Example 10

Using concentration as a conversion factor, how many liters of 2.35 M CuSO₄ are needed to obtain 4.88 mol of CuSO₄?

Solution

This is a one-step conversion, but the concentration must be written as the reciprocal for the units to work out:

$$4.88\cancel{{\text{ mol CuSO}}_{\text{4}}}\times \frac{1\text{ L}}{2.35\cancel{\text{ mol}}}=2.08\text{ L of solution}$$

Test Yourself

Using concentration as a conversion factor, how many liters of 0.0444 M CH₂O are needed to obtain 0.0773 mol of CH₂O?

Answer

1.74 L

Example 11

What mass of solute is present in 0.765 L of 1.93 M NaOH?

Solution

This is a two-step 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:

$$\text{0}\text{.765}\cancel{\text{ L}}\times \frac{1.93\cancel{\text{ mol NaOH}}}{\cancel{\text{ L solution}}}\times \frac{40.0\text{ g NaOH}}{1\cancel{\text{ mol NaOH}}}=59.1\text{ g NaOH}$$

Test Yourself

What mass of solute is present in 1.08 L of 0.0578 M H₂SO₄?

Answer

6.12 g

Example 12

What volume of 0.0995 M Al(NO₃)₃ will react with 3.66 g of Ag according to the following chemical equation?

3Ag(s) + Al(NO₃)₃(aq) → 3AgNO₃ + 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:

$$3.66\cancel{\text{ g Ag}}\times \frac{\text{1 }\cancel{\text{ mol Ag}}}{107.97\cancel{\text{ g Ag}}}\times \frac{\text{1}\cancel{{\text{ mol Al(NO}}_{\text{3}}{\text{)}}_{\text{3}}}}{3\cancel{\text{ mol Ag}}}\times \frac{\text{1 L solution}}{\text{0}\text{.0995}\cancel{{\text{ mol Al(NO}}_{\text{3}}{\text{)}}_{\text{3}}}}=0.114\text{ L}$$

The strikeouts show how the units cancel.

Test Yourself

What volume of 0.512 M NaOH will react with 17.9 g of H₂C₂O₄(s) according to the following chemical equation?

H₂C₂O₄(s) + 2NaOH(aq) → Na₂C₂O₄(aq) + 2H₂O(ℓ)

Answer

0.777 L

Example 13

A student takes a precisely measured sample, called an aliquot, of 10.00 mL of a solution of FeCl₃. The student carefully adds 0.1074 M Na₂C₂O₄ until all the Fe³⁺(aq) has precipitated as Fe₂(C₂O₄)₃(s). Using a precisely measured tube called a burette, the student finds that 9.04 mL of the Na₂C₂O₄ solution was added to completely precipitate the Fe³⁺(aq). What was the concentration of the FeCl₃ 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:

2FeCl₃(aq) + 3Na₂C₂O₄(aq) → Fe₂(C₂O₄)₃(s) + 6NaCl(aq)

Solution

First we need to determine the number of moles of Na₂C₂O₄ that reacted. We will convert the volume to liters and then use the concentration of the solution as a conversion factor:

$$\text{9}\text{.04}\cancel{\text{ mL}}\times \frac{\text{1}\cancel{\text{ L}}}{\text{1,000}\cancel{\text{ mL}}}\times \frac{0.1074{\text{ mol Na}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}}{\cancel{\text{ L}}}=0.000971{\text{ mol Na}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}$$

Now we will use the balanced chemical equation to determine the number of moles of Fe³⁺(aq) that were present in the initial aliquot:

$$0.000971\text{ mol}\cancel{{\text{ Na}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}}\times \frac{{\text{2 mol FeCl}}_{\text{3}}}{\text{3}\cancel{{\text{ mol Na}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}}}=0.000647{\text{ mol FeCl}}_{\text{3}}$$

Then we determine the concentration of FeCl₃ in the original solution. Converting 10.00 mL into liters (0.01000 L), we use the definition of molarity directly:

$$M=\frac{\text{mol}}{\text{L}}=\frac{0.000647{\text{ mol FeCl}}_{\text{3}}}{0.01000\text{ L}}=0.0647M{\text{ FeCl}}_{\text{3}}$$

Test Yourself

A student titrates 25.00 mL of H₃PO₄ with 0.0987 M KOH. She uses 54.06 mL to complete the chemical reaction. What is the concentration of H₃PO₄?

H₃PO₄(aq) + 3KOH(aq) → K₃PO₄(aq) + 3H₂O

Answer

0.0711 M

Example 14

H₂O₂ is used to determine the amount of Mn according to this balanced chemical equation:

2MnO₄⁻(aq) + 5H₂O₂(aq) + 6H⁺(aq) → 2Mn²⁺(aq) + 5O₂(g) + 8H₂O(ℓ)

What mass of 3.00% m/m H₂O₂ solution is needed to react with 0.355 mol of MnO₄⁻(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 H₂O₂ and then convert to find the mass of H₂O₂. Knowing that the H₂O₂ solution is 3.00% by mass, we can determine the mass of solution needed:

$$\text{0}\text{.355}\cancel{{\text{ mol MnO}}_{\text{4}}{}^{-}}\times \frac{\text{5}\cancel{{\text{ mol H}}_{\text{2}}{\text{O}}_{\text{2}}}}{\text{2}\cancel{{\text{ mol MnO}}_{\text{4}}{}^{-}}}\times \frac{34.02\cancel{{\text{ g H}}_{\text{2}}{\text{O}}_{\text{2}}}}{\cancel{{\text{ mol H}}_{\text{2}}{\text{O}}_{\text{2}}}}\times \frac{100\text{ g solution}}{\text{3}\cancel{{\text{ g H}}_{\text{2}}{\text{O}}_{\text{2}}}}=\mathrm{1,006}\text{ g solution}$$

The first conversion factor comes from the balanced chemical equation, the second conversion factor is the molar mass of H₂O₂, and the third conversion factor comes from the definition of percentage concentration by mass.

Test Yourself

Use the balanced chemical reaction for MnO₄⁻ and H₂O₂ to determine what mass of O₂ is produced if 258 g of 3.00% m/m H₂O₂ is reacted with MnO₄⁻.

Answer

7.28 g