Chapter 11. Solutions
Concentrations as Conversion Factors
Learning Objectives
1. Apply concentration units as conversion factors.
Concentration can be a conversion factor between the amount of solute and the amount of solution or solvent (depending on the definition of the concentration unit). As such, concentrations can be useful in a variety of stoichiometry problems. In many cases, it is best to use the original definition of the concentration unit; it is that definition that provides the conversion factor.
A simple example of using a concentration unit as a conversion factor is one in which we use the definition of the concentration unit and rearrange; we can do the calculation again as a unit conversion, rather than as a definition. For example, suppose we ask how many moles of solute are present in 0.108 L of a 0.887 M NaCl solution. Because 0.887 M means 0.887 mol/L, we can use this second expression for the concentration as a conversion factor:
0.108 L NaCl × (0.887 mol NaCl /(1 L NaCl)) = 0.0958 mol NaCl
(There is an understood 1 in the denominator of the conversion factor.) If we used the definition approach, we get the same answer, but now we are using conversion factor skills. Like any other conversion factor that relates two different types of units, the reciprocal of the concentration can be also used as a conversion factor.
Example 10
Using concentration as a conversion factor, how many liters of 2.35 M CuSO_{4} are needed to obtain 4.88 mol of CuSO_{4}?
Solution
This is a onestep conversion, but the concentration must be written as the reciprocal for the units to work out:
4.88 mol CuSO_{4}×(1 L / 2.35 mol) = 2.08 L of solution
Test Yourself
Using concentration as a conversion factor, how many liters of 0.0444 M CH_{2}O are needed to obtain 0.0773 mol of CH_{2}O?
Answer
1.74 L
Of course, once quantities in moles are available, another conversion can give the mass of the substance, using molar mass as a conversion factor.
Example 11
What mass of solute is present in 0.765 L of 1.93 M NaOH?
Solution
This is a twostep conversion, first using concentration as a conversion factor to determine the number of moles and then the molar mass of NaOH (40.0 g/mol) to convert to mass:
0.765 L × (1.93 mol NaOH / L solution) × (40.0 g NaOH / 1 mol NaOH) = 59.1 g NaOH
Test Yourself
What mass of solute is present in 1.08 L of 0.0578 M H_{2}SO_{4}?
Answer
6.12 g
More complex stoichiometry problems using balanced chemical reactions can also use concentrations as conversion factors. For example, suppose the following equation represents a chemical reaction:
2 AgNO_{3}(aq) + CaCl_{2}(aq) → 2 AgCl(s) + Ca(NO_{3})_{2}(aq)
If we wanted to know what volume of 0.555 M CaCl_{2} would react with 1.25 mol of AgNO_{3}, we first use the balanced chemical equation to determine the number of moles of CaCl_{2} that would react and then use concentration to convert to liters of solution:
1.25 mol AgNO_{3 }× (1 mol CaCl_{2 }/ 2 mol AgNO_{3}) × (1 L solution / 0.555 mol CaCl_{2}) =1.13 L CaCl_{2}
This can be extended by starting with the mass of one reactant, instead of moles of a reactant.
Example 12
What volume of 0.0995 M Al(NO_{3})_{3} will react with 3.66 g of Ag according to the following chemical equation?
3 Ag(s) + Al(NO_{3})_{3}(aq) → 3 AgNO_{3} + Al(s)
Solution
Here, we first must convert the mass of Ag to moles before using the balanced chemical equation and then the definition of molarity as a conversion factor:
3.66 g Ag × (1 mol Ag / 107.97 g Ag) × (1 mol Al(NO_{3})_{3 }/_{ }3 mol Ag) × (1 L solution / 0.0995 mol Al(NO_{3})_{3}) = 0.114 L
Test Yourself
What volume of 0.512 M NaOH will react with 17.9 g of H_{2}C_{2}O_{4}(s) according to the following chemical equation?
H_{2}C_{2}O_{4}(s) + 2 NaOH(aq) → Na_{2}C_{2}O_{4}(aq) + 2 H_{2}O(ℓ)
Answer
0.777 L
We can extend our skills even further by recognizing that we can relate quantities of one solution to quantities of another solution. Knowing the volume and concentration of a solution containing one reactant, we can determine how much of another solution of another reactant will be needed using the balanced chemical equation.
Example 13
A student takes a precisely measured sample, called an aliquot, of 10.00 mL of a solution of FeCl_{3}. The student carefully adds 0.1074 M Na_{2}C_{2}O_{4} until all the Fe^{3+}(aq) has precipitated as Fe_{2}(C_{2}O_{4})_{3}(s). Using a precisely measured tube called a burette, the student finds that 9.04 mL of the Na_{2}C_{2}O_{4} solution was added to completely precipitate the Fe^{3+}(aq). What was the concentration of the FeCl_{3} in the original solution? (A precisely measured experiment like this, which is meant to determine the amount of a substance in a sample, is called a titration.) The balanced chemical equation is as follows:
2 FeCl_{3}(aq) + 3 Na_{2}C_{2}O_{4}(aq) → Fe_{2}(C_{2}O_{4})_{3}(s) + 6 NaCl(aq)
Solution
First we need to determine the number of moles of Na_{2}C_{2}O_{4} that reacted. We will convert the volume to liters and then use the concentration of the solution as a conversion factor:
9.04 mL × (1 L / 1,000 mL) × (0.1074 mol Na_{2}C_{2}O_{4 }/_{ }L) = 0.000971 mol Na_{2}C_{2}O_{4}
Now we will use the balanced chemical equation to determine the number of moles of Fe^{3+}(aq) that were present in the initial aliquot:
0.000971 mol Na_{2}C_{2}O_{4} × (2 mol FeCl_{3 }/ 3 mol Na_{2}C_{2}O_{4}) = 0.000647 mol FeCl_{3}
Then we determine the concentration of FeCl_{3} in the original solution. Converting 10.00 mL into liters (0.01000 L), we use the definition of molarity directly:
M = mol / L = 0.000647 mol FeCl_{3 }/ 0.01000 L = 0.0647 M FeCl_{3}
Test Yourself
A student titrates 25.00 mL of H_{3}PO_{4} with 0.0987 M KOH. She uses 54.06 mL to complete the chemical reaction. What is the concentration of H_{3}PO_{4}?
H_{3}PO_{4}(aq) + 3 KOH(aq) → K_{3}PO_{4}(aq) + 3 H_{2}O
Answer
0.0711 M
We have used molarity exclusively as the concentration of interest, but that will not always be the case. The next example shows a different concentration unit being used.
Example 14
H_{2}O_{2} is used to determine the amount of Mn according to this balanced chemical equation:
2 MnO_{4}^{−}(aq) + 5 H_{2}O_{2}(aq) + 6H^{+}(aq) → 2 Mn^{2+}(aq) + 5 O_{2}(g) + 8 H_{2}O(ℓ)
What mass of 3.00% m/m H_{2}O_{2} solution is needed to react with 0.355 mol of MnO_{4}^{−}(aq)?
Solution
Because we are given an initial amount in moles, all we need to do is use the balanced chemical equation to determine the number of moles of H_{2}O_{2} and then convert to find the mass of H_{2}O_{2}. Knowing that the H_{2}O_{2} solution is 3.00% by mass, we can determine the mass of solution needed:
0.355 mol MnO_{4}^{− }× (5 mol H_{2}O_{2 }/_{ }2 mol MnO_{4}^{−}) × (34.02 g H_{2}O_{2} / mol H_{2}O_{2}) × (100 g solution / 3 g H_{2}O_{2}) = 1,006 g solution
The first conversion factor comes from the balanced chemical equation, the second conversion factor is the molar mass of H_{2}O_{2}, and the third conversion factor comes from the definition of percentage concentration by mass.
Test Yourself
Use the balanced chemical reaction for MnO_{4}^{−} and H_{2}O_{2} to determine what mass of O_{2} is produced if 258 g of 3.00% m/m H_{2}O_{2} is reacted with MnO_{4}^{−}.
Answer
7.28 g
Key Takeaways
 Know how to apply concentration units as conversion factors.
Exercises

Using concentration as a conversion factor, how many moles of solute are in 3.44 L of 0.753 M CaCl_{2}?

Using concentration as a conversion factor, how many moles of solute are in 844 mL of 2.09 M MgSO_{4}?

Using concentration as a conversion factor, how many liters are needed to provide 0.822 mol of NaBr from a 0.665 M solution?

Using concentration as a conversion factor, how many liters are needed to provide 2.500 mol of (NH_{2})_{2}CO from a 1.087 M solution?

What is the mass of solute in 24.5 mL of 0.755 M CoCl_{2}?

What is the mass of solute in 3.81 L of 0.0232 M Zn(NO_{3})_{2}?

What volume of solution is needed to provide 9.04 g of NiF_{2} from a 0.332 M solution?

What volume of solution is needed to provide 0.229 g of CH_{2}O from a 0.00560 M solution?

What volume of 3.44 M HCl will react with 5.33 mol of CaCO_{3}?
2 HCl + CaCO_{3} → CaCl_{2} + H_{2}O + CO_{2}

What volume of 0.779 M NaCl will react with 40.8 mol of Pb(NO_{3})_{2}?
Pb(NO_{3})_{2} + 2 NaCl → PbCl_{2} + 2 NaNO_{3}

What volume of 0.905 M H_{2}SO_{4} will react with 26.7 mL of 0.554 M NaOH?
H_{2}SO_{4} + 2 NaOH → Na_{2}SO_{4} + 2 H_{2}O

What volume of 1.000 M Na_{2}CO_{3} will react with 342 mL of 0.733 M H_{3}PO_{4}?
3 Na_{2}CO_{3} + 2 H_{3}PO_{4} → 2Na_{3}PO_{4} + 3 H_{2}O + 3 CO_{2}

It takes 23.77 mL of 0.1505 M HCl to titrate with 15.00 mL of Ca(OH)_{2}. What is the concentration of Ca(OH)_{2}? You will need to write the balanced chemical equation first.

It takes 97.62 mL of 0.0546 M NaOH to titrate a 25.00 mL sample of H_{2}SO_{4}. What is the concentration of H_{2}SO_{4}? You will need to write the balanced chemical equation first.

It takes 4.667 mL of 0.0997 M HNO_{3} to dissolve some solid Cu. What mass of Cu can be dissolved?
Cu + 4 HNO_{3}(aq) → Cu(NO_{3})_{2}(aq) + 2 NO_{2} + 2 H_{2}O

It takes 49.08 mL of 0.877 M NH_{3} to dissolve some solid AgCl. What mass of AgCl can be dissolved?
AgCl(s) + 4 NH_{3}(aq) → Ag(NH_{3})_{4}Cl(aq)

What mass of 3.00% H_{2}O_{2} is needed to produce 66.3 g of O_{2}(g)?
2 H_{2}O_{2}(aq) → 2 H_{2}O(ℓ) + O_{2}(g)

A 0.75% solution of Na_{2}CO_{3} is used to precipitate Ca^{2+} ions from solution. What mass of solution is needed to precipitate 40.7 L of solution with a concentration of 0.0225 M Ca^{2+}(aq)?
Na_{2}CO_{3}(aq) + Ca^{2+}(aq) → CaCO_{3}(s) + 2 Na^{+}(aq)
Answers
1.
2.59 mol
3.
1.24 L
5.
2.40 g
7.
0.282 L
9.
3.10 L
11.
8.17 mL
13.
0.1192 M
15.
7.39 mg
17.
4.70 kg