Question 3.5.1: When 1.827 g of a hydrocarbon, CxHy, was burned in a combust......
When 1.827 g of a hydrocarbon, C_{x}H_{y}, was burned in a combustion analysis apparatus, 6.373 g of CO_{2} and 0.7829 g of H_{2}O were produced. In a separate experiment, the molar mass of the compound was found to be 252.31 g/mol. Determine the empirical formula and molecular formula of the hydrocarbon.
You are asked to determine the empirical and molecular formulas of a compound.
You are given the mass of compound analyzed, data from a combustion analysis experiment (mass of H_{2}O and CO_{2}), and the molar mass of the compound.
Step 1. Use the mass of CO_{2} and H_{2}O produced in the combustion analysis to calculate moles of H and moles of C present in the original sample. Note that there are 2 mol of H present in 1 mol of H_{2}O.
6.373 g CO_{2} × \frac{1\text{ mol CO}_{2}}{44.010\text{ g}} \times \frac{1\text{ mol C}}{1\text{ mol CO}_{2}} = 0.1448 mol C
0.7829 g H_{2}O × \frac{1\text{ mol H}_{2}\text{O}}{18.015\text{ g}} \times \frac{2\text{ mol H}}{1\text{ mol H}_{2}\text{O}} = 0.08692 mol H
Step 2. Dividing each amount by the smallest value results in a whole-number ratio of elements.
\frac{0.1448\text{ mol C}}{0.08692} = 1.666 mol C
\frac{0.08692\text{ mol H}}{0.08692} = 1.000 mol H = 1 mol H
In this case, the ratio is not made up of two integers. Writing the relative amounts as a ratio and rewriting the ratio as a fraction results in a whole-number ratio of elements and the correct empirical formula.
\frac{1.666\text{ mol C}}{1\text{ mol H}} =\frac{1\frac{2}{3} \text{mol C}}{1\text{mol H}} = \frac{\frac{5}{3}\text{mol C} }{1\text{ mol H}} =\frac{5\text{ mol C}}{3\text{ mol H}}The empirical formula of the hydrocarbon is C_{5}H_{3}.
Step 3. Compare the molar mass of the compound to the molar mass of the empirical formula to determine the molecular formula.
\frac{\text{molar mass of compound}}{\text{molar mass of empirical formula}} =\frac{252.31\text{ g/mol}}{63.079\text{ g/mol}} = 4
The molecular formula is (C_{5}H_{3})_{4}, or C_{20}H_{12}.
Is your answer reasonable? The molar mass of the compound must be a whole-number multiple of the molar mass of the empirical formula, which is true in this case.