Question 3.9: Combustion of a 5.50-g sample of benzene produces 18.59 g CO...
Combustion of a 5.50-g sample of benzene produces 18.59 g CO2 and 3.81 g H2O. Determine the empirical formula and the molecular formula of benzene, given that its molar mass is approximately 78 g/mol.
Strategy From the product masses, determine the mass of C and the mass of H in the 5.50-g sample of benzene. Sum the masses of C and H; the difference between this sum and the original sample mass is the mass of O in the sample (if O is in fact present in benzene). Convert the mass of each element to moles, and use the results as subscripts in a chemical formula. Convert the subscripts to whole numbers by dividing each by the smallest subscript. This gives the empirical formula. To calculate the molecular formula, first divide the molar mass given in the problem statement by the empirical-formula mass. Then, multiply the subscripts in the empirical formula by the resulting number to obtain the subscripts in the molecular formula.
Setup The necessary molar masses are CO2, 44.01 g/mol; H2O, 18.02 g/mol; C, 12.01 g/mol; H, 1.008 g/mol; and O, 16.00 g/mol.
We calculate the mass of carbon and the mass of hydrogen in the products (and therefore in the original 5.50-g sample) as follows:
mass of C = 18.59 \cancel{g CO_{2}} × \frac{1 \cancel{mol CO_{2}}}{44.01 \cancel{g CO_{2}}} × \frac{1 \cancel{mol C}}{1 \cancel{mol CO_{2}}} × \frac{12.01 g C}{1 \cancel{mol C}} = 5.073 g C
mass of H = 3.81 \cancel{g H_{2}O} × \frac{1 \cancel{mol H_{2}O}}{18.02 \cancel{g H_{2}O}} × \frac{2 \cancel{mol H}}{1 \cancel{mol H_{2}O}} × \frac{1.008 g H}{1 \cancel{mol H}} = 0.426 g H
The total mass of products is 5.073 g + 0.426 g = 5.499 g. Because the combined masses of C and H account for the entire mass of the original sample (5.499 g ≈ 5.50 g), this compound must not contain O.
Converting mass to moles for each element present in the compound,
moles of C = 5.073 \cancel{g C} × \frac{1 mol C}{12.01 \cancel{g C}} = 0.4224 mol C
moles of H = 0.426 \cancel{g H} × \frac{1 mol H}{1.008 \cancel{g H}} = 0.423 mol H
gives the formula C0.4224H0.423. Converting the subscripts to whole numbers (0.4224/0.4224 = 1; 0.423/0.4224 ≈ 1) gives the empirical formula, CH.
Finally, dividing the approximate molar mass (78 g/mol) by the empirical-formula mass (12.01 g/mol + 1.008 g/mol = 13.02 g/mol) gives 78/13.02 ≈ 6. Then, multiplying both subscripts in the empirical formula by 6 gives the molecular formula, C6H6.