{"id":7335,"date":"2021-06-08T21:55:46","date_gmt":"2021-06-08T21:55:46","guid":{"rendered":"https:\/\/opentextbc.ca\/introductorychemistry\/chapter\/stoichiometry\/"},"modified":"2023-09-25T21:03:28","modified_gmt":"2023-09-25T21:03:28","slug":"stoichiometry","status":"publish","type":"chapter","link":"https:\/\/opentextbc.ca\/introductorychemistry\/chapter\/stoichiometry\/","title":{"raw":"Stoichiometry","rendered":"Stoichiometry"},"content":{"raw":"[latexpage]\r\n
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Learning Objectives<\/p>\r\n\r\n<\/header>\r\n

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  1. Define stoichiometry.<\/em><\/li>\r\n \t
  2. Relate quantities in a balanced chemical reaction on a molecular basis.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\nConsider a classic recipe for pound cake: 1 pound of eggs, 1 pound of butter, 1 pound of flour, and 1\u00a0pound of sugar. (That\u2019s why it\u2019s called \u201cpound cake.\u201d) 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.\r\n\r\nNow suppose you have 1.00 g H2<\/sub>. If the chemical reaction follows the balanced chemical equation 2H2<\/sub>(g) +\u00a0O2<\/sub>(g) \u2192\u00a02H2<\/sub>O(\u2113), then what mass of oxygen do you need to make water?\r\n\r\nCuriously, 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 [pb_glossary id=\"8002\"]stoichiometry[\/pb_glossary]. 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.\r\n\r\nLet us consider a more complicated example. A recipe for pancakes calls for 2 cups (c) of pancake mix, 1 egg, and \u00bd c of milk. We can write this in the form of a chemical equation:\r\n

    2 c mix +\u00a01 egg + \u00bd c milk \u2192\u00a01 batch of pancakes<\/p>\r\nIf 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 4\u00bd eggs and 2\u00bc c milk.\r\n\r\nHow 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\"<\/a>. Because one recipe\u2019s worth of pancakes requires 2 c of pancake mix, 1 egg, and \u00bd c of milk, we actually have the following mathematical relationships that relate these quantities:\r\n

    2 c pancake mix \u21d4 1 egg \u21d4 \u00bd c milk<\/p>\r\nwhere \u21d4 is the mathematical symbol for \u201cis equivalent to.\u201d This does not mean that 2 c of pancake mix equals 1 egg. However, as far as this recipe is concerned<\/em>, 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.\r\n\r\nWe 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:\r\n

    [latex]\\dfrac{1\\text{ egg}}{2\\text{ c pancake mix}}[\/latex]<\/p>\r\nThis 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, we get:\r\n

    [latex]9\\text{ \\cancel{c pancake mix}}\\times \\dfrac{1\\text{ egg}}{2\\text{ \\cancel{c pancake mix}}}=4.5\\text{ eggs}[\/latex]<\/p>\r\nNote how the units cups pancake mix<\/em> cancelled, leaving us with units of eggs<\/em>. 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:\r\n

    [latex]9\\text{ \\cancel{c pancake mix}}\\times \\dfrac{\\frac{1}{2}\\text{ c milk}}{2\\text{ \\cancel{c pancake mix}}}=2.25\\text{ c milk}[\/latex]<\/p>\r\nAgain, units cancel, and new units are introduced.\r\n\r\nA balanced chemical equation is nothing more than a recipe for a chemical reaction<\/em>. The difference is that a balanced chemical equation is written in terms of atoms and molecules, not cups, pounds, and eggs.\r\n\r\nFor example, consider the following chemical equation:\r\n

    2H2<\/sub>(g) +\u00a0O2<\/sub>(g) \u2192 2H2<\/sub>O(\u2113)<\/p>\r\nWe can interpret this as, literally, \u201ctwo hydrogen molecules react with one oxygen molecule to make two water molecules.\u201d That interpretation leads us directly to some equivalences, just as our pancake recipe did:\r\n

    2 H2<\/sub> molecules \u21d4 1 O2<\/sub> molecule \u21d4 2 H2<\/sub>O molecules<\/p>\r\nThese equivalences allow us to construct conversion factors:\r\n

    [latex]\\dfrac{2\\text{ molecules }\\ce{H2}}{1\\text{ molecule }\\ce{O2}}\\Leftrightarrow \\dfrac{2\\text{ molecules }\\ce{H2}}{2\\text{ molecules }\\ce{H2O}} \\Leftrightarrow \\dfrac{1\\text{ molecule }\\ce{O2}}{2\\text{ molecules }\\ce{H2O}}[\/latex]<\/p>\r\nand 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<\/sub>. As we did with converting units, we start with our given quantity and use the appropriate conversion factor:\r\n

    [latex]16\\text{ }\\cancel{\\text{molecules }\\ce{H2}}\\times \\dfrac{1\\text{ molecule }\\ce{O2}}{2\\text{ }\\cancel{\\text{molecules }\\ce{H2}}}=8\\text{ molecules }\\ce{O2}[\/latex]<\/p>\r\nNote how the unit molecules H<\/em>2<\/sub><\/em> 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.\r\n

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    Example 5.1<\/p>\r\n\r\n<\/header>\r\n

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    Problem<\/h1>\r\nHow many molecules of SO3<\/sub> are needed to react with 144 molecules of Fe2<\/sub>O3<\/sub> given the balanced chemical equation Fe2<\/sub>O3<\/sub>(s) + 3SO3<\/sub>(g) \u2192 Fe2<\/sub>(SO4<\/sub>)3<\/sub>?\r\n

    Solution<\/h2>\r\nWe use the balanced chemical equation to construct a conversion factor between Fe2<\/sub>O3<\/sub> and SO3<\/sub>. The number of molecules of Fe2<\/sub>O3<\/sub> goes on the bottom of our conversion factor so it cancels with our given amount, and the molecules of SO3<\/sub> go on the top. Thus, the appropriate conversion factor is:\r\n

    [latex]\\dfrac{3\\text{ molecules }\\ce{SO3}}{1\\text{ molecule }\\ce{Fe2O3}}[\/latex]<\/p>\r\nStarting with our given amount and applying the conversion factor, the result is:\r\n

    [latex]144\\text{ \\cancel{molecules \\ce{Fe2O3}}}\\times \\dfrac{3\\text{ molecules }\\ce{SO3}}{1\\cancel{\\text{ molecule }\\ce{Fe2O3}}}=432\\text{ molecules }\\ce{SO3}[\/latex]<\/p>\r\nWe need 432 molecules of SO3<\/sub> to react with 144 molecules of Fe2<\/sub>O3<\/sub>.\r\n

    Test Yourself<\/h1>\r\nHow many molecules of H2<\/sub> are needed to react with 29 molecules of N2<\/sub> to make ammonia if the balanced chemical equation is N2<\/sub> + 3H2<\/sub> \u2192 2NH3<\/sub>?\r\n

    Answer<\/h2>\r\n87 molecules\r\n\r\n<\/div>\r\n<\/div>\r\nChemical 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 H<\/em>2<\/em><\/sub>O<\/em> indicates that there are two hydrogen atoms and one oxygen atom in each molecule, and these relationships can be used to make conversion factors:\r\n

    [latex]\\dfrac{2\\text{ atoms H}}{1\\text{ molecule }\\ce{H2O}} \\Leftrightarrow \\dfrac{1\\text{ molecule }\\ce{H2O}}{1\\text{ atom O}}[\/latex]<\/p>\r\nConversion factors like this can also be used in stoichiometry calculations.\r\n

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    Example 5.2<\/p>\r\n\r\n<\/header>\r\n

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    Problem<\/h1>\r\nHow many molecules of NH3<\/sub> can you make if you have 228 atoms of H2<\/sub>?\r\n

    Solution<\/h2>\r\nFrom the formula, we know that one molecule of NH3<\/sub> has three H atoms. Use that fact as a conversion factor:\r\n

    [latex]228\\text{\\cancel{ atoms H}}\\times \\dfrac{1\\text{ molecule }\\ce{NH3}}{3\\cancel{\\text{ atoms H}}}=76\\text{ molecules }\\ce{NH3}[\/latex]<\/p>\r\n\r\n

    Test Yourself<\/h1>\r\nHow many molecules of Fe2<\/sub>(SO4<\/sub>)3<\/sub> can you make from 777 atoms of S?\r\n

    Answer<\/h2>\r\n259 molecules\r\n\r\n<\/div>\r\n<\/div>\r\nFor a video lecture on stoichiometry, view this video on stoichiometry<\/a> by Dr. Jessie A. Key.\r\n
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    Key Takeaways<\/p>\r\n\r\n<\/header>\r\n

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