Question 6.12: Calculating a Percent Composition from a Formula Glucose, or......
Calculating a Percent Composition from a Formula
Glucose, or blood sugar, has the molecular formula C_6H_{12}O_6. What is the empirical formula, and what is the percent composition of glucose?
STRATEGY AND SOLUTION
The empirical formula is found by reducing the subscripts in the molecular formula to their smallest whole-number values. In this case, dividing the subscripts by 6 reduces C_6H_{12}O_6 to CH_2O.
The percent composition of glucose can be calculated either from the molecular formula or from the empirical formula. Using the molecular formula, for instance, the C:H:O mole ratio of 6:12:6 can be converted into a mass ratio by assuming that we have 1 mol of compound and carrying out mole-to-gram conversions:
1 \ \cancel{ mol \ glucose} \times \frac{6 \ \cancel{mol \ C}}{1 \ \cancel{mol \ glucose}} \times \frac{12.0 \ g \ C}{1 \ \cancel{mol \ C}} = 72.0 \ g \ C \\ 1 \ \cancel{mol \ glucose} \times \frac{12 \ \cancel{mol \ H}}{1 \ \cancel{mol \ glucose}} \times \frac{1.01 \ g \ H}{1 \ \cancel{mol \ H}} = 12.1 \ g \ H \\ 1 \ \cancel{mol \ glucose} \times \frac{6 \ \cancel{mol \ O}}{1 \ \cancel{mol \ glucose}} \times \frac{16.0 \ g \ O}{1 \ \cancel{mol \ O}} = 96.0 \ g \ ODividing the mass of each element by the total mass, and multiplying by 100%, gives the percent composition. Note that the sum of the mass percentages is 100%.
Total mass of 1 mol glucose = 72.0 g + 12.1 g + 96.0 g = 180.1 g
\begin{aligned}& \% \mathrm{C}=\frac{72.0 \mathrm{~g} \mathrm{C}}{180.1 \mathrm{~g}} \times 100 \%=40.0 \% \\& \% \mathrm{H}=\frac{12.1 \mathrm{~g} \mathrm{H}}{180.1 \mathrm{~g}} \times 100 \%=6.72 \% \\& \% \mathrm{O}=\frac{96.0 \mathrm{~g} \mathrm{O}}{180.1 \mathrm{~g}} \times 100 \%=53.3 \%\end{aligned}