9.2 Mole-Mass and Mass-Mass Calculations
Learning Objectives
By the end of this section, you will be able to:
- From a given number of moles of a substance, calculate the mass of another substance involved using the balanced chemical equation.
- From a given mass of a substance, calculate the moles of another substance involved using the balanced chemical equation.
- 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:
2Al + 3Cl2 → 2AlCl3
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:
[latex]123.2 \cancel{\text{ g }\ce{Cl2}}\times \dfrac{1\text{ mol }\ce{Cl2}}{70.90\cancel{\text{ g }\ce{Cl2}}}=1.738\text{ mol }\ce{Cl2}[/latex]
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:
[latex]\dfrac{2\text{ mol }\ce{AlCl3}}{3\text{ mol }\ce{Cl2}}[/latex]
Using this conversion factor with the molar quantity we calculated above, we get:
[latex]1.738\cancel{\text{ mol }\ce{Cl2}}\times \dfrac{2\text{ mol }\ce{AlCl3}}{3\cancel{\text{ mol }\ce{Cl2}}}=1.159\text{ mol }\ce{AlCl3}[/latex]
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:
[latex]123.2\cancel{\text{ g }\ce{Cl2}}\times \dfrac{1\cancel{\text{ mol }\ce{Cl2}}}{70.90\cancel{\text{ g }\ce{Cl2}}}\times \dfrac{2\text{ mol }\ce{AlCl3}}{3\cancel{\text{ mol }\ce{Cl2}}}=1.159\text{ mol }\ce{AlCl3}[/latex]
The units still cancel appropriately, and we get the same numerical answer in the end. Sometimes the answer may be slightly different from