# 9.04: Mass to Mass Calculations

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It should be a trivial task now to extend the calculations to **mass-mass calculations**, 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 SO_{3} that can be produced by the reaction of 45.3 g of SO_{2} and O_{2}:

\[2\, SO_2 (g) + O_2 (g) → 2\,SO_3 (g) \]

First, we convert the given amount, 45.3 g of SO_{2}, to moles of SO_{2} using its molar mass (64.06 g/mol):

\[45.3\cancel{g\, SO_{2}}\times \frac{1\, mol\, SO_{2}}{64.06\cancel{g\, SO_{2}}}=0.707\, mol\, SO_{2}\]

Second, we use the balanced chemical reaction to convert from moles of SO_{2} to moles of SO_{3}:

\[0.707\cancel{mol\, SO_{2}}\times \frac{2\, mol\, SO_{3}}{2\cancel{mol\, SO_{2}}}=0.707\, mol\, SO_{3}\]

Finally, we use the molar mass of SO_{3} (80.06 g/mol) to convert to the mass of SO_{3}:

\[0.707\cancel{mol\, SO_{3}}\times \frac{80.06\, g\, SO_{3}}{1\cancel{mol\, SO_{3}}}=56.6\, g\, SO_{3}\]

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

\[45.3\cancel{g\, SO_{2}}\times \frac{1\, mol\, SO_{2}}{64.06\cancel{g\, SO_{2}}}\times \frac{2\, mol\, SO_{3}}{2\cancel{mol\, SO_{2}}}\times \frac{80.06\, g\, SO_{3}}{1\cancel{mol\, SO_{3}}}=56.6\, g\, SO_{3}\]

We get the same answer. Note how the initial and all the intermediate units cancel, leaving grams of SO_{3}, which is what we are looking for, as our final answer.

Example \(\PageIndex{3}\):

What mass of Mg will be produced when 86.4 g of K are reacted?

MgCl_{2}(s) + 2K(s) → Mg(s) + 2KCl(s)

**Solution**

We will simply follow the steps

mass K → mol K → mol Mg → mass Mg

In addition to the balanced chemical equation, we need the molar masses of K (39.09 g/mol) and Mg (24.31 g/mol). In one line,

\[86.4\cancel{g\, K}\times \frac{1\, mol\, K}{39.09\cancel{g\, K}}\times \frac{1\, \cancel{mol\, Mg}}{2\cancel{mol\, K}}\times \frac{24.31\, g\, Mg}{1\cancel{mol\, Mg}}=26.87\, g\, Mg\]

Exercise \(\PageIndex{3}\)

What mass of H_{2} will be produced when 122 g of Zn are reacted?

\[Zn (s) + 2\,HCl (aq) → ZnCl_2 (aq) + H_2 (g) \]

**Answer**

3.77 g