionic compound is aluminum fluoride.
  3. We know that cobalt can have more than one possible charge; we just need to determine what it is. Oxide always has a 2− charge, so with three oxide ions, we have a total negative charge of 6−. This means that the two cobalt ions have to contribute 6+, which for two cobalt ions means that each one is 3+. Therefore, the proper name for this ionic compound is cobalt(III) oxide.

Test Yourself

Name each ionic compound.

  1. Sc2O3
  2. AgCl

Answers

  1. scandium oxide
  2. silver chloride

How do you know whether a formula—and by extension, a name—is for a molecular compound or for an ionic compound? Molecular compounds form between nonmetals and nonmetals, while ionic compounds form between metals and nonmetals. The periodic table (Figure 3.2 “A Simple Periodic Table”) can be used to determine which elements are metals and nonmetals.

There also exists a group of ions that contain more than one atom. These are called polyatomic ions. Table 3.7 “Common Polyatomic Ions” lists the formulas, charges, and names of some common polyatomic ions. Only one of them, the ammonium ion, is a cation; the rest are anions. Most of them also contain oxygen atoms, so sometimes they are referred to as oxyanions. Some of them, such as nitrate and nitrite, and sulfate and sulfite, have very similar formulas and names, so care must be taken to get the formulas and names correct. Note that the -ite polyatomic ion has one less oxygen atom in its formula than the -ate ion but with the same ionic charge.

Table 3.7 Common Polyatomic Ions

Name Formula and Charge Name Formula and Charge
ammonium NH4+ hydroxide OH
acetate C2H3O2, or CH3COO nitrate NO3
bicarbonate (hydrogen carbonate) HCO3 nitrite NO2
bisulfate (hydrogen sulfate) HSO4 peroxide O22−
carbonate CO32− perchlorate ClO4
chlorate ClO3 phosphate PO43−
chromate CrO42− sulfate SO42−
cyanide CN sulfite SO32−
dichromate Cr2O72− triiodide I3

The naming of ionic compounds that contain polyatomic ions follows the same rules as the naming for other ionic compounds: simply combine the name of the cation and the name of the anion. Do not use numerical prefixes in the name if there is more than one polyatomic ion; the only exception to this is if the name of the ion itself contains a numerical prefix, such as dichromate or triiodide.

Writing the formulas of ionic compounds has one important difference. If more than one polyatomic ion is needed to balance the overall charge in the formula, enclose the formula of the polyatomic ion in parentheses and write the proper numerical subscript to the right and outside the parentheses. Thus, the formula between calcium ions, Ca2+, and nitrate ions, NO3, is properly written Ca(NO3)2, not CaNO32 or CaN2O6. Use parentheses where required. The name of this ionic compound is simply calcium nitrate.

Example 9

Write the proper formula and give the proper name for each ionic compound formed between the two listed ions.

  1. NH4+ and S2−
  2. Al3+ and PO43−
  3. Fe2+ and PO43−

Solution

  1. Because the ammonium ion has a 1+ charge and the sulfide ion has a 2− charge, we need two ammonium ions to balance the charge on a single sulfide ion. Enclosing the formula for the ammonium ion in parentheses, we have (NH4)2S. The compound’s name is ammonium sulfide.
  2. Because the ions have the same magnitude of charge, we need only one of each to balance the charges. The formula is AlPO4, and the name of the compound is aluminum phosphate.
  3. Neither charge is an exact multiple of the other, so we have to go to the least common multiple of 6. To get 6+, we need three iron(II) ions, and to get 6−, we need two phosphate ions. The proper formula is Fe3(PO4)2, and the compound’s name is iron(II) phosphate.

Test Yourself

Write the proper formula and give the proper name for each ionic compound formed between the two listed ions.

  1. NH4+ and PO43−
  2. Co3+ and NO2

Answers

  1. (NH4)3PO4, ammonium phosphate
  2. Co(NO2)3, cobalt(III) nitrite

Food and Drink App: Sodium in Your Food

The element sodium, at least in its ionic form as Na+, is a necessary nutrient for humans to live. In fact, the human body is approximately 0.15% sodium, with the average person having one-twentieth to one-tenth of a kilogram in their body at any given time, mostly in fluids outside cells and in other bodily fluids.

Sodium is also present in our diet. The common table salt we use on our foods is an ionic sodium compound. Many processed foods also contain significant amounts of sodium added to them as a variety of ionic compounds. Why are sodium compounds used so much? Usually sodium compounds are inexpensive, but, more importantly, most ionic sodium compounds dissolve easily. This allows processed food manufacturers to add sodium-containing substances to food mixtures and know that the compound will dissolve and distribute evenly throughout the food. Simple ionic compounds such as sodium nitrite (NaNO2) are added to cured meats, such as bacon and deli-style meats, while a compound called sodium benzoate is added to many packaged foods as a preservative. Table 3.8 “Some Sodium Compounds Added to Food” is a partial list of some sodium additives used in food. Some of them you may recognize after reading this chapter. Others you may not recognize, but they are all ionic sodium compounds with some negatively charged ion also present.

Table 3.8 Some Sodium Compounds Added to Food

Sodium Compound Use in Food
Sodium acetate preservative, acidity regulator
Sodium adipate food acid
Sodium alginate thickener, vegetable gum, stabilizer, gelling agent, emulsifier
Sodium aluminum phosphate acidity regulator, emulsifier
Sodium aluminosilicate anticaking agent
Sodium ascorbate antioxidant
Sodium benzoate preservative
Sodium bicarbonate mineral salt
Sodium bisulfite preservative, antioxidant
Sodium carbonate mineral salt
Sodium carboxymethylcellulose emulsifier
Sodium citrates food acid
Sodium dehydroacetate preservative
Sodium erythorbate antioxidant
Sodium erythorbin antioxidant
Sodium ethyl para-hydroxybenzoate preservative
Sodium ferrocyanide anticaking agent
Sodium formate preservative
Sodium fumarate food acid
Sodium gluconate stabilizer
Sodium hydrogen acetate preservative, acidity regulator
Sodium hydroxide mineral salt
Sodium lactate food acid
Sodium malate food acid
Sodium metabisulfite preservative, antioxidant, bleaching agent
Sodium methyl para-hydroxybenzoate preservative
Sodium nitrate preservative, color fixative
Sodium nitrite preservative, color fixative
Sodium orthophenyl phenol preservative
Sodium propionate preservative
Sodium propyl para-hydroxybenzoate preservative
Sodium sorbate preservative
Sodium stearoyl lactylate emulsifier
Sodium succinates acidity regulator, flavour enhancer
Sodium salts of fatty acids emulsifier, stabilizer, anticaking agent
Sodium sulfite mineral salt, preservative, antioxidant
Sodium sulfite preservative, antioxidant
Sodium tartrate food acid
Sodium tetraborate preservative

The use of so many sodium compounds in prepared and processed foods has alarmed some physicians and nutritionists. They argue that the average person consumes too much sodium from his or her diet. The average person needs only about 500 mg of sodium every day; most people consume more than this—up to 10 times as much. Some studies have implicated increased sodium intake with high blood pressure; newer studies suggest that the link is questionable. However, there has been a push to reduce the amount of sodium most people ingest every day: avoid processed and manufactured foods, read labels on packaged foods (which include an indication of the sodium content), don’t oversalt foods, and use other herbs and spices besides salt in cooking.

Nutrition Facts

 

Food labels include the amount of sodium per serving. This particular label shows that there are 75 mg of sodium in one serving of this particular food item.

Key Takeaways

  • Ions form when atoms lose or gain electrons.
  • Ionic compounds have positive ions and negative ions.
  • Ionic formulas balance the total positive and negative charges.
  • Ionic compounds have a simple system of naming.
  • Groups of atoms can have an overall charge and make ionic compounds.

 

Exercises

  1. Explain how cations form.

  2. Explain how anions form.

  3. Give the charge each atom takes when it forms an ion. If more than one charge is possible, list both.

a)  K

b)  O

c)  Co

 

4.  Give the charge each atom takes when it forms an ion. If more than one charge is possible, list both.

a)  Ca

b)  I

c)  Fe

 

5.  Give the charge each atom takes when it forms an ion. If more than one charge is possible, list both.

a)  Ag

b)  Au

c)  Br

 

6.  Give the charge each atom takes when it forms an ion. If more than one charge is possible, list both.

a)  S

b)  Na

c)  H

 

7.  Name the ions from Exercise 3.

 

8.  Name the ions from Exercise 4.

 

9  Name the ions from Exercise 5.

 

10.  Name the ions from Exercise 6.

 

11.  Give the formula and name for each ionic compound formed between the two listed ions.

a)  Mg2+ and Cl

b)  Fe2+ and O2−

c)  Fe3+ and O2−

 

12.  Give the formula and name for each ionic compound formed between the two listed ions.

a)  K+ and S2−

b)  Ag+ and Br

c)  Sr2+ and N3−

 

13.  Give the formula and name for each ionic compound formed between the two listed ions.

a)  Cu2+ and F

b)  Ca2+ and O2−

c)  K+ and P3−

 

14.  Give the formula and name for each ionic compound formed between the two listed ions.

a)  Na+ and N3−

b)  Co2+ and I

c)  Au3+ and S2−

 

15.  Give the formula and name for each ionic compound formed between the two listed ions.

a)  K+ and SO42−

b)  NH4+ and S2−

c)  NH4+ and PO43−

 

16.  Give the formula and name for each ionic compound formed between the two listed ions.

a)  Ca2+ and NO3

b)  Ca2+ and NO2

c)  Sc3+ and C2H3O2

 

17.  Give the formula and name for each ionic compound formed between the two listed ions.

a)  Pb4+ and SO42−

b)  Na+ and I3

c)  Li+ and Cr2O72−

 

18.  Give the formula and name for each ionic compound formed between the two listed ions.

a)  NH4+ and N3−

b)  Mg2+ and CO32−

c)  Al3+ and OH

 

19.  Give the formula and name for each ionic compound formed between the two listed ions.

a)  Ag+ and SO32−

b)  Na+ and HCO3

c)  Fe3+ and ClO3

 

20.  Give the formula and name for each ionic compound formed between the two listed ions.

a)  Rb+ and O22−

b)  Au3+ and HSO4

c)  Sr2+ and NO2

 

21.  What is the difference between SO3 and SO32−?

 

22.  What is the difference between NO2 and NO2?

Answers

1.

Cations form by losing electrons.

3.

a)  1+

b)  2−

c)  2+, 3+

5.

a)  1+

b)  1+, 3+

c)  1−

7.

a)  the potassium ion

b)  the oxide ion

c)  the cobalt(II) and cobalt(III) ions, respectively

9.

a)  the silver ion

b)  the gold(I) and gold(III) ions, respectively

c)  the bromide ion

11.

a)  magnesium chloride, MgCl2

b)  iron(II) oxide, FeO

c)  iron(III) oxide, Fe2O313.

a)  copper(II) fluoride, CuF2

b)  calcium oxide, CaO

c)  potassium phosphide, K3P

15.

a)  potassium sulfate, K2SO4

b)  ammonium sulfide, (NH4)2S

c)  ammonium phosphate, (NH4)3PO417.

a)  lead(IV) sulfate, Pb(SO4)2

b)  sodium triiodide, NaI3

c)  lithium dichromate, Li2Cr2O719.

a)  silver sulfite, Ag2SO3

b)  sodium hydrogen carbonate, NaHCO3

c)  iron(III) chlorate, Fe(ClO3)321.
SO3 is sulfur trioxide, while SO32− is the sulfite ion.

 

14

Masses of Atoms and Molecules

Learning Objective

1. Express the masses of atoms and molecules.

Because matter is defined as anything that has mass and takes up space, it should not be surprising to learn that atoms and molecules have mass.

Individual atoms and molecules, however, are very small, and the masses of individual atoms and molecules are also very small. For macroscopic objects, we use units such as grams and kilograms to state their masses, but these units are much too big to comfortably describe the masses of individual atoms and molecules. Another scale is needed.

The atomic mass unit (u; some texts use amu, but this older style is no longer accepted) is defined as one-twelfth of the mass of a carbon-12 atom, an isotope of carbon that has six protons and six neutrons in its nucleus. By this scale, the mass of a proton is 1.00728 u, the mass of a neutron is 1.00866 u, and the mass of an electron is 0.000549 u. There will not be much error if you estimate the mass of an atom by simply counting the total number of protons and neutrons in the nucleus (i.e., identify its mass number) and ignore the electrons. Thus, the mass of carbon-12 is about 12 u, the mass of oxygen-16 is about 16 u, and the mass of uranium-238 is about 238 u. More exact masses are found in scientific references—for example, the exact mass of uranium-238 is 238.050788 u, so you can see that we are not far off by using the whole-number value as the mass of the atom.

What is the mass of an element? This is somewhat more complicated because most elements exist as a mixture of isotopes, each of which has its own mass. Thus, although it is easy to speak of the mass of an atom, when talking about the mass of an element, we must take the isotopic mixture into account.

The atomic mass of an element is a weighted average of the masses of the isotopes that compose an element. What do we mean by a weighted average? Well, consider an element that consists of two isotopes, 50% with mass 10 u and 50% with mass 11 u. A weighted average is found by multiplying each mass by its fractional occurrence (in decimal form) and then adding all the products. The sum is the weighted average and serves as the formal atomic mass of the element. In this example, we have the following:

0.50 × 10 u = 5.0 u
0.50 × 11 u = 5.5 u
Sum = 10.5 u = the atomic mass of our element

Note that no atom in our hypothetical element has a mass of 10.5 u; rather, that is the average mass of the atoms, weighted by their percent occurrence.

This example is similar to a real element. Boron exists as about 20% boron-10 (five protons and five neutrons in the nuclei) and about 80% boron-11 (five protons and six neutrons in the nuclei). The atomic mass of boron is calculated similarly to what we did for our hypothetical example, but the percentages are different:

0.20 × 10 u = 2.0 u
0.80 × 11 u = 8.8 u
Sum = 10.8 u = the atomic mass of boron

Thus, we use 10.8 u for the atomic mass of boron.

Virtually all elements exist as mixtures of isotopes, so atomic masses may vary significantly from whole numbers. Table 3.5 “Selected Atomic Masses of Some Elements” lists the atomic masses of some elements; a more expansive table is in Chapter 17 “Appendix: Periodic Table of the Elements”. The atomic masses in Table 3.5 “Selected Atomic Masses of Some Elements” are listed to three decimal places where possible, but in most cases, only one or two decimal places are needed. Note that many of the atomic masses, especially the larger ones, are not very close to whole numbers. This is, in part, the effect of an increasing number of isotopes as the atoms increase in size. (The record number is 10 isotopes for tin.)

Table 3.5 Selected Atomic Masses of Some Elements

Element Name Atomic Mass (u) Element Name Atomic Mass (u)
Aluminum 26.981 Molybdenum 95.94
Argon 39.948 Neon 20.180
Arsenic 74.922 Nickel 58.693
Barium 137.327 Nitrogen 14.007
Beryllium 9.012 Oxygen 15.999
Bismuth 208.980 Palladium 106.42
Boron 10.811 Phosphorus 30.974
Bromine 79.904 Platinum 195.084
Calcium 40.078 Potassium 39.098
Carbon 12.011 Radium n/a
Chlorine 35.453 Radon n/a
Cobalt 58.933 Rubidium 85.468
Copper 63.546 Scandium 44.956
Fluorine 18.998 Selenium 78.96
Gallium 69.723 Silicon 28.086
Germanium 72.64 Silver 107.868
Gold 196.967 Sodium 22.990
Helium 4.003 Strontium 87.62
Hydrogen 1.008 Sulfur 32.065
Iodine 126.904 Tantalum 180.948
Iridium 192.217 Tin 118.710
Iron 55.845 Titanium 47.867
Krypton 83.798 Tungsten 183.84
Lead 207.2 Uranium 238.029
Lithium 6.941 Xenon 131.293
Magnesium 24.305 Zinc 65.409
Manganese 54.938 Zirconium 91.224
Mercury 200.59 Molybdenum 95.94
Note: Atomic mass is given to three decimal places, if known.

Now that we understand that atoms have mass, it is easy to extend the concept to the mass of molecules. The molecular mass is the sum of the masses of the atoms in a molecule. This may seem like a trivial extension of the concept, but it is important to count the number of each type of atom in the molecular formula. Also, although each atom in a molecule is a particular isotope, we use the weighted average, or atomic mass, for each atom in the molecule.

For example, if we were to determine the molecular mass of dinitrogen trioxide, N2O3, we would need to add the atomic mass of nitrogen two times with the atomic mass of oxygen three times:

2 N masses = 2 × 14.007 u = 28.014 u
3 O masses = 3 × 15.999 u = 47.997 u
Total = 76.011 u = the molecular mass of N2O3

We would not be far off if we limited our numbers to one or even two decimal places.

Example 5

What is the molecular mass of each substance?

  1. NBr3
  2. C2H6

Solution

  1. Add one atomic mass of nitrogen and three atomic masses of bromine:

    1 N mass = 14.007 u
    3 Br masses = 3 × 79.904 u = 239.712 u
    Total = 253.719 u = the molecular mass of NBr3
  2. Add two atomic masses of carbon and six atomic masses of hydrogen:

    2 C masses = 2 × 12.011 u = 24.022 u
    6 H masses = 6 × 1.008 u = 6.048 u
    Total = 30.070 u = the molecular mass of C2H6

    The compound C2H6 also has a common name—ethane.

Test Yourself

What is the molecular mass of each substance?

  1. SO2
  2. PF3

Answers

  1. 64.063 u
  2. 87.968 u

Chemistry Is Everywhere: Sulfur Hexafluoride

On March 20, 1995, the Japanese terrorist group Aum Shinrikyo (Sanskrit for “Supreme Truth”) released some sarin gas in the Tokyo subway system; twelve people were killed, and thousands were injured (part (a) in the accompanying figure). Sarin (molecular formula C4H10FPO2) is a nerve toxin that was first synthesized in 1938. It is regarded as one of the most deadly toxins known, estimated to be about 500 times more potent than cyanide. Scientists and engineers who study the spread of chemical weapons such as sarin (yes, there are such scientists) would like to have a less dangerous chemical, indeed one that is nontoxic, so they are not at risk themselves.

Sulfur hexafluoride is used as a model compound for sarin. SF6 (a molecular model of which is shown in part (b) in the accompanying figure) has a similar molecular mass (about 146 u) as sarin (about 140 u), so it has similar physical properties in the vapour phase. Sulfur hexafluoride is also very easy to accurately detect, even at low levels, and it is not a normal part of the atmosphere, so there is little potential for contamination from natural sources. Consequently, SF6 is also used as an aerial tracer for ventilation systems in buildings. It is nontoxic and very chemically inert, so workers do not have to take special precautions other than watching for asphyxiation.

Sulfur hexafluoride also has another interesting use: a spark suppressant in high-voltage electrical equipment. High-pressure SF6 gas is used in place of older oils that may have contaminants that are environmentally unfriendly (part (c) in the accompanying figure).

Key Takeaways

  • The atomic mass unit (u) is a unit that describes the masses of individual atoms and molecules.
  • The atomic mass is the weighted average of the masses of all isotopes of an element.
  • The molecular mass is the sum of the masses of the atoms in a molecule.

 

Exercises

Define atomic mass unit. What is its abbreviation?

Define atomic mass. What is its unit?

Estimate the mass, in whole numbers, of each isotope.

a)  hydrogen-1

b)  hydrogen-3

c)  iron-56

 

4.  Estimate the mass, in whole numbers, of each isotope.

a)  phosphorus-31

b)  carbon-14

c)  americium-241

 

5. Determine the atomic mass of each element, given the isotopic composition.

a)  lithium, which is 92.4% lithium-7 (mass 7.016 u) and 7.60% lithium-6 (mass 6.015 u)

b)  oxygen, which is 99.76% oxygen-16 (mass 15.995 u), 0.038% oxygen-17 (mass 16.999 u), and 0.205% oxygen-18 (mass 17.999 u)

 

6.  Determine the atomic mass of each element, given the isotopic composition.

a)  neon, which is 90.48% neon-20 (mass 19.992 u), 0.27% neon-21 (mass 20.994 u), and 9.25% neon-22 (mass 21.991 u)

b) uranium, which is 99.27% uranium-238 (mass 238.051 u) and 0.720% uranium-235 (mass 235.044 u)

 

7.  How far off would your answer be from Exercise 5a if you used whole-number masses for individual isotopes of lithium?

 

8.  How far off would your answer be from Exercise 6b if you used whole-number masses for individual isotopes of uranium?

 

9.  What is the atomic mass of an oxygen atom?

10.  What is the molecular mass of oxygen in its elemental form?

 

11.  What is the atomic mass of bromine?

12.  What is the molecular mass of bromine in its elemental form?

 

13.  Determine the mass of each substance.

a)  F2

b)  CO

c)  CO2

 

14.  Determine the mass of each substance.

a)  Kr

b)  KrF4

c)  PF5

 

15.  Determine the mass of each substance.

a)  Na

b)  B2O3

c)  S2Cl2

 

16.  Determine the mass of each substance.

a)  IBr3

b)  N2O5

c)  CCl4

 

17.  Determine the mass of each substance.

a)  GeO2

b)  IF3

c)  XeF6

 

18.  Determine the mass of each substance.

a)  NO

b)  N2O4

c)  Ca

Answers

1.

The atomic mass unit is defined as one-twelfth of the mass of a carbon-12 atom. Its abbreviation is u.

3.

a)  1

b)  3

c)  56

5.

a)  6.940 u

b)  16.000 u

7.

We would get 6.924 u.

9.

a)  15.999 u

b)  31.998 u

11.

a)  37.996 u

b)  28.010 u

c)  44.009 u

13.

a)  22.990 u

b)  69.619 u

c)  135.036 u

15.

a)  104.64 u

b)  183.898 u

c)  245.281 u

 

15

Molecules and Chemical Nomenclature

Learning Objectives

  1. Define molecule.
  2. Name simple molecules based on their formulas.
  3. Determine a formula of a molecule based on its name.

There are many substances that exist as two or more atoms connected together so strongly that they behave as a single particle. These multiatom combinations are called molecules. A molecule is the smallest part of a substance that has the physical and chemical properties of that substance. In some respects, a molecule is similar to an atom. A molecule, however, is composed of more than one atom.

Some elements exist naturally as molecules. For example, hydrogen and oxygen exist as two-atom molecules. Other elements also exist naturally as diatomic molecules (see Table 3.3 “Elements That Exist as Diatomic Molecules”). As with any molecule, these elements are labelled with a molecular formula, a formal listing of what and how many atoms are in a molecule. (Sometimes only the word formula is used, and its meaning is inferred from the context.) For example, the molecular formula for elemental hydrogen is H2, with H being the symbol for hydrogen and the subscript 2 implying that there are two atoms of this element in the molecule. Other diatomic elements have similar formulas: O2, N2, and so forth. Other elements exist as molecules—for example, sulfur normally exists as an eight-atom molecule, S8, while phosphorus exists as a four-atom molecule, P4 (see Figure 3.3 “Molecular Art of S”). Otherwise, we will assume that elements exist as individual atoms, rather than molecules. It is assumed that there is only one atom in a formula if there is no numerical subscript on the right side of an element’s symbol.

Table 3.3 Elements That Exist as Diatomic Molecules

Hydrogen
Oxygen
Nitrogen
Fluorine
Chlorine
Bromine
Iodine

Figure 3.3 Molecular Art of S8 and P4 Molecules

Molecular Art

 

If each green ball represents a sulfur atom, then the diagram on the left represents an S8 molecule. The molecule on the right shows that one form of elemental phosphorus exists, as a four-atom molecule.

Figure 3.3 “Molecular Art of S” shows two examples of how we will be representing molecules in this text. An atom is represented by a small ball or sphere, which generally indicates where the nucleus is in the molecule. A cylindrical line connecting the balls represents the connection between the atoms that make this collection of atoms a molecule. This connection is called a chemical bond. In Chapter 9 “Chemical Bonds”, we will explore the origin of chemical bonds. You will see other examples of this “ball and cylinder” representation of molecules throughout this book.

Many compounds exist as molecules. In particular, when nonmetals connect with other nonmetals, the compound typically exists as molecules. (Compounds between a metal and a nonmetal are different and will be considered in Section 3.4 “Ions and Ionic Compounds”.) Furthermore, in some cases there are many different kinds of molecules that can be formed between any given elements, with all the different molecules having different chemical and physical properties. How do we tell them apart?

The answer is a very specific system of naming compounds, called chemical nomenclature. By following the rules of nomenclature, each and every compound has its own unique name, and each name refers to one and only one compound. Here, we will start with relatively simple molecules that have only two elements in them, the so-called binary compounds:

  1. Identify the elements in the molecule from its formula. This is why you need to know the names and symbols of the elements in Table 3.2 “Names and Symbols of Common Elements”.
  2. Begin the name with the element name of the first element. If there is more than one atom of this element in the molecular formula, use a numerical prefix to indicate the number of atoms, as listed in Table 3.4 “Numerical Prefixes Used in Naming Molecular Compounds”. Do not use the prefix mono- if there is only one atom of the first element.

    Table 3.4 Numerical Prefixes Used in Naming Molecular Compounds

    The Number of Atoms of an Element Prefix
    1 mono-
    2 di-
    3 tri-
    4 tetra-
    5 penta-
    6 hexa-
    7 hepta-
    8 octa-
    9 nona-
    10 deca-
  3. Name the second element by using three pieces:

    1. a numerical prefix indicating the number of atoms of the second element, plus
    2. the stem of the element name (e.g., ox for oxygen, chlor for chlorine, etc.), plus
    3. the suffix -ide.
  4. Combine the two words, leaving a space between them.

Let us see how these steps work for a molecule whose molecular formula is SO2, which has one sulfur atom and two oxygen atoms—this completes step 1. According to step 2, we start with the name of the first element—sulfur. Remember, we don’t use the mono- prefix for the first element. Now for step 3, we combine the numerical prefix di- (see Table 3.4 “Numerical Prefixes Used in Naming Molecular Compounds”) with the stem ox- and the suffix -ide, to make dioxide. Bringing these two words together, we have the unique name for this compound—sulfur dioxide.

Why all this trouble? There is another common compound consisting of sulfur and oxygen whose molecular formula is SO3, so the compounds need to be distinguished. SO3 has three oxygen atoms in it, so it is a different compound with different chemical and physical properties. The system of chemical nomenclature is designed to give this compound its own unique name. Its name, if you go through all the steps, is sulfur trioxide. Different compounds have different names.

In some cases, when a prefix ends in a or o and the element name begins with o we drop the a or o on the prefix. So we see monoxide or pentoxide rather than monooxide or pentaoxide in molecule names.

One great thing about this system is that it works both ways. From the name of a compound, you should be able to determine its molecular formula. Simply list the element symbols, with a numerical subscript if there is more than one atom of that element, in the order of the name (we do not use a subscript 1 if there is only one atom of the element present; 1 is implied). From the name nitrogen trichloride, you should be able to get NCl3 as the formula for this molecule. From the name diphosphorus pentoxide, you should be able to get the formula P2O5 (note the numerical prefix on the first element, indicating there is more than one atom of phosphorus in the formula).

Example 3

Name each molecule.

  1. PF3
  2. CO
  3. Se2Br2

Solution

  1. A molecule with a single phosphorus atom and three fluorine atoms is called phosphorus trifluoride.
  2. A compound with one carbon atom and one oxygen atom is properly called carbon monoxide, not carbon monooxide.
  3. There are two atoms of each element, selenium and bromine. According to the rules, the proper name here is diselenium dibromide.

Test Yourself

Name each molecule.

  1. SF4
  2. P2S5

Answers

  1. sulfur tetrafluoride
  2. diphosphorus pentasulfide

Example 4

Give the formula for each molecule.

  1. carbon tetrachloride
  2. silicon dioxide
  3. trisilicon tetranitride

Solution

  1. The name carbon tetrachloride implies one carbon atom and four chlorine atoms, so the formula is CCl4.
  2. The name silicon dioxide implies one silicon atom and two oxygen atoms, so the formula is SiO2.
  3. We have a name that has numerical prefixes on both elements. Tri- means three, and tetra- means four, so the formula of this compound is Si3N4.

Test Yourself

Give the formula for each molecule.

  1. disulfur difluoride
  2. iodine pentabromide

Answers

  1. S2F2
  2. IBr5

Some simple molecules have common names that we use as part of the formal system of chemical nomenclature. For example, H2O is given the name water, not dihydrogen monoxide. NH3 is called ammonia, while CH4 is called methane. We will occasionally see other molecules that have common names; we will point them out as they occur.

Key Takeaways

  • Molecules are groups of atoms that behave as a single unit.
  • Some elements exist as molecules: hydrogen, oxygen, sulfur, and so forth.
  • There are rules that can express a unique name for any given molecule, and a unique formula for any given name.

 

Exercises

  1. Which of these formulas represent molecules? State how many atoms are in each molecule.

a)  Fe

b)  PCl3

c)  P4

d)  Ar

 

2.  Which of these formulas represent molecules? State how many atoms are in each molecule.

a)  I2

b)  He

c)  H2O

d)  Al

 

3.  What is the difference between CO and Co?

 

4.  What is the difference between H2O and H2O2 (hydrogen peroxide)?

 

5.  Give the proper formula for each diatomic element.

 

6.  In 1986, when Halley’s comet last passed the earth, astronomers detected the presence of S2 in their telescopes.  Why is sulfur not considered a diatomic element?

 

7.  What is the stem of fluorine used in molecule names? CF4 is one example.

 

8.  What is the stem of selenium used in molecule names? SiSe2 is an example.

 

9.  Give the proper name for each molecule.

a)  PF3

b)  TeCl2

c)  N2O3

 

10.  Give the proper name for each molecule.

a)  NO

b)  CS2

c)  As2O3

 

11.  Give the proper name for each molecule.

a)  XeF2

b)  O2F2

c)  SF6

 

12.  Give the proper name for each molecule.

a)  P4O10

b)  B2O3

c)  P2S3

 

13.  Give the proper name for each molecule.

a)  N2O

b)  N2O4

c)  N2O5

 

14.  Give the proper name for each molecule.

a)  SeO2

b)  Cl2O

c)  XeF6

 

15.  Give the proper formula for each name.

a)  dinitrogen pentoxide

b)  tetraboron tricarbide

c)  phosphorus pentachloride

 

16.  Give the proper formula for each name.

a)  nitrogen triiodide

b)  diarsenic trisulfide

c)  iodine trichloride

 

17.  Give the proper formula for each name.

a)  dioxygen dichloride

b)  dinitrogen trisulfide

c)  xenon tetrafluoride

 

18.  Give the proper formula for each name.

a)  chlorine dioxide

b)  selenium dibromide

c)  dinitrogen trioxide

 

19.  Give the proper formula for each name.

a)  iodine trifluoride

b)  xenon trioxide

c)  disulfur decafluoride

 

20.  Give the proper formula for each name.

a)  germanium dioxide

b)  carbon disulfide

c)  diselenium dibromide

Answers

1.

a)  not a molecule

b)  a molecule; four atoms total

c)  a molecule; four atoms total

d)  not a molecule

3.

CO is a compound of carbon and oxygen; Co is the element cobalt.

5.

H2, O2, N2, F2, Cl2, Br2, I27.

fluor-

9.

a)  phosphorus trifluoride

b)  tellurium dichloride

c)  dinitrogen trioxide

11.

a)  xenon difluoride

b)  dioxygen difluoride

c)  sulfur hexafluoride

13.

a)  dinitrogen monoxide

b)  dinitrogen tetroxide

c)  dinitrogen pentoxide

15.

a)  N2O5

b)  B4C3

c)  PCl517.

a)  O2Cl2

b)  N2S3

c)  XeF419.

a)  IF3

b)  XeO3

c)  S2F10

16

Atomic Theory

Learning Objectives

  1. State the modern atomic theory.
  2. Learn how atoms are constructed.

The smallest piece of an element that maintains the identity of that element is called an atom. Individual atoms are extremely small. It would take about fifty million atoms in a row to make a line that is 1 cm long. The period at the end of a printed sentence has several million atoms in it. Atoms are so small that it is difficult to believe that all matter is made from atoms—but it is.

The concept that atoms play a fundamental role in chemistry is formalized by the modern atomic theory, first stated by John Dalton, an English scientist, in 1808. It consists of three parts:

  1. All matter is composed of atoms.
  2. Atoms of the same element are the same; atoms of different elements are different.
  3. Atoms combine in whole-number ratios to form compounds.

These concepts form the basis of chemistry.

Although the word atom comes from a Greek word that means “indivisible,” we understand now that atoms themselves are composed of smaller parts called subatomic particles. The first part to be discovered was the electron, a tiny subatomic particle with a negative charge. It is often represented as e, with the right superscript showing the negative charge. Later, two larger particles were discovered. The proton is a more massive (but still tiny) subatomic particle with a positive charge, represented as p+. The neutron is a subatomic particle with about the same mass as a proton but no charge. It is represented as either n or n0. We now know that all atoms of all elements are composed of electrons, protons, and (with one exception) neutrons. Table 3.1 “Properties of the Three Subatomic Particles” summarizes the properties of these three subatomic particles.

Table 3.1 Properties of the Three Subatomic Particles

Name Symbol Mass (approx.; kg) Charge
Proton p+ 1.6 × 10−27 1+
Neutron n, n0 1.6 × 10−27 none
Electron e 9.1 × 10−31 1−

How are these particles arranged in atoms? They are not arranged at random. Experiments by Ernest Rutherford in England in the 1910s pointed to a nuclear model of the atom. The relatively massive protons and neutrons are collected in the center of an atom, in a region called the nucleus of the atom (plural nuclei). The electrons are outside the nucleus and spend their time orbiting in space about the nucleus. (See Figure 3.1 “The Structure of the Atom”.)

Figure 3.1 The Structure of the Atom

The Atom

 

Atoms have protons and neutrons in the center, making the nucleus, while the electrons orbit the nucleus.

The modern atomic theory states that atoms of one element are the same, while atoms of different elements are different. What makes atoms of different elements different? The fundamental characteristic that all atoms of the same element share is the number of protons. All atoms of hydrogen have one and only one proton in the nucleus; all atoms of iron have 26 protons in the nucleus. This number of protons is so important to the identity of an atom that it is called the atomic number of the element. Thus, hydrogen has an atomic number of 1, while iron has an atomic number of 26. Each element has its own characteristic atomic number.

Atoms of the same element can have different numbers of neutrons, however. Atoms of the same element (i.e., atoms with the same number of protons) with different numbers of neutrons are called isotopes. Most naturally occurring elements exist as isotopes. For example, most hydrogen atoms have a single proton in their nucleus. However, a small number (about one in a million) of hydrogen atoms have a proton and a neutron in their nuclei. This particular isotope of hydrogen is called deuterium. A very rare form of hydrogen has one proton and two neutrons in the nucleus; this isotope of hydrogen is called tritium. The sum of the number of protons and neutrons in the nucleus is called the mass number of the isotope.

Neutral atoms have the same number of electrons as they have protons, so their overall charge is zero. However, as we shall see later, this will not always be the case.

Example 1

  1. The most common carbon atoms have six protons and six neutrons in their nuclei. What are the atomic number and the mass number of these carbon atoms?
  2. An isotope of uranium has an atomic number of 92 and a mass number of 235. What are the number of protons and neutrons in the nucleus of this atom?

Solution

  1. If a carbon atom has six protons in its nucleus, its atomic number is 6. If it also has six neutrons in the nucleus, then the mass number is 6 + 6, or 12.
  2. If the atomic number of uranium is 92, then that is the number of protons in the nucleus. Because the mass number is 235, then the number of neutrons in the nucleus is 235 − 92, or 143.

Test Yourself

The number of protons in the nucleus of a tin atom is 50, while the number of neutrons in the nucleus is 68. What are the atomic number and the mass number of this isotope?

Answer

Atomic number = 50, mass number = 118

When referring to an atom, we simply use the element’s name: the term sodium refers to the element as well as an atom of sodium. But it can be unwieldy to use the name of elements all the time. Instead, chemistry defines a symbol for each element. The atomic symbol is a one- or two-letter abbreviation of the name of the element. By convention, the first letter of an element’s symbol is always capitalized, while the second letter (if present) is lowercase. Thus, the symbol for hydrogen is H, the symbol for sodium is Na, and the symbol for nickel is Ni. Most symbols come from the English name of the element, although some symbols come from an element’s Latin name. (The symbol for sodium, Na, comes from its Latin name, natrium.) Table 3.2 “Names and Symbols of Common Elements” lists some common elements and their symbols. You should memorize the symbols in Table 3.2 “Names and Symbols of Common Elements”, as this is how we will be representing elements throughout chemistry.

Table 3.2 Names and Symbols of Common Elements

Element Name Symbol Element Name Symbol
Aluminum Al Mercury Hg
Argon Ar Molybdenum Mo
Arsenic As Neon Ne
Barium Ba Nickel Ni
Beryllium Be Nitrogen N
Bismuth Bi Oxygen O
Boron B Palladium Pd
Bromine Br Phosphorus P
Calcium Ca Platinum Pt
Carbon C Potassium K
Chlorine Cl Radium Ra
Chromium Cr Radon Rn
Cobalt Co Rubidium Rb
Copper Cu Scandium Sc
Fluorine F Selenium Se
Gallium Ga Silicon Si
Germanium Ge Silver Ag
Gold Au Sodium Na
Helium He Strontium Sr
Hydrogen H Sulfur S
Iodine I Tantalum Ta
Iridium Ir Tin Sn
Iron Fe Titanium Ti
Krypton Kr Tungsten W
Lead Pb Uranium U
Lithium Li Xenon Xe
Magnesium Mg Zinc Zn
Manganese Mn Zirconium Zr

The elements are grouped together in a special chart called the periodic table. A simple periodic table is shown in Figure 3.2 “A Simple Periodic Table”, while a more extensive one is presented in Chapter 17 “Appendix: Periodic Table of the Elements”. The elements on the periodic table are listed in order of ascending atomic number. The periodic table has a special shape that will become important to us when we consider the organization of electrons in atoms (see Chapter 8 “Electronic Structure”). One immediate use of the periodic table helps us identify metals and nonmetals. Nonmetals are in the upper right corner of the periodic table, on one side of the heavy line splitting the right-hand part of the chart. All other elements are metals.

Figure 3.2 A Simple Periodic Table

Periodic Table

 

There is an easy way to represent isotopes using the atomic symbols. We use the construction

Isotopic generic symbol for atom X

where X is the symbol of the element, A is the mass number, and Z is the atomic number. Thus, for the isotope of carbon that has 6 protons and 6 neutrons, the symbol is

carbon 12 isotopic symbol

where C is the symbol for the element, 6 represents the atomic number, and 12 represents the mass number.

Example 2

  1. What is the symbol for an isotope of uranium that has an atomic number of 92 and a mass number of 235?
  2. How many protons and neutrons are in 26F?

Solution

  1. The symbol for this isotope is 92U.
  2. This iron atom has 26 protons and 56 − 26 = 30 neutrons.

Test Yourself

How many protons are in 11N?

Answer

11 protons

It is also common to state the mass number after the name of an element to indicate a particular isotope. Carbon-12 represents an isotope of carbon with 6 protons and 6 neutrons, while uranium-238 is an isotope of uranium that has 146 neutrons.

Key Takeaways

  • Chemistry is based on the modern atomic theory, which states that all matter is composed of atoms.
  • Atoms themselves are composed of protons, neutrons, and electrons.
  • Each element has its own atomic number, which is equal to the number of protons in its nucleus.
  • Isotopes of an element contain different numbers of neutrons.
  • Elements are represented by an atomic symbol.
  • The periodic table is a chart that organizes all the elements.

 

 

Exercises

 

List the three statements that make up the modern atomic theory.

 

Explain how atoms are composed.

 

Which is larger, a proton or an electron?

 

Which is larger, a neutron or an electron?

 

What are the charges for each of the three subatomic particles?

 

Where is most of the mass of an atom located?

 

Sketch a diagram of a boron atom, which has five protons and six neutrons in its nucleus.

 

Sketch a diagram of a helium atom, which has two protons and two neutrons in its nucleus.

 

Define atomic number. What is the atomic number for a boron atom?

 

What is the atomic number of helium?

 

Define isotope and give an example.

 

What is the difference between deuterium and tritium?

 

Which pair represents isotopes?

a)  13_a_question

b)  26F and 25M

c)  14S and 15P

 

14.  Which pair represents isotopes?

a)  20C and 19K

b)  26F and 26F

c)  92U and 92U

15.  Give complete symbols of each atom, including the atomic number and the mass number.

a)  an oxygen atom with 8 protons and 8 neutrons

b)  a potassium atom with 19 protons and 20 neutrons

c)  a lithium atom with 3 protons and 4 neutrons

 

16.  Give complete symbols of each atom, including the atomic number and the mass number.

a)  a magnesium atom with 12 protons and 12 neutrons

b)  a magnesium atom with 12 protons and 13 neutrons

c)  a xenon atom with 54 protons and 77 neutrons

 

17.    Americium-241 is an isotope used in smoke detectors. What is the complete symbol for this isotope?

 

18.    Carbon-14 is an isotope used to perform radioactive dating tests on previously living material. What is the complete symbol for this isotope?

 

19.  Give atomic symbols for each element.

a)  sodium

b)  argon

c)  nitrogen

d)  radon

 

20.  Give atomic symbols for each element.

a)  silver

b)  gold

c)  mercury

d)  iodine

 

21.  Give the name of the element.

a)  Si

b)  Mn

c)  Fe

d)  Cr

 

22.  Give the name of the element.

a)  F

b)  Cl

c)  Br

d)  I

Answers

 

1.

All matter is composed of atoms; atoms of the same element are the same, and atoms of different elements are different; atoms combine in whole-number ratios to form compounds.

3.

A proton is larger than an electron.

5.

proton: 1+; electron: 1−; neutron: 0

7.

Nucleus

9.

The atomic number is the number of protons in a nucleus. Boron has an atomic number of five.

11.

Isotopes are atoms of the same element but with different numbers of neutrons.  answer_11  are examples.

13.

a)  isotopes

b)  not isotopes

c)  not isotopes

15.

a)  5a_answer_oxygen_symbol

b)  19K

c)  15c_lithium_symbol

17.

95A

19.

a)  Na

b)  Ar

c)  N

d)  Rn

21.

a)  silicon

b)  manganese

c)  iron

d)  chromium

 

17

End-of-Chapter Material

Additional Exercises

  1. How many electrons does it take to make the mass of one proton?

  2. How many protons does it take to make the mass of a neutron?

  3. Dalton’s initial version of the modern atomic theory says that all atoms of the same element are the same. Is this actually correct? Why or why not?

  4. How are atoms of the same element the same? How are atoms of the same element different?

  5. Give complete atomic symbols for the three known isotopes of hydrogen.

  6. A rare isotope of helium has a single neutron in its nucleus. Write the complete atomic symbol of this isotope.

  7. Use its place on the periodic table to determine if indium, In, atomic number 49, is a metal or a nonmetal.

  8. Only a few atoms of astatine, At, atomic number 85, have been detected. On the basis of its position on the periodic table, would you expect it to be a metal or a nonmetal?

  9. Americium-241 is a crucial part of many smoke detectors. How many neutrons are present in its nucleus?

  10. Potassium-40 is a radioactive isotope of potassium that is present in the human body. How many neutrons are present in its nucleus?

  11. Determine the atomic mass of ruthenium from the given abundance and mass data.

    Ruthenium-96 5.54% 95.907 u
    Ruthenium-98 1.87% 97.905 u
    Ruthenium-99 12.76% 98.906 u
    Ruthenium-100 12.60% 99.904 u
    Ruthenium-101 17.06% 100.906 u
    Ruthenium-102 31.55% 101.904 u
    Ruthenium-104 18.62% 103.905 u
  12. Determine the atomic mass of tellurium from the given abundance and mass data.

    Tellurium-120 0.09% 119.904 u
    Tellurium-122 2.55% 121.903 u
    Tellurium-123 0.89% 122.904 u
    Tellurium-124 4.74% 123.903 u
    Tellurium-125 7.07% 124.904 u
    Tellurium-126 18.84% 125.903 u
    Tellurium-128 31.74% 127.904 u
    Tellurium-130 34.08% 129.906 u
  13. One atomic mass unit has a mass of 1.6605 × 10−24 g. What is the mass of one atom of sodium?

  14. One atomic mass unit has a mass of 1.6605 × 10−24 g. What is the mass of one atom of uranium?

  15. One atomic mass unit has a mass of 1.6605 × 10−24 g. What is the mass of one molecule of H2O?

  16. One atomic mass unit has a mass of 1.6605 × 10−24 g. What is the mass of one molecule of PF5?

  17. From their positions on the periodic table, will Cu and I form a molecular compound or an ionic compound?

  18. From their positions on the periodic table, will N and S form a molecular compound or an ionic compound?

  19. Mercury is an unusual element in that when it takes a 1+ charge as a cation, it always exists as the diatomic ion.

    a)  Propose a formula for the mercury(I) ion.  b)What is the formula of mercury(I) chloride?

  20. Propose a formula for hydrogen peroxide, a substance used as a bleaching agent. (Curiously, this compound does not behave as an acid, despite its formula. It behaves more like a classic nonmetal-nonmetal, molecular compound.)

  21. The uranyl cation has the formula UO22+. Propose formulas and names for the ionic compounds between the uranyl cation and F, SO42−, and PO43−.

  22. The permanganate anion has the formula MnO4. Propose formulas and names for the ionic compounds between the permanganate ion and K+, Ca2+, and Fe3+.

 

Answers

1.

about 1,800 electrons

3.

It is not strictly correct because of the existence of isotopes.

5.

H11, H12, and H13

7.

It is a metal.

9.

146 neutrons

11.

101.065 u

13.

3.817 × 10−23 g

15.

2.991 × 10−23 g

17.

ionic

19.

  1. Hg22+
  2. Hg2Cl2

21.

uranyl fluoride, UO2F2; uranyl sulfate, UO2SO4; uranyl phosphate, (UO2)3(PO4)2

 

IV

Chapter 4. Chemical Reactions and Equations

18

Introduction to Chemical Reactions and Equations

The space shuttle—and any other rocket-based system—uses chemical reactions to propel itself into space and maneuver itself when it gets into orbit. The rockets that lift the orbiter are of two different types. The three main engines are powered by reacting liquid hydrogen with liquid oxygen to generate water. Then there are the two solid rocket boosters, which use a solid fuel mixture that contains mainly ammonium perchlorate and powdered aluminum. The chemical reaction between these substances produces aluminum oxide, water, nitrogen gas, and hydrogen chloride. Although the solid rocket boosters each have a significantly lower mass than the liquid oxygen and liquid hydrogen tanks, they provide over 80% of the lift needed to put the shuttle into orbit—all because of chemical reactions.

Chemistry is largely about chemical changes. Indeed, if there were no chemical changes, chemistry as such would not exist! Chemical changes are a fundamental part of chemistry. Because chemical changes are so central, it may be no surprise that chemistry has developed some special ways of presenting them.

19

The Chemical Equation

Learning Objectives

  1. Define chemical equation.
  2. Identify the parts of a chemical equation.

A chemical reaction expresses a chemical change. For example, one chemical property of hydrogen is that it will react with oxygen to make water. We can write that as follows:

hydrogen reacts with oxygen to make water

We can represent this chemical change more succinctly as

hydrogen + oxygen → water

where the + sign means that the two substances interact chemically with each other and the → symbol implies that a chemical reaction takes place. But substances can also be represented by chemical formulas. Remembering that hydrogen and oxygen both exist as diatomic molecules, we can rewrite our chemical change as

H2 + O2 → H2O

This is an example of a chemical equation, which is a concise way of representing a chemical reaction. The initial substances are called reactants, and the final substances are called products.

Unfortunately, it is also an incomplete chemical equation. The law of conservation of matter says that matter cannot be created or destroyed. In chemical equations, the number of atoms of each element in the reactants must be the same as the number of atoms of each element in the products. If we count the number of hydrogen atoms in the reactants and products, we find two hydrogen atoms. But if we count the number of oxygen atoms in the reactants and products, we find that there are two oxygen atoms in the reactants but only one oxygen atom in the products.

What can we do? Can we change the subscripts in the formula for water so that it has two oxygen atoms in it? No; you cannot change the formulas of individual substances because the chemical formula for a given substance is characteristic of that substance. What you can do, however, is to change the number of molecules that react or are produced. We do this one element at a time, going from one side of the reaction to the other, changing the number of molecules of a substance until all elements have the same number of atoms on each side.

To accommodate the two oxygen atoms as reactants, let us assume that we have two water molecules as products:

H2 + O2 → 2 H2O

The 2 in front of the formula for water is called a coefficient. Now there is the same number of oxygen atoms in the reactants as there are in the product. But in satisfying the need for the same number of oxygen atoms on both sides of the reaction, we have also changed the number of hydrogen atoms on the product side, so the number of hydrogen atoms is no longer equal. No problem—simply go back to the reactant side of the equation and add a coefficient in front of the H2. The coefficient that works is 2:

2 H2 + O2 → 2 H2O

There are now four hydrogen atoms in the reactants and also four atoms of hydrogen in the product. There are two oxygen atoms in the reactants and two atoms of oxygen in the product. The law of conservation of matter has been satisfied. When the reactants and products of a chemical equation have the same number of atoms of all elements present, we say that an equation is balanced. All proper chemical equations are balanced. If a substance does not have a coefficient written in front of it, it is assumed to be 1. Also, the convention is to use all whole numbers when balancing chemical equations. This sometimes makes us do a bit more “back and forth” work when balancing a chemical equation.

Example 1

Write and balance the chemical equation for each given chemical reaction.

  1. Hydrogen and chlorine react to make HCl.
  2. Ethane, C2H6, reacts with oxygen to make carbon dioxide and water.

Solution

  1. Let us start by simply writing a chemical equation in terms of the formulas of the substances, remembering that both elemental hydrogen and chlorine are diatomic:

    H2 + Cl2 → HCl

    There are two hydrogen atoms and two chlorine atoms in the reactants and one of each atom in the product. We can fix this by including the coefficient 2 on the product side:

    H2 + Cl2 → 2 HCl

    Now there are two hydrogen atoms and two chlorine atoms on both sides of the chemical equation, so it is balanced.

  2. Start by writing the chemical equation in terms of the substances involved:

    C2H6 + O2 → CO2 + H2O

    We have two carbon atoms on the left, so we need two carbon dioxide molecules on the product side, so that each side has two carbon atoms; that element is balanced. We have six hydrogen atoms in the reactants, so we need six hydrogen atoms in the products. We can get this by having three water molecules:

    C2H6 + O2 → 2 CO2 + 3 H2O

    Now we have seven oxygen atoms in the products (four from the CO2 and three from the H2O). That means we need seven oxygen atoms in the reactants. However, because oxygen is a diatomic molecule, we can only get an even number of oxygen atoms at a time. We can achieve this by multiplying the other coefficients by 2:

    2 C2H6 + O2 → 4 CO2 + 6 H2O

    By multiplying everything else by 2, we don’t unbalance the other elements, and we now get an even number of oxygen atoms in the product—14. We can get 14 oxygen atoms on the reactant side by having 7 oxygen molecules:

    2 C2H6 + 7 O2 → 4 CO2 + 6 H2O

    As a check, recount everything to determine that each side has the same number of atoms of each element. This chemical equation is now balanced.

Test Yourself

Write and balance the chemical equation that represents nitrogen and hydrogen reacting to produce ammonia, NH3.

Answer

N2 + 3 H2 → 2 NH3

Many chemical equations also include phase labels for the substances: (s) for solid, (ℓ) for liquid, (g) for gas, and (aq) for aqueous (i.e., dissolved in water). Special conditions, such as temperature, may also be listed above the arrow. For example,

2 NaHCO3(s) →200°C Na2CO3(s) + CO2(g) + H2O(ℓ)

 

Key Takeaways

  • A chemical equation is a concise description of a chemical reaction.
  • Proper chemical equations are balanced.

 

Exercises

  1. From the statement “nitrogen and hydrogen react to produce ammonia,” identify the reactants and the products.

  2. From the statement “sodium metal reacts with water to produce sodium hydroxide and hydrogen,” identify the reactants and the products.

  3. From the statement “magnesium hydroxide reacts with nitric acid to produce magnesium nitrate and water,” identify the reactants and the products.

  4. From the statement “propane reacts with oxygen to produce carbon dioxide and water,” identify the reactants and the products.

  5. Write and balance the chemical equation described by Exercise 1.

  6. Write and balance the chemical equation described by Exercise 2.

  7. Write and balance the chemical equation described by Exercise 3.

  8. Write and balance the chemical equation described by Exercise 4. The formula for propane is C3H8.

  9. Balance: ___NaClO3 → ___NaCl + ___O2

  10. Balance: ___N2 + ___H2 → ___N2H4

  11. Balance: ___Al + ___O2 → ___Al2O3

  12. Balance: ___C2H4 + ___O2 → ___CO2 + ___H2O

  13. How would you write the balanced chemical equation in Exercise 10 if all substances were gases?

  14. How would you write the balanced chemical equation in Exercise 12 if all the substances except water were gases and water itself were a liquid?

Answers

1.

reactants: nitrogen and hydrogen; product: ammonia

3.

reactants: magnesium hydroxide and nitric acid; products: magnesium nitrate and water

5.

N2 + 3 H2 → 2 NH37.

Mg(OH)2 + 2 HNO3 → Mg(NO3)2 + 2 H2O

9.

2 NaClO3 → 2 NaCl + 3 O211.

4Al + 3O2 → 2Al2O313.

N2(g) + 3 H2(g) → 2 NH3(g)

 

20

Types of Chemical Reactions: Single- and Double-Displacement Reactions

Learning Objectives

  1. Recognize chemical reactions as single-replacement reactions and double-replacement reactions.
  2. Use the periodic table, an activity series, or solubility rules to predict whether single-replacement reactions or double-replacement reactions will occur.

Up to now, we have presented chemical reactions as a topic, but we have not discussed how the products of a chemical reaction can be predicted. Here we will begin our study of certain types of chemical reactions that allow us to predict what the products of the reaction will be.

A single-replacement reaction is a chemical reaction in which one element is substituted for another element in a compound, generating a new element and a new compound as products. For example,

2 HCl(aq) + Zn(s) → ZnCl2(aq) + H2(g)

is an example of a single-replacement reaction. The hydrogen atoms in HCl are replaced by Zn atoms, and in the process a new element—hydrogen—is formed. Another example of a single-replacement reaction is

2 NaCl(aq) + F2(g) → 2 NaF(s) + Cl2(g)

Here the negatively charged ion changes from chloride to fluoride. A typical characteristic of a single-replacement reaction is that there is one element as a reactant and another element as a product.

Not all proposed single-replacement reactions will occur between two given reactants. This is most easily demonstrated with fluorine, chlorine, bromine, and iodine. Collectively, these elements are called the halogens and are in the next-to-last column on the periodic table (see Figure 4.1 “Halogens on the Periodic Table”). The elements on top of the column will replace the elements below them on the periodic table but not the other way around. Thus, the reaction represented by

CaI2(s) + Cl2(g) → CaCl2(s) + I2(s)

will occur, but the reaction

CaF2(s) + Br2(ℓ) → CaBr2(s) + F2(g)

will not because bromine is below fluorine on the periodic table. This is just one of many ways the periodic table helps us understand chemistry.

Figure 4.1 Halogens on the Periodic Table

Halogens

 

The halogens are the elements in the next-to-last column on the periodic table.

Example 2

Will a single-replacement reaction occur? If so, identify the products.

  1. MgCl2 + I2 → ?
  2. CaBr2 + F2 → ?

Solution

  1. Because iodine is below chlorine on the periodic table, a single-replacement reaction will not occur.
  2. Because fluorine is above bromine on the periodic table, a single-replacement reaction will occur, and the products of the reaction will be CaF2 and Br2.

Test Yourself

Will a single-replacement reaction occur? If so, identify the products.

FeI2 + Cl2 → ?

Answer

Yes; FeCl2 and I2

Chemical reactivity trends are easy to predict when replacing anions in simple ionic compounds—simply use their relative positions on the periodic table. However, when replacing the cations, the trends are not as straightforward. This is partly because there are so many elements that can form cations; an element in one column on the periodic table may replace another element nearby, or it may not. A list called the activity series does the same thing the periodic table does for halogens: it lists the elements that will replace elements below them in single-replacement reactions. A simple activity series is shown below.

Activity Series for Cation Replacement in Single-Replacement Reactions

  • Li
  • K
  • Ba
  • Sr
  • Ca
  • Na
  • Mg
  • Al
  • Mn
  • Zn
  • Cr
  • Fe
  • Ni
  • Sn
  • Pb
  • H2
  • Cu
  • Hg
  • Ag
  • Pd
  • Pt
  • Au

Using the activity series is similar to using the positions of the halogens on the periodic table. An element on top will replace an element below it in compounds undergoing a single-replacement reaction. Elements will not replace elements above them in compounds.

Example 3

Use the activity series to predict the products, if any, of each equation.

  1. FeCl2 + Zn → ?
  2. HNO3 + Au → ?

Solution

  1. Because zinc is above iron in the activity series, it will replace iron in the compound. The products of this single-replacement reaction are ZnCl2 and Fe.
  2. Gold is below hydrogen in the activity series. As such, it will not replace hydrogen in a compound with the nitrate ion. No reaction is predicted.

Test Yourself

Use the activity series to predict the products, if any, of this equation.

AlPO4 + Mg → ?

Answer

Mg3(PO4)2 and Al

A double-replacement reaction occurs when parts of two ionic compounds are exchanged, making two new compounds. A characteristic of a double-replacement equation is that there are two compounds as reactants and two different compounds as products. An example is

CuCl2(aq) + 2 AgNO3(aq) → Cu(NO3)2(aq) + 2 AgCl(s)

There are two equivalent ways of considering a double-replacement equation: either the cations are swapped, or the anions are swapped. (You cannot swap both; you would end up with the same substances you started with.) Either perspective should allow you to predict the proper products, as long as you pair a cation with an anion and not a cation with a cation or an anion with an anion.

Example 4

Predict the products of this double-replacement equation: BaCl2 + Na2SO4 → ?

Solution

Thinking about the reaction as either switching the cations or switching the anions, we would expect the products to be BaSO4 and NaCl.

Test Yourself

Predict the products of this double-replacement equation: KBr + AgNO3 → ?

Answer

KNO3 and AgBr

Predicting whether a double-replacement reaction occurs is somewhat more difficult than predicting a single-replacement reaction. However, there is one type of double-replacement reaction that we can predict: the precipitation reaction. A precipitation reaction occurs when two ionic compounds are dissolved in water and form a new ionic compound that does not dissolve; this new compound falls out of solution as a solid precipitate. The formation of a solid precipitate is the driving force that makes the reaction proceed.

To judge whether double-replacement reactions will occur, we need to know what kinds of ionic compounds form precipitates. For this, we use solubility rules, which are general statements that predict which ionic compounds dissolve (are soluble) and which do not (are not soluble or insoluble). Table 4.1 “Some Useful Solubility Rules” lists some general solubility rules. We need to consider each ionic compound (both the reactants and the possible products) in light of the solubility rules in Table 4.1 “Some Useful Solubility Rules”. If a compound is soluble, we use the (aq) label with it, indicating it dissolves. If a compound is not soluble, we use the (s) label with it and assume that it will precipitate out of solution. If everything is soluble, then no reaction will be expected.

Table 4.1 Some Useful Solubility Rules

These compounds generally dissolve in water (are soluble): Exceptions:
All compounds of Li+, Na+, K+, Rb+, Cs+, and NH4+ None
All compounds of NO3 and C2H3O2 None
Compounds of Cl, Br, I Ag+, Hg22+, Pb2+
Compounds of SO42 Hg22+, Pb2+, Sr2+, Ba2+
These compounds generally do not dissolve in water (are insoluble): Exceptions:
Compounds of CO32− and PO43− Compounds of Li+, Na+, K+, Rb+, Cs+, and NH4+
Compounds of OH Compounds of Li+, Na+, K+, Rb+, Cs+, NH4+, Sr2+, and Ba2+

For example, consider the possible double-replacement reaction between Na2SO4 and SrCl2. The solubility rules say that all ionic sodium compounds are soluble and all ionic chloride compounds are soluble except for Ag+, Hg22+, and Pb2+, which are not being considered here. Therefore, Na2SO4 and SrCl2 are both soluble. The possible double-replacement reaction products are NaCl and SrSO4. Are these soluble? NaCl is (by the same rule we just quoted), but what about SrSO4? Compounds of the sulfate ion are generally soluble, but Sr2+ is an exception: we expect it to be insoluble—a precipitate. Therefore, we expect a reaction to occur, and the balanced chemical equation would be

Na2SO4(aq) + SrCl2(aq) → 2 NaCl(aq) + SrSO4(s)

You would expect to see a visual change corresponding to SrSO4 precipitating out of solution (Figure 4.2 “Double-Replacement Reactions”).

Figure 4.2 Double-Replacement Reactions

Double Replacement Reaction

 

Some double-replacement reactions are obvious because you can see a solid precipitate coming out of solution.

Example 5

Will a double-replacement reaction occur? If so, identify the products.

  1. Ca(NO3)2 + KBr → ?
  2. NaOH + FeCl2 → ?

Solution

  1. According to the solubility rules, both Ca(NO3)2 and KBr are soluble. Now we consider what the double-replacement products would be by switching the cations (or the anions)—namely, CaBr2 and KNO3. However, the solubility rules predict that these two substances would also be soluble, so no precipitate would form. Thus, we predict no reaction in this case.
  2. According to the solubility rules, both NaOH and FeCl2 are expected to be soluble. If we assume that a double-replacement reaction may occur, we need to consider the possible products, which would be NaCl and Fe(OH)2. NaCl is soluble, but, according to the solubility rules, Fe(OH)2 is not. Therefore, a reaction would occur, and Fe(OH)2(s) would precipitate out of solution. The balanced chemical equation is

    2NaOH(aq) + FeCl2(aq) → 2NaCl(aq) + Fe(OH)2(s)

Test Yourself

Will a double-replacement equation occur? If so, identify the products.

Sr(NO3)2 + KCl → ?

Answer

No reaction; all possible products are soluble.

Key Takeaways

  • A single-replacement reaction replaces one element for another in a compound.
  • The periodic table or an activity series can help predict whether single-replacement reactions occur.
  • A double-replacement reaction exchanges the cations (or the anions) of two ionic compounds.
  • A precipitation reaction is a double-replacement reaction in which one product is a solid precipitate.
  • Solubility rules are used to predict whether some double-replacement reactions will occur.

 

Exercises

What are the general characteristics that help you recognize single-replacement reactions?

What are the general characteristics that help you recognize double-replacement reactions?

Assuming that each single-replacement reaction occurs, predict the products and write each balanced chemical equation.

a)  Zn + Fe(NO3)2 → ?

b)  F2 + FeI3 → ?

4.  Assuming that each single-replacement reaction occurs, predict the products and write each balanced chemical equation.

a) Li + MgSO4 → ?

b)  NaBr + Cl2 → ?

5.  Assuming that each single-replacement reaction occurs, predict the products and write each balanced chemical equation.

a)  Sn + H2SO4 → ?

b)  Al + NiBr2 → ?

 

6.  Assuming that each single-replacement reaction occurs, predict the products and write each balanced chemical equation.

a)  Mg + HCl → ?

b)  HI + Br2 → ?

7.  Use the periodic table or the activity series to predict if each single-replacement reaction will occur and, if so, write a balanced chemical equation.

a)  FeCl2 + Br2 → ?

b)  Fe(NO3)3 + Al → ?

 

8.  Use the periodic table or the activity series to predict if each single-replacement reaction will occur and, if so, write a balanced chemical equation.

a)  Zn + Fe3(PO4)2 → ?

b)  Ag + HNO3 → ?

 

9.  Use the periodic table or the activity series to predict if each single-replacement reaction will occur and, if so, write a balanced chemical equation.

a)  NaI + Cl2 → ?

b)  AgCl + Au → ?

 

10.  Use the periodic table or the activity series to predict if each single-replacement reaction will occur and, if so, write a balanced chemical equation.

a)  Pt + H3PO4 → ?

b)  Li + H2O → ? (Hint: treat H2O as if it were composed of H+ and OH ions.)

 

11.  Assuming that each double-replacement reaction occurs, predict the products and write each balanced chemical equation.

a)  Zn(NO3)2 + NaOH → ?

b)  HCl + Na2S → ?

 

12.  Assuming that each double-replacement reaction occurs, predict the products and write each balanced chemical equation.

a)  Ca(C2H3O2)2 + HNO3 → ?

b)  Na2CO3 + Sr(NO2)2 → ?

 

13.  Assuming that each double-replacement reaction occurs, predict the products and write each balanced chemical equation.

a)  Pb(NO3)2 + KBr → ?

b)  K2O + MgCO3 → ?

 

14.  Assuming that each double-replacement reaction occurs, predict the products and write each balanced chemical equation.

a)  Sn(OH)2 + FeBr3 → ?

b)  CsNO3 + KCl → ?

 

15.  Use the solubility rules to predict if each double-replacement reaction will occur and, if so, write a balanced chemical equation.

a)  Pb(NO3)2 + KBr → ?

b)  K2O + Na2CO3 → ?

 

16.  Use the solubility rules to predict if each double-replacement reaction will occur and, if so, write a balanced chemical equation.

a)  Na2CO3 + Sr(NO2)2 → ?

b)  (NH4)2SO4 + Ba(NO3)2 → ?

 

17.  Use the solubility rules to predict if each double-replacement reaction will occur and, if so, write a balanced chemical equation.

a)  K3PO4 + SrCl2 → ?

b)  NaOH + MgCl2 → ?

 

18.  Use the solubility rules to predict if each double-replacement reaction will occur and, if so, write a balanced chemical equation.

a)  KC2H3O2 + Li2CO3 → ?

b)  KOH + AgNO3 → ?

Answers

1. One element replaces another element in a compound.

3.

a)  Zn + Fe(NO3)2 → Zn(NO3)2 + Fe

b)  3 F2 + 2 FeI3 → 3 I2 + 2 FeF35.

a)  Sn + H2SO4 → SnSO4 + H2

b)  2 Al + 3 NiBr2 → 2 AlBr3 + 3 Ni

7.

a)  No reaction occurs.

b)  Fe(NO3)3 + Al → Al(NO3)3 + Fe

9.

a)  2 NaI + Cl2 → 2 NaCl + I2

b)  No reaction occurs.

11.

a)  Zn(NO3)2 + 2 NaOH → Zn(OH)2 + 2 NaNO3

b)  2 HCl + Na2S → 2 NaCl + H2S

13.

a)  Pb(NO3)2 + 2 KBr → PbBr2 + 2 KNO3

b)  K2O + MgCO3 → K2CO3 + MgO

15.

a)  Pb(NO3)2 + 2 KBr → PbBr2(s) + 2 KNO3

b)  No reaction occurs.

17.

a)  2 K3PO4 + 3 SrCl2 → Sr3(PO4)2(s) + 6 KCl

b)  2 NaOH + MgCl2 → 2 NaCl + Mg(OH)2(s)

21

Ionic Equations: A Closer Look

Learning Objectives

  1. Write ionic equations for chemical reactions between ionic compounds.
  2. Write net ionic equations for chemical reactions between ionic compounds.

For single-replacement and double-replacement reactions, many of the reactions included ionic compounds: compounds between metals and nonmetals or compounds that contained recognizable polyatomic ions. Now we take a closer look at reactions that include ionic compounds.

One important aspect about ionic compounds that differs from molecular compounds has to do with dissolving in a liquid, such as water. When molecular compounds, such as sugar, dissolve in water, the individual molecules drift apart from each other. When ionic compounds dissolve, the ions physically separate from each other. We can use a chemical equation to represent this process—for example, with NaCl:

NaCl(s) → Na+(aq) + Cl(aq)

When NaCl dissolves in water, the ions separate and go their own way in solution; the ions are now written with their respective charges, and the (aq) phase label emphasizes that they are dissolved (Figure 4.3 “Ionic Solutions”). This process is called dissociation; we say that the ions dissociate.

Figure 4.3 Ionic Solutions

Ionic Compounds

 

When an ionic compound dissociates in water, water molecules surround each ion and separate it from the rest of the solid. Each ion goes its own way in solution.

All ionic compounds that dissolve behave this way. (This behaviour was first suggested by the Swedish chemist Svante August Arrhenius [1859–1927] as part of his PhD dissertation in 1884. Interestingly, his PhD examination team had a hard time believing that ionic compounds would behave like this, so they gave Arrhenius a barely passing grade. Later, this work was cited when Arrhenius was awarded the Nobel Prize in Chemistry.) Keep in mind that when the ions separate, all the ions separate. Thus, when CaCl2 dissolves, the one Ca2+ ion and the two Cl ions separate from each other:

CaCl2(s) →  Ca2+(aq) + Cl(aq) + Cl(aq) or CaCl2(s) →  Ca2+(aq) + 2Cl(aq)

That is, the two chloride ions go off on their own. They do not remain as Cl2 (that would be elemental chlorine; these are chloride ions); they do not stick together to make Cl2 or Cl22−. They become dissociated ions in their own right. Polyatomic ions also retain their overall identity when they are dissolved.

Example 6

Write the chemical equation that represents the dissociation of each ionic compound.

  1. KBr
  2. Na2SO4

Solution

  1. KBr(s) → K+(aq) + Br(aq)
  2. Not only do the two sodium ions go their own way, but the sulfate ion stays together as the sulfate ion. The dissolving equation is

    Na2SO4(s) → 2Na+(aq) + SO42−(aq)

Test Yourself

Write the chemical equation that represents the dissociation of (NH4)2S.

Answer

(NH4)2S(s) → 2NH4+(aq) + S2−(aq)

When chemicals in solution react, the proper way of writing the chemical formulas of the dissolved ionic compounds is in terms of the dissociated ions, not the complete ionic formula. A complete ionic equation is a chemical equation in which the dissolved ionic compounds are written as separated ions. Solubility rules are very useful in determining which ionic compounds are dissolved and which are not. For example, when NaCl(aq) reacts with AgNO3(aq) in a double-replacement reaction to precipitate AgCl(s) and form NaNO3(aq), the complete ionic equation includes NaCl, AgNO3, and NaNO3 written as separated ions:

Na+(aq) + Cl(aq) + Ag+(aq) + NO3(aq) → AgCl(s) + Na+(aq) + NO3(aq)

This is more representative of what is occurring in the solution.

Example 7

Write the complete ionic equation for each chemical reaction.

  1. KBr(aq) + AgC2H3O2(aq) → KC2H3O2(aq) + AgBr(s)
  2. MgSO4(aq) + Ba(NO3)2(aq) → Mg(NO3)2(aq) + BaSO4(s)

Solution

For any ionic compound that is aqueous, we will write the compound as separated ions.

  1. The complete ionic equation is

    K+(aq) + Br(aq) + Ag+(aq) + C2H3O2(aq) → K+(aq) + C2H3O2(aq) + AgBr(s)

  2. The complete ionic equation is

    Mg2+(aq) + SO42−(aq) + Ba2+(aq) + 2NO3(aq) → Mg2+(aq) + 2NO3(aq) + BaSO4(s)

Test Yourself

Write the complete ionic equation for

CaCl2(aq) + Pb(NO3)2(aq) → Ca(NO3)2(aq) + PbCl2(s)

Answer

Ca2+(aq) + 2Cl(aq) + Pb2+(aq) + 2NO3(aq) → Ca2+(aq) + 2NO3(aq) + PbCl2(s)

You may notice that in a complete ionic equation, some ions do not change their chemical form; they stay exactly the same on the reactant and product sides of the equation. For example, in

Na+(aq) + Cl(aq) + Ag+(aq) + NO3(aq) → AgCl(s) + Na+(aq) + NO3(aq)

the Ag+(aq) and Cl(aq) ions become AgCl(s), but the Na+(aq) ions and the NO3(aq) ions stay as Na+(aq) ions and NO3(aq) ions. These two ions are examples of spectator ions, ions that do nothing in the overall course of a chemical reaction. They are present, but they do not participate in the overall chemistry. It is common to cancel spectator ions (something also done with algebraic quantities) on the opposite sides of a chemical equation:

Na+(aq) + Cl(aq) + Ag+(aq) + NO3(aq) → AgCl(s) + Na+(aq) + NO3(aq)

What remains when the spectator ions are removed is called the net ionic equation, which represents the actual chemical change occurring between the ionic compounds:

Cl(aq) + Ag+(aq) → AgCl(s)

It is important to reiterate that the spectator ions are still present in solution, but they don’t experience any net chemical change, so they are not written in a net ionic equation.

Example 8

Write the net ionic equation for each chemical reaction.

  1. K+(aq) + Br(aq) + Ag+(aq) + C2H3O2(aq) → K+(aq) + C2H3O2(aq) + AgBr(s)
  2. Mg2+(aq) + SO42−(aq) + Ba2+(aq) + 2 NO3(aq) → Mg2+(aq) + 2 NO3(aq) + BaSO4(s)

Solution

  1. In the first equation, the K+(aq) and C2H3O2(aq) ions are spectator ions, so they are canceled:

     K+(aq) + Br(aq) + Ag+(aq) + C2H3O2(aq)→ K+(aq) + C2H3O2(aq) + AgBr(s)

    The net ionic equation is

    Br(aq) + Ag+(aq) → AgBr(s)

  2. In the second equation, the Mg2+(aq) and NO3(aq) ions are spectator ions, so they are canceled:

    Mg2+(aq) + SO42−(aq) + Ba2+(aq) + 2 NO3(aq) → Mg2+(aq) + 2 NO3(aq) + BaSO4(s)

    The net ionic equation is

    SO42−(aq) + Ba2+(aq) → BaSO4(s)

Test Yourself

Write the net ionic equation for

CaCl2(aq) + Pb(NO3)2(aq) → Ca(NO3)2(aq) + PbCl2(s)

Answer

Pb2+(aq) + 2Cl(aq) → PbCl2(s)

Chemistry Is Everywhere: Soluble and Insoluble Ionic Compounds

The concept of solubility versus insolubility in ionic compounds is a matter of degree. Some ionic compounds are very soluble, some are only moderately soluble, and some are soluble so little that they are considered insoluble. For most ionic compounds, there is also a limit to the amount of compound can be dissolved in a sample of water. For example, you can dissolve a maximum of 36.0 g of NaCl in 100 g of water at room temperature, but you can dissolve only 0.00019 g of AgCl in 100 g of water. We consider NaCl soluble but AgCl insoluble.

One place where solubility is important is in the tank-type water heater found in many homes in the United States. Domestic water frequently contains small amounts of dissolved ionic compounds, including calcium carbonate (CaCO3). However, CaCO3 has the relatively unusual property of being less soluble in hot water than in cold water. So as the water heater operates by heating water, CaCO3 can precipitate if there is enough of it in the water. This precipitate, called limescale, can also contain magnesium compounds, hydrogen carbonate compounds, and phosphate compounds. The problem is that too much limescale can impede the function of a water heater, requiring more energy to heat water to a specific temperature or even blocking water pipes into or out of the water heater, causing dysfunction.

Another place where solubility versus insolubility is an issue is the Grand Canyon. We usually think of rock as insoluble. But it is actually ever so slightly soluble. This means that over a period of about two billion years, the Colorado River carved rock from the surface by slowly dissolving it, eventually generating a spectacular series of gorges and canyons. And all because of solubility!

The Grand Canyon was formed by water running through rock for billions of years, very slowly dissolving it. Note the Colorado River is still present in the lower part of the photo. “Grand canyon yavapal point 2010′′ by chensiyuan is licensed under Creative Commons
The Grand Canyon was formed by water running through rock for billions of years, very slowly dissolving it. Note the Colorado River is still present in the lower part of the photo.
“Grand canyon yavapal point 2010′′ by chensiyuan is licensed under Creative Commons

Key Takeaways

  • Ionic compounds that dissolve separate into individual ions.
  • Complete ionic equations show dissolved ionic solids as separated ions.
  • Net ionic equations show only the ions and other substances that change in a chemical reaction.

Exercises

  1. Write a chemical equation that represents NaBr(s) dissociating in water.

  2. Write a chemical equation that represents SrCl2(s) dissociating in water.

  3. Write a chemical equation that represents (NH4)3PO4(s) dissociating in water.

  4. Write a chemical equation that represents Fe(C2H3O2)3(s) dissociating in water.

  5. Write the complete ionic equation for the reaction of FeCl2(aq) and AgNO3(aq). You may have to consult the solubility rules.

  6. Write the complete ionic equation for the reaction of BaCl2(aq) and Na2SO4(aq). You may have to consult the solubility rules.

  7. Write the complete ionic equation for the reaction of KCl(aq) and NaC2H3O2(aq). You may have to consult the solubility rules.

  8. Write the complete ionic equation for the reaction of Fe2(SO4)3(aq) and Sr(NO3)2(aq). You may have to consult the solubility rules.

  9. Write the net ionic equation for the reaction of FeCl2(aq) and AgNO3(aq). You may have to consult the solubility rules.

  10. Write the net ionic equation for the reaction of BaCl2(aq) and Na2SO4(aq). You may have to consult the solubility rules.

  11. Write the net ionic equation for the reaction of KCl(aq) and NaC2H3O2(aq). You may have to consult the solubility rules.

  12. Write the net ionic equation for the reaction of Fe2(SO4)3(aq) and Sr(NO3)2(aq). You may have to consult the solubility rules.

  13. Identify the spectator ions in Exercises 9 and 10.

  14. Identify the spectator ions in Exercises 11 and 12.

Answers

1.

NaBr(s)  Na+(aq) + Br(aq)

3.

(NH4)3PO4(s)  3 NH4+(aq) + PO43−(aq)

5.

Fe2+(aq) + 2 Cl(aq) + 2 Ag+(aq) + 2 NO3(aq) → Fe2+(aq) + 2 NO3(aq) + 2 AgCl(s)

7.

K+(aq) + Cl(aq) + Na+(aq) + C2H3O2(aq) → Na+(aq) + Cl(aq) + K+(aq) + C2H3O2(aq)

9.

2 Cl(aq) + 2 Ag+(aq) → 2 AgCl(s)

11.

There is no overall reaction.

13.

In Exercise 9, Fe2+(aq) and NO3(aq) are spectator ions; in Exercise 10, Na+(aq) and Cl(aq) are spectator ions.

22

Composition, Decomposition, and Combustion Reactions

Learning Objectives

  1. Recognize composition, decomposition, and combustion reactions.
  2. Predict the products of a combustion reaction.

Three classifications of chemical reactions will be reviewed in this section. Predicting the products in some of them may be difficult, but the reactions are still easy to recognize.

A composition reaction (sometimes also called a combination reaction or a synthesis reaction) produces a single substance from multiple reactants. A single substance as a product is the key characteristic of the composition reaction. There may be a coefficient other than one for the substance, but if the reaction has only a single substance as a product, it can be called a composition reaction. In the reaction

2 H2(g) + O2(g) → 2 H2O(ℓ)

water is produced from hydrogen and oxygen. Although there are two molecules of water being produced, there is only one substance—water—as a product. So this is a composition reaction.

A decomposition reaction starts from a single substance and produces more than one substance; that is, it decomposes. One substance as a reactant and more than one substance as the products is the key characteristic of a decomposition reaction. For example, in the decomposition of sodium hydrogen carbonate (also known as sodium bicarbonate),

2 NaHCO3(s) → Na2CO3(s) + CO2(g) + H2O(ℓ)

sodium carbonate, carbon dioxide, and water are produced from the single substance sodium hydrogen carbonate.

Composition and decomposition reactions are difficult to predict; however, they should be easy to recognize.

Example 9

Identify each equation as a composition reaction, a decomposition reaction, or neither.

  1. Fe2O3 + 3 SO3 → Fe2(SO4)3
  2. NaCl + AgNO3 → AgCl + NaNO3
  3. (NH4)2Cr2O7 → Cr2O3 + 4 H2O + N2

Solution

  1. In this equation, two substances combine to make a single substance. This is a composition reaction.
  2. Two different substances react to make two new substances. This does not fit the definition of either a composition reaction or a decomposition reaction, so it is neither. In fact, you may recognize this as a double-replacement reaction.
  3. A single substance reacts to make multiple substances. This is a decomposition reaction.

Test Yourself

Identify the equation as a composition reaction, a decomposition reaction, or neither.

C3H8 → C3H4 + 2 H2

Answer

decomposition

A combustion reaction occurs when a reactant combines with oxygen, many times from the atmosphere, to produce oxides of all other elements as products; any nitrogen in the reactant is converted to elemental nitrogen, N2. Many reactants, called fuels, contain mostly carbon and hydrogen atoms, reacting with oxygen to produce CO2 and H2O. For example, the balanced chemical equation for the combustion of methane, CH4, is as follows:

CH4 + 2 O2 → CO2 + 2 H2O

Kerosene can be approximated with the formula C12H26, and its combustion equation is

2 C12H26 + 37 O2 → 24 CO2 + 26 H2O

Sometimes fuels contain oxygen atoms, which must be counted when balancing the chemical equation. One common fuel is ethanol, C2H5OH, whose combustion equation is

C2H5OH + 3 O2 → 2 CO2 + 3 H2O

If nitrogen is present in the original fuel, it is converted to N2, not to a nitrogen-oxygen compound. Thus, for the combustion of the fuel dinitroethylene, whose formula is C2H2N2O4, we have

2 C2H2N2O4 + O2 → 4 CO2 + 2 H2O + 2 N2

Example 10

Complete and balance each combustion equation.

  1. the combustion of propane, C3H8
  2. the combustion of ammonia, NH3

Solution

  1. The products of the reaction are CO2 and H2O, so our unbalanced equation is

    C3H8 + O2 → CO2 + H2O

    Balancing (and you may have to go back and forth a few times to balance this), we get

    C3H8 + 5 O2 → 3 CO2 + 4 H2O

  2. The nitrogen atoms in ammonia will react to make N2, while the hydrogen atoms will react with O2 to make H2O:

    NH3 + O2 → N2 + H2O

    To balance this equation without fractions (which is the convention), we get

    4 NH3 + 3 O2 → 2 N2 + 6 H2O

Test Yourself

Complete and balance the combustion equation for cyclopropanol, C3H6O.

Answer

C3H6O + 4 O2 → 3 CO2 + 3 H2O

Propane is a fuel used to provide heat for some homes. Propane is stored in large tanks like that shown here. Source: “flowers and propane” by vistavision is licensed under the Creative Commons Attribution-NonCommercial-NoDerivs 2.0 Generic
Propane is a fuel used to provide heat for some homes. Propane is stored in large tanks like that shown here.
Source: “flowers and propane” by vistavision is licensed under the Creative Commons Attribution-NonCommercial-NoDerivs 2.0 Generic

Key Takeaways

  • A composition reaction produces a single substance from multiple reactants.
  • A decomposition reaction produces multiple products from a single reactant.
  • Combustion reactions are the combination of some compound with oxygen to make oxides of the other elements as products (although nitrogen atoms react to make N2).

 

Exercises

  1. Which is a composition reaction and which is not?

a)  NaCl + AgNO3 → AgCl + NaNO3

b)  CaO + CO2 → CaCO3

 

2.  Which is a composition reaction and which is not?

a)  H2 + Cl2 → 2 HCl

b)  2 HBr + Cl2 → 2 HCl + Br2

 

3.  Which is a composition reaction and which is not?

a)  2 SO2 + O2 → 2 SO3

b)  6 C + 3 H2 → C6H6

 

4.  Which is a composition reaction and which is not?

a)  4 Na + 2 C + 3 O2 → 2 Na2CO3

b)  Na2CO3 → Na2O + CO2

 

5.  Which is a decomposition reaction and which is not?

a)  HCl + NaOH → NaCl + H2O

b)  CaCO3 → CaO + CO2

 

6.  Which is a decomposition reaction and which is not?

a)  3 O2 → 2 O3

b)  2 KClO3 → 2 KCl + 3 O2

 

7.  Which is a decomposition reaction and which is not?

a)  Na2O + CO2 → Na2CO3

b)  H2SO3 → H2O + SO2

 

8.  Which is a decomposition reaction and which is not?

a)  2 C7H5N3O6 → 3 N2 + 5 H2O + 7 CO + 7 C

b)  C6H12O6 + 6 O2 → 6 CO2 + 6 H2O

 

9.  Which is a combustion reaction and which is not?

a)  C6H12O6 + 6 O2 → 6 CO2 + 6 H2O

b)  2 Fe2S3 + 9 O2 → 2 Fe2O3 + 6 SO2

 

10.  Which is a combustion reaction and which is not?

a)  CH4 + 2 F2 → CF4 + 2 H2

b)  2 H2 + O2 → 2 H2O

 

11.  Which is a combustion reaction and which is not?

a)  P4 + 5 O2 → 2 P2O5

b)  2 Al2S3 + 9 O2 → 2 Al2O3 + 6 SO2

 

12.  Which is a combustion reaction and which is not?

a)  C2H4 + O2 → C2H4O2

b)  C2H4 + Cl2 → C2H4Cl2

 

13.  Is it possible for a composition reaction to also be a combustion reaction? Give an example to support your case.

 

14.  Is it possible for a decomposition reaction to also be a combustion reaction? Give an example to support your case.

 

15.  Complete and balance each combustion equation.

a)  C4H9OH + O2 → ?

b)  CH3NO2 + O2 → ?

16.  Complete and balance each combustion equation.

a)  B2H6 + O2 → ? (The oxide of boron formed is B2O3.)

b)  Al2S3 + O2 → ? (The oxide of sulfur formed is SO2.)

c)  Al2S3 + O2 → ? (The oxide of sulfur formed is SO3.)

Answers

1.

a)  not composition

b)  composition

3.

a)  composition

b)  composition

5.

a)  not decomposition

b)  decomposition

7.

a)  not decomposition

b)  decomposition

9.

a)  combustion

b)  combustion

11.

a)  combustion

b)  combustion

13.

Yes; 2 H2 + O2 → 2 H2O (answers will vary)

15.

a)  C4H9OH + 6 O2 → 4 CO2 + 5 H2O

b)  4 CH3NO2 + 3 O2 → 4 CO2 + 6 H2O + 2 N2

 

23

Oxidation-Reduction Reactions

Learning Objectives

  1. Define oxidation and reduction.
  2. Assign oxidation numbers to atoms in simple compounds.
  3. Recognize a reaction as an oxidation-reduction reaction.

Consider this chemical reaction:

2 Na(s) + Cl2(g) → 2 NaCl

The reactants are elements, and it is assumed that they are electrically neutral; they have the same number of electrons as protons. The product, however, is ionic; it is composed of Na+ and Cl ions. Somehow, the individual sodium atoms as reactants had to lose an electron to make the Na+ ion, while the chlorine atoms as reactants had to each gain an electron to become the Cl ion. This reaction involves the transfer of electrons between atoms.

In reality, electrons are lost by some atoms and gained by other atoms simultaneously. However, mentally we can separate the two processes. Oxidation is defined as the loss of one or more electrons by an atom. Reduction is defined as the gain of one or more electrons by an atom. In reality, oxidation and reduction always occur together; it is only mentally that we can separate them. Chemical reactions that involve the transfer of electrons are called oxidation-reduction (or redox) reactions.

Redox reactions require that we keep track of the electrons assigned to each atom in a chemical reaction. How do we do that? We use an artificial count called the oxidation number to keep track of electrons in atoms. Oxidation numbers are assigned to atoms based on a series of rules. Oxidation numbers are not necessarily equal to the charge on the atom; we must keep the concepts of charge and oxidation numbers separate.

The rules for assigning oxidation numbers to atoms are as follows:

  1. Atoms in their elemental state are assigned an oxidation number of 0.
  2. Atoms in monatomic (i.e., one-atom) ions are assigned an oxidation number equal to their charge. Oxidation numbers are usually written with the sign first, then the magnitude, which differentiates them from charges.
  3. In compounds, fluorine is assigned a −1 oxidation number; oxygen is usually assigned a −2 oxidation number (except in peroxide compounds [where it is −1] and in binary compounds with fluorine [where it is positive]); and hydrogen is usually assigned a +1 oxidation number (except when it exists as the hydride ion, H, in which case rule 2 prevails).
  4. In compounds, all other atoms are assigned an oxidation number so that the sum of the oxidation numbers on all the atoms in the species equals the charge on the species (which is zero if the species is neutral).

Let us work through a few examples for practice. In H2, both hydrogen atoms have an oxidation number of 0, by rule 1. In NaCl, sodium has an oxidation number of +1, while chlorine has an oxidation number of −1, by rule 2. In H2O, the hydrogen atoms each have an oxidation number of +1, while the oxygen has an oxidation number of −2, even though hydrogen and oxygen do not exist as ions in this compound as per rule 3. By contrast, by rule 3 in hydrogen peroxide (H2O2), each hydrogen atom has an oxidation number of +1, while each oxygen atom has an oxidation number of −1. We can use rule 4 to determine oxidation numbers for the atoms in SO2. Each oxygen atom has an oxidation number of −2; for the sum of the oxidation numbers to equal the charge on the species (which is zero), the sulfur atom is assigned an oxidation number of +4. Does this mean that the sulfur atom has a 4+ charge on it? No, it only means that the sulfur atom is assigned a +4 oxidation number by our rules of apportioning electrons among the atoms in a compound.

Example 13

Assign oxidation numbers to the atoms in each substance.

  1. Br2
  2. SiO2
  3. Ba(NO3)2

Solution

  1. Br2 is the elemental form of bromine. Therefore, by rule 1, each atom has an oxidation number of 0.
  2. By rule 3, oxygen is normally assigned an oxidation number of −2. For the sum of the oxidation numbers to equal the charge on the species (which is zero), the silicon atom is assigned an oxidation number of +4.
  3. The compound barium nitrate can be separated into two parts: the Ba2+ ion and the nitrate ion. Considering these separately, the Ba2+ ion has an oxidation number of +2 by rule 2. Now consider the NO3 ion. Oxygen is assigned an oxidation number of −2, and there are three oxygens. According to rule 4, the sum of the oxidation number on all atoms must equal the charge on the species, so we have the simple algebraic equation

    x + 3(−2) = −1

    where x is the oxidation number of the nitrogen atom and −1 represents the charge on the species. Evaluating,

    x + (−6) = −1
    x = +5

    Thus, the oxidation number on the N atom in the nitrate ion is +5.

Test Yourself

Assign oxidation numbers to the atoms in H3PO4.

Answer

H = +1, O = −2, P = +5

All redox reactions occur with a simultaneous change in the oxidation numbers of some atoms. At least two elements must change their oxidation numbers. When an oxidation number of an atom is increased in the course of a redox reaction, that atom is being oxidized. When an oxidation number of an atom is decreased in the course of a redox reaction, that atom is being reduced. Oxidation and reduction are thus also defined in terms of increasing or decreasing oxidation numbers, respectively.

Example 14

Identify what is being oxidized and reduced in this redox equation.

2 Na + Cl2 → 2 NaCl

Solution

Consider the reactants. Because both reactants are the elemental forms of their atoms, the Na and Cl atoms as reactants have oxidation numbers of 0. In the ionic product, the sodium ions have an oxidation number of +1, while the chloride ions have an oxidation number of −1:

NaCl-1

We note that the sodium is increasing its oxidation number from 0 to +1, so it is being oxidized; chlorine is decreasing its oxidation number from 0 to −1, so it is being reduced:

NaCl-2

Because oxidation numbers are changing, this is a redox reaction. Note that the total number of electrons lost by the sodium (two, one lost from each atom) is gained by the chlorine atoms (two, one gained for each atom).

Test Yourself

Identify what is being oxidized and reduced in this redox equation.

C + O2 → CO2

Answer

C is being oxidized from 0 to +4; O is being reduced from 0 to −2.

In this introduction to oxidation-reduction reactions, we are using straightforward examples. However, oxidation reactions can become quite complex; the following equation represents a redox reaction:

Redox Reaction

To demonstrate that this is a redox reaction, the oxidation numbers of the species being oxidized and reduced are listed; can you determine what is being oxidized and what is being reduced? This is also an example of a net ionic reaction; spectator ions that do not change oxidation numbers are not displayed in the equation.

Food and Drink App: Acids in Foods

Many foods and beverages contain acids. Acids impart a sour note to the taste of foods, which may add some pleasantness to the food. For example, orange juice contains citric acid, H3C6H5O7. Note how this formula shows hydrogen atoms in two places; the first hydrogen atoms written are the hydrogen atoms that can form H+ ions, while the second hydrogen atoms written are part of the citrate ion, C6H5O73−. Lemons and limes contain much more citric acid—about 60 times as much—which accounts for these citrus fruits being more sour than most oranges. Vinegar is essentially a ~5% solution of acetic acid (HC2H3O2) in water. Apples contain malic acid (H2C4H4O5; the name malic acid comes from the apple’s botanical genus name, malus), while lactic acid (HC3H5O3) is found in wine and sour milk products, such as yogurt and some cottage cheeses.

Table 4.2 “Various Acids Found in Food and Beverages” lists some acids found in foods, either naturally or as an additive. Frequently, the salts of acid anions are used as additives, such as monosodium glutamate (MSG), which is the sodium salt derived from glutamic acid. As you read the list, you should come to the inescapable conclusion that it is impossible to avoid acids in food and beverages.

Table 4.2 Various Acids Found in Food and Beverages

Acid Name Acid Formula Use and Appearance
acetic acid HC2H3O2 flavouring; found in vinegar
adipic acid H2C6H8O4 flavouring; found in processed foods and some antacids
alginic acid various thickener; found in drinks, ice cream, and weight loss products
ascorbic acid HC6H7O6 antioxidant, also known as vitamin C; found in fruits and vegetables
benzoic acid HC6H5CO2 preservative; found in processed foods
citric acid H3C6H5O7 flavouring; found in citrus fruits
dehydroacetic acid HC8H7O4 preservative, especially for strawberries and squash
erythrobic acid HC6H7O6 antioxidant; found in processed foods
fatty acids various thickener and emulsifier; found in processed foods
fumaric acid H2C4H2O4 flavouring; acid reactant in some baking powders
glutamic acid H2C5H7NO4 flavouring; found in processed foods and in tomatoes, some cheeses, and soy products
lactic acid HC3H5O3 flavouring; found in wine, yogurt, cottage cheese, and other sour milk products
malic acid H2C4H4O5 flavouring; found in apples and unripe fruit
phosphoric acid H3PO4 flavouring; found in some colas
propionic acid HC3H5O2 preservative; found in baked goods
sorbic acid HC6H7O2 preservative; found in processed foods
stearic acid HC18H35O2 anticaking agent; found in hard candies
succinic acid H2C4H4O4 flavouring; found in wine and beer
tartaric acid H2C4H4O6 flavouring; found in grapes, bananas, and tamarinds

Key Takeaways

  • Oxidation-reduction (redox) reactions involve the transfer of electrons from one atom to another.
  • Oxidation numbers are used to keep track of electrons in atoms.
  • There are rules for assigning oxidation numbers to atoms.
  • Oxidation is an increase of oxidation number (a loss of electrons); reduction is a decrease in oxidation number (a gain of electrons).

 

Exercises

  1. Is the reaction

    2 K(s) + Br2(ℓ) → 2 KBr(s)

    an oxidation-reduction reaction? Explain your answer.

  2. Is the reaction

    NaCl(aq) + AgNO3(aq) → NaNO3(aq) + AgCl(s)

    an oxidation-reduction reaction? Explain your answer.

  3. In the reaction

    2 Ca(s) + O2(g) → 2 CaO

    indicate what has lost electrons and what has gained electrons.

  4. In the reaction

    16 Fe(s) + 3 S8(s) → 8 Fe2S3(s)

    indicate what has lost electrons and what has gained electrons.

  5. In the reaction

    2 Li(s) + O2(g) → Li2O2(s)

    indicate what has been oxidized and what has been reduced.

  6. In the reaction

    2 Ni(s) + 3 I2(s) → 2 NiI3(s)

    indicate what has been oxidized and what has been reduced.

  7. What are two different definitions of oxidation?

  8. What are two different definitions of reduction?

  9. Assign oxidation numbers to each atom in each substance.

a)  P4

b)  SO2

c)  SO22−

d)  Ca(NO3)2

10.  Assign oxidation numbers to each atom in each substance.

a)  PF5

b)  (NH4)2S

c)  Hg

d)  Li2O2 (lithium peroxide)

 

11.  Assign oxidation numbers to each atom in each substance.

a)  CO

b)  CO2

c)  NiCl2

d)  NiCl3

 

12.  Assign oxidation numbers to each atom in each substance.

a)  NaH (sodium hydride)

b)  NO2

c)  NO2

d)  AgNO3

 

13.  Assign oxidation numbers to each atom in each substance.

a)  CH2O

b)  NH3

c)  Rb2SO4

d)  Zn(C2H3O2)2

 

14.  Assign oxidation numbers to each atom in each substance.

a)  C6H6

b)  B(OH)3

c)  Li2S

d)  Au

 

15.  Identify what is being oxidized and reduced in this redox equation by assigning oxidation numbers to the atoms.

2 NO + Cl2 → 2 NOCl

 

16.  Identify what is being oxidized and reduced in this redox equation by assigning oxidation numbers to the atoms.

Fe + SO3 → FeSO3

 

17.  Identify what is being oxidized and reduced in this redox equation by assigning oxidation numbers to the atoms.

2 KrF2 + 2 H2O → 2 Kr + 4 HF + O2

 

18.  Identify what is being oxidized and reduced in this redox equation by assigning oxidation numbers to the atoms.

SO3 + SCl2 → SOCl2 + SO2

 

19.  Identify what is being oxidized and reduced in this redox equation by assigning oxidation numbers to the atoms.

2 K + MgCl2 → 2 KCl + Mg

 

20.  Identify what is being oxidized and reduced in this redox equation by assigning oxidation numbers to the atoms.

C7H16 + 11 O2 → 7 CO2 + 8 H2O

Answers

1.

Yes; both K and Br are changing oxidation numbers.

3.

Ca has lost electrons, and O has gained electrons.

5.

Li has been oxidized, and O has been reduced.

7.

loss of electrons; increase in oxidation number

9.

a)  P: 0

b)  S: +4; O: −2

c)  S: +2; O: −2

d)  Ca: 2+; N: +5; O: −2

11.

a)  C: +2; O: −2

b)  C: +4; O: −2

c)  Ni: +2; Cl: −1

d)  Ni: +3; Cl: −1

13.

a)  C: 0; H: +1; O: −2

b)  N: −3; H: +1

c)  Rb: +1; S: +6; O: −2

d)  Zn: +2; C: 0; H: +1; O: −2

15.

N is being oxidized, and Cl is being reduced.

17.

O is being oxidized, and Kr is being reduced.

19.

K is being oxidized, and Mg is being reduced.

 

24

Neutralization Reactions

Learning Objectives

  1. Identify an acid and a base.
  2. Identify a neutralization reaction and predict its products.

In Chapter 3 “Atoms, Molecules, and Ions”, Section 3.5 “Acids”, we defined an acid as an ionic compound that contains H+ as the cation. This is slightly incorrect, but until additional concepts were developed, a better definition needed to wait. Now we can redefine an acid: an acid is any compound that increases the amount of hydrogen ion (H+) in an aqueous solution. The chemical opposite of an acid is a base. The equivalent definition of a base is that a base is a compound that increases the amount of hydroxide ion (OH) in an aqueous solution. These original definitions were proposed by Arrhenius (the same person who proposed ion dissociation) in 1884, so they are referred to as the Arrhenius definition of an acid and a base, respectively.

You may recognize that, based on the description of a hydrogen atom, an H+ ion is a hydrogen atom that has lost its lone electron; that is, H+ is simply a proton. Do we really have bare protons moving about in aqueous solution? No. What is more likely is that the H+ ion has attached itself to one (or more) water molecule(s). To represent this chemically, we define the hydronium ion H3O+(aq), a water molecule with an extra hydrogen ion attached to it. as H3O+, which represents an additional proton attached to a water molecule. We use the hydronium ion as the more logical way a hydrogen ion appears in an aqueous solution, although in many chemical reactions H+ and H3O+ are treated equivalently.

The reaction of an acid and a base is called a neutralization reaction. Although acids and bases have their own unique chemistries, the acid and base cancel each other’s chemistry to produce a rather innocuous substance—water. In fact, the general reaction between an acid and a base is

acid + base → water + salt

where the term salt is generally used to define any ionic compound (soluble or insoluble) that is formed from a reaction between an acid and a base. (In chemistry, the word salt refers to more than just table salt.) For example, the balanced chemical equation for the reaction between HCl(aq) and KOH(aq) is

HCl(aq) + KOH(aq) → H2O(ℓ) + KCl(aq)

where the salt is KCl. By counting the number of atoms of each element, we find that only one water molecule is formed as a product. However, in the reaction between HCl(aq) and Mg(OH)2(aq), additional molecules of HCl and H2O are required to balance the chemical equation:

2 HCl(aq) + Mg(OH)2(aq) → 2 H2O(ℓ) + MgCl2(aq)

Here, the salt is MgCl2. (This is one of several reactions that take place when a type of antacid—a base—is used to treat stomach acid.)

Example 11

Write the neutralization reactions between each acid and base.

  1. HNO3(aq) and Ba(OH)2(aq)
  2. H3PO4(aq) and Ca(OH)2(aq)

Solution

First, we will write the chemical equation with the formulas of the reactants and the expected products; then we will balance the equation.

  1. The expected products are water and barium nitrate, so the initial chemical reaction is

    HNO3(aq) + Ba(OH)2(aq) → H2O(ℓ) + Ba(NO3)2(aq)

    To balance the equation, we need to realize that there will be two H2O molecules, so two HNO3 molecules are required:

    2HNO3(aq) + Ba(OH)2(aq) → 2H2O(ℓ) + Ba(NO3)2(aq)

    This chemical equation is now balanced.

  2. The expected products are water and calcium phosphate, so the initial chemical equation is

    H3PO4(aq) + Ca(OH)2(aq) → H2O(ℓ) + Ca3(PO4)2(s)

    According to the solubility rules, Ca3(PO4)2 is insoluble, so it has an (s) phase label. To balance this equation, we need two phosphate ions and three calcium ions; we end up with six water molecules to balance the equation:

    2 H3PO4(aq) + 3 Ca(OH)2(aq) → 6 H2O(ℓ) + Ca3(PO4)2(s)

    This chemical equation is now balanced.

Test Yourself

Write the neutralization reaction between H2SO4(aq) and Sr(OH)2(aq).

Answer

H2SO4(aq) + Sr(OH)2(aq) → 2 H2O(ℓ) + SrSO4(aq)

Neutralization reactions are one type of chemical reaction that proceeds even if one reactant is not in the aqueous phase. For example, the chemical reaction between HCl(aq) and Fe(OH)3(s) still proceeds according to the equation

3 HCl(aq) + Fe(OH)3(s) → 3 H2O(ℓ) + FeCl3(aq)

even though Fe(OH)3 is not soluble. When one realizes that Fe(OH)3(s) is a component of rust, this explains why some cleaning solutions for rust stains contain acids—the neutralization reaction produces products that are soluble and wash away. (Washing with acids like HCl is one way to remove rust and rust stains, but HCl must be used with caution!)

Complete and net ionic reactions for neutralization reactions will depend on whether the reactants and products are soluble, even if the acid and base react. For example, in the reaction of HCl(aq) and NaOH(aq),

HCl(aq) + NaOH(aq) → H2O(ℓ) + NaCl(aq)

the complete ionic reaction is

H+(aq) + Cl(aq) + Na+(aq) + OH(aq) → H2O(ℓ) + Na+(aq) + Cl(aq)

The Na+(aq) and Cl(aq) ions are spectator ions, so we can remove them to have

H+(aq) + OH(aq) → H2O(ℓ)

as the net ionic equation. If we wanted to write this in terms of the hydronium ion, H3O+(aq), we would write it as

H3O+(aq) + OH(aq) → 2H2O(ℓ)

With the exception of the introduction of an extra water molecule, these two net ionic equations are equivalent.

However, for the reaction between HCl(aq) and Cr(OH)2(s), because chromium(II) hydroxide is insoluble, we cannot separate it into ions for the complete ionic equation:

2 H+(aq) + 2 Cl(aq) + Cr(OH)2(s) → 2 H2O(ℓ) + Cr2+(aq) + 2 Cl(aq)

The chloride ions are the only spectator ions here, so the net ionic equation is

2 H+(aq) + Cr(OH)2(s) → 2 H2O(ℓ) + Cr2+(aq)

Example 12

Oxalic acid, H2C2O4(s), and Ca(OH)2(s) react very slowly. What is the net ionic equation between these two substances if the salt formed is insoluble? (The anion in oxalic acid is the oxalate ion, C2O42−.)

Solution

The products of the neutralization reaction will be water and calcium oxalate:

H2C2O4(s) + Ca(OH)2(s) → 2 H2O(ℓ) + CaC2O4(s)

Because nothing is dissolved, there are no substances to separate into ions, so the net ionic equation is the equation of the three solids and one liquid.

Test Yourself

What is the net ionic equation between HNO3(aq) and Ti(OH)4(s)?

Answer

4 H+(aq) + Ti(OH)4(s) → 4 H2O(ℓ) + Ti4+(aq)

Key Takeaways

  • The Arrhenius definition of an acid is a substance that increases the amount of H+ in an aqueous solution.
  • The Arrhenius definition of a base is a substance that increases the amount of OH in an aqueous solution.
  • Neutralization is the reaction of an acid and a base, which forms water and a salt.
  • Net ionic equations for neutralization reactions may include solid acids, solid bases, solid salts, and water.

 

Exercises

  1. What is the Arrhenius definition of an acid?

  2. What is the Arrhenius definition of a base?

  3. Predict the products of each acid-base combination listed. Assume that a neutralization reaction occurs.

a)  HCl and KOH

b)  H2SO4 and KOH

c)  H3PO4 and Ni(OH)2

 

4.  Predict the products of each acid-base combination listed. Assume that a neutralization reaction occurs.

a)  HBr and Fe(OH)3

b)  HNO2 and Al(OH)3

c)  HClO3 and Mg(OH)2

 

5.  Write a balanced chemical equation for each neutralization reaction in Exercise 3.

 

6.  Write a balanced chemical equation for each neutralization reaction in Exercise 4.

 

7.  Write a balanced chemical equation for the neutralization reaction between each given acid and base. Include the proper phase labels.

a)  HI(aq) + KOH(aq) → ?

b)  H2SO4(aq) + Ba(OH)2(aq) → ?

 

8.  Write a balanced chemical equation for the neutralization reaction between each given acid and base. Include the proper phase labels.

a)  HNO3(aq) + Fe(OH)3(s) → ?

b)  H3PO4(aq) + CsOH(aq) → ?

 

9.  Write the net ionic equation for each neutralization reaction in Exercise 7.

 

10.  Write the net ionic equation for each neutralization reaction in Exercise 8.

 

11.  Write the complete and net ionic equations for the neutralization reaction between HClO3(aq) and Zn(OH)2(s). Assume the salt is soluble.

 

12.  Write the complete and net ionic equations for the neutralization reaction between H2C2O4(s) and Sr(OH)2(aq). Assume the salt is insoluble.

 

13.  Explain why the net ionic equation for the neutralization reaction between HCl(aq) and KOH(aq) is the same as the net ionic equation for the neutralization reaction between HNO3(aq) and RbOH.

 

14.  Explain why the net ionic equation for the neutralization reaction between HCl(aq) and KOH(aq) is different from the net ionic equation for the neutralization reaction between HCl(aq) and AgOH.

 

15.  Write the complete and net ionic equations for the neutralization reaction between HCl(aq) and KOH(aq) using the hydronium ion in place of H+. What difference does it make when using the hydronium ion?

 

16.  Write the complete and net ionic equations for the neutralization reaction between HClO3(aq) and Zn(OH)2(s) using the hydronium ion in place of H+. Assume the salt is soluble. What difference does it make when using the hydronium ion?

Answers

  1. An Arrhenius acid increases the amount of H+ ions in an aqueous solution.

3.

a)  KCl and H2O

b)  K2SO4 and H2O

c)  Ni3(PO4)2 and H2O

5.

a)    HCl + KOH → KCl + H2O

b)  H2SO4 + 2 KOH → K2SO4 + 2 H2O

c)  2 H3PO4 + 3 Ni(OH)2 → Ni3(PO4)2 + 6 H2O

 

7.

a)  HI(aq) + KOH(aq) → KCl(aq) + H2O(ℓ)

b)  H2SO4(aq) + Ba(OH)2(aq) → BaSO4(s) + 2 H2O(ℓ)

 

9.

a)  H+(aq) + OH(aq) → H2O(ℓ)

b)  2 H+(aq) + SO42−(aq) + Ba2+(aq) + 2 OH(aq) → BaSO4(s) + 2 H2O(ℓ)

11.  Complete ionic equation:

 

2 H+(aq) + 2 ClO3(aq) + Zn2+(aq) + 2 OH(aq) → Zn2+(aq) + 2 ClO3(aq) + 2 H2O(ℓ)

Net ionic equation:

2 H+(aq) + 2 OH(aq) → 2 H2O(ℓ)

 

13.  Because the salts are soluble in both cases, the net ionic reaction is just H+(aq) + OH(aq) → H2O(ℓ).

 

15.  Complete ionic equation:

H3O+(aq) + Cl(aq) + K+(aq) + OH(aq) → 2 H2O(ℓ) + K+(aq) + Cl(aq)

Net ionic equation:

H3O+(aq) + OH(aq) → 2 H2O(ℓ)

The difference is simply the presence of an extra water molecule as a product.

 

25

End-of-Chapter Material

Additional Exercises

  1. Chemical equations can also be used to represent physical processes. Write a chemical reaction for the boiling of water, including the proper phase labels.

  2. Chemical equations can also be used to represent physical processes. Write a chemical reaction for the freezing of water, including the proper phase labels.

  3. Explain why

    4 Na(s) + 2 Cl2(g) → 4 NaCl(s)

    should not be considered a proper chemical equation.

  4. Explain why

    H2(g) + 1/2 O2(g) → H2O(ℓ)

    should not be considered a proper chemical equation.

  5. Does the chemical reaction represented by

    3 Zn(s) + 2 Al(NO3)3(aq) → 3 Zn(NO3)2(aq) + 2 Al(s)

    proceed as written? Why or why not?

  6. Does the chemical reaction represented by

    2 Au(s) + 2 HNO3(aq) → 2 AuNO3(aq) + H2(g)

    proceed as written? Gold is a relatively useful metal for certain applications, such as jewelry and electronics. Does your answer suggest why this is so?

  7. Explain what is wrong with this double-replacement reaction.

    NaCl(aq) + KBr(aq) → NaK(aq) + ClBr(aq)

  8. Predict the products of and balance this double-replacement reaction.

    Ag2SO4(aq) + SrCl2(aq) → ?

  9. Write the complete and net ionic equations for this double-replacement reaction.

    BaCl2(aq) + Ag2SO4(aq) → ?

  10. Write the complete and net ionic equations for this double-replacement reaction.

    Ag2SO4(aq) + SrCl2(aq) → ?

  11. Identify the spectator ions in this reaction. What is the net ionic equation?

    NaCl(aq) + KBr(aq) → NaBr(aq) + KCl(aq)

  12. Complete this reaction and identify the spectator ions. What is the net ionic equation?

    3 H2SO4(aq) + 2 Al(OH)3(s) → ?

  13. Can a reaction be a composition reaction and a redox reaction at the same time? Give an example to support your answer.

  14. Can a reaction be a combustion reaction and a redox reaction at the same time? Give an example to support your answer.

  15. Can a reaction be a decomposition reaction and a redox reaction at the same time? Give an example to support your answer.

  16. Can a reaction be a combustion reaction and a double-replacement reaction at the same time? Give an example to support your answer.

  17. Why is CH4 not normally considered an acid?

  18. Methyl alcohol has the formula CH3OH. Why would methyl alcohol not normally be considered a base?

  19. What are the oxidation numbers of the nitrogen atoms in these substances?

a)  N2

b)  NH3

c)  NO

d)  N2O

e)  NO2

f)  N2O4

g)  N2O5

h)  NaNO3

 

20.  What are the oxidation numbers of the sulfur atoms in these substances?

a)  SF6

b)  Na2SO4

c)  K2SO3

d)  SO3

e)  SO2

f)  S8

g)  Na2S

 

21.  Disproportion is a type of redox reaction in which the same substance is both oxidized and reduced. Identify the element that is disproportionating and indicate the initial and final oxidation numbers of that element.

2 CuCl(aq) → CuCl2(aq) + Cu(s)

 

22. Disproportion is a type of redox reaction in which the same substance is both oxidized and reduced. Identify the element that is disproportionating and indicate the initial and final oxidation numbers of that element.

3 Cl2(g) + 6 OH(aq) → 5 Cl(aq) + ClO3(aq) + 3 H2O(ℓ)

 

Answers

1.

H2O(ℓ) → H2O(g)

3.

The coefficients are not in their lowest whole-number ratio.

5.

No; zinc is lower in the activity series than aluminum.

7.

In the products, the cation is pairing with the cation, and the anion is pairing with the anion.

9.

Complete ionic equation: Ba2+(aq) + 2 Cl(aq) + 2 Ag+(aq) + SO42−(aq) → BaSO4(s) + 2 AgCl(s)

Net ionic equation: The net ionic equation is the same as the complete ionic equation.

11.

Each ion is a spectator ion; there is no overall net ionic equation.

13.

Yes; H2 + Cl2 → 2 HCl (answers will vary)

15.

Yes; 2 HCl → H2 + Cl2 (answers will vary)

17.

It does not increase the H+ ion concentration; it is not a compound of H+.

19.

a)  0

b)  −3

c)  +2

d)  +1

e)  +4

f) +4

g)  +5

h)  +5

21.

Copper is disproportionating. Initially, its oxidation number is +1; in the products, its oxidation numbers are +2 and 0, respectively.

V

Chapter 5. Stoichiometry and the Mole

26

Introduction to Stoichiometry and the Mole

At Contrived State University in Anytown, Ohio, a new building was dedicated in March 2010 to house the College of Education. The 100,000-square-foot building has enough office space to accommodate 86 full-time faculty members and 167 full-time staff.

In a fit of monetary excess, the university administration offered to buy new furniture (desks and chairs) and computer workstations for all faculty and staff members moving into the new building. However, to save on long-term energy and materials costs, the university offered to buy only 1 laser printer per 10 employees, with the plan to network the printers together.

How many laser printers did the administration have to buy? It is rather simple to show that 26 laser printers are needed for all the employees. However, what if a chemist was calculating quantities for a chemical reaction? Interestingly enough, similar calculations can be performed for chemicals as well as laser printers.

In filling a new office building with furniture and equipment, managers do calculations similar to those performed by scientists doing chemical reactions.

We have already established that quantities are important in science, especially in chemistry. It is important to make accurate measurements of a variety of quantities when performing experiments. However, it is also important to be able to relate one measured quantity to another, unmeasured quantity. In this chapter, we will consider how we manipulate quantities to relate them to each other.

27

Stoichiometry

Learning Objectives

  1. Define stoichiometry.
  2. Relate quantities in a balanced chemical reaction on a molecular basis.

Consider a classic recipe for pound cake: 1 pound of eggs, 1 pound of butter, 1 pound of flour, and 1 pound of sugar. (That’s why it’s called “pound cake.”) If you have 4 pounds of butter, how many pounds of sugar, flour, and eggs do you need? You would need 4 pounds each of sugar, flour, and eggs.

Now suppose you have 1.00 g H2. If the chemical reaction follows the balanced chemical equation

2 H2(g) + O2(g) → 2 H2O(ℓ)

then what mass of oxygen do you need to make water?

Curiously, this chemical reaction question is very similar to the pound cake question. Both of them involve relating a quantity of one substance to a quantity of another substance or substances. The relating of one chemical substance to another using a balanced chemical reaction is called stoichiometry. Using stoichiometry is a fundamental skill in chemistry; it greatly broadens your ability to predict what will occur and, more importantly, how much is produced.

Let us consider a more complicated example. A recipe for pancakes calls for 2 cups (c) of pancake mix, 1 egg, and 1/2 c of milk. We can write this in the form of a chemical equation:

2 c mix + 1 egg + 1/2 c milk → 1 batch of pancakes

If you have 9 c of pancake mix, how many eggs and how much milk do you need? It might take a little bit of work, but eventually you will find you need 4½ eggs and 2¼ c milk.

How can we formalize this? We can make a conversion factor using our original recipe and use that conversion factor to convert from a quantity of one substance to a quantity of another substance, similar to the way we constructed a conversion factor between feet and yards in Chapter 2 “Measurements”. Because one recipe’s worth of pancakes requires 2 c of pancake mix, 1 egg, and 1/2 c of milk, we actually have the following mathematical relationships that relate these quantities:

2 c pancake mix ⇔ 1 egg ⇔ 1/2 c milk

where ⇔ is the mathematical symbol for “is equivalent to.” This does not mean that 2 c of pancake mix equal 1 egg. However, as far as this recipe is concerned, these are the equivalent quantities needed for a single recipe of pancakes. So, any possible quantities of two or more ingredients must have the same numerical ratio as the ratios in the equivalence.

We can deal with these equivalences in the same way we deal with equalities in unit conversions: we can make conversion factors that essentially equal 1. For example, to determine how many eggs we need for 9 c of pancake mix, we construct the conversion factor:

Screen Shot 2014-07-17 at 10.44.33 AM

This conversion factor is, in a strange way, equivalent to 1 because the recipe relates the two quantities. Starting with our initial quantity and multiplying by our conversion factor,

Screen Shot 2014-07-17 at 10.45.13 AM

Note how the units cups pancake mix canceled, leaving us with units of eggs. This is the formal, mathematical way of getting our amounts to mix with 9 c of pancake mix. We can use a similar conversion factor for the amount of milk:

Screen Shot 2014-07-17 at 10.45.40 AM

Again, units cancel, and new units are introduced.

A balanced chemical equation is nothing more than a recipe for a chemical reaction. The difference is that a balanced chemical equation is written in terms of atoms and molecules, not cups, pounds, and eggs.

For example, consider the following chemical equation:

2 H2(g) + O2(g) → 2 H2O(ℓ)

We can interpret this as, literally, “two hydrogen molecules react with one oxygen molecule to make two water molecules.” That interpretation leads us directly to some equivalences, just as our pancake recipe did:

2 H2 molecules ⇔ 1 O2 molecule ⇔ 2 H2O molecules

These equivalences allow us to construct conversion factors:

Screen Shot 2014-07-17 at 10.46.06 AM

and so forth. These conversions can be used to relate quantities of one substance to quantities of another. For example, suppose we need to know how many molecules of oxygen are needed to react with 16 molecules of H2. As we did with converting units, we start with our given quantity and use the appropriate conversion factor:

Screen Shot 2014-07-17 at 10.46.31 AM

Note how the unit molecules H2 cancels algebraically, just as any unit does in a conversion like this. The conversion factor came directly from the coefficients in the balanced chemical equation. This is another reason why a properly balanced chemical equation is important.

Example 1

How many molecules of SO3 are needed to react with 144 molecules of Fe2O3 given this balanced chemical equation?

Fe2O3(s) + 3SO3(g) → Fe2(SO4)3

Solution

We use the balanced chemical equation to construct a conversion factor between Fe2O3 and SO3. The number of molecules of Fe2O3 goes on the bottom of our conversion factor so it cancels with our given amount, and the molecules of SO3 go on the top. Thus, the appropriate conversion factor is

ss2

Starting with our given amount and applying the conversion factor, the result is

ss1

We need 432 molecules of SO3 to react with 144 molecules of Fe2O3.

Test Yourself

How many molecules of H2 are needed to react with 29 molecules of N2 to make ammonia if the balanced chemical equation is N2 + 3H2 → 2NH3?

Answer

87 molecules

Chemical equations also allow us to make conversions regarding the number of atoms in a chemical reaction because a chemical formula lists the number of atoms of each element in a compound. The formula H2O indicates that there are two hydrogen atoms and one oxygen atom in each molecule, and these relationships can be used to make conversion factors:

Screen Shot 2014-07-17 at 10.49.14 AM

Conversion factors like this can also be used in stoichiometry calculations.

Example 2

How many molecules of NH3 can you make if you have 228 atoms of H2?

Solution

From the formula, we know that one molecule of NH3 has three H atoms. Use that fact as a conversion factor:

Screen-Shot-2014-07-17-at-10.49.40-AM

Test Yourself

How many molecules of Fe2(SO4)3 can you make from 777 atoms of S?

Answer

259 molecules

qrcode.23437574

Video Source: Stiochiometry by keyj (https://viuvideos.viu.ca/media/Stoichiometry/0_b6t2h6o2)

Key Takeaway

  • Quantities of substances can be related to each other using balanced chemical equations.

Exercises

  1. Think back to the pound cake recipe. What possible conversion factors can you construct relating the components of the recipe?

  2. Think back to the pancake recipe. What possible conversion factors can you construct relating the components of the recipe?

  3. What are all the conversion factors that can be constructed from the balanced chemical reaction 2H2(g) + O2(g) → 2H2O(ℓ)?

  4. What are all the conversion factors that can be constructed from the balanced chemical reaction N2(g) + 3H2(g) → 2NH3(g)?

  5. Given the chemical equation

    Na(s) + H2O(ℓ) → NaOH(aq) + H2(g)

a)  Balance the equation.

b)  How many molecules of H2 are produced when 332 atoms of Na react?

6.  Given the chemical equation

S(s) + O2(g) → SO3(g)

a)  Balance the equation.

b)  How many molecules of O2 are needed when 38 atoms of S react?

7.  For the balanced chemical equation

6 H+(aq) + 2 MnO4(aq) + 5 H2O2(ℓ) → 2 Mn2+(aq) + 5 O2(g) + 8 H2O(ℓ)

how many molecules of H2O are produced when 75 molecules of H2O2 react?

8.  For the balanced chemical reaction

2 C6H6(ℓ) + 15 O2(g) → 12 CO2(g) + 6 H2O(ℓ)

how many molecules of CO2 are produced when 56 molecules of C6H6 react?

9.  Given the balanced chemical equation

Fe2O3(s) + 3SO3(g) → Fe2(SO4)3

how many molecules of Fe2(SO4)3 are produced if 321 atoms of S are reacted?

10.  For the balanced chemical equation

CuO(s) + H2S(g) → CuS + H2O(ℓ)

how many molecules of CuS are formed if 9,044 atoms of H react?

11.  For the balanced chemical equation

Fe2O3(s) + 3 SO3(g) → Fe2(SO4)3

suppose we need to make 145,000 molecules of Fe2(SO4)3. How many molecules of SO3 do we need?

12.  One way to make sulfur hexafluoride is to react thioformaldehyde, CH2S, with elemental fluorine:

CH2S + 6 F2 → CF4 + 2 HF + SF6

If 45,750 molecules of SF6 are needed, how many molecules of F2 are required?

13.  Construct the three independent conversion factors possible for these two reactions:

2 H2 + O2 → 2 H2O

H2 + O2 → H2O2

Why are the ratios between H2 and O2 different?

The conversion factors are different because the stoichiometries of the balanced chemical reactions are different.

14.  Construct the three independent conversion factors possible for these two reactions:

2 Na + Cl2 → 2 NaCl

4 Na + 2 Cl2 → 4 NaCl

What similarities, if any, exist in the conversion factors from these two reactions?

Answers

1.

Screen Shot 2014-07-17 at 10.50.06 AM are two conversion factors that can be constructed from the pound cake recipe. Other conversion factors are also possible.

3.

Screen Shot 2014-07-17 at 10.51.35 AM and their reciprocals are the conversion factors that can be constructed.

5.

2Na(s) + 2H2O(ℓ) → 2NaOH(aq) + H2(g) and 166 molecules

7.
120 molecules
9.
107 molecules
11.
435,000 molecules
13.
Screen Shot 2014-07-17 at 10.52.09 AM

28

The Mole

Learning Objectives

  1. Describe the unit mole.
  2. Relate the mole quantity of substance to its mass.

So far, we have been talking about chemical substances in terms of individual atoms and molecules. Yet we don’t typically deal with substances an atom or a molecule at a time; we work with millions, billions, and trillions of atoms and molecules at a time. What we need is a way to deal with macroscopic, rather than microscopic, amounts of matter. We need a unit of amount that relates quantities of substances on a scale that we can interact with.

Chemistry uses a unit called mole. A mole (mol) is a number of things equal to the number of atoms in exactly 12 g of carbon-12. Experimental measurements have determined that this number is very large:

1 mol = 6.02214179 × 1023 things

Understand that a mole means a number of things, just like a dozen means a certain number of things—twelve, in the case of a dozen. But a mole is a much larger number of things. These things can be atoms, or molecules, or eggs; however, in chemistry, we usually use the mole to refer to the amounts of atoms or molecules. Although the number of things in a mole is known to eight decimal places, it is usually fine to use only two or three decimal places in calculations. The numerical value of things in a mole is often called Avogadro’s number (NA), which is also known as the Avogadro constant, after Amadeo Avogadro, an Italian chemist who first proposed its importance.

Example 3

How many molecules are present in 2.76 mol of H2O? How many atoms is this?

Solution

The definition of a mole is an equality that can be used to construct a conversion factor. Also, because we know that there are three atoms in each molecule of H2O, we can also determine the number of atoms in the sample.

ss3

 

To determine the total number of atoms, we have

ss4

Test Yourself

How many molecules are present in 4.61 × 10−2 mol of O2?

Answer

2.78 × 1022 molecules

How big is a mole? It is very large. Suppose you had a mole of dollar bills that need to be counted. If everyone on earth (about 6 billion people) counted one bill per second, it would take about 3.2 million years to count all the bills. A mole of sand would fill a cube about 32 km on a side. A mole of pennies stacked on top of each other would have about the same diameter as our galaxy, the Milky Way. A mole is a lot of things—but atoms and molecules are very tiny. One mole of carbon atoms would make a cube that is 1.74 cm on a side, small enough to carry in your pocket.

Why is the mole unit so important? It represents the link between the microscopic and the macroscopic, especially in terms of mass. A mole of a substance has the same mass in grams as one unit (atom or molecules) has in atomic mass units. The mole unit allows us to express amounts of atoms and molecules in visible amounts that we can understand.

For example, we already know that, by definition, a mole of carbon has a mass of exactly 12 g. This means that exactly 12 g of C has 6.022 × 1023 atoms:

12 g C = 6.022 × 1023 atoms C

We can use this equality as a conversion factor between the number of atoms of carbon and the number of grams of carbon. How many grams are there, say, in 1.50 × 1025 atoms of carbon? This is a one-step conversion:

Screen Shot 2014-07-20 at 3.09.54 PM

But it also goes beyond carbon. Previously we defined atomic and molecular masses as the number of atomic mass units per atom or molecule. Now we can do so in terms of grams. The atomic mass of an element is the number of grams in 1 mol of atoms of that element, while the molecular mass of a compound is the number of grams in 1 mol of molecules of that compound. Sometimes these masses are called molar masses to emphasize the fact that they are the mass for 1 mol of things. (The term molar is the adjective form of mole and has nothing to do with teeth.)

Here are some examples. The mass of a hydrogen atom is 1.0079 u; the mass of 1 mol of hydrogen atoms is 1.0079 g. Elemental hydrogen exists as a diatomic molecule, H2. One molecule has a mass of 1.0079 + 1.0079 = 2.0158 u, while 1 mol H2 has a mass of 2.0158 g. A molecule of H2O has a mass of about 18.01 u; 1 mol H2O has a mass of 18.01 g. A single unit of NaCl has a mass of 58.45 u; NaCl has a molar mass of 58.45 g. In each of these moles of substances, there are 6.022 × 1023 units: 6.022 × 1023 atoms of H, 6.022 × 1023 molecules of H2 and H2O, 6.022 × 1023 units of NaCl ions. These relationships give us plenty of opportunities to construct conversion factors for simple calculations.

Example 4

What is the molar mass of C6H12O6?

Solution

To determine the molar mass, we simply add the atomic masses of the atoms in the molecular formula but express the total in grams per mole, not atomic mass units. The masses of the atoms can be taken from the periodic table or the list of elements in Chapter 17 “Appendix: Periodic Table of the Elements”:

6 C = 6 × 12.011 = 72.066
12 H = 12 × 1.0079 = 12.0948
6 O = 6 × 15.999 = 95.994
TOTAL = 180.155 g/mol

Per convention, the unit grams per mole is written as a fraction.

Test Yourself

What is the molar mass of AgNO3?

Answer

169.87 g/mol

Knowing the molar mass of a substance, we can calculate the number of moles in a certain mass of a substance and vice versa, as these examples illustrate. The molar mass is used as the conversion factor.

Example 5

What is the mass of 3.56 mol of HgCl2? The molar mass of HgCl2 is 271.49 g/mol.

Solution

Use the molar mass as a conversion factor between moles and grams. Because we want to cancel the mole unit and introduce the gram unit, we can use the molar mass as given:

ss5

Test Yourself

What is the mass of 33.7 mol of H2O?

Answer

607 g

Example 6

How many moles of H2O are present in 240.0 g of water (about the mass of a cup of water)?

Solution

Use the molar mass of H2O as a conversion factor from mass to moles. The molar mass of water is (1.0079 + 1.0079 + 15.999) = 18.015 g/mol. However, because we want to cancel the gram unit and introduce moles, we need to take the reciprocal of this quantity, or 1 mol/18.015 g:

ss6

Test Yourself

How many moles are present in 35.6 g of H2SO4 (molar mass = 98.08 g/mol)?

Answer

0.363 mol

Other conversion factors can be combined with the definition of mole—density, for example.

Example 7

The density of ethanol is 0.789 g/mL. How many moles are in 100.0 mL of ethanol? The molar mass of ethanol is 46.08 g/mol.

Solution

Here, we use density to convert from volume to mass and then use the molar mass to determine the number of moles.

ss7

Test Yourself

If the density of benzene, C6H6, is 0.879 g/mL, how many moles are present in 17.9 mL of benzene?

Answer

0.201 mol

Key Takeaways

  • The mole is a key unit in chemistry.
  • The molar mass of a substance, in grams, is numerically equal to one atom’s or molecule’s mass in atomic mass units.

 

Exercises

  1. How many atoms are present in 4.55 mol of Fe?

  2. How many atoms are present in 0.0665 mol of K?

  3. How many molecules are present in 2.509 mol of H2S?

  4. How many molecules are present in 0.336 mol of acetylene (C2H2)?

  5. How many moles are present in 3.55 × 1024 Pb atoms?

  6. How many moles are present in 2.09 × 1022 Ti atoms?

  7. How many moles are present in 1.00 × 1023 PF3 molecules?

  8. How many moles are present in 5.52 × 1025 penicillin molecules?

  9. Determine the molar mass of each substance.

a)  Si

b)  SiH4

c)  K2O

 

10.  Determine the molar mass of each substance.

a)  Cl2

b)  SeCl2

c)  Ca(C2H3O2)2

11.  Determine the molar mass of each substance.

a)  Al

b)  Al2O3

c)  CoCl3

12.  Determine the molar mass of each substance.

a)  O3

b)  NaI

c)  C12H22O11

 

13.  What is the mass of 4.44 mol of Rb?

 

14.  What is the mass of 0.311 mol of Xe?

 

15.  What is the mass of 12.34 mol of Al2(SO4)3?

 

16.  What is the mass of 0.0656 mol of PbCl2?

 

17.  How many moles are present in 45.6 g of CO?

 

18.  How many moles are present in 0.00339 g of LiF?

 

19.  How many moles are present in 1.223 g of SF6?

 

20.  How many moles are present in 48.8 g of BaCO3?

 

21.  How many moles are present in 54.8 mL of mercury if the density of mercury is 13.6 g/mL?

 

22.  How many moles are present in 56.83 mL of O2 if the density of O2 is 0.00133 g/mL?

Answers

1.

2.74 × 1024 atoms
3.  1.511 × 1024 molecules
5.  90 mol

7.  
0.166 mol

9.

a)  28.086 g

b)  32.118 g

c)  94.195 g

11.

a)  26.981 g

b)  101.959 g

c)  165.292 g

13.  379 g
15.  4,222 g
17.  1.63 mol
19.  0.008374 mol
21.  3.72 mol

29

Mole-Mass and Mass-Mass Calculations

 Learning Objectives

  1. From a given number of moles of a substance, calculate the mass of another substance involved using the balanced chemical equation.
  2. From a given mass of a substance, calculate the moles of another substance involved using the balanced chemical equation.
  3. From a given mass of a substance, calculate the mass of another substance involved using the balanced chemical equation.

Mole-mole calculations are not the only type of calculations that can be performed using balanced chemical equations. Recall that the molar mass can be determined from a chemical formula and used as a conversion factor. We can add that conversion factor as another step in a calculation to make a mole-mass calculation, where we start with a given number of moles of a substance and calculate the mass of another substance involved in the chemical equation, or vice versa.

For example, suppose we have the balanced chemical equation

2 Al + 3 Cl2 → 2 AlCl3

Suppose we know we have 123.2 g of Cl2. How can we determine how many moles of AlCl3 we will get when the reaction is complete? First and foremost, chemical equations are not balanced in terms of grams; they are balanced in terms of moles. So to use the balanced chemical equation to relate an amount of Cl2 to an amount of AlCl3, we need to convert the given amount of Cl2 into moles. We know how to do this by simply using the molar mass of Cl2 as a conversion factor. The molar mass of Cl2 (which we get from the atomic mass of Cl from the periodic table) is 70.90 g/mol. We must invert this fraction so that the units cancel properly:

Screen Shot 2014-07-21 at 5.16.11 PM

Now that we have the quantity in moles, we can use the balanced chemical equation to construct a conversion factor that relates the number of moles of Cl2 to the number of moles of AlCl3. The numbers in the conversion factor come from the coefficients in the balanced chemical equation:

Screen Shot 2014-07-21 at 5.16.18 PM

Using this conversion factor with the molar quantity we calculated above, we get

Screen Shot 2014-07-21 at 5.16.21 PM

So, we will get 1.159 mol of AlCl3 if we react 123.2 g of Cl2.

In this last example, we did the calculation in two steps. However, it is mathematically equivalent to perform the two calculations sequentially on one line:

Screen Shot 2014-07-21 at 5.16.27 PM

The units still cancel appropriately, and we get the same numerical answer in the end. Sometimes the answer may be slightly different from doing it one step at a time because of rounding of the intermediate answers, but the final answers should be effectively the same.

Example 10

How many moles of HCl will be produced when 249 g of AlCl3 are reacted according to this chemical equation?

2 AlCl3 + 3 H2O(ℓ) → Al2O3 + 6 HCl(g)

Solution

We will do this in two steps: convert the mass of AlCl3 to moles and then use the balanced chemical equation to find the number of moles of HCl formed. The molar mass of AlCl3 is 133.33 g/mol, which we have to invert to get the appropriate conversion factor:

187mol

Now we can use this quantity to determine the number of moles of HCl that will form. From the balanced chemical equation, we construct a conversion factor between the number of moles of AlCl3 and the number of moles of HCl:

6molhcl

Applying this conversion factor to the quantity of AlCl3, we get

187mol

Alternatively, we could have done this in one line:

249galcl3

The last digit in our final answer is slightly different because of rounding differences, but the answer is essentially the same.

Test Yourself

How many moles of Al2O3 will be produced when 23.9 g of H2O are reacted according to this chemical equation?

2 AlCl3 + 3 H2O(ℓ) → Al2O3 + 6 HCl(g)

Answer

0.442 mol

A variation of the mole-mass calculation is to start with an amount in moles and then determine an amount of another substance in grams. The steps are the same but are performed in reverse order.

Example 11

How many grams of NH3 will be produced when 33.9 mol of H2 are reacted according to this chemical equation?

N2(g) + 3 H2(g) → 2 NH3(g)

Solution

The conversions are the same, but they are applied in a different order. Start by using the balanced chemical equation to convert to moles of another substance and then use its molar mass to determine the mass of the final substance. In two steps, we have

339molh2

Now, using the molar mass of NH3, which is 17.03 g/mol, we get

226molnh3

Test Yourself

How many grams of N2 are needed to produce 2.17 mol of NH3 when reacted according to this chemical equation?

N2(g) + 3 H2(g) → 2 NH3(g)

Answer

30.4 g (Note: here we go from a product to a reactant, showing that mole-mass problems can begin and end with any substance in the chemical equation.)

It should be a trivial task now to extend the calculations to mass-mass calculations, in which we start with a mass of some substance and end with the mass of another substance in the chemical reaction. For this type of calculation, the molar masses of two different substances must be used—be sure to keep track of which is which. Again, however, it is important to emphasize that before the balanced chemical reaction is used, the mass quantity must first be converted to moles. Then the coefficients of the balanced chemical reaction can be used to convert to moles of another substance, which can then be converted to a mass.

For example, let us determine the number of grams of SO3 that can be produced by the reaction of 45.3 g of SO2 and O2:

2 SO2(g) + O2(g) → 2 SO3(g)

First, we convert the given amount, 45.3 g of SO2, to moles of SO2 using its molar mass (64.06 g/mol):

Screen Shot 2014-07-21 at 5.17.00 PM

Second, we use the balanced chemical reaction to convert from moles of SO2 to moles of SO3:

Screen Shot 2014-07-21 at 5.17.03 PM

Finally, we use the molar mass of SO3 (80.06 g/mol) to convert to the mass of SO3:

Screen Shot 2014-07-21 at 5.17.07 PM

We can also perform all three steps sequentially, writing them on one line as

Screen Shot 2014-07-21 at 5.17.13 PM

We get the same answer. Note how the initial and all the intermediate units cancel, leaving grams of SO3, which is what we are looking for, as our final answer.

Example 12

What mass of Mg will be produced when 86.4 g of K are reacted?

MgCl2(s) + 2 K(s) → Mg(s) + 2 KCl(s)

Solution

We will simply follow the steps

mass K → mol K → mol Mg → mass Mg

In addition to the balanced chemical equation, we need the molar masses of K (39.09 g/mol) and Mg (24.31 g/mol). In one line,

864gk

Test Yourself

What mass of H2 will be produced when 122 g of Zn are reacted?

Zn(s) + 2 HCl(aq) → ZnCl2(aq) + H2(g)

Answer

3.77 g

Key Takeaways

  • Mole quantities of one substance can be related to mass quantities using a balanced chemical equation.
  • Mass quantities of one substance can be related to mass quantities using a balanced chemical equation.
  • In all cases, quantities of a substance must be converted to moles before the balanced chemical equation can be used to convert to moles of another substance.

 

Exercises

  1. What mass of CO2 is produced by the combustion of 1.00 mol of CH4?

    CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(ℓ)

  2. What mass of H2O is produced by the combustion of 1.00 mol of CH4?

    CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(ℓ)

  3. What mass of HgO is required to produce 0.692 mol of O2?

    2 HgO(s) → 2 Hg(ℓ) + O2(g)

  4. What mass of NaHCO3 is needed to produce 2.659 mol of CO2?

    2 NaHCO3(s) → Na2CO3(s) + H2O(ℓ) + CO2(g)

  5. How many moles of Al can be produced from 10.87 g of Ag?

    Al(NO3) 3(s) + 3 Ag → Al + 3 AgNO3

  6. How many moles of HCl can be produced from 0.226 g of SOCl2?

    SOCl2(ℓ) + H2O(ℓ) → SO2(g) + 2HCl(g)

  7. How many moles of O2 are needed to prepare 1.00 g of Ca(NO3)2?

    Ca(s) + N2(g) + 3 O2(g) → Ca(NO3) 2(s)

  8. How many moles of C2H5OH are needed to generate 106.7 g of H2O?

    C2H5OH(ℓ) + 3 O2(g) → 2 CO2(g) + 3 H2O(ℓ)

  9. What mass of O2 can be generated by the decomposition of 100.0 g of NaClO3?

    2 NaClO3 → 2 NaCl(s) + 3 O2(g)

  10. What mass of Li2O is needed to react with 1,060 g of CO2?

    Li2O(aq) + CO2(g) → Li2CO3(aq)

  11. What mass of Fe2O3 must be reacted to generate 324 g of Al2O3?

    Fe2O3(s) + 2 Al(s) → 2 Fe(s) + Al2O3(s)

  12. What mass of Fe is generated when 100.0 g of Al are reacted?

    Fe2O3(s) + 2 Al(s) → 2 Fe(s) + Al2O3(s)

  13. What mass of MnO2 is produced when 445 g of H2O are reacted?

    H2O(ℓ) + 2 MnO4(aq) + Br(aq) → BrO3(aq) + 2 MnO2(s) + 2 OH(aq)

  14. What mass of PbSO4 is produced when 29.6 g of H2SO4 are reacted?

    Pb(s) + PbO2(s) + 2 H2SO4(aq) → 2 PbSO4(s) + 2 H2O(ℓ)

  15. If 83.9 g of ZnO are formed, what mass of Mn2O3 is formed with it?

    Zn(s) + 2 MnO2(s) → ZnO(s) + Mn2O3(s)

  16. If 14.7 g of NO2 are reacted, what mass of H2O is reacted with it?

    3 NO2(g) + H2O(ℓ) → 2 HNO3(aq) + NO(g)

  17. If 88.4 g of CH2S are reacted, what mass of HF is produced?

    CH2S + 6 F2 → CF4 + 2 HF + SF6

  18. If 100.0 g of Cl2 are needed, what mass of NaOCl must be reacted?

    NaOCl + HCl → NaOH + Cl2

Answers

1.

44.0 g

3.

3.00 × 102 g

5.

0.0336 mol

7.

0.0183 mol

9.

45.1 g

11.

507 g

13.

4.30 × 103 g

15.

163 g

17.

76.7 g

 

30

Limiting Reagents

Learning Objectives

  1. Identify a limiting reagent from a set of reactants.
  2. Calculate how much product will be produced from the limiting reagent.
  3. Calculate how much reactant(s) remains when the reaction is complete.

One additional assumption we have made about chemical reactions—in addition to the assumption that reactions proceed all the way to completion—is that all the reactants are present in the proper quantities to react to products. This is not always the case.

Consider Figure 5.2 “Making Water”. Here we are taking hydrogen atoms and oxygen atoms (left) to make water molecules (right). However, there are not enough oxygen atoms to use up all the hydrogen atoms. We run out of oxygen atoms and cannot make any more water molecules, so the process stops when we run out of oxygen atoms.

Figure 5.2 Making Water

Making Water

 

In this scenario for making water molecules, we run out of O atoms before we use up all the H atoms. Similar situations exist for many chemical reactions when one reactant runs out before the other.

A similar situation exists for many chemical reactions: you usually run out of one reactant before all of the other reactant has reacted. The reactant you run out of is called the limiting reagent; the other reactant or reactants are considered to be in excess. A crucial skill in evaluating the conditions of a chemical process is to determine which reactant is the limiting reagent and which is in excess.

The key to recognizing which reactant is the limiting reagent is based on a mole-mass or mass-mass calculation: whichever reactant gives the lesser amount of product is the limiting reagent. What we need to do is determine an amount of one product (either moles or mass) assuming all of each reactant reacts. Whichever reactant gives the least amount of that particular product is the limiting reagent. It does not matter which product we use, as long as we use the same one each time. It does not matter whether we determine the number of moles or grams of that product; however, we will see shortly that knowing the final mass of product can be useful.

For example, consider this reaction:

4 As(s) + 3 O2(g) → 2 As2O3(s)

Suppose we start a reaction with 50.0 g of As and 50.0 g of O2. Which one is the limiting reagent? We need to perform two mole-mass calculations, each assuming that each reactant reacts completely. Then we compare the amount of the product produced by each and determine which is less.

The calculations are as follows:

Screen Shot 2014-07-22 at 1.09.10 PM
Screen Shot 2014-07-22 at 1.09.16 PM

Comparing these two answers, it is clear that 0.334 mol of As2O3 is less than 1.04 mol of As2O3, so arsenic is the limiting reagent. If this reaction is performed under these initial conditions, the arsenic will run out before the oxygen runs out. We say that the oxygen is “in excess.”

Identifying the limiting reagent, then, is straightforward. However, there are usually two associated questions: (1) what mass of product (or products) is then actually formed? and (2) what mass of what reactant is left over? The first question is straightforward to answer: simply perform a conversion from the number of moles of product formed to its mass, using its molar mass. For As2O3, the molar mass is 197.84 g/mol; knowing that we will form 0.334 mol of As2O3 under the given conditions, we will get

Screen Shot 2014-07-22 at 1.14.42 PM

The second question is somewhat more convoluted to answer. First, we must do a mass-mass calculation relating the limiting reagent (here, As) to the other reagent (O2). Once we determine the mass of O2 that reacted, we subtract that from the original amount to determine the amount left over. According to the mass-mass calculation,

Screen Shot 2014-07-22 at 1.09.26 PM

Because we reacted 16.0 g of our original O2, we subtract that from the original amount, 50.0 g, to get the mass of O2 remaining:

50.0 g O2 − 16.0 g O2 reacted = 34.0 g O2 left over

You must remember to perform this final subtraction to determine the amount remaining; a common error is to report the 16.0 g as the amount remaining.

Example 14

A 5.00 g quantity of Rb are combined with 3.44 g of MgCl2 according to this chemical reaction:

2 Rb(s) + MgCl2(s) → Mg(s) + 2 RbCl(s)

What mass of Mg is formed, and what mass of what reactant is left over?

Solution

Because the question asks what mass of magnesium is formed, we can perform two mass-mass calculations and determine which amount is less.

5OOGRB

The 0.711 g of Mg is the lesser quantity, so the associated reactant—5.00 g of Rb—is the limiting reagent. To determine how much of the other reactant is left, we have to do one more mass-mass calculation to determine what mass of MgCl2 reacted with the 5.00 g of Rb and then subtract the amount reacted from the original amount.

500GRB

Because we started with 3.44 g of MgCl2, we have

3.44 g MgCl2 − 2.78 g MgCl2 reacted = 0.66 g MgCl2 left

Test Yourself

Given the initial amounts listed, what is the limiting reagent, and what is the mass of the leftover reagent?

MGOS

Answer

H2S is the limiting reagent; 1.5 g of MgO are left over.

Key Takeaways

  • The limiting reagent is that reactant that produces the least amount of product.
  • Mass-mass calculations can determine how much product is produced and how much of the other reactants remain.

 

Exercises

The box below shows a group of nitrogen and hydrogen molecules that will react to produce ammonia, NH3. What is the limiting reagent? Limiting Reagent

The box below shows a group of hydrogen and oxygen molecules that will react to produce water, H2O. What is the limiting reagent?

Making Water 2

Given the statement “20.0 g of methane is burned in excess oxygen,” is it obvious which reactant is the limiting reagent?

Given the statement “the metal is heated in the presence of excess hydrogen,” is it obvious which substance is the limiting reagent despite not specifying any quantity of reactant?

Acetylene (C2H2) is formed by reacting 7.08 g of C and 4.92 g of H2.

2 C(s) + H2(g) → C2H2(g)

What is the limiting reagent? How much of the other reactant is in excess?

Ethane (C2H6) is formed by reacting 7.08 g of C and 4.92 g of H2.

2 C(s) + 3 H2(g) → C2H6(g)

What is the limiting reagent? How much of the other reactant is in excess?

Given the initial amounts listed, what is the limiting reagent, and how much of the other reactant is in excess?

Screen Shot 2014-07-22 at 1.49.34 PM

Given the initial amounts listed, what is the limiting reagent, and how much of the other reactant is in excess?

Screen Shot 2014-07-22 at 1.49.39 PM

To form the precipitate PbCl2, 2.88 g of NaCl and 7.21 g of Pb(NO3)2 are mixed in solution. How much precipitate is formed? How much of which reactant is in excess?

In a neutralization reaction, 18.06 g of KOH are reacted with 13.43 g of HNO3. What mass of H2O is produced, and what mass of which reactant is in excess?

Answers

1.

Nitrogen is the limiting reagent.

3.

Yes; methane is the limiting reagent.

5.

C is the limiting reagent; 4.33 g of H2 are left over.

7.

H2O is the limiting reagent; 25.9 g of P4O6 are left over.

9.

6.06 g of PbCl2 are formed; 0.33 g of NaCl is left over.

 

31

The Mole in Chemical Reactions

Learning Objectives

  1. Balance a chemical equation in terms of moles.
  2. Use the balanced equation to construct conversion factors in terms of moles.
  3. Calculate moles of one substance from moles of another substance using a balanced chemical equation.

Consider this balanced chemical equation:

2 H2 + O2 → 2 H2O

We interpret this as “two molecules of hydrogen react with one molecule of oxygen to make two molecules of water.” The chemical equation is balanced as long as the coefficients are in the ratio 2:1:2. For instance, this chemical equation is also balanced:

100 H2 + 50 O2 → 100 H2O

This equation is not conventional—because convention says that we use the lowest ratio of coefficients—but it is balanced. So is this chemical equation:

5,000 H2 + 2,500 O2 → 5,000 H2O

Again, this is not conventional, but it is still balanced. Suppose we use a much larger number:

12.044 × 1023 H2 + 6.022 × 1023 O2 → 12.044 × 1023 H2O

These coefficients are also in the ratio of 2:1:2. But these numbers are related to the number of things in a mole: the first and last numbers are two times Avogadro’s number, while the second number is Avogadro’s number. That means that the first and last numbers represent 2 mol, while the middle number is just 1 mol. Well, why not just use the number of moles in balancing the chemical equation?

2 H2 + O2 → 2 H2O

is the same balanced chemical equation we started with! What this means is that chemical equations are not just balanced in terms of molecules; they are also balanced in terms of moles. We can just as easily read this chemical equation as “two moles of hydrogen react with one mole of oxygen to make two moles of water.” All balanced chemical reactions are balanced in terms of moles.

Example 8

Interpret this balanced chemical equation in terms of moles.

P4 + 5 O2 → P4O10

Solution

The coefficients represent the number of moles that react, not just molecules. We would speak of this equation as “one mole of molecular phosphorus reacts with five moles of elemental oxygen to make one mole of tetraphosphorus decoxide.”

Test Yourself

Interpret this balanced chemical equation in terms of moles.

N2 + 3 H2 → 2 NH3

Answer

One mole of elemental nitrogen reacts with three moles of elemental hydrogen to produce two moles of ammonia.

In Chapter 4 “Chemical Reactions and Equations”, Section 4.1 “The Chemical Equation”, we stated that a chemical equation is simply a recipe for a chemical reaction. As such, chemical equations also give us equivalences—equivalences between the reactants and the products. However, now we understand that these equivalences are expressed in terms of moles. Consider the chemical equation

2 H2 + O2 → 2 H2O

This chemical reaction gives us the following equivalences:

2 mol H2 ⇔ 1 mol O2 ⇔ 2 mol H2O

Any two of these quantities can be used to construct a conversion factor that lets us relate the number of moles of one substance to an equivalent number of moles of another substance. If, for example, we want to know how many moles of oxygen will react with 17.6 mol of hydrogen, we construct a conversion factor between 2 mol of H2 and 1 mol of O2 and use it to convert from moles of one substance to moles of another:

Screen Shot 2014-07-20 at 3.16.01 PM

Note how the mol H2 unit cancels, and mol O2 is the new unit introduced. This is an example of a mole-mole calculation, when you start with moles of one substance and convert to moles of another substance by using the balanced chemical equation. The example may seem simple because the numbers are small, but numbers won’t always be so simple!

Example 9

For the balanced chemical equation

2 C4H10(g) + 13 O2 → 8 CO2(g) + 10 H2O(ℓ)

if 154 mol of O2 are reacted, how many moles of CO2 are produced?

Solution

We are relating an amount of oxygen to an amount of carbon dioxide, so we need the equivalence between these two substances. According to the balanced chemical equation, the equivalence is

13 mol O2 ⇔ 8 mol CO2

We can use this equivalence to construct the proper conversion factor. We start with what we are given and apply the conversion factor:

ss8

The mol O2 unit is in the denominator of the conversion factor so it cancels. Both the 8 and the 13 are exact numbers, so they don’t contribute to the number of significant figures in the final answer.

Test Yourself

Using the above equation, how many moles of H2O are produced when 154 mol of O2 react?

Answer

118 mol

It is important to reiterate that balanced chemical equations are balanced in terms of moles. Not grams, kilograms, or liters—but moles. Any stoichiometry problem will likely need to work through the mole unit at some point, especially if you are working with a balanced chemical reaction.

Key Takeaways

  • Balanced chemical reactions are balanced in terms of moles.
  • A balanced chemical reaction gives equivalences in moles that allow stoichiometry calculations to be performed.

Exercises

  1. Express in mole terms what this chemical equation means.

    CH4 + 2O2 → CO2 + 2H2O

  2. Express in mole terms what this chemical equation means.

    Na2CO3 + 2HCl → 2NaCl + H2O + CO2

  3. How many molecules of each substance are involved in the equation in Exercise 1 if it is interpreted in terms of moles?

  4. How many molecules of each substance are involved in the equation in Exercise 2 if it is interpreted in terms of moles?

  5. For the chemical equation

    2 C2H6 + 7 O2 → 4 CO2 + 6 H2O

    what equivalences can you write in terms of moles? Use the ⇔ sign.

  6. For the chemical equation

    2 Al + 3 Cl2 → 2 AlCl3

    what equivalences can you write in terms of moles? Use the ⇔ sign.

  7. Write the balanced chemical reaction for the combustion of C5H12 (the products are CO2 and H2O) and determine how many moles of H2O are formed when 5.8 mol of O2 are reacted.

  8. Write the balanced chemical reaction for the formation of Fe2(SO4)3 from Fe2O3 and SO3 and determine how many moles of Fe2(SO4)3 are formed when 12.7 mol of SO3 are reacted.

  9. For the balanced chemical equation

    3 Cu(s) + 2 NO3(aq) + 8 H+(aq) → 3 Cu2+(aq) + 4 H2O(ℓ) + 2 NO(g)

    how many moles of Cu2+ are formed when 55.7 mol of H+ are reacted?

  10. For the balanced chemical equation

    Al(s) + 3 Ag+(aq) → Al3+(aq) + 3 Ag(s)

    how many moles of Ag are produced when 0.661 mol of Al are reacted?

  11. For the balanced chemical reaction

    4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(ℓ)

    how many moles of H2O are produced when 0.669 mol of NH3 react?

  12. For the balanced chemical reaction

    4 NaOH(aq) + 2 S(s) + 3 O2(g) → 2 Na2SO4(aq) + 2 H2O(ℓ)

    how many moles of Na2SO4 are formed when 1.22 mol of O2 react?

  13. For the balanced chemical reaction

    4 KO2(s) + 2 CO2(g) → 2 K2CO3(s) + 3 O2(g)

    determine the number of moles of both products formed when 6.88 mol of KO2 react.

  14. For the balanced chemical reaction

    2 AlCl3 + 3 H2O(ℓ) → Al2O3 + 6 HCl(g)

    determine the number of moles of both products formed when 0.0552 mol of AlCl3 react.

Answers

1.

One mole of CH4 reacts with 2 mol of O2 to make 1 mol of CO2 and 2 mol of H2O.

3.

6.022 × 1023 molecules of CH4, 1.2044 × 1024 molecules of O2, 6.022 × 1023 molecules of CO2, and 1.2044 × 1024 molecules of H2O

5.

2 mol of C2H6 ⇔ 7 mol of O2 ⇔ 4 mol of CO2 ⇔ 6 mol of H2O

7.

C5H12 + 😯2 → 5CO2 + 6H2O; 4.4 mol

9.

20.9 mol

11.

1.00 mol

13.

3.44 mol of K2CO3; 5.16 mol of O2

32

Yields

Learning Objectives

1. Define and determine theoretical yields, actual yields, and percent yields.

In all the previous calculations we have performed involving balanced chemical equations, we made two assumptions: (1) the reaction goes exactly as written, and (2) the reaction proceeds completely. In reality, such things as side reactions occur that make some chemical reactions rather messy. For example, in the actual combustion of some carbon-containing compounds, such as methane, some CO is produced as well as CO2. However, we will continue to ignore side reactions, unless otherwise noted.

The second assumption, that the reaction proceeds completely, is more troublesome. Many chemical reactions do not proceed to completion as written, for a variety of reasons (some of which we will consider in Chapter 13 “Chemical Equilibrium”). When we calculate an amount of product assuming that all the reactant reacts, we calculate the theoretical yield, an amount that is theoretically produced as calculated using the balanced chemical reaction.

In many cases, however, this is not what really happens. In many cases, less—sometimes much less—of a product is made during the course of a chemical reaction. The amount that is actually produced in a reaction is called the actual yield. By definition, the actual yield is less than or equal to the theoretical yield. If it is not, then an error has been made.

Both theoretical yields and actual yields are expressed in units of moles or grams. It is also common to see something called a percent yield. The percent yield is a comparison between the actual yield and the theoretical yield and is defined as

Screen Shot 2014-07-22 at 12.54.26 PM

It does not matter whether the actual and theoretical yields are expressed in moles or grams, as long as they are expressed in the same units. However, the percent yield always has units of percent. Proper percent yields are between 0% and 100%—again, if percent yield is greater than 100%, an error has been made.

Example 13

A worker reacts 30.5 g of Zn with nitric acid and evapourates the remaining water to obtain 65.2 g of Zn(NO3)2. What are the theoretical yield, the actual yield, and the percent yield?

Zn(s) + 2 HNO3(aq) → Zn(NO3)2(aq) + H2(g)

Solution

A mass-mass calculation can be performed to determine the theoretical yield. We need the molar masses of Zn (65.39 g/mol) and Zn(NO3)2 (189.41 g/mol). In three steps, the mass-mass calculation is

305GZN

Thus, the theoretical yield is 88.3 g of Zn(NO3)2. The actual yield is the amount that was actually made, which was 65.2 g of Zn(NO3)2. To calculate the percent yield, we take the actual yield and divide it by the theoretical yield and multiply by 100:

642GZN

The worker achieved almost three-fourths of the possible yield.

Test Yourself

A synthesis produced 2.05 g of NH3 from 16.5 g of N2. What is the theoretical yield and the percent yield?

N2(g) + 3 H2(g) → 2 NH3(g)

Answer

theoretical yield = 20.1 g; percent yield = 10.2%

Chemistry Is Everywhere: Actual Yields in Drug Synthesis and Purification

Many drugs are the product of several steps of chemical synthesis. Each step typically occurs with less than 100% yield, so the overall percent yield might be very small. The general rule is that the overall percent yield is the product of the percent yields of the individual synthesis steps. For a drug synthesis that has many steps, the overall percent yield can be very tiny, which is one factor in the huge cost of some drugs. For example, if a 10-step synthesis has a percent yield of 90% for each step, the overall yield for the entire synthesis is only 35%. Many scientists work every day trying to improve percent yields of the steps in the synthesis to decrease costs, improve profits, and minimize waste.

Even purifications of complex molecules into drug-quality purity are subject to percent yields. Consider the purification of impure albuterol. Albuterol (C13H21NO2; accompanying figure) is an inhaled drug used to treat asthma, bronchitis, and other obstructive pulmonary diseases. It is synthesized from norepinephrine, a naturally occurring hormone and neurotransmitter. Its initial synthesis makes very impure albuterol that is purified in five chemical steps. The details of the steps do not concern us; only the percent yields do:

impure albuterol → intermediate A percent yield = 70%
intermediate A → intermediate B percent yield = 100%
intermediate B → intermediate C percent yield = 40%
intermediate C → intermediate D percent yield = 72%
intermediate D → purified albuterol percent yield = 35%
overall percent yield = 70% × 100% × 40% × 72% × 35% = 7.5%

That is, only about one-fourteenth of the original material was turned into the purified drug. This gives you one reason why some drugs are so expensive; a lot of material is lost in making a high-purity pharmaceutical.

Key Takeaways

  • Theoretical yield is what you calculate the yield will be using the balanced chemical reaction.
  • Actual yield is what you actually get in a chemical reaction.
  • Percent yield is a comparison of the actual yield with the theoretical yield.

 

Exercises

  1. What is the difference between the theoretical yield and the actual yield?

  2. What is the difference between the actual yield and the percent yield?

  3. A worker isolates 2.675 g of SiF4 after reacting 2.339 g of SiO2 with HF. What are the theoretical yield and the actual yield?

    SiO2(s) + 4 HF(g) → SiF4(g) + 2 H2O(ℓ)

  4. A worker synthesizes aspirin, C9H8O4, according to this chemical equation. If 12.66 g of C7H6O3 are reacted and 12.03 g of aspirin are isolated, what are the theoretical yield and the actual yield?

    C7H6O3 + C4H6O3 → C9H8O4 + HC2H3O2

  5. A chemist decomposes 1.006 g of NaHCO3 and obtains 0.0334 g of Na2CO3. What are the theoretical yield and the actual yield?

    2 NaHCO3(s) → Na2CO3(s) + H2O(ℓ) + CO2(g)

  6. A chemist combusts a 3.009 g sample of C5H12 and obtains 3.774 g of H2O. What are the theoretical yield and the actual yield?

    C5H12(ℓ) + 8 O2(g) → 5 CO2 + 6 H2O(ℓ)

  7. What is the percent yield in Exercise 3?

  8. What is the percent yield in Exercise 4?

  9. What is the percent yield in Exercise 5?

  10. What is the percent yield in Exercise 6?

Answers

1.

Theoretical yield is what you expect stoichiometrically from a chemical reaction; actual yield is what you actually get from a chemical reaction.

3.

theoretical yield = 4.052 g; actual yield = 2.675 g

5.

theoretical yield = 0.635 g; actual yield = 0.0334 g

7.

66.02%

9.

5.26%

 

33

End-of-Chapter Material

Additional Exercises

  1. How many molecules of O2 will react with 6.022 × 1023 molecules of H2 to make water? The reaction is 2 H2(g) + O2(g) → 2 H2O(ℓ).

  2. How many molecules of H2 will react with 6.022 × 1023 molecules of N2 to make ammonia? The reaction is N2(g) + 3 H2(g) → 2 NH3(g).

  3. How many moles are present in 6.411 kg of CO2? How many molecules is this?

  4. How many moles are present in 2.998 mg of SCl4? How many molecules is this?

  5. What is the mass in milligrams of 7.22 × 1020 molecules of CO2?

  6. What is the mass in kilograms of 3.408 × 1025 molecules of SiS2?

  7. What is the mass in grams of 1 molecule of H2O?

  8. What is the mass in grams of 1 atom of Al?

  9. What is the volume of 3.44 mol of Ga if the density of Ga is 6.08 g/mL?

  10. What is the volume of 0.662 mol of He if the density of He is 0.1785 g/L?

  11. For the chemical reaction

    2 C4H10(g) + 13 O2(g) → 8 CO2(g) + 10 H2O(ℓ)

    assume that 13.4 g of C4H10 reacts completely to products. The density of CO2 is 1.96 g/L. What volume in liters of CO2 is produced?

  12. For the chemical reaction

    2 GaCl3(s) + 3 H2(g) → 2 Ga(ℓ) + 6 HCl(g)

    if 223 g of GaCl3 reacts completely to products and the density of Ga is 6.08 g/mL, what volume in milliliters of Ga is produced?

  13. Calculate the mass of each product when 100.0 g of CuCl react according to the reaction

    2 CuCl(aq) → CuCl2(aq) + Cu(s)

    What do you notice about the sum of the masses of the products? What concept is being illustrated here?

  14. Calculate the mass of each product when 500.0 g of SnCl2 react according to the reaction

    2 SnCl2(aq) → SnCl4(aq) + Sn(s)

    What do you notice about the sum of the masses of the products? What concept is being illustrated here?

  15. What mass of CO2 is produced from the combustion of 1 gal of gasoline? The chemical formula of gasoline can be approximated as C8H18. Assume that there are 2,801 g of gasoline per gallon.

  16. What mass of H2O is produced from the combustion of 1 gal of gasoline? The chemical formula of gasoline can be approximated as C8H18. Assume that there are 2,801 g of gasoline per gallon.

  17. A chemical reaction has a theoretical yield of 19.98 g and a percent yield of 88.40%. What is the actual yield?

  18. A chemical reaction has an actual yield of 19.98 g and a percent yield of 88.40%. What is the theoretical yield?

  19. Given the initial amounts listed, what is the limiting reagent, and how much of the other reactants are in excess?

    Screen Shot 2014-07-22 at 2.18.07 PM

  20. Given the initial amounts listed, what is the limiting reagent, and how much of the other reactants are in excess?

    Screen Shot 2014-07-22 at 2.18.11 PM

  21. Verify that it does not matter which product you use to predict the limiting reagent by using both products in this combustion reaction to determine the limiting reagent and the amount of the reactant in excess. Initial amounts of each reactant are given.

    Screen Shot 2014-07-22 at 2.18.16 PM

  22. Just in case you suspect Exercise 21 is rigged, do it for another chemical reaction and verify that it does not matter which product you use to predict the limiting reagent by using both products in this combustion reaction to determine the limiting reagent and the amount of the reactant in excess. Initial amounts of each reactant are given.

    Screen Shot 2014-07-22 at 2.18.22 PM

 

Answers

1.

1.2044 × 1024 molecules

3.

145.7 mol; 8.77 × 1025 molecules

5.

52.8 mg

7.

2.99 × 10−23 g

9.

39.4 mL

11.

20.7 L

13.

67.91 g of CuCl2; 32.09 g of Cu. The two masses add to 100.0 g, the initial amount of starting material, demonstrating the law of conservation of matter.

15.

8,632 g

17.

17.66 g

19.

The limiting reagent is NaOH; 21.9 g of P4 and 3.61 g of H2O are left over.

21.

Both products predict that O2 is the limiting reagent; 20.3 g of C3H8 are left over.

 

VI

Chapter 6. Gases

34

Introduction to Gases

Opening Essay

Perhaps one of the most spectacular chemical reactions involving a gas occurred on May 6, 1937, when the German airship Hindenburg exploded on approach to the Naval Air Station in Lakehurst, New Jersey. The actual cause of the explosion is still unknown, but the entire volume of hydrogen gas used to float the airship, about 200,000 m3, burned in less than a minute. Thirty-six people, including one on the ground, were killed.

The German airship Hindenburg (left) was one of the largest airships ever built. However, it was filled with hydrogen gas and exploded in Lakehurst, New Jersey, at the end of a transatlantic voyage in May 1937 (right). Source: “Hindenburg” by James Vaughan is licensed under the Creative Commons Attribution-NonCommercial-ShareAlike 2.0 Generic. “Hindenburg burning” by Gus Pasquerella is in the public domain.
The German airship Hindenburg (left) was one of the largest airships ever built. However, it was filled with hydrogen gas and exploded in Lakehurst, New Jersey, at the end of a transatlantic voyage in May 1937 (right).
Source: “Hindenburg” by James Vaughan is licensed under the Creative Commons Attribution-NonCommercial-ShareAlike 2.0 Generic. “Hindenburg burning” by Gus Pasquerella is in the public domain.

Hydrogen is the lightest known gas. Any balloon filled with hydrogen gas will float in air if its mass is not too great. This makes hydrogen an obvious choice for flying machines based on balloons—airships, dirigibles, and blimps. However, hydrogen also has one obvious drawback: it burns in air according to the well-known chemical equation

2 H2(g) + O2(g) → 2 H2O(ℓ)

So although hydrogen is an obvious choice, it is also a dangerous choice.

Helium gas is also lighter than air and has 92% of the lifting power of hydrogen. Why, then, was helium not used in the Hindenburg? In the 1930s, helium was much more expensive. In addition, the best source of helium at the time was the United States, which banned helium exports to pre–World War II Germany. Today all airships use helium, a legacy of the Hindenburg disaster.

Of the three basic phases of matter—solids, liquids, and gases—only one of them has predictable physical properties: gases. In fact, the study of the properties of gases was the beginning of the development of modern chemistry from its alchemical roots. The interesting thing about some of these properties is that they are independent of the identity of the gas. That is, it doesn’t matter if the gas is helium gas, oxygen gas, or sulfur vapours; some of their behaviour is predictable and, as we will find, very similar. In this chapter, we will review some of the common behaviours of gases.

Let us start by reviewing some properties of gases. Gases have no definite shape or volume; they tend to fill whatever container they are in. They can compress and expand, sometimes to a great extent. Gases have extremely low densities, one-thousandth or less the density of a liquid or solid. Combinations of gases tend to mix together spontaneously; that is, they form solutions. Air, for example, is a solution of mostly nitrogen and oxygen. Any understanding of the properties of gases must be able to explain these characteristics.

35

Pressure

Learning Objectives

  1. Define pressure.
  2. Learn the units of pressure and how to convert between them.

The kinetic theory of gases indicates that gas particles are always in motion and are colliding with other particles and the walls of the container holding them. Although collisions with container walls are elastic (i.e., there is no net energy gain or loss because of the collision), a gas particle does exert a force on the wall during the collision. The accumulation of all these forces distributed over the area of the walls of the container causes something we call pressure. Pressure (P) is defined as the force of all the gas particle/wall collisions divided by the area of the wall:

pressure=force/area

All gases exert pressure; it is one of the fundamental measurable quantities of this phase of matter. Even our atmosphere exerts pressure—in this case, the gas is being “held in” by the earth’s gravity, rather than the gas being in a container. The pressure of the atmosphere is 101,325 Pa.

Pressure has a variety of units. The formal, SI-approved unit of pressure is the pascal (Pa), which is defined as 1 N/m2 (one newton of force over an area of one square meter). However, this is usually too small in magnitude to be useful. A common unit of pressure is the atmosphere (atm), which was originally defined as the average atmospheric pressure at sea level.

However, “average atmospheric pressure at sea level” is difficult to pinpoint because of atmospheric pressure variations. A more reliable and common unit is millimeters of mercury (mmHg), which is the amount of pressure exerted by a column of mercury exactly 1 mm high. An equivalent unit is the torr, which equals 1 mmHg (The torr is named after Evangelista Torricelli, a seventeenth-century Italian scientist who invented the mercury barometer.) With these definitions of pressure, the atmosphere unit is redefined: 1 atm is defined as exactly 760 mmHg, or 760 torr. We thus have the following equivalences:

1 atm = 760 mmHg = 760 torr

We can use these equivalences as with any equivalences—to perform conversions from one unit to another. Relating these to the formal SI unit of pressure, 1 atm = 101,325 Pa.

Example 1

How many atmospheres are there in 595 torr?

Solution

Using the pressure equivalences, we construct a conversion factor between torr and atmospheres: 1 atm = 760 torr. Thus,

595TORR

Because the numbers in the conversion factor are exact, the number of significant figures in the final answer is determined by the initial value of pressure.

Test Yourself

How many atmospheres are there in 1,022 torr?

Answer

1.345 atm

Example 2

The atmosphere on Mars is largely CO2 at a pressure of 6.01 mmHg. What is this pressure in atmospheres?

Solution

Use the pressure equivalences to construct the proper conversion factor between millimeters of mercury and atmospheres.

601MMHG

At the end, we expressed the answer in scientific notation.

Test Yourself

Atmospheric pressure is low in the eye of a hurricane. In a 1979 hurricane in the Pacific Ocean, a pressure of 0.859 atm was reported inside the eye. What is this pressure in torr?

Answer

652 torr

Key Takeaways

  • Pressure is a force exerted over an area.
  • Pressure has several common units that can be converted.

 

Exercises

  1. Define pressure. What causes it?

  2. Define and relate three units of pressure.

  3. If a force of 16.7 N is pressed against an area of 2.44 m2, what is the pressure in pascals?

  4. If a force of 2,546 N is pressed against an area of 0.0332 m2, what is the pressure in pascals?

  5. Explain why the original definition of atmosphere did not work well.

  6. What units of pressure are equal to each other?

  7. How many atmospheres are in 889 mmHg?

  8. How many atmospheres are in 223 torr?

  9. How many torr are in 2.443 atm?

  10. How many millimeters of mercury are in 0.334 atm?

  11. How many millimeters of mercury are in 334 torr?

  12. How many torr are in 0.777 mmHg?

  13. How many pascals are in 1 torr?

  14. A pressure of 0.887 atm equals how many pascals?

Answers

1.

Pressure is force per unit area. It is caused by gas particles hitting the walls of their container.

3.

6.84 Pa

5.

Because the atmospheric pressure at sea level is variable, it is not a consistent unit of pressure.

7.

1.17 atm

9.

1,857 torr

11.

334 mmHg

13.

133 Pa

 

36

Gas Laws

Learning Objectives

  1. Learn what is meant by the term gas laws.
  2. Learn and apply Boyle’s law.
  3. Learn and apply Charles’s law.

When seventeenth-century scientists began studying the physical properties of gases, they noticed simple relationships between some of the measurable properties of gases. Take pressure (P) and volume (V), for example. Scientists noted that for a given amount of a gas (usually expressed in units of moles [n]), if the temperature (T) of the gas is kept constant, pressure and volume are related: as one increases, the other decreases. As one decreases, the other increases. We say that pressure and volume are inversely related.

There is more to it, however: pressure and volume of a given amount of gas at a constant temperature are numerically related. If you take the pressure value and multiply it by the volume value, the product is a constant for a given amount of gas at a constant temperature:

P × V = constant at constant n and T

If either volume or pressure changes while the amount and temperature stay the same, then the other property must change so that the product of the two properties still equals that same constant. That is, if the original conditions are labelled P1 and V1 and the new conditions are labelled P2 and V2, we have

P1V1 = constant = P2V2

where the properties are assumed to be multiplied together. Leaving out the middle part, we have simply

P1V1 = P2V2 at constant n and T

This equation is an example of a gas law. A gas law is a simple mathematical formula that allows you to model, or predict, the behaviour of a gas. This particular gas law is called Boyle’s law, after the English scientist Robert Boyle, who first announced it in 1662. Figure 6.1 Boyle’s Law shows two representations of what Boyle’s law describes.

Figure 6.1 Boyle’s Law

Figure 6-1

 

A piston having a certain pressure and volume (left piston) will have half the volume when its pressure is twice as much (right piston). One can also plot P versus V for a given amount of gas at a certain temperature; such a plot will look like the graph on the right.

Boyle’s law is an example of a second type of mathematical problem we see in chemistry—one based on a mathematical formula. Tactics for working with mathematical formulas are different from tactics for working with conversion factors. First, most of the questions you will have to answer using formulas are word-type questions, so the first step is to identify what quantities are known and assign them to variables. Second, in most formulas, some mathematical rearrangements (i.e., algebra) must be performed to solve for an unknown variable. The rule is that to find the value of the unknown variable, you must mathematically isolate the unknown variable by itself and in the numerator of one side of the equation. Finally, units must be consistent. For example, in Boyle’s law there are two pressure variables, and they must have the same unit. There are also two volume variables; they also must have the same unit. In most cases, it won’t matter what the unit is, but the unit must be the same on both sides of the equation.

Example 3

A sample of gas has an initial pressure of 2.44 atm and an initial volume of 4.01 L. Its pressure changes to 1.93 atm. What is the new volume if temperature and amount are kept constant?

Solution

First, determine what quantities we are given. We are given an initial pressure and an initial volume, so let these values be P1 and V1:

P1 = 2.44 atm and V1 = 4.01 L

We are given another quantity, final pressure of 1.93 atm, but not a final volume. This final volume is the variable we will solve for.

P2 = 1.93 atm and V2 = ? L

Substituting these values into Boyle’s law, we get

(2.44 atm)(4.01 L) = (1.93 atm)V2

To solve for the unknown variable, we isolate it by dividing both sides of the equation by 1.93 atm—both the number and the unit:

244ATM

Note that, on the left side of the equation, the unit atm is in the numerator and the denominator of the fraction. They cancel algebraically, just as a number would. On the right side, the unit atm and the number 1.93 are in the numerator and the denominator, so the entire quantity cancels:

244ATM2

What we have left is

244401

Now we simply multiply and divide the numbers together and combine the answer with the L unit, which is a unit of volume. Doing so, we get

V2 = 5.07 L

Does this answer make sense? We know that pressure and volume are inversely related; as one decreases, the other increases. Pressure is decreasing (from 2.44 atm to 1.93 atm), so volume should be increasing to compensate, and it is (from 4.01 L to 5.07 L). So the answer makes sense based on Boyle’s law.

Test Yourself

If P1 = 334 torr, V1 = 37.8 mL, and P2 = 102 torr, what is V2?

Answer

124 mL

As mentioned, you can use any units for pressure or volume, but both pressures must be expressed in the same units, and both volumes must be expressed in the same units.

Example 4

A sample of gas has an initial pressure of 722 torr and an initial volume of 88.8 mL. Its volume changes to 0.663 L. What is the new pressure?

Solution

We can still use Boyle’s law to answer this, but now the two volume quantities have different units. It does not matter which unit we change, as long as we perform the conversion correctly. Let us change the 0.663 L to milliliters:

0633L

Now that both volume quantities have the same units, we can substitute into Boyle’s law:

722TORR

The mL units cancel, and we multiply and divide the numbers to get

P2 = 96.7 torr

The volume is increasing, and the pressure is decreasing, which is as expected for Boyle’s law.

Test Yourself

If V1 = 456 mL, P1 = 308 torr, and P2 = 1.55 atm, what is V2?

Answer

119 mL

There are other measurable characteristics of a gas. One of them is temperature (T). One could vary the temperature of a gas sample and note what effect it has on the other properties of the gas. Early scientists did just this, discovering that if the amount of a gas and its pressure are kept constant, then changing the temperature changes the volume (V). As temperature increases, volume increases; as temperature decreases, volume decreases. We say that these two characteristics are directly related.

A mathematical relationship between V and T should be possible except for one question: What temperature scale should we use? We know from Chapter 2 “Measurements” that science uses several possible temperature scales. Experiments show that the volume of a gas is related to its absolute temperature in Kelvin, not its temperature in degrees Celsius. If the temperature of a gas is expressed in kelvins, then experiments show that the ratio of volume to temperature is a constant:

Screen Shot 2014-07-22 at 2.35.38 PM

We can modify this equation as we modified Boyle’s law: the initial conditions V1 and T1 have a certain value, and the value must be the same when the conditions of the gas are changed to some new conditions V2 and T2, as long as pressure and the amount of the gas remain constant. Thus, we have another gas law:

Screen Shot 2014-07-22 at 2.35.42 PM

This gas law is commonly referred to as Charles’s law, after the French scientist Jacques Charles, who performed experiments on gases in the 1780s. The tactics for using this mathematical formula are similar to those for Boyle’s law. To determine an unknown quantity, use algebra to isolate the unknown variable by itself and in the numerator; the units of similar variables must be the same. But we add one more tactic: all temperatures must be expressed in the absolute temperature scale (Kelvin). As a reminder, we review the conversion between the absolute temperature scale and the Celsius temperature scale:

K = °C + 273

where K represents the temperature in kelvins, and °C represents the temperature in degrees Celsius.

Figure 6.2 “Charles’s Law” shows two representations of how Charles’s law works.

Figure 6.2 Charles’s Law

Figure 6-2

 

A piston having a certain volume and temperature (left piston) will have twice the volume when its temperature is twice as much (right piston). One can also plot V versus T for a given amount of gas at a certain pressure; such a plot will look like the graph on the right.

Example 5

A sample of gas has an initial volume of 34.8 mL and an initial temperature of 315 K. What is the new volume if the temperature is increased to 559 K? Assume constant pressure and amount for the gas.

Solution

First, we assign the given values to their variables. The initial volume is V1, so V1 = 34.8 mL, and the initial temperature is T1, so T1 = 315 K. The temperature is increased to 559 K, so the final temperature T2 = 559 K. We note that the temperatures are already given in kelvins, so we do not need to convert the temperatures. Substituting into the expression for Charles’s law yields

348ml

We solve for V2 by algebraically isolating the V2 variable on one side of the equation. We do this by multiplying both sides of the equation by 559 K (number and unit). When we do this, the temperature unit cancels on the left side, while the entire 559 K cancels on the right side:

559k

The expression simplifies to

559k2

By multiplying and dividing the numbers, we see that the only remaining unit is mL, so our final answer is

V2 = 61.8 mL

Does this answer make sense? We know that as temperature increases, volume increases. Here, the temperature is increasing from 315 K to 559 K, so the volume should also increase, which it does.

Test Yourself

If V1 = 3.77 L and T1 = 255 K, what is V2 if T2 = 123 K?

Answer

1.82 L

It is more mathematically complicated if a final temperature must be calculated because the T variable is in the denominator of Charles’s law. There are several mathematical ways to work this, but perhaps the simplest way is to take the reciprocal of Charles’s law. That is, rather than write it as

Screen Shot 2014-07-22 at 2.36.09 PM

write the equation as

Screen Shot 2014-07-22 at 2.36.13 PM

It is still an equality and a correct form of Charles’s law, but now the temperature variable is in the numerator, and the algebra required to predict a final temperature is simpler.

Example 6

A sample of a gas has an initial volume of 34.8 L and an initial temperature of −67°C. What must be the temperature of the gas for its volume to be 25.0 L?

Solution

Here, we are looking for a final temperature, so we will use the reciprocal form of Charles’s law. However, the initial temperature is given in degrees Celsius, not kelvins. We must convert the initial temperature to kelvins:

−67°C + 273 = 206 K

In using the gas law, we must use T1 = 206 K as the temperature. Substituting into the reciprocal form of Charles’s law, we get

206k

Bringing the 25.0 L quantity over to the other side of the equation, we get

250l

The L units cancel, so our final answer is

T2 = 148 K

This is also equal to −125°C. As temperature decreases, volume decreases, which it does in this example.

Test Yourself

If V1 = 623 mL, T1 = 255°C, and V2 = 277 mL, what is T2?

Answer

235 K, or −38°C

Key Takeaways

  • The behaviour of gases can be modelled with gas laws.
  • Boyle’s law relates a gas’s pressure and volume at constant temperature and amount.
  • Charles’s law relates a gas’s volume and temperature at constant pressure and amount.
  • In gas laws, temperatures must always be expressed in kelvins.

 

Exercises

  1. Define gas law. What restrictions are there on the units that can be used for the physical properties?

  2. What unit of temperature must be used for gas laws?

  3. Boyle’s law relates the _____________ of a gas inversely with the ___________ of that gas.

  4. Charles’s law relates the _____________ of a gas directly with the ___________ of that gas.

  5. What properties must be held constant when applying Boyle’s law?

  6. What properties must be held constant when applying Charles’s law?

  7. A gas has an initial pressure of 1.445 atm and an initial volume of 1.009 L. What is its new pressure if volume is changed to 0.556 L? Assume temperature and amount are held constant.

  8. A gas has an initial pressure of 633 torr and an initial volume of 87.3 mL. What is its new pressure if volume is changed to 45.0 mL? Assume temperature and amount are held constant.

  9. A gas has an initial pressure of 4.33 atm and an initial volume of 5.88 L. What is its new volume if pressure is changed to 0.506 atm? Assume temperature and amount are held constant.

  10. A gas has an initial pressure of 87.0 torr and an initial volume of 28.5 mL. What is its new volume if pressure is changed to 206 torr? Assume temperature and amount are held constant.

  11. A gas has an initial volume of 638 mL and an initial pressure of 779 torr. What is its final volume in liters if its pressure is changed to 0.335 atm? Assume temperature and amount are held constant.

  12. A gas has an initial volume of 0.966 L and an initial pressure of 3.07 atm. What is its final pressure in torr if its volume is changed to 3,450 mL? Assume temperature and amount are held constant.

  13. A gas has an initial volume of 67.5 mL and an initial temperature of 315 K. What is its new volume if temperature is changed to 244 K? Assume pressure and amount are held constant.

  14. A gas has an initial volume of 2.033 L and an initial temperature of 89.3 K. What is its volume if temperature is changed to 184 K? Assume pressure and amount are held constant.

  15. A gas has an initial volume of 655 mL and an initial temperature of 295 K. What is its new temperature if volume is changed to 577 mL? Assume pressure and amount are held constant.

  16. A gas has an initial volume of 14.98 L and an initial temperature of 238 K. What is its new temperature if volume is changed to 12.33 L? Assume pressure and amount are held constant.

  17. A gas has an initial volume of 685 mL and an initial temperature of 29°C. What is its new temperature if volume is changed to 1.006 L? Assume pressure and amount are held constant.

  18. A gas has an initial volume of 3.08 L and an initial temperature of −73°C. What is its new volume if its temperature is changed to 104°C? Assume pressure and amount are held constant.

Answers

1.

A gas law is a simple mathematical formula that allows one to predict the physical properties of a gas. The units of changing properties (volume, pressure, etc.) must be the same.

3.

pressure; volume

5.

amount of gas and temperature

7.

2.62 atm

9.

50.3 L

11.

1.95 L

13.

52.3 mL

15.

260 K

17.

444 K, or 171°C

 

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This work (Gas Laws by Jessie A. Key) is free of known copyright restrictions.

37

Other Gas Laws

Learning Objectives

  1. Review other simple gas laws.
  2. Learn and apply the combined gas law.

You may notice in Boyle’s law and Charles’s law that we actually refer to four physical properties of a gas: pressure (P), volume (V), temperature (T), and amount (in moles, n). We do this because these are the only four independent physical properties of a gas. There are other physical properties, but they are all related to one (or more) of these four properties.

Boyle’s law is written in terms of two of these properties, with the other two being held constant. Charles’s law is written in terms of two different properties, with the other two being held constant. It may not be surprising to learn that there are other gas laws that relate other pairs of properties—as long as the other two are held constant. Here we will mention a few.

Gay-Lussac’s law relates pressure with absolute temperature. In terms of two sets of data, Gay-Lussac’s law is

Screen Shot 2014-07-22 at 5.14.07 PM

Note that it has a structure very similar to that of Charles’s law, only with different variables—pressure instead of volume. Avogadro’s law introduces the last variable for amount. The original statement of Avogadro’s law states that equal volumes of different gases at the same temperature and pressure contain the same number of particles of gas. Because the number of particles is related to the number of moles (1 mol = 6.022 × 1023 particles), Avogadro’s law essentially states that equal volumes of different gases at the same temperature and pressure contain the same amount (moles, particles) of gas. Put mathematically into a gas law, Avogadro’s law is

Screen Shot 2014-07-22 at 5.14.11 PM

(First announced in 1811, it was Avogadro’s proposal that volume is related to the number of particles that eventually led to naming the number of things in a mole as Avogadro’s number.) Avogadro’s law is useful because for the first time we are seeing amount, in terms of the number of moles, as a variable in a gas law.

Example 7

A 2.45 L volume of gas contains 4.5 × 1021 gas particles. How many gas particles are there in 3.87 L if the gas is at constant pressure and temperature?

Solution

We can set up Avogadro’s law as follows:

245L

We algebraically rearrange to solve for n2:

n2

The L units cancel, so we solve for n2:

n2 = 7.1 × 1021 particles

Test Yourself

A 12.8 L volume of gas contains 3.00 × 1020 gas particles. At constant temperature and pressure, what volume do 8.22 × 1018 gas particles fill?

Answer

0.351 L

The variable n in Avogadro’s law can also stand for the number of moles of gas in addition to the number of particles.

One thing we notice about all the gas laws is that, collectively, volume and pressure are always in the numerator, and temperature is always in the denominator. This suggests that we can propose a gas law that combines pressure, volume, and temperature. This gas law is known as the combined gas law, and its mathematical form is

Screen Shot 2014-07-22 at 5.14.27 PM

This allows us to follow changes in all three major properties of a gas. Again, the usual warnings apply about how to solve for an unknown algebraically (isolate it on one side of the equation in the numerator), units (they must be the same for the two similar variables of each type), and units of temperature (must be in kelvins).

Example 8

A sample of gas at an initial volume of 8.33 L, an initial pressure of 1.82 atm, and an initial temperature of 286 K simultaneously changes its temperature to 355 K and its volume to 5.72 L. What is the final pressure of the gas?

Solution

We can use the combined gas law directly; all the units are consistent with each other, and the temperatures are given in Kelvin. Substituting,

182

We rearrange this to isolate the P2 variable all by itself. When we do so, certain units cancel:

182ATM2

Multiplying and dividing all the numbers, we get

P2 = 3.29 atm

Ultimately, the pressure increased, which would have been difficult to predict because two properties of the gas were changing.

Test Yourself

If P1 = 662 torr, V1 = 46.7 mL, T1 = 266 K, P2 = 409 torr, and T2 = 371 K, what is V2?

Answer

105 mL

As with other gas laws, if you need to determine the value of a variable in the denominator of the combined gas law, you can either cross-multiply all the terms or just take the reciprocal of the combined gas law. Remember, the variable you are solving for must be in the numerator and all by itself on one side of the equation.

Key Takeaways

  • There are other gas laws that relate any two physical properties of a gas.
  • The combined gas law relates pressure, volume, and temperature of a gas.

 

Exercises

  1. State Gay-Lussac’s law.

  2. State Avogadro’s law.

  3. Use Gay-Lussac’s law to determine the final pressure of a gas whose initial pressure is 602 torr, initial temperature is 356 K, and final temperature is 277 K. Assume volume and amount are held constant.

  4. Use Gay-Lussac’s law to determine the final temperature of a gas whose initial pressure is 1.88 atm, initial temperature is 76.3 K, and final pressure is 6.29 atm. Assume volume and amount are held constant.

  5. If 3.45 × 1022 atoms of Ar have a volume of 1.55 L at a certain temperature and pressure, what volume do 6.00 × 1023 atoms of Ar have at the same temperature and pressure?

  6. If 5.55 × 1022 atoms of He occupy a volume of 2.06 L at 0°C at 1.00 atm pressure, what volume do 2.08 × 1023 atoms of He occupy under the same conditions?

  7. Use Avogadro’s law to determine the final volume of a gas whose initial volume is 6.72 L, initial amount is 3.88 mol, and final amount is 6.10 mol. Assume pressure and temperature are held constant.

  8. Use Avogadro’s law to determine the final amount of a gas whose initial volume is 885 mL, initial amount is 0.552 mol, and final volume is 1,477 mL. Assume pressure and temperature are held constant.

  9. Use the combined gas law to complete this table. Assume that the amount remains constant in all cases.

    V1 = P1 = T1 = V2 = P2 = T2 =
    56.9 mL 334 torr 266 K 722 torr 334 K
    0.976 L 2.33 atm 443 K 1.223 L 355 K
    3.66 L 889 torr 23°C 2.19 L 739 torr
  10. Use the combined gas law to complete this table. Assume that the amount remains constant in all cases.

    V1 = P1 = T1 = V2 = P2 = T2 =
    56.7 mL 1.07 atm −34°C 998 torr 375 K
    3.49 L 338 torr 45°C 1,236 mL 392 K
    2.09 mL 776 torr 45°C 0.461 mL 0.668 atm
  11. A gas starts at the conditions 78.9 mL, 3.008 atm, and 56°C. Its conditions change to 35.6 mL and 2.55 atm. What is its final temperature?

  12. The initial conditions of a sample of gas are 319 K, 3.087 L, and 591 torr. What is its final pressure if volume is changed to 2.222 L and temperature is changed to 299 K?

  13. A gas starts with initial pressure of 7.11 atm, initial temperature of 66°C, and initial volume of 90.7 mL. If its conditions change to 33°C and 14.33 atm, what is its final volume?

  14. A sample of gas doubles its pressure and doubles its absolute temperature. By what amount does the volume change?

Answers

1.

The pressure of a gas is proportional to its absolute temperature.

3.

468 torr

5.

27.0 L

7.

10.6 L

9.

V1 = P1 = T1 = V2 = P2 = T2 =
56.9 mL 334 torr 266 K 33.1 mL 722 torr 334 K
0.976 L 2.33 atm 443 K 1.223 L 1.49 atm 355 K
3.66 L 889 torr 23°C 2.19 L 739 torr 147 K, or −126°C

11.

126 K, or −147°C

13.

40.6 mL

 

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This work (Other Gas Laws by Jessie A. Key) is free of known copyright restrictions.

38

The Ideal Gas Law and Some Applications

Learning Objectives

  1. Learn the ideal gas law.
  2. Apply the ideal gas law to any set of conditions of a gas.
  3. Apply the ideal gas law to molar volumes, density, and stoichiometry problems.

So far, the gas laws we have considered have all required that the gas change its conditions; then we predict a resulting change in one of its properties. Are there any gas laws that relate the physical properties of a gas at any given time?

Consider a further extension of the combined gas law to include n. By analogy to Avogadro’s law, n is positioned in the denominator of the fraction, opposite the volume. So

Screen Shot 2014-07-22 at 5.23.00 PM

Because pressure, volume, temperature, and amount are the only four independent physical properties of a gas, the constant in the above equation is truly a constant; indeed, because we do not need to specify the identity of a gas to apply the gas laws, this constant is the same for all gases. We define this constant with the symbol R, so the previous equation is written as

Screen Shot 2014-07-22 at 5.23.04 PM

which is usually rearranged as

PV = nRT

This equation is called the ideal gas law. It relates the four independent properties of a gas at any time. The constant R is called the ideal gas law constant. Its value depends on the units used to express pressure and volume. Table 6.1 “Values of the Ideal Gas Law Constant R lists the numerical values of R.

Table 6.1 Values of the Ideal Gas Law Constant R

Numerical Value Units
0.08205  L·atm/mol·K
62.36  L·torr/mol·K = L·mmHg/mol·K
8.314  J/mol·K

The ideal gas law is used like any other gas law, with attention paid to the units and making sure that temperature is expressed in kelvins. However, the ideal gas law does not require a change in the conditions of a gas sample. The ideal gas law implies that if you know any three of the physical properties of a gas, you can calculate the fourth property.

Example 9

A 4.22 mol sample of Ar has a pressure of 1.21 atm and a temperature of 34°C. What is its volume?

Solution

The first step is to convert temperature to kelvins:

34 + 273 = 307 K

Now we can substitute the conditions into the ideal gas law:

equations-01

The atm unit is in the numerator of both sides, so it cancels. On the right side of the equation, the mol and K units appear in the numerator and the denominator, so they cancel as well. The only unit remaining is L, which is the unit of volume we are looking for. We isolate the volume variable by dividing both sides of the equation by 1.21:

equations-05

Then solving for volume, we get

V = 87.9 L

Test Yourself

A 0.0997 mol sample of O2 has a pressure of 0.692 atm and a temperature of 333 K. What is its volume?

Answer

3.94 L

Example 10

At a given temperature, 0.00332 g of Hg in the gas phase has a pressure of 0.00120 mmHg and a volume of 435 L. What is its temperature?

Solution

We are not given the number of moles of Hg directly, but we are given a mass. We can use the molar mass of Hg to convert to the number of moles.

equations-07

Pressure is given in units of millimeters of mercury. We can either convert this to atmospheres or use the value of the ideal gas constant that includes the mmHg unit. We will take the second option. Substituting into the ideal gas law,

example10eq2

The mmHg, L, and mol units cancel, leaving the K unit, the unit of temperature. Isolating T all by itself on one side, we get

example10eq3

Then solving for K, we get

T = 507 K

Test Yourself

For a 0.00554 mol sample of H2, P = 23.44 torr and T = 557 K. What is its volume?

Answer

8.21 L

The ideal gas law can also be used in stoichiometry problems.

Example 11

What volume of H2 is produced at 299 K and 1.07 atm when 55.8 g of Zn metal react with excess HCl?

Zn(s) + 2 HCl(aq) → ZnCl2(aq) + H2(g)

Solution

Here we have a stoichiometry problem where we need to find the number of moles of H2 produced. Then we can use the ideal gas law, with the given temperature and pressure, to determine the volume of gas produced. First, the number of moles of H2 is calculated:

equations-06

Now that we know the number of moles of gas, we can use the ideal gas law to determine the volume, given the other conditions:

107atm

All the units cancel except for L, for volume, which means

V = 19.6 L

Test Yourself

What pressure of HCl is generated if 3.44 g of Cl2 are reacted in 4.55 L at 455 K?

H2(g) + Cl2(g) → 2HCl(g)

Answer

0.796 atm

It should be obvious by now that some physical properties of gases depend strongly on the conditions. What we need is a set of standard conditions so that properties of gases can be properly compared to each other. Standard temperature and pressure (STP) is defined as exactly 100 kPa of pressure (0.986 atm) and 273 K (0°C). For simplicity, we will use 1 atm as standard pressure. Defining STP allows us to compare more directly the properties of gases that differ from each other.

One property shared among gases is a molar volume. The molar volume is the volume of 1 mol of a gas. At STP, the molar volume of a gas can be easily determined by using the ideal gas law:

Screen Shot 2014-07-22 at 5.23.15 PM

All the units cancel except for L, the unit of volume. So

V = 22.4 L

Note that we have not specified the identity of the gas; we have specified only that the pressure is 1 atm and the temperature is 273 K. This makes for a very useful approximation: any gas at STP has a volume of 22.4 L per mole of gas; that is, the molar volume at STP is 22.4 L/mol (Figure 6.3 “Molar Volume”). This molar volume makes a useful conversion factor in stoichiometry problems if the conditions are at STP. If the conditions are not at STP, a molar volume of 22.4 L/mol is not applicable. However, if the conditions are not at STP, the combined gas law can be used to calculate the volume of the gas at STP; then the 22.4 L/mol molar volume can be used.

Figure 6.3 Molar Volume

Figure 6-3

A mole of gas at STP occupies 22.4 L, the volume of a cube that is 28.2 cm on a side.

Example 12

How many moles of Ar are present in 38.7 L at STP?

Solution

We can use the molar volume, 22.4 L/mol, as a conversion factor, but we need to reverse the fraction so that the L units cancel and mol units are introduced. It is a one-step conversion:

equations-11

Test Yourself

What volume does 4.87 mol of Kr have at STP?

Answer

109 L

Example 13

What volume of H2 is produced at STP when 55.8 g of Zn metal react with excess HCl?

Zn(s) + 2 HCl(aq) → ZnCl2(aq) + H2(g)

Solution

This is a stoichiometry problem with a twist: we need to use the molar volume of a gas at STP to determine the final answer. The first part of the calculation is the same as in a previous example:

equations-04

Now we can use the molar volume, 22.4 L/mol, because the gas is at STP:

equations-08

Alternatively, we could have applied the molar volume as a third conversion factor in the original stoichiometry calculation.

Test Yourself

What volume of HCl is generated if 3.44 g of Cl2 are reacted at STP?

H2(g) + Cl2(g) → 2 HCl(g)

Answer

2.17 L

The ideal gas law can also be used to determine the densities of gases. Recall that density is defined as the mass of a substance divided by its volume:

Screen Shot 2014-07-22 at 5.23.22 PM

Assume that you have exactly 1 mol of a gas. If you know the identity of the gas, you can determine the molar mass of the substance. Using the ideal gas law, you can also determine the volume of that mole of gas, using whatever the temperature and pressure conditions are. Then you can calculate the density of the gas by using

Screen Shot 2014-07-22 at 5.23.24 PM

Example 14

What is the density of N2 at 25°C and 0.955 atm?

Solution

First, we must convert the temperature into kelvins:

25 + 273 = 298 K

If we assume exactly 1 mol of N2, then we know its mass: 28.0 g. Using the ideal gas law, we can calculate the volume:

equations-02

All the units cancel except for L, the unit of volume. So

V = 25.6 L

Knowing the molar mass and the molar volume, we can determine the density of N2 under these conditions:

equations-09

Test Yourself

What is the density of CO2 at a pressure of 0.0079 atm and 227 K? (These are the approximate atmospheric conditions on Mars.)

Answer

0.019 g/L

Chemistry Is Everywhere: Breathing

Breathing (more properly called respiration) is the process by which we draw air into our lungs so that our bodies can take up oxygen from the air. Let us apply the gas laws to breathing.

Start by considering pressure. We draw air into our lungs because the diaphragm, a muscle underneath the lungs, moves down to reduce pressure in the lungs, causing external air to rush in to fill the lower-pressure volume. We expel air by the diaphragm pushing against the lungs, increasing pressure inside the lungs and forcing the high-pressure air out. What are the pressure changes involved? A quarter of an atmosphere? A tenth of an atmosphere? Actually, under normal conditions, a pressure difference of  only 1 or 2 torr makes us breathe in and out.

Figure 6.4 Breathing Mechanics

Figure 6-4

Breathing involves pressure differences between the inside of the lungs and the air outside. The pressure differences are only a few torr.

A normal breath is about 0.50 L. If room temperature is about 22°C, then the air has a temperature of about 295 K. With normal pressure being 1.0 atm, how many moles of air do we take in for every breath? The ideal gas law gives us an answer:

Screen Shot 2014-07-22 at 5.23.34 PM

Solving for the number of moles, we get

n = 0.021 mol air

This ends up being about 0.6 g of air per breath—not much but enough to keep us alive.

Key Takeaways

  • The ideal gas law relates the four independent physical properties of a gas at any time.
  • The ideal gas law can be used in stoichiometry problems in which chemical reactions involve gases.
  • Standard temperature and pressure (STP) are a useful set of benchmark conditions to compare other properties of gases.
  • At STP, gases have a volume of 22.4 L per mole.
  • The ideal gas law can be used to determine densities of gases.

Exercises

  1. What is the ideal gas law? What is the significance of R?

  2. Why does R have different numerical values (see Table 6.1 “Values of the Ideal Gas Law Constant R)?

  3. A sample of gas has a volume of 3.91 L, a temperature of 305 K, and a pressure of 2.09 atm. How many moles of gas are present?

  4. A 3.88 mol sample of gas has a temperature of 28°C and a pressure of 885 torr. What is its volume?

  5. A 0.0555 mol sample of Kr has a temperature of 188°C and a volume of 0.577 L. What pressure does it have?

  6. If 1.000 mol of gas has a volume of 5.00 L and a pressure of 5.00 atm, what is its temperature?

  7. A sample of 7.55 g of He has a volume of 5,520 mL and a temperature of 123°C. What is its pressure in torr?

  8. A sample of 87.4 g of Cl2 has a temperature of −22°C and a pressure of 993 torr. What is its volume in milliliters?

  9. A sample of Ne has a pressure of 0.772 atm and a volume of 18.95 L. If its temperature is 295 K, what mass is present in the sample?

  10. A mercury lamp contains 0.0055 g of Hg vapor in a volume of 15.0 mL. If the operating temperature is 2,800 K, what is the pressure of the mercury vapor?

  11. Oxygen is a product of the decomposition of mercury(II) oxide:

    2 HgO(s) → 2 Hg(ℓ) + O2(g)

    What volume of O2 is formed from the decomposition of 3.009 g of HgO if the gas has a pressure of 744 torr and a temperature of 122°C?

  12. Lithium oxide is used to absorb carbon dioxide:

    Li2O(s) + CO2(g) → Li2CO3(s)

    What volume of CO2 can 6.77 g of Li2O absorb if the CO2 pressure is 3.5 × 10−4 atm and the temperature is 295 K?

  13. What is the volume of 17.88 mol of Ar at STP?

  14. How many moles are present in 334 L of H2 at STP?

  15. How many liters of CO2 at STP are produced from 100.0 g of C8H18, the approximate formula of gasoline?

    2 C8H18(ℓ) + 25 O2(g) → 16 CO2(g) + 18 H2O(ℓ)

  16. How many liters of O2 at STP are required to burn 3.77 g of butane from a disposable lighter?

    2 C4H10(g) + 13 O2(g) → 8 CO2(g) + 10 H2O(ℓ)

  17. What is the density of each gas at STP?

a)  He

b)  Ne

c)  Ar

d)  Kr

18.  What is the density of each gas at STP?

a)  H2

b)  O2

c)  N2

19.  What is the density of SF6 at 335 K and 788 torr?

20.  What is the density of He at −200°C and 33.9 torr?

Answers

1. The ideal gas law is PV = nRT. R is the ideal gas law constant, which relates the other four variables.

3. 0.327 mol

5. 3.64 atm

7. 8,440 torr

9. 12.2 g

11. 0.230 L

13. 401 L

15. 157 L

17.

a)   0.179 g/L
b)   0.901 g/L
c)   1.78 g/L
d)   3.74 g/L

19. 5.51 g/L

39

Gas Mixtures

Learning Objective

  1. Learn Dalton’s law of partial pressures.

One of the properties of gases is that they mix with each other. When they do so, they become a solution—a homogeneous mixture. Some of the properties of gas mixtures are easy to determine if we know the composition of the gases in the mix.

In gas mixtures, each component in the gas phase can be treated separately. Each component of the mixture shares the same temperature and volume. (Remember that gases expand to fill the volume of their container; gases in a mixture do that as well.) However, each gas has its own pressure. The partial pressure of a gas, Pi, is the pressure that an individual gas in a mixture has. Partial pressures are expressed in torr, millimeters of mercury, or atmospheres like any other gas pressure; however, we use the term pressure when talking about pure gases and the term partial pressure when we are talking about the individual gas components in a mixture.

Dalton’s law of partial pressures states that the total pressure of a gas mixture, Ptot, is equal to the sum of the partial pressures of the components, Pi:

Screen Shot 2014-07-22 at 5.51.15 PM

Although this may seem to be a trivial law, it reinforces the idea that gases behave independently of each other.

Example 15

A mixture of H2 at 2.33 atm and N2 at 0.77 atm is in a container. What is the total pressure in the container?

Solution

Dalton’s law of partial pressures states that the total pressure is equal to the sum of the partial pressures. We simply add the two pressures together:

Ptot = 2.33 atm + 0.77 atm = 3.10 atm

Test Yourself

Air can be thought of as a mixture of N2 and O2. In 760 torr of air, the partial pressure of N2 is 608 torr. What is the partial pressure of O2?

Answer

152 torr

Example 16

A 2.00 L container with 2.50 atm of H2 is connected to a 5.00 L container with 1.90 atm of O2 inside. The containers are opened, and the gases mix. What is the final pressure inside the containers?

Solution

Because gases act independently of each other, we can determine the resulting final pressures using Boyle’s law and then add the two resulting pressures together to get the final pressure. The total final volume is 2.00 L + 5.00 L = 7.00 L. First, we use Boyle’s law to determine the final pressure of H2:

(2.50 atm)(2.00 L) = P2(7.00 L)

Solving for P2, we get

P2 = 0.714 atm = partial pressure of H2

Now we do that same thing for the O2:

(1.90 atm)(5.00 L) = P2(7.00 L)
P2 = 1.36 atm = partial pressure of O2

The total pressure is the sum of the two resulting partial pressures:

Ptot = 0.714 atm + 1.36 atm = 2.07 atm

Test Yourself

If 0.75 atm of He in a 2.00 L container is connected to a 3.00 L container with 0.35 atm of Ne and the connection between the containers is opened, what is the resulting total pressure?

Answer

0.51 atm

One of the reasons we have to deal with Dalton’s law of partial pressures is that gases are frequently collected by bubbling through water. As we will see in Chapter 10 “Solids and Liquids,” liquids are constantly evapourating into a vapour until the vapour achieves a partial pressure characteristic of the substance and the temperature. This partial pressure is called a vapour pressure. Table 6.2 “Vapor Pressure of Water versus Temperature” lists the vapour pressures of H2O versus temperature. Note that if a substance is normally a gas under a given set of conditions, the term partial pressure is used; the term vapour pressure is reserved for the partial pressure of a vapour when the liquid is the normal phase under a given set of conditions.

Table 6.2 Vapor Pressure of Water versus Temperature

Temperature (°C) Vapor Pressure (torr) Temperature (°C) Vapor Pressure (torr)
5 6.54 30 31.84
10 9.21 35 42.20
15 12.79 40 55.36
20 17.54 50 92.59
21 18.66 60 149.5
22 19.84 70 233.8
23 21.08 80 355.3
24 22.39 90 525.9
25 23.77 100 760.0

Any time a gas is collected over water, the total pressure is equal to the partial pressure of the gas plus the vapour pressure of water. This means that the amount of gas collected will be less than the total pressure suggests.

Example 17

Hydrogen gas is generated by the reaction of nitric acid and elemental iron. The gas is collected in an inverted 2.00 L container immersed in a pool of water at 22°C. At the end of the collection, the partial pressure inside the container is 733 torr. How many moles of H2 gas were generated?

Solution

We need to take into account that the total pressure includes the vapour pressure of water. According to Table 6.2 “Vapor Pressure of Water versus Temperature,” the vapour pressure of water at 22°C is 19.84 torr. According to Dalton’s law of partial pressures, the total pressure equals the sum of the pressures of the individual gases, so

733 torr = PH2 + PH2O = PH2 + 19.84 torr

We solve by subtracting:

PH2 = 713 torr

Now we can use the ideal gas law to determine the number of moles (remembering to convert the temperature to kelvins, making it 295 K):

713torr

All the units cancel except for mol, which is what we are looking for. So

n = 0.0775 mol H2 collected

Test Yourself

CO2, generated by the decomposition of CaCO3, is collected in a 3.50 L container over water. If the temperature is 50°C and the total pressure inside the container is 833 torr, how many moles of CO2 were generated?

Answer

0.129 mol

Finally, we introduce a new unit that can be useful, especially for gases. The mole fraction, χi, is the ratio of the number of moles of component i in a mixture divided by the total number of moles in the sample:

Screen Shot 2014-07-22 at 5.51.24 PM

(χ is the lowercase Greek letter chi.) Note that mole fraction is not a percentage; its values range from 0 to 1. For example, consider the combination of 4.00 g of He and 5.0 g of Ne. Converting both to moles, we get

Screen Shot 2014-07-22 at 5.51.28 PM

The total number of moles is the sum of the two mole amounts:

total moles = 1.00 mol + 0.025 mol = 1.25 mol

The mole fractions are simply the ratio of each mole amount to the total number of moles, 1.25 mol:

Screen Shot 2014-07-22 at 5.51.35 PM

The sum of the mole fractions equals exactly 1.

For gases, there is another way to determine the mole fraction. When gases have the same volume and temperature (as they would in a mixture of gases), the number of moles is proportional to partial pressure, so the mole fractions for a gas mixture can be determined by taking the ratio of partial pressure to total pressure:

Screen Shot 2014-07-22 at 5.51.40 PM

This expression allows us to determine mole fractions without calculating the moles of each component directly.

Example 18

A container has a mixture of He at 0.80 atm and Ne at 0.60 atm. What is the mole fraction of each component?

Solution

According to Dalton’s law, the total pressure is the sum of the partial pressures:

Ptot = 0.80 atm + 0.60 atm = 1.40 atm

The mole fractions are the ratios of the partial pressure of each component to the total pressure:

xhe

Again, the sum of the mole fractions is exactly 1.

Test Yourself

What are the mole fractions when 0.65 atm of O2 and 1.30 atm of N2 are mixed in a container?

Answer

χO2=0.33; χN2=0.67

Food and Drink App: Carbonated Beverages

Carbonated beverages—sodas, beer, sparkling wines—have one thing in common: they have CO2 gas dissolved in them in such sufficient quantities that it affects the drinking experience. Most people find the drinking experience pleasant—indeed, in the United States alone, over 1.5 × 109 gal of soda are consumed each year, which is almost 50 gal per person! This figure does not include other types of carbonated beverages, so the total consumption is probably significantly higher.

All carbonated beverages are made in one of two ways. First, the flat beverage is subjected to a high pressure of CO2 gas, which forces the gas into solution. The carbonated beverage is then packaged in a tightly sealed package (usually a bottle or a can) and sold. When the container is opened, the CO2 pressure is released, resulting in the well-known hiss, and CO2 bubbles come out of solution (Figure 6.5 (Opening a Carbonated Beverage). This must be done with care: if the CO2 comes out too violently, a mess can occur!

Figure 6.5 Opening a Carbonated Beverage


image
If you are not careful opening a container of a carbonated beverage, you can make a mess as the CO2 comes out of solution suddenly. Source: “Champagne uncorking” by Niels Noordhoek s licensed under the Creative Commons Attribution-Share Alike 3.0 Unported license.

The second way a beverage can become carbonated is by the ingestion of sugar by yeast, which then generates CO2 as a digestion product. This process is called fermentation. The overall reaction is

C6H12O6(aq) → 2 C2H5OH(aq) + 2 CO2(aq)

When this process occurs in a closed container, the CO2 produced dissolves in the liquid, only to be released from solution when the container is opened. Most fine sparkling wines and champagnes are turned into carbonated beverages this way. Less-expensive sparkling wines are made like sodas and beer, with exposure to high pressures of CO2 gas.

Key Takeaways

  • The pressure of a gas in a gas mixture is termed the partial pressure.
  • Dalton’s law of partial pressure states that the total pressure in a gas mixture is the sum of the individual partial pressures.
  • Collecting gases over water requires that we take the vapour pressure of water into account.
  • Mole fraction is another way to express the amounts of components in a mixture.

 

Exercises

  1. What is the total pressure of a gas mixture containing these partial pressures: PN2 = 0.78 atm, PH2 = 0.33 atm, and PO2 = 1.59 atm?

  2. What is the total pressure of a gas mixture containing these partial pressures: PNe = 312 torr, PHe = 799 torr, and PAr = 831 torr?

  3. In a gas mixture of He and Ne, the total pressure is 335 torr and the partial pressure of He is 0.228 atm. What is the partial pressure of Ne?

  4. In a gas mixture of O2 and N2, the total pressure is 2.66 atm and the partial pressure of O2 is 888 torr. What is the partial pressure of N2?

  5. A 3.55 L container has a mixture of 56.7 g of Ar and 33.9 g of He at 33°C. What are the partial pressures of the gases and the total pressure inside the container?

  6. A 772 mL container has a mixture of 2.99 g of H2 and 44.2 g of Xe at 388 K. What are the partial pressures of the gases and the total pressure inside the container?

  7. A sample of O2 is collected over water in a 5.00 L container at 20°C. If the total pressure is 688 torr, how many moles of O2 are collected?

  8. A sample of H2 is collected over water in a 3.55 L container at 50°C. If the total pressure is 445 torr, how many moles of H2 are collected?

  9. A sample of CO is collected over water in a 25.00 L container at 5°C. If the total pressure is 0.112 atm, how many moles of CO are collected?

  10. A sample of NO2 is collected over water in a 775 mL container at 25°C. If the total pressure is 0.990 atm, how many moles of NO2 are collected?

  11. A sample of NO is collected over water in a 75.0 mL container at 25°C. If the total pressure is 0.495 atm, how many grams of NO are collected?

  12. A sample of ClO2 is collected over water in a 0.800 L container at 15°C. If the total pressure is 1.002 atm, how many grams of ClO2 are collected?

  13. Determine the mole fractions of each component when 44.5 g of He is mixed with 8.83 g of H2.

  14. Determine the mole fractions of each component when 9.33 g of SO2 is mixed with 13.29 g of SO3.

  15. In a container, 4.56 atm of F2 is combined with 2.66 atm of Cl2. What is the mole fraction of each component?

  16. In a container, 77.3 atm of SiF4 are mixed with 33.9 atm of O2. What is the mole fraction of each component?

Answers

1.

2.70 atm

3.

162 torr, or 0.213 atm

5.

PAr = 10.0 atm; PHe = 59.9 atm; Ptot = 69.9 atm

7.

0.183 mol

9.

0.113 mol

11.

0.0440 g

13.

χHe = 0.718; χH2 = 0.282

15.

χF2 = 0.632; χCl3 = 0.368

 

40

Kinetic Molecular Theory of Gases

Jessie A. Key and David W. Ball

Learning Objectives

  1. State the major concepts behind the kinetic molecular theory of gases.
  2. Demonstrate the relationship between kinetic energy and molecular speed.
  3. Apply the kinetic molecular theory to explain and predict the gas laws.

Gases were among the first substances studied using the modern scientific method, which was developed in the 1600s. It did not take long to recognize that gases all shared certain physical behaviours, suggesting that gases could be described by one all-encompassing theory. The kinetic molecular theory of gases is a model that helps us understand the physical properties of gases at the molecular level. It is based on the following concepts:

  1. Gases consist of particles (molecules or atoms) that are in constant random motion.
  2. Gas particles are constantly colliding with each other and the walls of their container. These collisions are elastic; that is, there is no net loss of energy from the collisions.
  3. Gas particles are small and the total volume occupied by gas molecules is negligible relative to the total volume of their container.
  4. There are no interactive forces (i.e., attraction or repulsion) between the particles of a gas.
  5. The average kinetic energy of gas particles is proportional to the absolute temperature of the gas, and all gases at the same temperature have the same average kinetic energy.

Figure 6.6 “The Kinetic Molecular Theory of Gases” shows a representation of how we mentally picture the gas phase.

Figure 6-6

The kinetic molecular theory of gases describes this state of matter as composed of tiny particles in constant motion with a lot of distance between the particles. Because most of the volume occupied by a gas is empty space, a gas has a low density and can expand or contract under the appropriate influence. The fact that gas particles are in constant motion means that two or more gases will always mix as the particles from the individual gases move and collide with each other. The number of collisions the gas particles make with the walls of their container and the force with which they collide determine the magnitude of the gas pressure.

 

Kinetic Energy and Molecular Speed

Gas particles are in constant motion, and any object in motion has kinetic energy (Ek). Kinetic energy, for an individual atom, can be calculated by the following equation where m is the mass, and u is the speed.

Ek = 1/2 mu2  

Overall the molecules in a sample of a gas share an average kinetic energy; however, individual molecules exhibit a distribution of kinetic energies because of having a distribution of speeds (Figure 6.7 “Stylized molecular speed distribution”). This distribution of speeds arises from the collisions that occur between molecules in the gas phase. Although these collisions are elastic (there is no net loss of energy), the individual speeds of each molecule involved in the collision may change. For example, in the collision of two molecules, one molecule may be deflected at a slightly higher speed and the other at a slightly lower speed, but the average kinetic energy does not change.

Figure 6.## Stylized molecular speed distribution
Figure 6.7 Stylized molecular speed distribution

When analyzing a diagram of the distribution of molecular speeds, there are several commonly used terms to be familiar with. The most probable speed (ump) is the speed of the largest number of molecules, and corresponds to the peak of the distribution. The average speed (uav) is the mean speed of all gas molecules in the sample. The root-mean-square (rms) speed (urms) corresponds to the speed of molecules having exactly the same kinetic energy as the average kinetic energy of the sample. 

 

Figure 6.# Distribution of molecular speeds, Oxygen gas at -100, 20 and 600 oC.

Figure 6.8 Distribution of molecular speeds, oxygen gas at -100, 20, and 600°CAdapted from Maxwell-Boltzmann distribution 1.png by Superborsuk/CC-BY-SA-3.0.

 

According to the kinetic molecular theory, the average kinetic energy of gas particles is proportional to the absolute temperature of the gas. This can be expressed with the following equation where k represents the Boltzmann constant. The Boltzmann constant is simply the gas constant R divided by the Avogadro’s constant (NA). The bar above certain terms indicates they are average values.

\overline{E_k} = \frac{3}{2} \textit{kT}

Since average kinetic energy is related both to the absolute temperature and the molecular speed, we can combine the equation above with the previous one to determine the rms speed.

\overline{E_k} \textit{ = }\frac{1}{2} \textit{m} \overline{u^2} \textit{ = } \frac{3}{2} \textit{kT}

\sqrt{\overline{u^2}} = \sqrt{\frac{3kT}{m}}

This demonstrates that the rms speed is related to the temperature. We can further manipulate this equation by multiplying the numerator and denominator by Avogadro’s constant  (NA) to give us a form using the gas constant (R) and molar mass (M).

\sqrt{\overline{u^2}}\ = \sqrt{\frac{3RT}{M}}

This form of the equation demonstrates that the rms speed of gas molecules is also related to the molar mass of the substance. Comparing two gases of different molar mass at the same temperature, we see that despite having the same average kinetic energy, the gas with the smaller molar mass will have a higher rms speed.

Figure 6.# Molecular Speed Distribution of Noble Gases

Figure 6.9 Molecular Speed Distribution of Noble GasesAdapted from MaxwellBoltzmann-en.svg by Pdbailey/Public Domain

 

Example 19
Calculate the rms speed of nitrogen molecules at 25ºC.Solution\sqrt{\overline{u^2}}=\ \sqrt{\frac{3RT}{M}}=\ \sqrt{\frac{3(8.3145\ J{\ K}^{-1}{\ mol}^{-1})(298.15\ K)}{28.02\ x\ {10}^{-3}\ kg{\ mol}^{-1}}}\sqrt{\overline{u^2}}=\ \sqrt{2.654\ x\ {10}^{5\ }J\ {kg}^{-1}}Knowing that 1 J = 1 kg ms-2 we can convert to metres per second:  \sqrt{\overline{u^2}}=\ \sqrt{2.654\ x\ {10}^{5\ }J\ {kg}^{-1}\ x\ \frac{1\ kg\ m^2{\ s}^{-2}}{1\ J}}\ \sqrt{\overline{u^2}}= \sqrt{2.654\ x\ {10}^{5\ }m^2{\ s}^{-2}}\ \textit{ = 515.2 m} {\ s}^{-1} \textit{}

Applying the Kinetic Molecular Theory to the Gas Laws

The kinetic molecular theory can be used to explain or predict the experimental trends that were used to generate the gas laws. Let’s work through a few scenarios to demonstrate this point.

What will happen to the pressure of a system where the volume is decreased at constant temperature? 

This problem can be approached in two ways:

1. The ideal gas law can be rearranged to solve for pressure and estimate the change in pressure:

\textit{PV = nRT}

\textit{P = }\frac{nRT}{V}\textit{}

Volume is located in the denominator of the equation, and it is being decreased. This means the rest of the equation is being divided by a smaller number, so that should make the pressure larger.

2. The kinetic molecular theory can be used. Since the temperature is remaining constant, the average kinetic energy and the rms speed remain the same as well. The volume of the container has decreased, which means that the gas molecules have to move a shorter distance to have a collision. There will therefore be more collisions per second, causing an increase in pressure.

What will happen to the pressure of a system where the temperature is increased and the volume remains constant?

Again, this type of problem can be approached in two ways:

1. The ideal gas law can be rearranged to solve for pressure  and estimate the change in pressure.

\textit{p = }\frac{nRT}{V}\textit{}

Temperature is located in the numerator; there is a direct relationship between temperature and pressure. Therefore an increase in temperature should cause an increase in pressure.

2. The kinetic molecular theory can be used. Temperature is increased, so the average kinetic energy and the rms speed should also increase. This means that the gas molecules will hit the container walls more frequently and with greater force because they are all moving faster. This should increase the pressure.

 

Key Takeaways

  • The physical behaviour of gases is explained by the kinetic molecular theory of gases.
  • The number of collisions that gas particles make with the walls of their container and the force at which they collide determine the magnitude of the gas pressure.
  • Temperature is proportional to average kinetic energy.

 

Exercises

  1. State the ideas of the kinetic molecular theory of gases.
  2. Calculate the rms speed of  CO2 at 40°C.
  3. Using the kinetic molecular theory, explain how an increase in the number of moles of gas at constant volume and temperature affects the pressure.

Answers

  1. Gases consist of tiny particles of matter that are in constant motion. Gas particles are constantly colliding with each other and the walls of a container. These collisions are elastic; that is, there is no net loss of energy from the collisions. Gas particles are separated by large distances. The size of gas particles is tiny compared to the distances that separate them and the volume of the container. There are no interactive forces (i.e., attraction or repulsion) between the particles of a gas. The average kinetic energy of gas particles is dependent on the temperature of the gas.
  2. 421 m/s
  3. Temperature remains the same, so the average kinetic energy and the rms speed should remain the same. Increasing the number of moles of gas means there are more molecules of gas available to collide with the walls of the container at any given time. Therefore pressure should increase.

 

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Kinetic Molecular Theory of Gases by Jessie A. Key and David W. Ball is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.

41

Molecular Effusion and Diffusion

Jessie A. Key

Learning Objectives

  1. Explore the nature of gas movement: molecular effusion and diffusion.
  2. Examine and apply Graham’s law of effusion.

Effusion

The movement of gas molecules can be divided into a few different types. Effusion is the movement of gas molecules from one container to another via a tiny hole. Typically the container to which the gas is moving is kept under lower pressure.

Figure 6.## Molecular Effusion
Figure 6.10 Molecular Effusion

In 1846, the Scottish chemist Thomas Graham found that the rate of effusion of a gas (the amount of gas transferred between containers in a certain amount of time) is inversely proportional to the square root of its molar mass. This means that gases with a lighter molecular weight have higher effusion rates.

Figure 6.## Thomas Graham who proposed Graham's law of effusion in 1846.

Figure 6.11 Thomas Graham who proposed his law of effusion in 1846.Thomas Graham Lithogram Public Domain

This finding is summarized in Graham’s law of effusion:

\frac{{\rm Rate\ of\ Effusion\ (Gas\ 1)}}{{\rm Rate\ of\ Effusion\ (Gas\ 2)}} =\sqrt{\frac{M_2}{M_1}}\

This finding can be rationalized by thinking through the process of effusion on the molecular level. For a gas molecule to successfully move from one container to another, it must hit and pass through the tiny hole present in the container. Gases with higher rms speed are more likely to hit and pass through the hole so effusion is dependent on rms speed:

\frac{{\rm Rate\ of\ Effusion\ (Gas\ 1)}}{{\rm Rate\ of\ Effusion\ (Gas\ 2)}} =\frac{\sqrt{\frac{3RT}{M_1}}}{\sqrt{\frac{3RT}{M_2}}} =\sqrt{\frac{M_2}{M_1}}\

 

Example 20

An unknown halogen (diatomic) gas effuses at a rate that is approximately 1.89 times the rate of I2 gas at the same temperature. Determine the molar mass and identity of this unknown gas.

Solution

\frac{{\rm Rate\ of\ Effusion\ (Gas\ 1)}}{{\rm Rate\ of\ Effusion\ (Gas\ 2)}}=\sqrt{\frac{M_2}{M_1}}\

\frac{{\rm Rate\ of\ Effusion\ (Gas\ }{{\rm I}}_2{\rm )}}{{\rm Rate\ of\ Effusion\ (Gas\ 2)}} = \sqrt{\frac{M_2}{M{{\rm I}}_2}}\
\frac{{\rm 1}}{1.89} = \sqrt{\frac{M_2}{253.80\ \text{g/mol}}}\

0.279 =\frac{M_2}{253.80\ \text{g/mol}}\

M2 = 71.1 g/mol. Therefore the unknown gas is Cl2

Diffusion

Another type of gas movement is called diffusion; it is the movement of gas molecules through one or more additional types of gas via random molecular motion. Similar to effusion, gases with lower molecular weights (which have a higher rms speed) diffuse faster than gases with higher molecular weights. However, in diffusion, movement is much more complicated as collisions occur between molecules that change the direction and speed of the molecules. As a result of these collisions, the path a molecule travels in diffusion is made up of numerous straight, short segments. The term mean free path is used to describe the average distance travelled by a molecule between collisions.

Figure 6.## Molecular Diffusion
Figure 6.12 Molecular Diffusion
Figure 6.## Stylized depiction of the path travelled by a gas particle during diffusion. Other particles omitted for clarity.
Figure 6.13 Stylized depiction of the path travelled by a gas particle during diffusion. Other particles have been omitted for clarity.

qrcode.23437599

Video Source: Diffusion by keyj (https://viuvideos.viu.ca/media/Diffusion+/0_zel419p5)
Image: ParticleMeanFreePath.png by DaisyDaisy/Public Domain

Key Takeaways

  • Effusion is the movement of gas molecules from one container to another through a tiny hole.
  • Rates of effusion can be compared at the same temperature using Graham’s law.
  • Diffusion  is the movement of gas molecules through one or more other types of gas via random molecular motion.
  • Both the rates of effusion and diffusion are influenced by the molecular weight of the gas particle.

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Molecular Effusion and Diffusion by Jessie A. Key is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.

42

Real Gases

Jessie A. Key

Learning Objectives

  1. To examine the differences between ideal and real gases.
  2. To explore the effects of high pressure and low temperature on real gas samples.
  3. To apply van der Waal’s equation to correct the ideal gas law for real gases using the experimentally determined constants a and b.

An ideal gas is one that conforms exactly to the tenets of the kinetic molecular theory, where the volume occupied by the gas particles is negligible relative to the total volume of the container, and there are no appreciable intermolecular attractions or repulsions.

Real gases can deviate from ideal behaviour, especially at high pressures and low temperatures. The extent of deviation is measured using the compressibility factor. The compressibility factor is obtained by solving for n in the ideal gas law: dividing the product of pressure and volume by the product of the gas constant and temperature (PV/RT) for one mole of a given substance. Under ideal conditions, this ratio of PV/RT should be exactly equal to 1.

 

Real Gases at High Pressure

At higher pressures, gas molecules are closer together in a space. As a result of this crowding, gas molecules experience greater attractive intermolecular forces. Intermolecular forces hold molecules together more, lessening the force and frequency of collisions with the container wall and thus lowering the pressure below ideal values. As well, at higher pressure, molecules occupy a larger proportion of the volume of the container. With other gas molecules taking up a larger proportion of the volume of the container, the unoccupied volume of the container available to any one molecule is smaller than in ideal conditions. This decrease in available volume causes an increase in pressure beyond ideal conditions.

Figure #.#. Approximated compressibility factors of three gases at 250 K
Figure 6.14. Approximate compressibility factors of three gases at 250 K.

 

Real Gases at Low Temperature

Temperature also influences deviations from ideal gas behaviour (Figure 6.15). As temperature decreases, the average kinetic energy of the gas particles decreases. A larger proportion of gas molecules therefore have insufficient kinetic energy to overcome attractive intermolecular forces from neighbouring atoms. This means that gas molecules become “stickier” to each other, and collide with the walls of the container with less frequency and force, decreasing pressure below that of ideal values.

Figure #.#. Approximate compressibility factor of nitrogen at different temperatures.
Figure 6.15. Approximate compressibility factor of nitrogen at different temperatures.

 

The van der Waals Equation

Figure 6.## Johannes Diderik van der Waals

Figure 6.16 Johannes Diderik van der WaalsJohannes Diderik van der Waals from Public Domain

In 1873, Dutch scientist Johannes van der Waals developed an equation that compensates for deviations from ideal gas behaviour. The van der Waals equation uses two additional experimentally determined constants: a, which is a term to correct for intermolecular forces, and b, which corrects for the volume of the gas molecules (Table 6.3 “Selected van der Waals Constants for Gas Molecules”).

It should be noted that if the new terms a and b are equal to zero (under ideal conditions), the equation simplifies back to the ideal gas law: PV = nRT.

Table 6.3 Selected van der Waals Constants for Gas Molecules. Adapted from van der Waals constants by Wikipedia/CC-BY-SA-3.0

 

a (L2atm/mol2)

b (L/mol)

Helium 0.03457 0.0237
Neon 0.2135 0.01709
Hydrogen 0.2476 0.02661
Argon 1.355 0.0320
Nitric oxide 1.358 0.02789
Oxygen 1.378 0.03183
Nitrogen 1.408 0.03913
Carbon monoxide 1.505 0.03985
Methane 2.283 0.04278
Krypton 2.349 0.03978
Carbon dioxide 3.640 0.04267
Hydrogen chloride 3.716 0.04081
Nitrous oxide 3.832 0.04415
Ammonia 4.225 0.0371
Xenon 4.250 0.05105
 

Example 21

Use the van der Waals equation and Table 6.3 to determine the pressure, in atmospheres, of 2.00 moles of oxygen gas in a 30.00 L flask at 25.0oC.

Solution

\left(P+a\ {\left[\frac{n}{V}\right]}^2\right)\\left(V-nb\right)=nRT\

\left(P+a\ {\left[\frac{n}{V}\right]}^2\right)\\left(V-nb\right)=nRT\
\left(P+{\rm 1.378}\frac{L^2atm}{{mol}^2}\ {\left[\frac{2.00\ mol}{30.00\ L}\right]}^2\right)\\left(30.00\ L-2.00\ mol\ ({\rm 0.03183}\frac{{\rm L}}{{\rm mol}})\right)=(2.00\ mol)\\left(0.08206\ L\ atm\ K^{-1}\ \ {mol}^{-1}\right)(\ 298.15\ K)\

P =1.63 atm

 

Key Takeaways

  • An ideal gas is one that conforms exactly to the tenets of the kinetic molecular theory, where the volume occupied by the gas particles is negligible relative to the total volume of the container and there are no appreciable intermolecular attractions or repulsions.
  • A real gas is one that deviates from ideal behaviour, due to the effects of gas particles occupying a finite volume and the strength of intermolecular forces.
  • The van der Waal’s equation compensates for deviations from ideal gas behaviour.

 

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Real Gases by Jessie A. Key is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.

43

End-of-Chapter Material

David W. Ball

Additional Exercises

  1. What is the pressure in pascals if a force of 4.88 kN is pressed against an area of 235 cm2?
  2. What is the pressure in pascals if a force of 3.44 ×104 MN is pressed against an area of 1.09 km2?
  3. What is the final temperature of a gas whose initial conditions are 667 mL, 822 torr, and 67°C, and whose final volume and pressure are 1.334 L and 2.98 atm, respectively? Assume the amount remains constant.
  4. What is the final pressure of a gas whose initial conditions are 1.407 L, 2.06 atm, and −67°C, and whose final volume and temperature are 608 mL and 449 K, respectively? Assume the amount remains constant.
  5. Propose a combined gas law that relates volume, pressure, and amount at constant temperature.
  6. Propose a combined gas law that relates amount, pressure, and temperature at constant volume.
  7. A sample of 6.022 ×1023 particles of gas has a volume of 22.4 L at 0°C and a pressure of 1.000 atm. Although it may seem silly to contemplate, what volume would one particle of gas occupy?
  8. One mole of liquid N2 has a volume of 34.65 mL at −196°C. At that temperature, 1 mol of N2 gas has a volume of 6.318 L if the pressure is 1.000 atm. What pressure is needed to compress the N2 gas to 34.65 mL?
  9. Use two values of R to determine the ratio between an atmosphere and a torr. Does the number make sense?
  10. Use two values of R to determine how many joules are in a liter·atmosphere.

11.  At an altitude of 40 km above the earth’s surface, the atmospheric pressure is 5.00 torr, and the surrounding temperature is −20°C. If a weather balloon is filled with 1.000 mol of He at 760 torr and 22°C, what is its

a)  initial volume before ascent?

b)  final volume when it reaches 40 km in altitude? (Assume the pressure of the gas equals the surrounding pressure.)

12.  If a balloon is filled with 1.000 mol of He at 760 torr and 22°C, what is its

a)  initial volume before ascent?

b)  final volume if it descends to the bottom of the Mariana Trench, where the surrounding temperature is 1.4°C and the pressure is 1,060 atm?

13.  Air, a mixture of mostly N2 and O2, can be approximated as having a molar mass of 28.8 g/mol. What is the density of air at 1.00 atm and 22°C? (This is approximately sea level.)

14.  Air, a mixture of mostly N2 and O2, can be approximated as having a molar mass of 28.8 g/mol. What is the density of air at 0.26 atm and −26°C? (This is approximately the atmospheric condition at the summit of Mount Everest.)

15.  On the surface of Venus, the atmospheric pressure is 91.8 atm, and the temperature is 460°C. What is the density of CO2 under these conditions? (The Venusian atmosphere is composed largely of CO2.)

16.  On the surface of Mars, the atmospheric pressure is 4.50 torr, and the temperature is −87°C. What is the density of CO2 under these conditions? (The Martian atmosphere, similar to its Venusian counterpart, is composed largely of CO2.)

17.  HNO3 reacts with iron metal according to Fe(s) + 2HNO3(aq) → Fe(NO3)2(aq) + H2(g) In a reaction vessel, 23.8 g of Fe are reacted but only 446 mL of H2 are collected over water at 25°C and a pressure of 733 torr. What is the percent yield of the reaction?

18.  NaHCO3 is decomposed by heat according to 2NaHCO3(s) → Na2CO3(s) + H2O(ℓ) + CO2(g) If you start with 100.0 g of NaHCO3 and collect 10.06 L of CO2 over water at 20°C and 0.977 atm, what is the percent yield of the decomposition reaction?

19.  Determine if the following actions will cause the pressure of a particular gas sample to increase, decrease, or remain the same:

a)  decreasing the temperature

b)  decreasing the molar mass of the gas

c)  decreasing the volume of the container

20.  Under what conditions do gases deviate most from ideal gas behaviour? Explain your answer.

21.  Place the following gases in order from lowest to highest average molecular speed at 25°C: He, Ar, O2, I2.

22.  The effusion rate of an unknown noble gas sample is 0.35 times that of neon, at the same temperature. Determine the molecular weight and identity of the unknown noble gas.

23.  Use the van der Waal’s equation to determine the pressure of 2.00 moles of helium in a 5.00 L balloon at 300.00 K. How does this value compare to what you would obtain with the ideal gas law?

 

Answers

1. 208,000 Pa

3. 1,874 K

5.

Screen Shot 2014-07-22 at 5.59.06 PM

7. 3.72 × 10-23 L

9. 1 atm = 760 torr

11.

11.1. 24.2 L
11.2. 3155 L

13. 1.19 g/L

15. 67.2 g/L

17. 3.99%

19.

19.1. decreased pressure
19.2. increased pressure
19.3. increased pressure

21. I< Ar < O< He

23. van der Waals: 9.94 atm, ideal: 9.85 atm

 

VII

Chapter 7. Energy and Chemistry

44

Introduction to Energy and Chemistry

It takes energy to launch a spaceship into space. If it takes 1 energy unit to warm 0.25 g of water by 1°C, then it takes over 15,100 energy units to put that 0.25 g of water into earth orbit. The most powerful engines designed to lift rockets into space were part of the Saturn V rocket, that was built by the National Aeronautics and Space Administration (NASA). The rocket had three stages, with the first stage having the capability of launching about 3.5 million kg of mass. About 2.3 million kg was the actual fuel for the first stage; rockets in space have the unpleasant task of having to take their own chemicals with them to provide thrust.

It takes a lot of energy to launch a rocket into space. The Saturn V rocket used five of the most powerful engines ever built to take its initial step into orbit. Source: “Soyuz TMA-5 launch” s in the public domain because it was solely created by NASA.
It takes a lot of energy to launch a rocket into space. The Saturn V rocket used five of the most powerful engines ever built to take its initial step into orbit.
Source: “Soyuz TMA-5 launch” s in the public domain because it was solely created by NASA.

Having to carry its own fuel puts a lot of mass burden on an engine in space. This is why NASA is developing other types of engines to minimize fuel mass. An ion thruster uses xenon atoms that have had at least one electron removed from their atoms. The resulting ions can be accelerated by electric fields, causing a thrust. Because xenon atoms are very large for atoms, the thrusting efficiency is high even though the actual thrust is low. Because of this, ion engines are useful only in space.

Ion Engine

Ion drives have low thrust but high efficiency. They have already been used on several space missions, including NASA’s Deep Space 1 spacecraft and Japan’s Hayabusa asteroid sampling probe.

Energy is a very important quantity in science and the world around us. Although most of our energy ultimately comes from the sun, much of the energy we use on a daily basis is rooted in chemical reactions. The gasoline in your car, the electricity in your house, the food in your diet—all provide substances for chemical reactions to provide energy (gasoline, food) or are produced from chemical reactions (electricity, about 50% of which is generated by burning coal). As such, it is only natural that the study of chemistry involves energy.

45

Formation Reactions

Learning Objectives

  1. Define a formation reaction and be able to recognize one.
  2. Use enthalpies of formation to determine the enthalpy of reaction.

Hess’s law allows us to construct new chemical reactions and predict what their enthalpies of reaction will be. This is a very useful tool because now we don’t have to measure the enthalpy changes of every possible reaction. We need measure only the enthalpy changes of certain benchmark reactions and then use these reactions to algebraically construct any possible reaction and combine the enthalpies of the benchmark reactions accordingly.

But what are the benchmark reactions? We need to have some agreed-on sets of reactions that provide the central data for any thermochemical equation.

Formation reactions are chemical reactions that form one mole of a substance from its constituent elements in their standard states. By standard states we mean as a diatomic molecule if that is how the element exists and the proper phase at normal temperatures (typically room temperature). The product is one mole of substance, which may require that coefficients on the reactant side be fractional (a change from our normal insistence that all coefficients be whole numbers). For example, the formation reaction for methane (CH4) is

C(s) + 2 H2(g) → CH4(g)

The formation reaction for carbon dioxide (CO2) is

C(s) + O2(g) → CO2(g)

In both cases, one of the elements is a diatomic molecule because that is the standard state for that particular element. The formation reaction for H2O—

2 H2(g) + O2(g) → 2 H2O(ℓ)

—is not in a standard state because the coefficient on the product is 2; for a proper formation reaction, only one mole of product is formed. Thus, we have to divide all coefficients by 2:

H2(g) + 1/2 O2(g) → H2O(ℓ)

On a molecular scale, we are using half of an oxygen molecule, which may be problematic to visualize. However, on a molar level, it implies that we are reacting only half of a mole of oxygen molecules, which should be an easy concept for us to understand.

Example 11

Which of the following are proper formation reactions?

  1. H2(g) + Cl2(g) → 2 HCl(g)
  2. Si(s) + 2 F2(g) → SiF4(g)
  3. CaO(s) + CO2 → CaCO3(s)

Solution

  1. In this reaction, two moles of product are produced, so this is not a proper formation reaction.
  2. In this reaction, one mole of a substance is produced from its elements in their standard states, so this is a proper formation reaction.
  3. One mole of a substance is produced, but it is produced from two other compounds, not its elements. So this is not a proper formation reaction.

Test Yourself

Is this a proper formation reaction? Explain why or why not.

2 Fe(s) + 3 P(s) + 12 O(g) → Fe2(PO4)3(s)

Answer

This is not a proper formation reaction because oxygen is not written as a diatomic molecule.

Given the formula of any substance, you should be able to write the proper formation reaction for that substance.

Example 12

Write formation reactions for each of the following.

  1. FeO(s)
  2. C2H6(g)

Solution

In both cases, there is one mole of the substance as product, and the coefficients of the reactants may have to be fractional to balance the reaction.

  1. Fe(s) + 1/2 O2(g) → FeO(s)
  2. 2 C(s) + 3 H2(g) → C2H6(g)

Test Yourself

Write the equation for the formation of CaCO3(s).

Answer

Ca(s) + C(s) + 3/2 O2(g) → CaCO3(s)

The enthalpy change for a formation reaction is called the enthalpy of formation The subscript f is the clue that the reaction of interest is a formation reaction. Thus, for the formation of FeO(s),

Screen Shot 2014-07-22 at 7.57.34 PM

Note that now we are using kJ/mol as the unit because it is understood that the enthalpy change is for one mole of substance. Note, too, by definition, that the enthalpy of formation of an element is exactly zero because making an element from an element is no change. For example,

H2(g) → H2(g) ΔHf = 0

Formation reactions and their enthalpies are important because these are the thermochemical data that are tabulated for any chemical reaction. Table 7.2 “Enthalpies of Formation for Various Substances” lists some enthalpies of formation for a variety of substances; in some cases, however, phases can be important (e.g., for H2O).

It is easy to show that any general chemical equation can be written in terms of the formation reactions of its reactants and products, some of them reversed (which means the sign must change in accordance with Hess’s law). For example, consider

2 NO2(g) → N2O4(g)

We can write it in terms of the (reverse) formation reaction of NO2 and the formation reaction of N2O4:

Formation Reaction

We must multiply the first reaction by 2 to get the correct overall balanced equation. We are simply using Hess’s law in combining the ΔHf values of the formation reactions.

Table 7.2 Enthalpies of Formation for Various Substances

Compound ΔHf (kJ/mol) Compound ΔHf (kJ/mol) Compound ΔHf (kJ/mol) Compound ΔHf (kJ/mol)
Ag(s) 0 Ca(s) 0 Hg2Cl2(s) −265.37 NaHCO3(s) −950.81
AgBr(s) −100.37 CaCl2(s) −795.80 I2(s) 0 NaN3(s) 21.71
AgCl(s) −127.01 CaCO3(s, arag) −1,207.1 K(s) 0 Na2CO3(s) −1,130.77
Al(s) 0 CaCO3(s, calc) −1,206.9 KBr(s) −393.8 Na2O(s) −417.98
Al2O3(s) −1,675.7 Cl2(g) 0 KCl(s) −436.5 Na2SO4(s) −331.64
Ar(g) 0 Cr(s) 0 KF(s) −567.3 Ne(g) 0
Au(s) 0 Cr2O3(s) −1,134.70 KI(s) −327.9 Ni(s) 0
BaSO4(s) −1,473.19 Cs(s) 0 Li(s) 0 O2(g) 0
Br2(ℓ) 0 Cu(s) 0 LiBr(s) −351.2 O3(g) 142.67
C(s, dia) 1.897 F2(g) 0 LiCl(s) −408.27 PH3(g) 22.89
C(s, gra) 0 Fe(s) 0 LiF(s) −616.0 Pb(s) 0
CCl4(ℓ) −128.4 Fe2(SO4)3(s) −2,583.00 LiI(s) −270.4 PbCl2(s) −359.41
CH2O(g) −115.90 Fe2O3(s) −825.5 Mg(s) 0 PbO2(s) −274.47
CH3COOH(ℓ) −483.52 Ga(s) 0 MgO(s) −601.60 PbSO4(s) −919.97
CH3OH(ℓ) −238.4 HBr(g) −36.29 NH3(g) −45.94 Pt(s) 0
CH4(g) −74.87 HCl(g) −92.31 NO(g) 90.29 S(s) 0
CO(g) −110.5 HF(g) −273.30 NO2(g) 33.10 SO2(g) −296.81
CO2(g) −393.51 HI(g) 26.5 N2(g) 0 SO3(g) −395.77
C2H5OH(ℓ) −277.0 HNO2(g) −76.73 N2O(g) 82.05 SO3(ℓ) −438
C2H6(g) −83.8 HNO3(g) −134.31 N2O4(g) 9.08 Si(s) 0
C6H12(ℓ) −157.7 H2(g) 0 N2O5(g) 11.30 U(s) 0
C6H12O6(s) −1277 H2O(g) −241.8 Na(s) 0 UF6(s) −2,197.0
C6H14(ℓ) −198.7 H2O(ℓ) −285.83 NaBr(s) −361.1 UO2(s) −1,085.0
C6H5CH3(ℓ) 12.0 H2O(s) −292.72 NaCl(s) −385.9 Xe(g) 0
C6H6(ℓ) 48.95 He(g) 0 NaF(s) −576.6 Zn(s) 0
C10H8(s) 77.0 Hg(ℓ) 0 NaI(s) −287.8 ZnCl2(s) −415.05
C12H22O11(s) −2,221.2

Example 13

Show that the reaction

Fe2O3(s) + 3 SO3(g) → Fe2(SO4)3(s)

can be written as a combination of formation reactions.

Solution

There will be three formation reactions. The one for the products will be written as a formation reaction, while the ones for the reactants will be written in reverse. Furthermore, the formation reaction for SO3 will be multiplied by 3 because there are three moles of SO3 in the balanced chemical equation. The formation reactions are as follows:

fe2o3
2 Fe(s) + 3 S(s) + 6 O2(g) → Fe2(SO4)3(s)

When these three equations are combined and simplified, the overall reaction is

Fe2O3(s) + 3 SO3(s) → Fe2(SO4)3(s)

Test Yourself

Write the formation reactions that will yield

2 SO2(g) + O2(g) → 2 SO3(g).

Answer

2so2

Now that we have established formation reactions as the major type of thermochemical reaction we will be interested in, do we always need to write all the formation reactions when we want to determine the enthalpy change of any random chemical reaction? No. There is an easier way. You may have noticed in all our examples that we change the signs on all the enthalpies of formation of the reactants, and we don’t change the signs on the enthalpies of formation of the products. We also multiply the enthalpies of formation of any substance by its coefficient—technically, even when it is just 1. This allows us to make the following statement: the enthalpy change of any chemical reaction is equal to the sum of the enthalpies of formation of the products minus the sum of the enthalpies of formation of the reactants. In mathematical terms,

ahrxb

where np and nr are the number of moles of products and reactants, respectively (even if they are just 1 mol), and ΔHf,p and ΔHf,r are the enthalpies of formation of the product and reactant species, respectively. This products-minus-reactants scheme is very useful in determining the enthalpy change of any chemical reaction, if the enthalpy of formation data are available. Because the mol units cancel when multiplying the amount by the enthalpy of formation, the enthalpy change of the chemical reaction has units of energy (joules or kilojoules) only.

Example 14

Use the products-minus-reactants approach to determine the enthalpy of reaction for

2hbr

Solution

The enthalpies of formation are multiplied by the number of moles of each substance in the chemical equation, and the total enthalpy of formation for reactants is subtracted from the total enthalpy of formation of the products:

ΔHrxn=[(2 mol)(−92.3 kJ/ mol)+(1 mol)(0 kJ/ mol)]−[(2 mol)(−36.3 kJ/ mol)+(1 mol)(0 kJ/ mol)]

All the mol units cancel. Multiplying and combining all the values, we get

ΔHrxn = −112.0 kJ

Test Yourself

What is the enthalpy of reaction for this chemical equation?

cog

Answer

+2.8 kJ

Food and Drink App: Calories and Nutrition

Section 7.1 “Energy” mentioned the connection between the calorie unit and nutrition: the calorie is the common unit of energy used in nutrition, but we really consider the kilocalorie (spelled Calorie with a capital C). A daily diet of 2,000 Cal is actually 2,000,000 cal, or over 8,000,000 J, of energy.

Nutritionists typically generalize the Calorie content of foods by separating it into the three main food types: proteins, carbohydrates, and fats. The general rule of thumb is as follows:

If the food is It has this energy content
protein 4 Cal/g
carbohydrate 4 Cal/g
fat 9 Cal/g

This table is very useful. Assuming a 2,000 Cal daily diet, if our diet consists solely of proteins and carbohydrates, we need only about 500 g of food for sustenance—a little more than a pound. If our diet consists solely of fats, we need only about 220 g of food—less than a half pound. Of course, most of us have a mixture of proteins, carbohydrates, and fats in our diets. Water has no caloric value in the diet, so any water in the diet is calorically useless. (However, it is important for hydration; also, many forms of water in our diet are highly flavoured and sweetened, which bring other nutritional issues to bear.)

When your body works, it uses calories provided by the diet as its energy source. If we eat more calories than our body uses, we gain weight—about 1 lb of weight for every additional 3,500 Cal we ingest. Similarly, if we want to lose weight, we need to expend an extra 3,500 Cal than we ingest to lose 1 lb of weight. No fancy or fad diets are needed; maintaining an ideal body weight is a straightforward matter of thermochemistry—pure and simple.

Key Takeaways

  • A formation reaction is the formation of one mole of a substance from its constituent elements.
  • Enthalpies of formation are used to determine the enthalpy change of any given reaction.

 

Exercises

  1. Define formation reaction and give an example.

  2. Explain the importance of formation reactions in thermochemical equations.

  3. Which of the following reactions is a formation reaction? If it is not a formation reaction, explain why.

a)  H2(g) + S(s) → H2S(g)

b)  2 HBr(g) + Cl2(g) → 2 HCl(g) + Br2(ℓ)

 

4.  Which of the following reactions is a formation reaction? If it is not a formation reaction, explain why.

a)  Fe(g) + 1/2 O2(g) → FeO(s)

b)  Hg(ℓ) + 1/2 O2(g) → HgO(s)

 

5.  Which of the following reactions is a formation reaction? If it is not a formation reaction, explain why.

a)  H2(g) + S(s) + 2 O2(g) → H2SO4(ℓ)

b)  C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(ℓ)

 

6.  Which of the following reactions is a formation reaction? If it is not a formation reaction, explain why.

a)  Zn(s) + 2 HCl(aq) → ZnCl2(aq) + H2(g)

b)  2 Na(s) + C(s) + 3/2 O2(g) → Na2CO3(s)

 

7.  Write a proper formation reaction for each substance.

a)  H3PO4(s)

b)  Na2O(s)

c)  C3H7OH(ℓ)

 

8.  Write a proper formation reaction for each substance.

a)  N2O5(g)

b)  BaSO4(s)

c)  Fe(OH)3(s)

 

9.  Write a proper formation reaction for each substance.

a)  C12H22O11(s)

b)  Zn(NO3)2(s)

c)  Al(OH)3(s)

 

10.  Write a proper formation reaction for each substance.

a)  O3(g)

b)  Na2O2(s)

c)  PCl5(g)

 

11.  Write this reaction in terms of formation reactions.

MgCO3(s) → MgO(s) + CO2(g)

 

12.  Write this reaction in terms of formation reactions.

2 NO + 4 NO2 → 2 N2O5 + N2

 

13. Write this reaction in terms of formation reactions.

2 CuCl(s) → Cu(s) + CuCl2(s)

 

14.  Write this reaction in terms of formation reactions.

SiH4 + 4 F2 → SiF4 + 4 HF

 

15.  Determine the enthalpy change of this reaction. Data can be found in Table 7.2 “Enthalpies of Formation for Various Substances”.

CH2O(g) + O2 → CO2(g) + H2O(ℓ)

 

16.  Determine the enthalpy change of this reaction. Data can be found in Table 7.2 “Enthalpies of Formation for Various Substances”.

2 AgBr(s) + Cl2(g) → 2 AgCl(s) + Br2(ℓ)

 

17.  Determine the enthalpy change of this reaction. Data can be found in Table 7.2 “Enthalpies of Formation for Various Substances”.

Mg(s) + N2O5(g) → MgO(s) + 2NO2(g)

 

18.  Determine the enthalpy change of this reaction. Data can be found in Table 7.2 “Enthalpies of Formation for Various Substances”.

2C6H6(ℓ) + 15O2(g) → 12CO2(g) + 6H2O(ℓ)

Answers

1.

A formation reaction is a reaction that produces one mole of a substance from its elements. Example: C(s) + O2(g) → CO2(g)

3.

a)  formation reaction

b)  It is not the formation of a single substance, so it is not a formation reaction.

5.

a)  formation reaction

b)  It is not the formation of a single substance, so it is not a formation reaction.

7.

a)  3/2 H2(g) + P(s) + 2 O2(g) → H3PO4(s)

b)  2 Na(s) + 1/2 O2(g) → Na2O(s)

c)  3 C(s) + 1/2 O2(g) + 4 H2(g) → C3H7OH(ℓ)

9.

a)  12 C(s) + 11 H2(g) + 11/2 O2(g) → C12H22O11(s)

b)  Zn(s) + N2(g) + 3 O2(g) → Zn(NO3)2

c)  Al(s) + 3/2 O2(g) + 3/2 H2(g) → Al(OH)3(s)

11.

a)  MgCO3(s) → Mg(s) + C(s) + 3/2 O2(g)

b)  Mg(s) + 1/2 O2(g) → MgO(s)

c)  C(s) + O2(g) → CO2(g)

13.

a)  2 × [CuCl(s) → Cu(s) + 1/2 Cl2(g)]

b)  Cu(s) → Cu(s)

c)  Cu(s) + Cl2(g) → CuCl2(s)

15.

ΔH = −563.44 kJ

17.

ΔH = −546.7 kJ

 

46

Energy

Learning Objectives

  1. Define energy.
  2. Know the units of energy.
  3. Understand the law of conservation of energy.

Energy: is the ability to do work. Think about it: when you have a lot of energy, you can do a lot of work; but if you’re low on energy, you don’t want to do much work. Work (w) itself is defined as a force (F) operating over a distance (Δx):

w = F × Δx

In SI, force has units of newtons (N), while distance has units of meters. Therefore, work has units of N·m. This compound unit is redefined as a joule (J):

1 joule = 1 newton·meter
1 J = 1 N·m

Because energy is the ability to do work, energy is also measured in joules. This is the primary unit of energy we will use here.

How much is 1 J? It is enough to warm up about one-fourth of a gram of water by 1°C. It takes about 12,000 J to warm a cup of coffee from room temperature to 50°C. So a joule is not a lot of energy. It will not be uncommon to measure energies in thousands of joules, so the kilojoule (kJ) is a common unit of energy, with 1 kJ equal to 1,000 J.

An older—but still common—unit of energy is the calorie. The calorie (cal) was originally defined in terms of warming up a given quantity of water. The modern definition of calorie equates it to joules:

1 cal = 4.184 J

One area where the calorie is used is in nutrition. Energy contents of foods are often expressed in calories. However, the calorie unit used for foods is actually the kilocalorie (kcal). Most foods indicate this by spelling the word with a capital C—Calorie. Figure 7.1 “Calories on Food Labels” shows one example. So be careful counting calories when you eat!

Figure 7.1 Calories on Food Labels

 

This label expresses the energy content of the food, but in Calories (which are actually kilocalories). Source: “FDA Nutrition Facts Label 2014′′ by U.S. Food and Drug Administration is in the public domain.
This label expresses the energy content of the food, but in Calories (which are actually kilocalories).
Source: “FDA Nutrition Facts Label 2014′′ by U.S. Food and Drug Administration is in the public domain.

Example 1

The label in Figure 7.1 “Calories on Food Labels” states that the serving has 38 Cal. How many joules is this?

Solution

We recognize that with a capital C, the Calories unit is actually kilocalories. To determine the number of joules, we convert first from kilocalories to calories (using the definition of the kilo- prefix) and then from calories to joules (using the relationship between calories and joules). So

38kcal

Test Yourself

A serving of breakfast cereal usually has 110 Cal. How many joules of energy is this?

Answer

460,000 J

In the study of energy, we use the term system to describe the part of the universe under study: a beaker, a flask, or a container whose contents are being observed and measured. An isolated system is a system that does not allow a transfer of energy or matter into or out of the system. A good approximation of an isolated system is a closed, insulated thermos-type bottle. The fact that the thermos-type bottle is closed keeps matter from moving in or out, and the fact that it is insulated keeps energy from moving in or out.

One of the fundamental ideas about the total energy of an isolated system is that is does not increase or decrease. When this happens to a quantity, we say that the quantity is conserved. The statement that the total energy of an isolated system does not change is called the law of conservation of energy. As a scientific law, this concept occupies the highest level of understanding we have about the natural universe.

Key Takeaways

  • Energy is the ability to do work and uses the unit joule.
  • The law of conservation of energy states that the total energy of an isolated system does not increase or decrease.

 

Exercises

  1. Define energy. How is work related to energy?

  2. Give two units of energy and indicate which one is preferred.

  3. Express the quantity of 422 J in calories.

  4. Express the quantity of 3.225 kJ in calories.

  5. Express the quantity 55.69 cal in joules.

  6. Express the quantity 965.33 kcal in joules.

  7. How does a Calorie differ from a calorie?

  8. Express the quantity 965.33 Cal in joules.

  9. What is the law of conservation of energy?

  10. What does the word conserved mean as applied to the law of conservation of energy?

Answers

1.

Energy is the ability to do work. Work is a form of energy.

3.

101 cal

5.

233.0 J

7.

A Calorie is actually a kilocalorie, or 1,000 calories.

9.

The total energy of an isolated system does not increase or decrease.

 

47

Stoichiometry Calculations Using Enthalpy

Learning Objective

1. Perform stoichiometry calculations using energy changes from thermochemical equations.

In Chapter 5 “Stoichiometry and the Mole”, we related quantities of one substance to another in a chemical equation by performing calculations that used the balanced chemical equation; the balanced chemical equation provided equivalences that we used to construct conversion factors. For example, in the balanced chemical equation

2 H2(g) + O2(g) → 2 H2O(ℓ)

we recognized the equivalences

2 mol H2 ⇔ 1 mol O2 ⇔ 2 mol H2O

where ⇔ is the mathematical symbol for “is equivalent to.” In our thermochemical equation, however, we have another quantity—energy change:

2H2(g) + O2(g) → 2H2O(ℓ) ΔH = −570 kJ

This new quantity allows us to add another equivalence to our list:

2 mol H2 ⇔ 1 mol O2 ⇔ 2 mol H2O ⇔ −570 kJ

That is, we can now add an energy amount to the equivalences—the enthalpy change of a balanced chemical reaction. This equivalence can also be used to construct conversion factors so that we can relate enthalpy change to amounts of substances reacted or produced.

Note that these equivalences address a concern. When an amount of energy is listed for a balanced chemical reaction, what amount(s) of reactants or products does it refer to? The answer is that relates to the number of moles of the substance as indicated by its coefficient in the balanced chemical reaction. Thus, 2 mol of H2 are related to −570 kJ, while 1 mol of O2 is related to −570 kJ. This is why the unit on the energy change is kJ, not kJ/mol.

For example, consider the thermochemical equation

H2(g) + Cl2(g) → 2 HCl(g) ΔH = −184.6 kJ

The equivalences for this thermochemical equation are

1 mol H2 ⇔ 1 mol Cl2 ⇔ 2 mol HCl ⇔ −184.6 kJ

Suppose we asked how much energy is given off when 8.22 mol of H2 react. We would construct a conversion factor between the number of moles of H2 and the energy given off, −184.6 kJ:

Screen Shot 2014-07-22 at 7.37.43 PM

The negative sign means that this much energy is given off.

Example 8

Given the thermochemical equation

N2(g) + 3 H2(g) → 2 NH3(g) ΔH = −91.8 kJ

how much energy is given off when 222.4 g of N2 reacts?

Solution

The balanced thermochemical equation relates the energy change to moles, not grams, so we first convert the amount of N2 to moles and then use the thermochemical equation to determine the energy change:

2224g

Test Yourself

Given the thermochemical equation

N2(g) + 3 H2(g) → 2 NH3(g) ΔH = −91.8 kJ

how much heat is given off when 1.00 g of H2 reacts?

Answer

−15.1 kJ

Like any stoichiometric quantity, we can start with energy and determine an amount, rather than the other way around.

Example 9

Given the thermochemical equation

N2(g) + O2(g) → 2 NO(g) ΔH = 180.6 kJ

if 558 kJ of energy are supplied, what mass of NO can be made?

Solution

This time, we start with an amount of energy:

558kj

Test Yourself

How many grams of N2 will react if 100.0 kJ of energy are supplied?

N2(g) + O2(g) → 2 NO(g) ΔH = 180.6 kJ

Answer

15.5 g

Chemistry Is Everywhere: Welding with Chemical Reactions

One very energetic reaction is called the thermite reaction. Its classic reactants are aluminum metal and iron(III) oxide; the reaction produces iron metal and aluminum oxide:

2 Al(s) + Fe2O3(s) → Al2O3(s) + 2 Fe(s) ΔH = −850.2 kJ

When properly done, the reaction gives off so much energy that the iron product comes off as a liquid. (Iron normally melts at 1,536°C.) If carefully directed, the liquid iron can fill spaces between two or more metal parts and, after it quickly cools, can weld the metal parts together.

Thermite reactions are used for this purpose even today. For civilian purposes, they are used to reweld broken locomotive axles that cannot be easily removed for repair. They are used to weld railroad tracks together. Thermite reactions can also be used to separate thin pieces of metal if, for whatever reason, a torch doesn’t work.

image

A small clay pot contains a thermite mixture. It is reacting at high temperature in the photo and will eventually produce molten metal to join the railroad tracks below it.

Thermite reactions are also used for military purposes. Thermite mixtures are frequently used with additional components as incendiary devices—devices that start fires. Thermite reactions are also useful in disabling enemy weapons: a piece of artillery doesn’t work so well when it has a hole melted into its barrel because of a thermite reaction!

Key Takeaway

  • The energy change of a chemical reaction can be used in stoichiometry calculations.

 

Exercises

  1. Write the equivalences that this balanced thermochemical equation implies.

    PCl3(g) + Cl2(g) → PCl5(g) ΔH = −87.9 kJ

  2. Write the equivalences that this balanced thermochemical equation implies.

    2 SO3(g) → 2 SO2(g) + O2(g) ΔH = 197.9 kJ

  3. How many kilojoules are given off when 17.8 mol of CH4(g) react?

    CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(ℓ) ΔH = −890.1 kJ

  4. How many kilojoules are absorbed when 0.772 mol of N2(g) reacts?

    N2(g) + 2 NO(g) → 2 N2O(g) ΔH = 73.8 kJ

  5. How many kilojoules are absorbed when 23.09 mol of C6H6(ℓ) are formed?

    6 C(s) + 3 H2(g) → C6H6(ℓ) ΔH = 49.0 kJ

  6. How many kilojoules are given off when 8.32 mol of Mg react?

    2 Mg(s) + O2(g) → 2 MgO(s) ΔH = −1,213 kJ

  7. Glucose is the main fuel metabolized in animal cells:

    C6H12O6 + 6 O2 → 6 CO2 + 6 H2O ΔH = −2,799 kJ

    How much energy is given off when 100.0 g of C6H12O6 react?

  8. Given the thermochemical equation

    2 Al(s) + Fe2O3(s) → Al2O3(s) + 2 Fe(s) ΔH = −850.2 kJ

    how much energy is given off when 288 g of Fe are produced?

  9. Given the thermochemical equation

    2 CO2(g) → 2 CO(g) + O2(g) ΔH = 566 kJ

    how much energy is absorbed when 85.2 g of CO2 are reacted?

  10. Given the thermochemical equation

    2 Na+(aq) + SO42−(aq) → Na2SO4(s) ΔH = 819.8 kJ

    how much energy is absorbed when 55.9 g of Na+(aq) are reacted?

  11. NaHCO3 decomposes when exposed to heat:

    2 NaHCO3(s) → Na2CO3(s) + CO2(g) + H2O(ℓ) ΔH = 91.5 kJ

    What mass of NaHCO3 is decomposed by 256 kJ?

  12. HgO decomposes when exposed to heat:

    2 HgO(s) → 2 Hg(ℓ) + O2(g) ΔH = 181.6 kJ

    What mass of O2 can be made with 100.0 kJ?

  13. For the thermochemical equation

    Fe2O3(s) + 3 SO3(g) → Fe2(SO4)3 (s) ΔH = −570.2 kJ

    what mass of SO3 is needed to generate 1,566 kJ?

  14. For the thermochemical equation

    H2(g) + Br2(ℓ) → 2 HBr(g) ΔH = −72.6 kJ

    what mass of HBr will be formed when 553 kJ of energy are given off?

Answers

1.

1 mol of PCl3 ⇔ 1 mol of Cl2 ⇔ 1 mol of PCl5 ⇔ −87.9 kJ

3.

15,800 kJ

5.

1,130 kJ

7.

1,554 kJ

9.

548 kJ

11.

470 g

13.

6.60 × 102 g

 

48

Enthalpy and Chemical Reactions

Learning Objectives

  1. Define enthalpy.
  2. Properly express the enthalpy change of chemical reactions.
  3. Explain how enthalpy changes are measured experimentally.

Now that we have shown how energy, work, and heat are related, we are ready to consider energy changes in chemical reactions. A fundamental concept is that every chemical reaction occurs with a concurrent change in energy. Now we need to learn how to properly express these energy changes.

Our study of gases in Chapter 6 “Gases” and our definition of work in Section 7.2 “Work and Heat” indicate that conditions like pressure, volume, and temperature affect the energy content of a system. What we need is a definition of energy that holds when some of these conditions are specified (somewhat similar to our definition of standard temperature and pressure in our study of gases). We define the enthalpy change (ΔH) as the heat of a process when pressure is held constant:

ΔH ≡ q at constant pressure

The letter H stands for “enthalpy,” a kind of energy, while the Δ implies a change in the quantity. We will always be interested in the change in H, rather than the absolute value of H itself.

When a chemical reaction occurs, there is a characteristic change in enthalpy. The enthalpy change for a reaction is typically written after a balanced chemical equation and on the same line. For example, when two moles of hydrogen react with one mole of oxygen to make two moles of water, the characteristic enthalpy change is 570 kJ. We write the equation as

2 H2(g) + O2(g) → 2 H2O(ℓ) ΔH = −570 kJ

A chemical equation that includes an enthalpy change is called a thermochemical equation. A thermochemical equation is assumed to refer to the equation in molar quantities, which means it must be interpreted in terms of moles, not individual molecules.

Example 5

Write the thermochemical equation for the reaction of PCl3(g) with Cl2(g) to make PCl5(g), which has an enthalpy change of −88 kJ.

Solution

The thermochemical equation is

PCl3(g) + Cl2(g) → PCl5(g) ΔH = −88 kJ

Test Yourself

Write the thermochemical equation for the reaction of N2(g) with O2(g) to make 2NO(g), which has an enthalpy change of 181 kJ.

Answer

N2(g) + O2(g) → 2 NO(g) ΔH = 181 kJ

You may have noticed that the ΔH for a chemical reaction may be positive or negative. The number is assumed to be positive if it has no sign; a + sign can be added explicitly to avoid confusion. A chemical reaction that has a positive ΔH is said to be endothermic, while a chemical reaction that has a negative ΔH is said to be exothermic.

What does it mean if the ΔH of a process is positive? It means that the system in which the chemical reaction is occurring is gaining energy. If one considers the energy of a system as being represented as a height on a vertical energy plot, the enthalpy change that accompanies the reaction can be diagrammed as in part (a) in Figure 7.3 “Reaction Energy”: the energy of the reactants has some energy, and the system increases its energy as it goes to products. The products are higher on the vertical scale than the reactants. Endothermic, then, implies that the system gains, or absorbs, energy.

An opposite situation exists for an exothermic process, as shown in part (b) in Figure 7.3 “Reaction Energy”. If the enthalpy change of a reaction is negative, the system is losing energy, so the products have less energy than the reactants, and the products are lower on the vertical energy scale than the reactants are. Exothermic, then, implies that the system loses, or gives off, energy.

Figure 7.3 Reaction Energy

Reaction Energy

 

(a) In an endothermic reaction, the energy of the system increases (i.e., moves higher on the vertical scale of energy). (b) In an exothermic reaction, the energy of the system decreases (i.e., moves lower on the vertical scale of energy).

Example 6

Consider this thermochemical equation.

2 CO(g) + O2(g) → 2 CO2(g) ΔH = −565 kJ

Is it exothermic or endothermic? How much energy is given off or absorbed?

Solution

By definition, a chemical reaction that has a negative ΔH is exothermic, meaning that this much energy—in this case, 565 kJ—is given off by the reaction.

Test Yourself

Consider this thermochemical equation.

CO2(g) + H2(g) → CO(g) + H2O(g) ΔH = 42 kJ

Is it exothermic or endothermic? How much energy is given off or absorbed?

Answer

Endothermic; 42 kJ are absorbed.

How are ΔH values measured experimentally? Actually, ΔH is not measured; q is measured. But the measurements are performed under conditions of constant pressure, so ΔH is equal to the q measured.

Experimentally, q is measured by taking advantage of the equation

q = mcΔT

We premeasure the mass of the chemicals in a system. Then we let the chemical reaction occur and measure the change in temperature (ΔT) of the system. If we know the specific heat of the materials in the system (typically, we do), we can calculate q. That value of q is numerically equal to the ΔH of the process, which we can scale up to a molar scale. The container in which the system resides is typically insulated, so any energy change goes into changing the temperature of the system, rather than being leaked from the system. The container is referred to as a calorimeter, and the process of measuring changes in enthalpy is called calorimetry.

Figure 7.4 Calorimeters

Calorimeter

 

A simple calorimeter can be constructed from some nested foam coffee cups, a cover, a thermometer, and a stirrer.

For example, suppose 4.0 g of NaOH, or 0.10 mol of NaOH, are dissolved to make 100.0 mL of aqueous solution, while 3.65 g of HCl, or 0.10 mol of HCl, are dissolved to make another 100.0 mL of aqueous solution. The two solutions are mixed in an insulated calorimeter, a thermometer is inserted, and the calorimeter is covered (see Figure 7.4 “Calorimeters” for an example setup). The thermometer measures the temperature change as the following chemical reaction occurs:

NaOH (aq) + HCl(aq) → NaCl(aq) + H2O(ℓ)

An observer notes that the temperature increases from 22.4°C to 29.1°C. Assuming that the heat capacities and densities of the solutions are the same as those of pure water, we now have the information we need to determine the enthalpy change of the chemical reaction. The total amount of solution is 200.0 mL, and with a density of 1.00 g/mL, we thus have 200.0 g of solution. Using the equation for q, we substitute for our experimental measurements and the specific heat of water (Table 7.1 “Specific Heats of Various Substances”):

Screen Shot 2014-07-22 at 7.32.26 PM

Solving for q, we get

q = 5,600 J ≡ ΔH for the reaction

The heat q is equal to the ΔH for the reaction because the chemical reaction occurs at constant pressure. However, the reaction is giving off this amount of energy, so the actual sign on ΔH is negative:

ΔH = −5,600 J for the reaction

Thus, we have the following thermochemical equation for the chemical reaction that occurred in the calorimeter:

Screen Shot 2014-07-22 at 7.32.31 PM

The 1/10 coefficients are present to remind us that we started with one-tenth of a mole of each reactant, so we make one-tenth of a mole of each product. Typically, however, we report thermochemical equations in terms of moles, not one-tenth of a mole. To scale up to molar quantities, we must multiply the coefficients by 10. However, when we do this, we get 10 times as much energy. Thus, we have

NaOH (aq) + HCl(aq) → NaCl(aq) + H2O(ℓ) ΔH = −56,000 J

The ΔH can be converted into kJ units, so our final thermochemical equation is

NaOH (aq) + HCl(aq) → NaCl(aq) + H2O(ℓ) ΔH = −56 kJ

We have just taken our experimental data from calorimetry and determined the enthalpy change of a chemical reaction. Similar measurements on other chemical reactions can determine the ΔH values of any chemical reaction you want to study.

Example 7

A 100 mL solution of 0.25 mol of Ca2+(aq) was mixed with 0.50 mol of F(aq) ions, and CaF2 was precipitated:

Ca2+(aq) + 2 F(aq) → CaF2(s)

The temperature of the solution increased by 10.5°C. What was the enthalpy change for the chemical reaction? What was the enthalpy change for the production of 1 mol of CaF2? Assume that the solution has the same density and specific heat as water.

Solution

Because we are given ΔT directly, we can determine the heat of the reaction, which is equal to ΔH:

q100g

Solving for q, we get

q = 4,400 J

Therefore, ΔH = −4,400 J.

According to the stoichiometry of the reaction, exactly 0.25 mol of CaF2 will form, so this quantity of heat is for 0.25 mol. For 1 mol of CaF2, we need to scale up the heat by a factor of four:

q = 4,400 J × 4 = 17,600 J for 1 mol CaF2

On a molar basis, the change in enthalpy is

ΔH = −17,600 J = −17.6 kJ

Test Yourself

In a calorimeter at constant pressure, 0.10 mol of CH4(g) and 0.20 mol of O2(g) are reacted.

CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(ℓ)

The reaction warms 750.0 g of H2O by 28.4°C. What is ΔH for the reaction on a molar scale?

Answer

−891 kJ

Key Takeaways

  • Every chemical reaction occurs with a concurrent change in energy.
  • The change in enthalpy equals heat at constant pressure.
  • Enthalpy changes can be expressed by using thermochemical equations.
  • Enthalpy changes are measured by using calorimetry.

 

Exercises

  1. Under what circumstances are q and ΔH the same?

  2. Under what circumstances are q and ΔH different?

  3. Hydrogen gas and chlorine gas react to make hydrogen chloride gas with an accompanying enthalpy change of −184 kJ. Write a properly balanced thermochemical equation for this process.

  4. Propane (C3H8) reacts with elemental oxygen gas to produce carbon dioxide and liquid water with an accompanying enthalpy change of −2,220 kJ. Write a properly balanced thermochemical equation for this process.

  5. Nitrogen gas reacts with oxygen gas to make NO(g) while absorbing 180 kJ. Write a properly balanced thermochemical equation for this process.

  6. Solid sodium reacts with chlorine gas to make solid sodium chloride while giving off 772 kJ. Write a properly balanced thermochemical equation for this process.

  7. Hydrogen gas and chlorine gas react to make hydrogen chloride gas with an accompanying enthalpy change of −184 kJ. Is this process endothermic or exothermic?

  8. Propane (C3H8) reacts with elemental oxygen gas to produce carbon dioxide while giving off 2,220 kJ of energy. Is this process endothermic or exothermic?

  9. Nitrogen gas reacts with oxygen gas to make NO(g) while absorbing 180 kJ. Is this process exothermic or endothermic?

  10. Sodium metal can react with nitrogen to make sodium azide (NaN3) with a ΔH of 21.72 kJ. Is this process exothermic or endothermic?

  11. Draw an energy level diagram for the chemical reaction in Exercise 8. (See Figure 7.3 “Reaction Energy” for an example.)

  12. Draw an energy level diagram for the chemical reaction in Exercise 9. (See Figure 7.3 “Reaction Energy” for an example.)

  13. In a 250 mL solution, 0.25 mol of KOH(aq) and 0.25 mol of HNO3(aq) are combined. The temperature of the solution increases from 22.5°C to 35.9°C. Assume the solution has the same density and heat capacity of water. What is the heat of the reaction, and what is the ΔH of the reaction on a molar basis?

  14. In a 600 mL solution, 0.50 mol of Ca(OH)2(aq) and 0.50 mol of H2SO4(aq) are combined. The temperature of the solution increases by 22.3°C. What is the heat of the reaction, and what is the ΔH of the reaction on a molar basis? Assume the solution has the same density and heat capacity of water.

  15. To warm 400.0 g of H2O, 0.050 mol of ethanol (C2H5OH) is burned. The water warms from 24.6°C to 65.6°C. What is the heat of the reaction, and what is the ΔH of the reaction on a molar basis?

  16. To warm 100.0 g of H2O, 0.066 mol beeswax is burned. The water warms from 21.4°C to 25.5°C. What is the heat of the reaction, and what is the ΔH of the reaction on a molar basis?

Answers

1.

under conditions of constant pressure

3.

H2(g) + Cl2(g) → 2HCl(g) ΔH = −184 kJ

5.

N2(g) + O2(g) → 2NO(g) ΔH = 180 kJ

7.

exothermic

9.

endothermic

11.

Energy Graph

 

13.

heat of reaction = −14.0 kJ; ΔH = −56.0 kJ/mol of reactants

15.

heat of reaction = −68.6 kJ; ΔH = −1,370 kJ/mole of ethanol

 

49

Work and Heat

Learning Objectives

  1. Define a type of work in terms of pressure and volume.
  2. Define heat.
  3. Relate the amount of heat to a temperature change.

We have already defined work as a force acting through a distance. It turns out that there are other equivalent definitions of work that are also important in chemistry.

When a certain volume of a gas expands, it works against an external pressure to expand (Figure 7.2 “Volume versus Pressure”). That is, the gas must perform work. Assuming that the external pressure Pext is constant, the amount of work done by the gas is given by the equation

w = −Pext × ΔV

where ΔV is the change in volume of the gas. This term is always the final volume minus the initial volume,

ΔV = VfinalVinitial

and can be positive or negative, depending on whether Vfinal is larger (is expanding) or smaller (is contracting) than Vinitial. The negative sign in the equation for work is important and implies that as volume expands (ΔV is positive), the gas in the system is losing energy as work. On the other hand, if the gas is contracting, ΔV is negative, and the two negative signs make the work positive, so energy is being added to the system.

Figure 7.2 Volume versus Pressure

Volume vs Pressure

 

When a gas expands against an external pressure, the gas does work.

Finally, let us consider units. Volume changes are usually expressed in units like liters, while pressures are usually expressed in atmospheres. When we use the equation to determine work, the unit for work comes out as liter·atmospheres, or L·atm. This is not a very common unit for work. However, there is a conversion factor between L·atm and the common unit of work, joules:

1 L·atm = 101.32 J

Using this conversion factor and the previous equation for work, we can calculate the work performed when a gas expands or contracts.

Example 2

What is the work performed by a gas if it expands from 3.44 L to 6.19 L against a constant external pressure of 1.26 atm? Express the final answer in joules.

Solution

First we need to determine the change in volume, ΔV. A change is always the final value minus the initial value:

ΔV = VfinalVinitial = 6.19 L − 3.44 L = 2.75 L

Now we can use the definition of work to determine the work done:

w = −Pext · ΔV = −(1.26 atm)(2.75 L) = −3.47 L·atm

Now we construct a conversion factor from the relationship between liter·atmospheres and joules:

_347

We limit the final answer to three significant figures, as appropriate.

Test Yourself

What is the work performed when a gas expands from 0.66 L to 1.33 L against an external pressure of 0.775 atm?

Answer

−53 J

Heat is another aspect of energy. Heat is the transfer of energy from one body to another due to a difference in temperature. For example, when we touch something with our hands, we interpret that object as either hot or cold depending on how energy is transferred: If energy is transferred into your hands, the object feels hot. If energy is transferred from your hands to the object, your hands feel cold. Because heat is a measure of energy transfer, heat is also measured in joules.

For a given object, the amount of heat (q) involved is proportional to two things: the mass of the object (m) and the temperature change (ΔT) evoked by the energy transfer. We can write this mathematically as

q∝m × ΔT

where ∝ means “is proportional to.” To make a proportionality an equality, we include a proportionality constant. In this case, the proportionality constant is labelled c and is called the specific heat capacity, or, more succinctly, specific heat:

q = mcΔT

where the mass, specific heat, and change in temperature are multiplied together. Specific heat is a measure of how much energy is needed to change the temperature of a substance; the larger the specific heat, the more energy is needed to change the temperature. The units for specific heat are Jg⋅°C or Jg⋅K, depending on what the unit of ΔT is. You may note a departure from the insistence that temperature be expressed in Kelvin. That is because a change in temperature has the same value whether the temperatures are expressed in degrees Celsius or kelvins.

Example 3

Calculate the heat involved when 25.0 g of Fe increase temperature from 22°C to 76°C. The specific heat of Fe is 0.449 J/g·°C.

Solution

First we need to determine ΔT. A change is always the final value minus the initial value:

ΔT = 76°C − 22°C = 54°C

Now we can use the expression for q, substitute for all variables, and solve for heat:

q250

Note how the g and °C units cancel, leaving J, a unit of heat. Also note that this value of q is inherently positive, meaning that energy is going into the system.

Test Yourself

Calculate the heat involved when 76.5 g of Ag increase temperature from 17.8°C to 144.5°C. The specific heat of Ag is 0.233 J/g·°C.

Answer

2,260 J

As with any equation, when you know all but one variable in the expression for q, you can determine the remaining variable by using algebra.

Example 4

It takes 5,408 J of heat to raise the temperature of 373 g of Hg by 104°C. What is the specific heat of Hg?

Solution

We can start with the equation for q, but now different values are given, and we need to solve for specific heat. Note that ΔT is given directly as 104°C. Substituting,

5,408 J = (373 g)c(104°C)

We divide both sides of the equation by 373 g and 104°C:

c5408

Combining the numbers and bringing together all the units, we get

c0139

Test Yourself

Gold has a specific heat of 0.129 J/g·°C. If 1,377 J are needed to increase the temperature of a sample of gold by 99.9°C, what is the mass of the gold?

Answer

107 g

Table 7.1 “Specific Heats of Various Substances” lists the specific heats of some substances. Specific heat is a physical property of substances, so it is a characteristic of the substance. The general idea is that the lower the specific heat, the less energy is required to change the temperature of the substance by a certain amount.

Table 7.1 Specific Heats of Various Substances

Substance Specific Heat (J/g·°C)
water 4.184
iron 0.449
gold 0.129
mercury 0.139
aluminum 0.900
ethyl alcohol 2.419
magnesium 1.03
helium 5.171
oxygen 0.918

Key Takeaways

  • Work can be defined as a gas changing volume against a constant external pressure.
  • Heat is the transfer of energy due to temperature differences.
  • Heat can be calculated in terms of mass, temperature change, and specific heat.

 

Exercises

  1. Give two definitions of work.

  2. What is the sign on work when a sample of gas increases its volume? Explain why work has that sign.

  3. What is the work when a gas expands from 3.00 L to 12.60 L against an external pressure of 0.888 atm?

  4. What is the work when a gas expands from 0.666 L to 2.334 L against an external pressure of 2.07 atm?

  5. What is the work when a gas contracts from 3.45 L to 0.97 L under an external pressure of 0.985 atm?

  6. What is the work when a gas contracts from 4.66 L to 1.22 L under an external pressure of 3.97 atm?

  7. Like work, the sign on heat can be positive or negative. What is happening to the total energy of a system if heat is positive?

  8. Like work, the sign on heat can be positive or negative. What is happening to the total energy of a system if heat is negative?

  9. What is the heat when 55.6 g of Fe increase temperature from 25.6°C to 177.9°C? The heat capacity of Fe is in Table 7.1 “Specific Heats of Various Substances”.

  10. What is the heat when 0.444 g of Au increases temperature from 17.8°C to 222.5°C? The heat capacity of Au is in Table 7.1 “Specific Heats of Various Substances”.

  11. What is the heat when 245 g of H2O cool from 355 K to 298 K? The heat capacity of H2O is in Table 7.1 “Specific Heats of Various Substances”.

  12. What is the heat when 100.0 g of Mg cool from 725 K to 552 K? The heat capacity of Mg is in Table 7.1 “Specific Heats of Various Substances”.

  13. It takes 452 J of heat to raise the temperature of a 36.8 g sample of a metal from 22.9°C to 98.2°C. What is the heat capacity of the metal?

  14. It takes 2,267 J of heat to raise the temperature of a 44.5 g sample of a metal from 33.9°C to 288.3°C. What is the heat capacity of the metal?

  15. An experimenter adds 336 J of heat to a 56.2 g sample of Hg. What is its change in temperature? The heat capacity of Hg is in Table 7.1 “Specific Heats of Various Substances”.

  16. To a 0.444 g sample of H2O, 23.4 J of heat are added. What is its change in temperature? The heat capacity of H2O is in Table 7.1 “Specific Heats of Various Substances”.

  17. An unknown mass of Al absorbs 187.9 J of heat and increases its temperature from 23.5°C to 35.6°C. What is the mass of the aluminum? How many moles of aluminum is this?

  18. A sample of He goes from 19.4°C to 55.9°C when 448 J of energy are added. What is the mass of the helium? How many moles of helium is this?

Answers

1.

Work is a force acting through a distance or a volume changing against some pressure.

3.

−864 J

5.

248 J

7.

When heat is positive, the total energy of the system is increasing.

9.

3.80 × 103 J

11.

−58,400 J

13.

0.163 J/g·°C

15.

43.0°C

17.

17.3 g; 0.640 mol

 

50

Hess’s Law

Learning Objective

  1. Learn how to combine chemical equations and their enthalpy changes.

Now that we understand that chemical reactions occur with a simultaneous change in energy, we can apply the concept more broadly. To start, remember that some chemical reactions are rather difficult to perform. For example, consider the combustion of carbon to make carbon monoxide:

2 C(s) + O2(g) → 2 CO(g) ΔH = ?

In reality, this is extremely difficult to do; given the opportunity, carbon will react to make another compound, carbon dioxide:

2 C(s) + O2(g) → 2 CO2(g) ΔH = −393.5 kJ

Is there a way around this? Yes. It comes from the understanding that chemical equations can be treated like algebraic equations, with the arrow acting like the equals sign. Like algebraic equations, chemical equations can be combined, and if the same substance appears on both sides of the arrow, it can be canceled out (much like a spectator ion in ionic equations). For example, consider these two reactions:

2 C(s) + 2 O2(g) → 2 CO2(g)
2 CO2(g) → 2 CO(g) + O2(g)

If we added these two equations by combining all the reactants together and all the products together, we would get

2 C(s) + 2 O2(g) + 2 CO2(g) → 2 CO2(g) + 2 CO(g) + O2(g)

We note that 2 CO2(g) appears on both sides of the arrow, so they cancel:

Screen Shot 2014-07-22 at 7.42.02 PM

We also note that there are 2 mol of O2 on the reactant side, and 1 mol of O2 on the product side. We can cancel 1 mol of O2 from both sides:

Screen Shot 2014-07-22 at 7.42.06 PM

What do we have left?

2 C(s) + O2(g) → 2 CO(g)

This is the reaction we are looking for! So by algebraically combining chemical equations, we can generate new chemical equations that may not be feasible to perform.

What about the enthalpy changes? Hess’s law states that when chemical equations are combined algebraically, their enthalpies can be combined in exactly the same way. Two corollaries immediately present themselves:

  1. If a chemical reaction is reversed, the sign on ΔH is changed.
  2. If a multiple of a chemical reaction is taken, the same multiple of the ΔH is taken as well.

What are the equations being combined? The first chemical equation is the combustion of C, which produces CO2:

2 C(s) + 2 O2(g) → 2 CO2(g)

This reaction is two times the reaction to make CO2 from C(s) and O2(g), whose enthalpy change is known:

C(s) + O2(g) → CO2(g) ΔH = −393.5 kJ

According to the first corollary, the first reaction has an energy change of two times −393.5 kJ, or −787.0 kJ:

2 C(s) + 2 O2(g) → 2 CO2(g) ΔH = −787.0 kJ

The second reaction in the combination is related to the combustion of CO(g):

2 CO(g) + O2(g) → 2 CO2(g) ΔH = −566.0 kJ

The second reaction in our combination is the reverse of the combustion of CO. When we reverse the reaction, we change the sign on the ΔH:

2 CO2(g) → 2 CO(g) + O2(g) ΔH = +566.0 kJ

Now that we have identified the enthalpy changes of the two component chemical equations, we can combine the ΔH values and add them:

Enthalpy Changes

Hess’s law is very powerful. It allows us to combine equations to generate new chemical reactions whose enthalpy changes can be calculated, rather than directly measured.

Example 10

Determine the enthalpy change of

C2H4 + 3 O2 → 2 CO2 + 2 H2O ΔH = ?

from these reactions:

C2H2 + H2 → C2H4 ΔH = −174.5 kJ
2 C2H2 + 5 O2 → 4 CO2 + 2 H2O ΔH = −1,692.2 kJ
2 CO2 + H2 → 2 O2 + C2H2 ΔH = −167.5 kJ

Solution

We will start by writing chemical reactions that put the correct number of moles of the correct substance on the proper side. For example, our desired reaction has C2H4 as a reactant, and only one reaction from our data has C2H4. However, it has C2H4 as a product. To make it a reactant, we need to reverse the reaction, changing the sign on the ΔH:

C2H4 → C2H2 + H2 ΔH = +174.5 kJ

We need CO2 and H2O as products. The second reaction has them on the proper side, so let us include one of these reactions (with the hope that the coefficients will work out when all our reactions are added):

2 C2H2 + 5 O2 → 4 CO2 + 2 H2O ΔH = −1,692.2 kJ

We note that we now have 4 mol of CO2 as products; we need to get rid of 2 mol of CO2. The last reaction has 2CO2 as a reactant. Let us use it as written:

2 CO2 + H2 → 2 O2 + C2H2 ΔH = −167.5 kJ

We combine these three reactions, modified as stated:

Enthalpy Changes 2

What cancels? 2 C2H2, H2, 2 O2, and 2 CO2. What is left is

C2H4 + 3 O2 → 2 CO2 + 2H2O

which is the reaction we are looking for. The ΔH of this reaction is the sum of the three ΔH values:

ΔH = +174.5 − 1,692.2 − 167.5 = −1,685.2 kJ

Test Yourself

Given the thermochemical equations

Pb + Cl2 → PbCl2 ΔH = −223 kJ
PbCl2 + Cl2 → PbCl4 ΔH = −87 kJ

determine ΔH for

2 bCl2 → Pb + PbCl4

Answer

+136 kJ

Key Takeaway

  • Hess’s law allows us to combine reactions algebraically and then combine their enthalpy changes the same way.

 

Exercises

  1. Define Hess’s law.

  2. What does Hess’s law require us to do to the ΔH of a thermochemical equation if we reverse the equation?

  3. If the ΔH for

    C2H4 + H2 → C2H6

    is −65.6 kJ, what is the ΔH for this reaction?

    C2H6 → C2H4 + H2

  4. If the ΔH for

    2 Na + Cl2 → 2NaCl

    is −772 kJ, what is the ΔH for this reaction:

    2 NaCl → 2 Na + Cl2

  5. If the ΔH for

    C2H4 + H2 → C2H6

    is −65.6 kJ, what is the ΔH for this reaction?

    2 C2H4 + 2 H2 → 2 C2H6

  6. If the ΔH for

    2 C2H6 + 7 O2 → 4 CO2 + 6 H2O

    is −2,650 kJ, what is the ΔH for this reaction?

    6 C2H6 + 21 O2 → 12 CO2 + 18 H2O

  7. The ΔH for

    C2H4 + H2O → C2H5OH

    is −44 kJ. What is the ΔH for this reaction?

    2 C2H5OH → 2 C2H4 + 2 H2O

  8. The ΔH for

    N2 + O2 → 2 NO

    is 181 kJ. What is the ΔH for this reaction?

    NO → 1/2 N2 + 1/2 O2

  9. Determine the ΔH for the reaction

    Cu + Cl2 → CuCl2

    given these data:

    2 Cu + Cl2 → 2 CuCl ΔH = −274 kJ
    2 CuCl + Cl2 → 2 CuCl2 ΔH = −166 kJ

  10. Determine ΔH for the reaction

    2 CH4 → 2 H2 + C2H4

    given these data:

    CH4 + 2 O2 → CO2 + 2 H2O ΔH = −891 kJ
    C2H4 + 3 O2 → 2 CO2 + 2 H2O ΔH = −1,411 kJ
    2 H2 + O2 → 2 H2O ΔH = −571 kJ

  11. Determine ΔH for the reaction

    Fe2(SO4)3 → Fe2O3 + 3SO3

    given these data:

    4 Fe + 3O2 → 2 Fe2O3 ΔH = −1,650 kJ
    2 S + 3 O2 → 2 SO3 ΔH = −792 kJ
    2 Fe + 3 S + 6 O2 → Fe2(SO4)3 ΔH = −2,583 kJ

  12. Determine ΔH for the reaction

    CaCO3 → CaO + CO2

    given these data:

    2 Ca + 2 C + 3 O2 → 2 CaCO3 ΔH = −2,414 kJ
    C + O2 → CO2 ΔH = −393.5 kJ
    2 Ca + O2 → 2 CaO ΔH = −1,270 kJ

Answers

1.

If chemical equations are combined, their energy changes are also combined.

3.

ΔH = 65.6 kJ

5.

ΔH = −131.2 kJ

7.

ΔH = 88 kJ

9.

ΔH = −220 kJ

11.

ΔH = 570 kJ

 

51

End-of-Chapter Material

Additional Exercises

 

  1. What is the work when 124 mL of gas contract to 72.0 mL under an external pressure of 822 torr?

  2. What is the work when 2,345 mL of gas contract to 887 mL under an external pressure of 348 torr?

  3. A 3.77 L volume of gas is exposed to an external pressure of 1.67 atm. As the gas contracts, 156 J of work are added to the gas. What is the final volume of the gas?

  4. A 457 mL volume of gas contracts when 773 torr of external pressure act on it. If 27.4 J of work are added to the gas, what is its final volume?

  5. What is the heat when 1,744 g of Hg increase in temperature by 334°C? Express your final answer in kJ.

  6. What is the heat when 13.66 kg of Fe cool by 622°C? Express your final answer in kJ.

  7. What is final temperature when a 45.6 g sample of Al at 87.3°C gains 188 J of heat?

  8. What is final temperature when 967 g of Au at 557°C lose 559 J of heat?

  9. Plants take CO2 and H2O and make glucose (C6H12O6) and O2. Write a balanced thermochemical equation for this process. Use data in Table 7.2 “Enthalpies of Formation for Various Substances”.

  10. Exercise 9 described the formation of glucose in plants, which take in CO2 and H2O and give off O2. Is this process exothermic or endothermic? If exothermic, where does the energy go? If endothermic, where does the energy come from?

  11. The basic reaction in the refining of aluminum is to take Al2O3(s) and turn it into Al(s) and O2(g). Write the balanced thermochemical equation for this process. Use data in Table 7.2 “Enthalpies of Formation for Various Substances”.

  12. Is the enthalpy change of the reaction

    H2O(ℓ) → H2O(g)

    zero or nonzero? Use data in Table 7.2 “Enthalpies of Formation for Various Substances” to determine the answer.

  13. What mass of H2O can be heated from 22°C to 80°C in the combustion of 1 mol of CH4? You will need the balanced thermochemical equation for the combustion of CH4. Use data in Table 7.2 “Enthalpies of Formation for Various Substances”.

  14. What mass of H2O can be heated from 22°C to 80°C in the combustion of 1 mol of C2H6? You will need the balanced thermochemical equation for the combustion of C2H6. Use data in Table 7.2 “Enthalpies of Formation for Various Substances”. Compare your answer to Exercise 13.

  15. What is the enthalpy change for the unknown reaction?

    Pb(s) + Cl2(g) → PbCl2(s) ΔH = −359 kJ
    PbCl2(s) + Cl2(g) → PbCl4(ℓ) ΔH = ?
    Pb(s) + 2 Cl2(g) → PbCl4(ℓ) ΔH = −329 kJ

  16. What is the enthalpy change for the unknown reaction?

    P(s) + 3/2 Br2(ℓ) → PBr3(ℓ) ΔH = −185 kJ
    PI3(s) → P(s) + 3/2 I2(s) ΔH = ?
    PI3(s) + 3/2 Br2(ℓ) → PBr3(ℓ) + 3/2 I2(s) ΔH = −139 kJ

  17. What is the ΔH for this reaction? The label gra means graphite, and the label dia means diamond. What does your answer mean?

    C(s, gra) → C(s, dia)

  18. Without consulting any tables, determine the ΔH for this reaction. Explain your answer.

    H2O(ℓ, 25°C) → H2O(ℓ, 25°C)

Answers

1.

5.70 J

3.

4.69 L

5.

80.97 kJ

7.

91.9°C

9.

6 CO2(g) + 6 H2O(ℓ) → C6H12O6(s) + 6 O2(g) ΔH = 2,799 kJ

11.

2 Al2O3(s) → 4 Al(s) + 3 O2(g) ΔH = 3351.4 kJ

13.

3,668 g

15.

ΔH = 30 kJ

17.

ΔH = 1.897 kJ; the reaction is endothermic.

 

VIII

Chapter 8. Electronic Structure

52

Introduction to Electronic Structure

Normal light microscopes can magnify objects up to about 1,500 times. Electron microscopes can magnify objects up to 1,000,000 times. Why can electron microscopes magnify images so much?

A microscope’s resolution depends on the wavelength of light used. The smaller the wavelength, the more a microscope can magnify. Light is a wave, and, as such, it has a wavelength associated with it. The wavelength of visible light, which is detected by the eyes, varies from about 700 nm to about 400 nm.

One of the startling conclusions about modern science is that electrons also act as waves. However, the wavelength of electrons is much, much shorter—about 0.5 to 1 nm. This allows electron microscopes to magnify 600–700 times more than light microscopes. This allows us to see even smaller features in a world that is invisible to the naked eye.

Atoms act the way they do because of their structure. We already know that atoms are composed of protons, neutrons, and electrons. Protons and neutrons are located in the nucleus, and electrons orbit around the nucleus. But we need to know the structural details to understand why atoms react the way they do.

Virtually everything we know about atoms ultimately comes from light. Before we can understand the composition of atoms (especially electrons), we need to understand the properties of light.

53

Organization of Electrons in Atoms

David W. Ball and Jessie A. Key

Learning Objectives

  1. Learn how electrons are organized in atoms.
  2. Represent the organization of electrons by an electron configuration.

Now that you know that electrons have quantum numbers, how are they arranged in atoms? The key to understanding electronic arrangement is summarized in the Pauli exclusion principle: no two electrons in an atom can have the same set of four quantum numbers. This dramatically limits the number of electrons that can exist in a shell or a subshell.

Electrons are typically organized around an atom by starting at the lowest possible quantum numbers first, which are the shells-subshells with lower energies. Consider H, an atom with a single electron only. Under normal conditions, the single electron would go into the n = 1 shell, which has only a single s subshell with one orbital (because m can equal only 0). The convention is to label the shell-subshell combination with the number of the shell and the letter that represents the subshell. Thus, the electron goes in the 1s shell-subshell combination. It is usually not necessary to specify the m or ms quantum numbers, but for the H atom, the electron has m = 0 (the only possible value) and an ms of either +1/2 or −1/2.

The He atom has two electrons. The second electron can also go into the 1s shell-subshell combination but only if its spin quantum number is different from the first electron’s spin quantum number. Thus, the sets of quantum numbers for the two electrons are {1, 0, 0, +1/2} and {1, 0, 0, −1/2}. Notice that the overall set is different for the two electrons, as required by the Pauli exclusion principle.

The next atom is Li, with three electrons. However, now the Pauli exclusion principle implies that we cannot put that electron in the 1s shell-subshell because no matter how we try, this third electron would have the same set of four quantum numbers as one of the first two electrons. So this third electron must be assigned to a different shell-subshell combination. However, the n = 1 shell doesn’t have another subshell; it is restricted to having just ℓ = 0, or an s subshell. Therefore, this third electron has to be assigned to the n = 2 shell, which has an s (ℓ = 0) subshell and a p (ℓ = 1) subshell. Again, we usually start with the lowest quantum number, so this third electron is assigned to the 2s shell-subshell combination of quantum numbers.

The Pauli exclusion principle has the net effect of limiting the number of electrons that can be assigned a shell-subshell combination of quantum numbers. For example, in any s subshell, no matter what the shell number, there can be a maximum of only two electrons. Once the s subshell is filled up, any additional electrons must go to another subshell in the shell (if it exists) or to higher-numbered shell. A similar analysis shows that a p subshell can hold a maximum of six electrons. A d subshell can hold a maximum of 10 electrons, while an f subshell can have a maximum of 14 electrons. By limiting subshells to these maxima, we can distribute the available electrons to their shells and subshells.

Example 4

How would the six electrons for C be assigned to the n and ℓ quantum numbers?

Solution

The first two electrons go into the 1s shell-subshell combination. Two additional electrons can go into the 2s shell-subshell, but now this subshell is filled with the maximum number of electrons. The n = 2 shell also has a p subshell, so the remaining two electrons can go into the 2p subshell. The 2p subshell is not completely filled because it can hold a maximum of six electrons.

Test Yourself

How would the 11 electrons for Na be assigned to the n and ℓ quantum numbers?

Answer

two 1s electrons, two 2s electrons, six 2p electrons, and one 3s electron

Now that we see how electrons are partitioned among the shells and subshells, we need a more concise way of communicating this partitioning. Chemists use an electron configuration, to represent the organization of electrons in shells and subshells in an atom. An electron configuration simply lists the shell and subshell labels, with a right superscript giving the number of electrons in that subshell. The shells and subshells are listed in the order of filling.

For example, an H atom has a single electron in the 1s subshell. Its electron configuration is

H: 1s1

He has two electrons in the 1s subshell. Its electron configuration is

He: 1s2

The three electrons for Li are arranged in the 1s subshell (two electrons) and the 2s subshell (one electron). The electron configuration of Li is

Li: 1s22s1

Be has four electrons, two in the 1s subshell and two in the 2s subshell. Its electron configuration is

Be: 1s22s2

Now that the 2s subshell is filled, electrons in larger atoms must go into the 2p subshell, which can hold a maximum of six electrons. The next six elements progressively fill up the 2p subshell:

B: 1s22s22p1
C: 1s22s22p2
N: 1s22s22p3
O: 1s22s22p4
F: 1s22s22p5
Ne: 1s22s22p6

Now that the 2p subshell is filled (all possible subshells in the n = 2 shell), the next electron for the next-larger atom must go into the n = 3 shell, s subshell.

Example 5

What is the electron configuration for Na, which has 11 electrons?

Solution

The first two electrons occupy the 1s subshell. The next two occupy the 2s subshell, while the next six electrons occupy the 2p subshell. This gives us 10 electrons so far, with 1 electron left. This last electron goes into the n = 3 shell, s subshell. Thus, the electron configuration of Na is 1s22s22p63s1.

Test Yourself

What is the electron configuration for Mg, which has 12 electrons?

Answer

1s22s22p63s2

For larger atoms, the electron arrangement becomes more complicated. This is because after the 3p subshell is filled, filling the 4s subshell first actually leads to a lesser overall energy than filling the 3d subshell. Recall that while the principal quantum number largely dictates the energy of an electron, the angular momentum quantum number also has an impact on energy; by the time we get to the 3d and 4s subshells, we see overlap in the filling of the shells. Thus, after the 3p subshell is completely filled (which occurs for Ar), the next electron for K occupies the 4s subshell, not the 3d subshell:

K: 1s22s22p63s23p64s1, not 1s22s22p63s23p63d1

For larger and larger atoms, the order of filling the shells and subshells seems to become even more complicated. There are some useful ways to remember the order, like that shown in Figure 8.7 “Electron Shell Filling Order”. If you follow the arrows in order, they pass through the subshells in the order that they are filled with electrons in larger atoms. Initially, the order is the same as the expected shell-subshell order, but for larger atoms, there is some shifting around of the principal quantum numbers. However, Figure 8.7 “Electron Shell Filling Order” gives a valid ordering of filling subshells with electrons for most atoms.

Figure 8.7 Electron Shell Filling Order

Electron Shells

 

Starting with the top arrow, follow each arrow. The subshells you reach along each arrow give the ordering of filling of subshells in larger atoms. The n = 5 and higher shells have more subshells, but only those subshells that are needed to accommodate the electrons of the known elements are given.

Example 6

What is the predicted electron configuration for Sn, which has 50 electrons?

Solution

We will follow the chart in Figure 8.7 “Electron Shell Filling Order” until we can accommodate 50 electrons in the subshells in the proper order:

Sn: 1s22s22p63s23p64s23d104p65s24d105p2

Verify by adding the superscripts, which indicate the number of electrons: 2 + 2 + 6 + 2 + 6 + 2 + 10 + 6 + 2 + 10 + 2 = 50, so we have placed all 50 electrons in subshells in the proper order.

Test Yourself

What is the electron configuration for Ba, which has 56 electrons?

Answer

1s22s22p63s23p64s23d104p65s24d105p66s2

As the previous example demonstrated, electron configurations can get fairly long. An abbreviated electron configuration uses one of the elements from the last column of the periodic table, which contains what are called the noble gases, to represent the core of electrons up to that element. Then the remaining electrons are listed explicitly. For example, the abbreviated electron configuration for Li, which has three electrons, would be

Li: [He]2s1

where [He] represents the two-electron core that is equivalent to He’s electron configuration. The square brackets represent the electron configuration of a noble gas. This is not much of an abbreviation. However, consider the abbreviated electron configuration for W, which has 74 electrons:

W: [Xe]6s24f145d4

This is a significant simplification over an explicit listing of all 74 electrons. So for larger elements, the abbreviated electron configuration can be a very useful shorthand.

Example 7

What is the abbreviated electron configuration for P, which has 15 electrons?

Solution

With 15 electrons, the electron configuration of P is

P: 1s22s22p63s23p3

The first immediate noble gas is Ne, which has an electron configuration of 1s22s22p6. Using the electron configuration of Ne to represent the first 10 electrons, the abbreviated electron configuration of P is

P: [Ne]3s23p3

Test Yourself

What is the abbreviated electron configuration for Rb, which has 37 electrons?

Answer

[Kr]5s1

There are some exceptions to the rigorous filling of subshells by electrons. In many cases, an electron goes from a higher-numbered shell to a lower-numbered but later-filled subshell to fill the later-filled subshell. One example is Ag. With 47 electrons, its electron configuration is predicted to be

Ag: [Kr]5s24d9

However, experiments have shown that the electron configuration is actually

Ag: [Kr]5s14d10

This, then, qualifies as an exception to our expectations. At this point, you do not need to memorize the exceptions; but if you come across one, understand that it is an exception to the normal rules of filling subshells with electrons, which can happen.

Electron Configuration Energy Diagrams

We have just seen that electrons fill orbitals in shells and subshells in a regular pattern, but why does it follow this pattern? There are three principles which should be followed to properly fill electron orbital energy diagrams:

  1. The Aufbau principle
  2. The Pauli exclusion principle
  3. Hund’s rule

The overall pattern of the electron shell filling order emerges from the Aufbau principle (German for “building up”):  electrons fill the lowest energy orbitals first. Increasing the principle quantum number, n, increases orbital energy levels, as the electron density becomes more spread out away from the nucleus. In many-electron atoms (all atoms except hydrogen), the energy levels of subshells varies due to electron-electron repulsions. The trend that emerges is that energy levels increase with value of the angular momentum quantum number, l, for orbitals sharing the same principle quantum number, n. This is demonstrated in Figure 8.8, where each line represents an orbital, and each set of lines at the same energy represents a subshell of orbitals.

Figure 8.# Generic energy diagram of orbitals in a multi-electron atom.
Figure 8.8. Generic energy diagram of orbitals in a multi-electron atom.

As previously discussed, the Pauli exclusion principle states that we can only fill each orbital with a maximum of two electrons of opposite spin. But how should we fill multiple orbitals of the same energy level within a subshell (eg. The three orbitals in the 2p subshell)? Orbitals of the same energy level are known as degenerate orbitals, and we fill them using Hund’s rule: place one electron into each degenerate orbital first, before pairing them in the same orbital.

 

Let’s examine a few examples to demonstrate the use of the three principles.

Boron is atomic number 5, and therefore has 5 electrons. First fill the lowest energy 1s orbital with two electrons of opposite spin, then the 2s orbital with 2 electrons of opposite spin and finally place the last electron into any of the three degenerate 2p orbitals (Figure 8.9).

Figure 8.#. Boron electron configuration energy diagram
Figure 8.9. Boron electron configuration energy diagram

Moving across the periodic table, we follow Hund’s rule and add an additional electron to each degenerate 2p orbital for each subsequent element (Figure 8.10). At oxygen we can finally pair up and fill one of the degenerate 2p orbitals.

 

Figure 8.#. Electron configuration energy diagrams for carbon, nitrogen and oxygen.
Figure 8.10. Electron configuration energy diagrams for carbon, nitrogen and oxygen.

Key Takeaways

  • The Pauli exclusion principle limits the number of electrons in the subshells and shells.
  • Electrons in larger atoms fill shells and subshells in a regular pattern that we can follow.
  • Electron configurations are a shorthand method of indicating what subshells electrons occupy in atoms.
  • Abbreviated electron configurations are a simpler way of representing electron configurations for larger atoms.
  • Exceptions to the strict filling of subshells with electrons occur.
  • Electron configurations are assigned from lowest to highest energy following the Aufbau principle
  • One electron is placed in each degenerate orbital before pairing electrons following Hund’s rule.
  • Electron configuration energy diagrams follow three principles: the Aufbau principle, the Pauli exclusion principle and Hund’s rule.

Exercises

  1. Give two possible sets of four quantum numbers for the electron in an H atom.

  2. Give the possible sets of four quantum numbers for the electrons in a Li atom.

  3. How many subshells are completely filled with electrons for Na? How many subshells are unfilled?

  4. How many subshells are completely filled with electrons for Mg? How many subshells are unfilled?

  5. What is the maximum number of electrons in the entire n = 2 shell?

  6. What is the maximum number of electrons in the entire n = 4 shell?

  7. Write the complete electron configuration for each atom.

a)  Si, 14 electrons

b)  Sc, 21 electrons

 

8.  Write the complete electron configuration for each atom.

a)  Br, 35 electrons

b)  Be, 4 electrons

 

9.  Write the complete electron configuration for each atom.

a)  Cd, 48 electrons

b)  Mg, 12 electrons

 

10.  Write the complete electron configuration for each atom.

a)  Cs, 55 electrons

b)  Ar, 18 electrons

11.  Write the abbreviated electron configuration for each atom in Exercise 7.

 

12.  Write the abbreviated electron configuration for each atom in Exercise 8.

 

13.  Write the abbreviated electron configuration for each atom in Exercise 9.

14.  Write the abbreviated electron configuration for each atom in Exercise 10.

15.   Draw electron configuration energy diagrams for potassium, and bromine.

Answers

1.

{1, 0, 0, 1/2} and [1, 0, 0, −1/2}

3.

Three subshells (1s, 2s, 2p) are completely filled, and one shell (3s) is partially filled.

5.

8 electrons

7.

a)  1s22s22p63s23p2

b)  1s22s22p63s23p64s23d1

9.

a)  1s22s22p63s23p64s23d104p65s24d10

b)  1s22s22p63s2

11.

a)  [Ne]3s23p2

b)  [Ar]4s23d1

13.

a)  [Kr]5s24d10

b)  [Ne]3s2

15.

Electron_configuration_potassium.svg

Orbital representation diagram for potassium, depicting each orbital as a line. Adrignola\Public domain

1000px-Electron_configuration_bromine.svg
  1. Orbital representation diagram for bromine, depicting each orbital as a line. Adrignola\Public domain

 

54

Electronic Structure and the Periodic Table

Learning Objectives

  1. Relate the electron configurations of the elements to the shape of the periodic table.
  2. Determine the expected electron configuration of an element by its place on the periodic table.

In Chapter 3 “Atoms, Molecules, and Ions”, we introduced the periodic table as a tool for organizing the known chemical elements. A periodic table is shown in Figure 8.11 “The Periodic Table”. The elements are listed by atomic number (the number of protons in the nucleus), and elements with similar chemical properties are grouped together in columns.

Figure 8.11 The Periodic Table

Periodic Table

 

Why does the periodic table have the structure it does? The answer is rather simple, if you understand electron configurations: the shape of the periodic table mimics the filling of the subshells with electrons.

Let us start with H and He. Their electron configurations are 1s1 and 1s2, respectively; with He, the n = 1 shell is filled. These two elements make up the first row of the periodic table (see Figure 8.12 “The 1”).

Figure 8.12 The 1s Subshell

1s Subshell

 

H and He represent the filling of the 1s subshell.

The next two electrons, for Li and Be, would go into the 2s subshell. Figure 8.13 “The 2” shows that these two elements are adjacent on the periodic table.

Figure 8.13 The 2s Subshell

2s Subshell

 

In Li and Be, the 2s subshell is being filled.

For the next six elements, the 2p subshell is being occupied with electrons. On the right side of the periodic table, these six elements (B through Ne) are grouped together (Figure 8.14 “The 2”).

Figure 8.14 The 2p Subshell

2p Subshell

 

For B through Ne, the 2p subshell is being occupied.

The next subshell to be filled is the 3s subshell. The elements when this subshell is being filled, Na and Mg, are back on the left side of the periodic table (Figure 8.15 “The 3”).

Figure 8.15 The 3s Subshell

3s Subshell

 

Now the 3s subshell is being occupied.

Next, the 3p subshell is filled with the next six elements (Figure 8.16 “The 3”).

Figure 8.16 The 3p Subshell

3p Subshell

 

Next, the 3p subshell is filled with electrons.

Instead of filling the 3d subshell next, electrons go into the 4s subshell (Figure 8.17 “The 4”).

Figure 8.17 The 4s Subshell

4s Subshell

 

The 4s subshell is filled before the 3d subshell. This is reflected in the structure of the periodic table.

After the 4s subshell is filled, the 3d subshell is filled with up to 10 electrons. This explains the section of 10 elements in the middle of the periodic table (Figure 8.18 “The 3”).

Figure 8.18 The 3d Subshell

3d Subshell

 

The 3d subshell is filled in the middle section of the periodic table.

And so forth. As we go across the rows of the periodic table, the overall shape of the table outlines how the electrons are occupying the shells and subshells.

The first two columns on the left side of the periodic table are where the s subshells are being occupied. Because of this, the first two rows of the periodic table are labelled the s block. Similarly, the p block are the right-most six columns of the periodic table, the d block is the middle 10 columns of the periodic table, while the block is the 14-column section that is normally depicted as detached from the main body of the periodic table. It could be part of the main body, but then the periodic table would be rather long and cumbersome. Figure 8.19 “Blocks on the Periodic Table” shows the blocks of the periodic table.

Figure 8.19 Blocks on the Periodic Table

Periodic Table Blocks

 

The periodic table is separated into blocks depending on which subshell is being filled for the atoms that belong in that section.

The electrons in the highest-numbered shell, plus any electrons in the last unfilled subshell, are called valence electrons; the highest-numbered shell is called the valence shell. (The inner electrons are called core electrons.) The valence electrons largely control the chemistry of an atom. If we look at just the valence shell’s electron configuration, we find that in each column, the valence shell’s electron configuration is the same. For example, take the elements in the first column of the periodic table: H, Li, Na, K, Rb, and Cs. Their electron configurations (abbreviated for the larger atoms) are as follows, with the valence shell electron configuration highlighted:

H: 1s1
Li: 1s22s1
Na: [Ne]3s1
K: [Ar]4s1
Rb: [Kr]5s1
Cs: [Xe]6s1

They all have a similar electron configuration in their valence shells: a single s electron. Because much of the chemistry of an element is influenced by valence electrons, we would expect that these elements would have similar chemistry—and they do. The organization of electrons in atoms explains not only the shape of the periodic table but also the fact that elements in the same column of the periodic table have similar chemistry.

The same concept applies to the other columns of the periodic table. Elements in each column have the same valence shell electron configurations, and the elements have some similar chemical properties. This is strictly true for all elements in the s and p blocks. In the d and f blocks, because there are exceptions to the order of filling of subshells with electrons, similar valence shells are not absolute in these blocks. However, many similarities do exist in these blocks, so a similarity in chemical properties is expected.

Similarity of valence shell electron configuration implies that we can determine the electron configuration of an atom solely by its position on the periodic table. Consider Se, as shown in Figure 8.20 “Selenium on the Periodic Table”. It is in the fourth column of the p block. This means that its electron configuration should end in a p4 electron configuration. Indeed, the electron configuration of Se is [Ar]4s23d104p4, as expected.

Figure 8.20 Selenium on the Periodic Table

Periodic Table Selenium

 

Example 8

From the element’s position on the periodic table, predict the valence shell electron configuration for each atom. See Figure 8.21 “Various Elements on the Periodic Table”.

  1. Ca
  2. Sn

Solution

  1. Ca is located in the second column of the s block. We would expect that its electron configuration should end with s2. Calcium’s electron configuration is [Ar]4s2.
  2. Sn is located in the second column of the p block, so we expect that its electron configuration would end in p2. Tin’s electron configuration is [Kr]5s24d105p2.

Test Yourself

From the element’s position on the periodic table, predict the valence shell electron configuration for each atom. See Figure 8.21 “Various Elements on the Periodic Table”.

  1. Ti
  2. Cl

Answer

  1. [Ar]4s23d2
  2. [Ne]3s23p5

Figure 8.21 Various Elements on the Periodic Table

Periodic Table Various Elements

 

Food and Drink App: Artificial Colors

The color of objects comes from a different mechanism than the colors of neon and other discharge lights. Although colored lights produce their colors, objects are colored because they preferentially reflect a certain color from the white light that shines on them. A red tomato, for example, is bright red because it reflects red light while absorbing all the other colors of the rainbow.

Many foods, such as tomatoes, are highly colored; in fact, the common statement “you eat with your eyes first” is an implicit recognition that the visual appeal of food is just as important as its taste. But what about processed foods?

Many processed foods have food colorings added to them. There are two types of food colorings: natural and artificial. Natural food colorings include caramelized sugar for brown; annatto, turmeric, and saffron for various shades of orange or yellow; betanin from beets for purple; and even carmine, a deep red dye that is extracted from the cochineal, a small insect that is a parasite on cacti in Central and South America. (That’s right: you may be eating bug juice!)

Some colorings are artificial. In the United States, the Food and Drug Administration currently approves only seven compounds as artificial colorings in food, beverages, and cosmetics:

  1. FD&C Blue #1: Brilliant Blue FCF
  2. FD&C Blue #2: Indigotine
  3. FD&C Green #3: Fast Green FCF
  4. RD&C Red #3: Erythrosine
  5. FD&C Red #40: Allura Red AC
  6. FD&C Yellow #5: Tartrazine
  7. FD&C Yellow #6: Sunset Yellow FCF

Lower-numbered colors are no longer on the market or have been removed for various reasons. Typically, these artificial colorings are large molecules that absorb certain colors of light very strongly, making them useful even at very low concentrations in foods and cosmetics. Even at such low amounts, some critics claim that a small portion of the population (especially children) is sensitive to artificial colorings and urge that their use be curtailed or halted. However, formal studies of artificial colorings and their effects on behaviour have been inconclusive or contradictory. Despite this, most people continue to enjoy processed foods with artificial coloring (like those shown in the accompanying figure).

Food Colouring

 

Artificial food colorings are found in a variety of food products, such as processed foods, candies, and egg dyes. Even pet foods have artificial food coloring in them, although it’s likely that the animal doesn’t care!

Key Takeaways

  • The arrangement of electrons in atoms is responsible for the shape of the periodic table.
  • Electron configurations can be predicted by the position of an atom on the periodic table.

 

Exercises

  1. Where on the periodic table are s subshells being occupied by electrons?

  2. Where on the periodic table are d subshells being occupied by electrons?

  3. In what block is Ra found?

  4. In what block is Br found?

  5. What are the valence shell electron configurations of the elements in the second column of the periodic table?

  6. What are the valence shell electron configurations of the elements in the next-to-last column of the periodic table?

  7. What are the valence shell electron configurations of the elements in the first column of the p block?

  8. What are the valence shell electron configurations of the elements in the last column of the p block?

  9. From the element’s position on the periodic table, predict the electron configuration of each atom.

a)  Sr

b)  S

 

10.  From the element’s position on the periodic table, predict the electron configuration of each atom.

a)  Fe

b)  Ba

 

11.  From the element’s position on the periodic table, predict the electron configuration of each atom.

a)  V

b)  Ar

 

12.  From the element’s position on the periodic table, predict the electron configuration of each atom.

a)  Cl

b)  K

 

13.  From the element’s position on the periodic table, predict the electron configuration of each atom.

a)  Ge

b)  C

 

14.  From the element’s position on the periodic table, predict the electron configuration of each atom.

a)  Mg

b)  I

Answers

1.

the first two columns

3.

the s block

5.

ns27.

ns2np19.

a)  1s22s22p63s23p64s23d104p65s2

b)  1s22s22p63s23p411.

a)  1s22s22p63s23p64s23d3

b)  1s22s22p63s23p613.

a)  1s22s22p63s23p64s23d104p2

b)  1s22s22p2

55

Quantum Numbers for Electrons

Learning Objectives

  1. Explain what spectra are.
  2. Learn the quantum numbers that are assigned to electrons.

There are two fundamental ways of generating light: either heat an object up so hot it glows or pass an electrical current through a sample of matter (usually a gas). Incandescent lights and fluorescent lights generate light via these two methods, respectively.

A hot object gives off a continuum of light. We notice this when the visible portion of the electromagnetic spectrum is passed through a prism: the prism separates light into its constituent colors, and all colors are present in a continuous rainbow (part (a) in Figure 8.3 “Prisms and Light”). This image is known as a continuous spectrum. However, when electricity is passed through a gas and light is emitted and this light is passed though a prism, we see only certain lines of light in the image (part (b) in Figure 8.3 “Prisms and Light”). This image is called a line spectrum. It turns out that every element has its own unique, characteristic line spectrum.

Figure 8.3 Prisms and Light

Prisms and Light

 

(a) A glowing object gives off a full rainbow of colors, which are noticed only when light is passed through a prism to make a continuous spectrum. (b) However, when electricity is passed through a gas, only certain colors of light are emitted. Here are the colors of light in the line spectrum of Hg.

Why does the light emitted from an electrically excited gas have only certain colors, while light given off by hot objects has a continuous spectrum? For a long time, it was not well explained. Particularly simple was the spectrum of hydrogen gas, which could be described easily by an equation; no other element has a spectrum that is so predictable (Figure 8.4 “Hydrogen Spectrum”). Late-nineteenth-century scientists found that the positions of the lines obeyed a pattern given by the equation

Screen Shot 2014-07-22 at 8.04.37 PM

where n = 3, 4, 5, 6,…, but they could not explain why this was so.

Figure 8.4 Hydrogen Spectrum

Hydrogen Spectrum

 

The spectrum of hydrogen was particularly simple and could be predicted by a simple mathematical expression.

In 1913, the Danish scientist Niels Bohr suggested a reason why the hydrogen atom spectrum looked this way. He suggested that the electron in a hydrogen atom could not have any random energy, having only certain fixed values of energy that were indexed by the number n (the same n in the equation above and now called a quantum number). Quantities that have certain specific values are called quantized. Bohr suggested that the energy of the electron in hydrogen was quantized because it was in a specific orbit. Because the energies of the electron can have only certain values, the changes in energies can have only certain values (somewhat similar to a staircase: not only are the stair steps set at specific heights but the height between steps is fixed). Finally, Bohr suggested that the energy of light emitted from electrified hydrogen gas was equal to the energy difference of the electron’s energy states:

Elight = hν = ΔEelectron

This means that only certain frequencies (and thus, certain wavelengths) of light are emitted. Figure 8.5 “Bohr’s Model of the Hydrogen Atom” shows a model of the hydrogen atom based on Bohr’s ideas.

Figure 8.5 Bohr’s Model of the Hydrogen Atom

Bohr's Hydrogen Atom

 

Bohr’s description of the hydrogen atom had specific orbits for the electron, which had quantized energies.

Bohr’s ideas were useful but were applied only to the hydrogen atom. However, later researchers generalized Bohr’s ideas into a new theory called quantum mechanics, which explains the behaviour of electrons as if they were acting as a wave, not as particles. Quantum mechanics predicts two major things: quantized energies for electrons of all atoms (not just hydrogen) and an organization of electrons within atoms. Electrons are no longer thought of as being randomly distributed around a nucleus or restricted to certain orbits (in that regard, Bohr was wrong). Instead, electrons are collected into groups and subgroups that explain much about the chemical behaviour of the atom.

In the quantum-mechanical model of an atom, the state of an electron is described by four quantum numbers, not just the one predicted by Bohr. The first quantum number is called the principal quantum number. Represented by n. (n). The principal quantum number largely determines the energy of an electron. Electrons in the same atom that have the same principal quantum number are said to occupy an electron shell of the atom. The principal quantum number can be any nonzero positive integer: 1, 2, 3, 4,….

Within a shell, there may be multiple possible values of the next quantum number, the angular momentum quantum number. Represented by ℓ. (ℓ). The ℓ quantum number has a minor effect on the energy of the electron but also affects the spatial distribution of the electron in three-dimensional space—that is, the shape of an electron’s distribution in space. The value of the ℓ quantum number can be any integer between 0 and n − 1:

ℓ = 0, 1, 2,…, n − 1

Thus, for a given value of n, there are different possible values of ℓ:

If n equals ℓ can be
1 0
2 0 or 1
3 0, 1, or 2
4 0, 1, 2, or 3

and so forth. Electrons within a shell that have the same value of ℓ are said to occupy a subshell in the atom. Commonly, instead of referring to the numerical value of ℓ, a letter represents the value of ℓ (to help distinguish it from the principal quantum number):

If ℓ equals The letter is
0 s
1 p
2 d
3 f

The next quantum number is called the magnetic quantum number, represented by mℓ. (m). For any value of ℓ, there are 2ℓ + 1 possible values of m, ranging from −ℓ to ℓ:

−ℓ ≤ m ≤ ℓ

or

|mℓ|≤ ℓ

The following explicitly lists the possible values of m for the possible values of ℓ:

If ℓ equals The mℓ values can be
0 0
1 −1, 0, or 1
2 −2, −1, 0, 1, or 2
3 −3, −2, −1, 0, 1, 2, or 3

The particular value of m dictates the orientation of an electron’s distribution in space. When ℓ is zero, m can be only zero, so there is only one possible orientation. When ℓ is 1, there are three possible orientations for an electron’s distribution. When ℓ is 2, there are five possible orientations of electron distribution. This goes on and on for other values of ℓ, but we need not consider any higher values of ℓ here. Each value of m designates a certain orbital. Thus, there is only one orbital when ℓ is zero, three orbitals when ℓ is 1, five orbitals when ℓ is 2, and so forth. The m quantum number has no effect on the energy of an electron unless the electrons are subjected to a magnetic field—hence its name.

The ℓ quantum number dictates the general shape of electron distribution in space (Figure 8.6 “Electron Orbitals”). Any s orbital is spherically symmetric (part (a) in Figure 8.6 “Electron Orbitals”), and there is only one orbital in any s subshell. Any p orbital has a two-lobed, dumbbell-like shape (part (b) in Figure 8.6 “Electron Orbitals”); because there are three of them, we normally represent them as pointing along the x-, y-, and z-axes of Cartesian space. The d orbitals are four-lobed rosettes (part (c) in Figure 8.6 “Electron Orbitals”); they are oriented differently in space (the one labelled dz2 has two lobes and a torus instead of four lobes, but it is equivalent to the other orbitals). When there is more than one possible value of m, each orbital is labelled with one of the possible values. It should be noted that the diagrams in Figure 8.6 “Electron Orbitals” are estimates of the electron distribution in space, not surfaces electrons are fixed on.

Figure 8.6 Electron Orbitals

Electron Orbitals

 

(a) The lone s orbital is spherical in distribution. (b) The three p orbitals are shaped like dumbbells, and each one points in a different direction. (c) The five d orbitals are rosette in shape, except for the dz2 orbital, which is a “dumbbell + torus” combination. They are all oriented in different directions.

The final quantum number is the spin quantum number. Represented by ms. (ms). Electrons and other subatomic particles behave as if they are spinning (we cannot tell if they really are, but they behave as if they are). Electrons themselves have two possible spin states, and because of mathematics, they are assigned the quantum numbers +1/2 and −1/2. These are the only two possible choices for the spin quantum number of an electron.

Example 3

Of the set of quantum numbers {n, ℓ, m, ms}, which are possible and which are not allowed?

  1. {3, 2, 1, +1/2}
  2. {2, 2, 0, −1/2}
  3. {3, −1, 0, +1/2}

Solution

  1. The principal quantum number n must be an integer, which it is here. The quantum number ℓ must be less than n, which it is. The m quantum number must be between −ℓ and ℓ, which it is. The spin quantum number is +1/2, which is allowed. Because this set of quantum numbers follows all restrictions, it is possible.
  2. The quantum number n is an integer, but the quantum number ℓ must be less than n, which it is not. Thus, this is not an allowed set of quantum numbers.
  3. The principal quantum number n is an integer, but ℓ is not allowed to be negative. Therefore this is not an allowed set of quantum numbers.

Test Yourself

Of the set of quantum numbers {n, ℓ, m, ms}, which are possible and which are not allowed?

  1. {4, 2, −2, 1}
  2. {3, 1, 0, −1/2}

Answers

  1. Spin must be either +1/2 or −1/2, so this set of quantum number is not allowed.
  2. allowed

Chemistry Is Everywhere: Neon Lights

A neon light is basically an electrified tube with a small amount of gas in it. Electricity excites electrons in the gas atoms, which then give off light as the electrons go back into a lower energy state. However, many so-called “neon” lights don’t contain neon!

Although we know now that a gas discharge gives off only certain colors of light, without a prism or other component to separate the individual light colors, we see a composite of all the colors emitted. It is not unusual for a certain color to predominate. True neon lights, with neon gas in them, have a reddish-orange light due to the large amount of red-, orange-, and yellow-colored light emitted. However, if you use krypton instead of neon, you get a whitish light, while using argon yields a blue-purple light. A light filled with nitrogen gas glows purple, as does a helium lamp. Other gases—and mixtures of gases—emit other colors of light. Ironically, despite its importance in the development of modern electronic theory, hydrogen lamps emit little visible light and are rarely used for illumination purposes.

The different colors of these “neon” lights are caused by gases other than neon in the discharge tubes. Source: “Neon Internet Cafe open 24 hours” by JustinC is licensed under the Creative Commons Attribution- Share Alike 2.0 Generic license.
The different colors of these “neon” lights are caused by gases other than neon in the discharge tubes. Source: “Neon Internet Cafe open 24 hours” by JustinC is licensed under the Creative Commons Attribution- Share Alike 2.0 Generic license.

Key Takeaways

  • Electrons in atoms have quantized energies.
  • The state of electrons in atoms is described by four quantum numbers.

 

Exercises

Differentiate between a continuous spectrum and a line spectrum.

Under what circumstances is a continuous spectrum formed? Under what circumstances is a line spectrum formed?

What is the wavelength of light from the hydrogen atom spectrum when n = 3?

What is the wavelength of light from the hydrogen atom spectrum when n = 5?

What are the restrictions on the principal quantum number?

What are the restrictions on the angular momentum quantum number?

What are the restrictions on the magnetic quantum number?

What are the restrictions on the spin quantum number?

What are the possible values for ℓ when n = 5?

What are the possible values for ℓ when n = 1?

What are the possible values for m when ℓ = 3?

What are the possible values for m when ℓ = 6?

Describe the shape of an s orbital.

Describe the shape of a p orbital.

Which of these sets of quantum numbers is allowed? If it is not, explain why.

a)  {4, 1, −2, +1/2}

b)  {2, 0, 0, −1/2}

 

16.  Which of these sets of quantum numbers is allowed? If it is not, explain why.

a)  {5, 2, −1, −1/2}

b)  {3, −1, −1, −1/2}

Answers

1.

A continuous spectrum is a range of light frequencies or wavelengths; a line spectrum shows only certain frequencies or wavelengths.

3.

6.56 × 10−7 m, or 656 nm

5.

The principal quantum number is restricted to being a positive whole number.

7.

The absolute value of m must be less than or equal to ℓ: |m| ≤ ℓ.

9.

ℓ can be 0, 1, 2, 3, or 4.

11.

m can be −3, −2, −1, 0, 1, 2, or 3.

13.

An s orbital is spherical in shape.

15.

a)  Because |m| must be less than ℓ, this set of quantum numbers is not allowed.

b)  allowed

 

56

Light

Learning Objectives

  1. Describe light with its frequency and wavelength.
  2. Describe light as a particle of energy.

What we know as light is more properly called electromagnetic radiation. We know from experiments that light acts as a wave. As such, it can be described as having a frequency and a wavelength. The wavelength of light is the distance between corresponding points in two adjacent light cycles, and the frequency of light is the number of cycles of light that pass a given point in one second. Wavelength is typically represented by λ, the lowercase Greek letter lambda, while frequency is represented by ν, the lowercase Greek letter nu (although it looks like a Roman “vee,” it is actually the Greek equivalent of the letter “en”). Wavelength has units of length (meters, centimeters, etc.), while frequency has units of per second, written as s−1 and sometimes called a hertz (Hz). Figure 8.1 “Characteristics of Light Waves” shows how these two characteristics are defined.

Figure 8.1 Characteristics of Light Waves

Light Waves

 

Light acts as a wave and can be described by a wavelength λ and a frequency ν.

One property of waves is that their speed is equal to their wavelength times their frequency. That means we have

speed = λν

For light, however, speed is actually a universal constant when light is traveling through a vacuum (or, to a very good approximation, air). The measured speed of light (c) in a vacuum is 2.9979 × 108 m/s, or about 3.00 × 108 m/s. Thus, we have

c = λν<