5.4 Mole-Mass and Mass-Mass Calculations

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 calculationA calculation in which you 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., 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:

$$\text{123}\text{.2}\cancel{{\text{ g Cl}}_2}\times \frac{1{\text{ mol Cl}}_2}{70.90\cancel{{\text{ g Cl}}_2}}=1.738{\text{ mol Cl}}_2$$

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:

$$\frac{{\text{2 mol AlCl}}_3}{{\text{3 mol Cl}}_2}$$

Using this conversion factor with the molar quantity we calculated above, we get

$$1.738\cancel{{\text{ mol Cl}}_2}\times \frac{2{\text{ mol AlCl}}_3}{\text{3}\cancel{{\text{ mol Cl}}_2}}=1.159{\text{ mol AlCl}}_3$$

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:

$$\text{123}\text{.2}\cancel{{\text{ g Cl}}_2}\times \frac{1\cancel{{\text{ mol Cl}}_2}}{70.90\cancel{{\text{ g Cl}}_2}}\times \frac{2{\text{ mol AlCl}}_3}{3\cancel{{\text{ mol Cl}}_2}}=1.159{\text{ mol AlCl}}_3$$

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.

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.

It should be a trivial task now to extend the calculations to mass-mass calculationsA calculation in which you start with a given mass of a substance and calculate the mass of another substance involved in the chemical equation., 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:

2SO₂(g) + O₂(g) → 2SO₃(g)

First, we convert the given amount, 45.3 g of SO₂, to moles of SO₂ using its molar mass (64.06 g/mol):

$$\text{45}\text{.3}\cancel{{\text{ g SO}}_2}\times \frac{{\text{1 mol SO}}_2}{\text{64}\text{.06}\cancel{{\text{ g SO}}_2}}=0.707{\text{ mol SO}}_2$$

Second, we use the balanced chemical reaction to convert from moles of SO₂ to moles of SO₃:

$$\text{0}\text{.707}\cancel{{\text{ mol SO}}_2}\times \frac{{\text{2 mol SO}}_3}{\text{2}\cancel{{\text{ mol SO}}_2}}=\text{0}{\text{.707 mol SO}}_3$$

Finally, we use the molar mass of SO₃ (80.06 g/mol) to convert to the mass of SO₃:

$$\text{0}\text{.707}\cancel{{\text{ mol SO}}_3}\times \frac{\text{80}{\text{.06 g SO}}_3}{1\cancel{{\text{ mol SO}}_3}}=\text{56}{\text{.6 g SO}}_3$$

We can also perform all three steps sequentially, writing them on one line as

$$\text{45}\text{.3}\cancel{{\text{ g SO}}_2}\times \frac{\text{1}\cancel{{\text{ mol SO}}_2}}{\text{64}\text{.06 }\cancel{{\text{ g SO}}_2}}\times \frac{\text{2}\cancel{{\text{ mol SO}}_3}}{\text{2}\cancel{{\text{ mol SO}}_2}}\times \frac{\text{80}{\text{.06 g SO}}_3}{1\cancel{{\text{ mol SO}}_3}}=\text{56}{\text{.6 g SO}}_3$$

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.