A \(0.1888 \mathrm{g}\) sample of a hydrocarbon produces \(0.6260 \mathrm{g}\) \(\mathrm{CO}_{2}\) and \(0.1602 \mathrm{g} \mathrm{H}_{2} \mathrm{O}\) in combustion analysis. Its molecular mass is found to be 106 u. For this hydrocarbon, determine its (a) mass percent composition; (b) empirical formula; (c) molecular formula.

To determine the composition, refer to the fact that the carbon in the original compound ends up as carbon in the carbon dioxide and the hydrogen in the original compound ends up as hydrogen in water. In carbon dioxide, there are 1 mole of carbon for each mole of CO2. This is represented as 12 g carbon / 44 g CO2. In water, there are 2 moles of hydrogen for each mole of H2O. This is represented as 2 g hydrogen / 18 g H2O. To find the amount of carbon in the original compound, multiply the weight of the CO2 by the mole ratio of carbon in CO2. \(0.6260 \mathrm{g} \mathrm{CO}_{2} * \frac{12 \mathrm{g} \mathrm{C}}{44 \mathrm{g} \mathrm{CO}_{2}} = 0.17 \mathrm{g} \mathrm{C}\). Similarly, to find the amount of hydrogen in the original compound, multiply the weight of the H2O by the mole ratio of hydrogen in H2O.\(0.1602 \mathrm{g} \mathrm{H}_{2} \mathrm{O} * \frac{2 \mathrm{g} \mathrm{H}}{18 \mathrm{g} \mathrm{H}_{2} \mathrm{O}} = 0.018 \mathrm{g} \mathrm{H}\)

The mass percent composition can be calculated by dividing the mass of each element by the total mass of the compound and multiplying by 100. For Carbon: \(\frac{0.17 \mathrm{g} \mathrm{C}}{0.1888 \mathrm{g} \mathrm{Compound}} * 100 = 89.9\%\). For Hydrogen: \(\frac{0.018 \mathrm{g} \mathrm{H}}{0.1888 \mathrm{g} \mathrm{Compound}} * 100 = 9.5 \%\)

To find the empirical formula, convert the mass of each component to moles. For Carbon: \(0.17 \mathrm{g} \mathrm{C} * \frac{1 \mathrm{mole}}{12 \mathrm{g}} = 0.014 \mathrm{moles} \mathrm{C}\). For Hydrogen: \(0.018 \mathrm{g} \mathrm{H} * \frac{1 \mathrm{mole}}{1 \mathrm{g}} = 0.018 \mathrm{moles} \mathrm{H}\). The empirical formula is the simplest whole number ratio of these moles, which is CH3.

To find the molecular formula, you must know the molar mass of the compound. The molar mass can be calculated from the empirical formula, which was found to be CH3. The molar mass of CH3 is \(((12 * 1) + (3 * 1)) = 15\ g/mol\). The molecular formula is then found by comparing the empirical formula mass with the given molar mass , \( \frac{106}{15} \approx 7\). This number, multiplied with the subscript in the empirical formula, will give the molecular formula. So, the molecular formula is \(C_{7}H_{21}\).