When hydrocarbons are burned in a limited amount of air, both \(\mathrm{CO}\) and \(\mathrm{CO}_{2}\) form. When \(0.450 \mathrm{~g}\) of a particular hydrocarbon was burned in air, \(0.467 \mathrm{~g}\) of \(\mathrm{CO}, 0.733 \mathrm{~g}\) of \(\mathrm{CO}_{2},\) and \(0.450 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) were formed. (a) What is the empirical formula of the compound? (b) How many grams of \(\mathrm{O}_{2}\) were used in the reaction? (c) How many grams would have been required for complete combustion?

First, we need to convert the given mass of each product into moles. We will use the molar mass of each component (C = 12.01 g/mol, O = 16.00 g/mol, H = 1.008 g/mol). Moles of CO: \(\frac{0.467 \mathrm{~g}}{28.01 \mathrm{~g/mol}} = 0.0167 \mathrm{~mol}\) Moles of CO2: \(\frac{0.733 \mathrm{~g}}{44.01 \mathrm{~g/mol}} = 0.0167 \mathrm{~mol}\) Moles of H2O: \(\frac{0.450 \mathrm{~g}}{18.02 \mathrm{~g/mol}} = 0.0250 \mathrm{~mol}\) Since the hydrocarbon contains only carbon and hydrogen, we can calculate the moles of carbon (C) and hydrogen (H) in the hydrocarbon based on the moles of CO and CO2 and moles of H2O, respectively. Moles of C = moles of CO + moles of CO2 = 0.0167 + 0.0167 = 0.0334 moles Moles of H = 2 * moles of H2O = 2 * 0.0250 = 0.0500 moles Oxygen is present in CO, CO2, and H2O. Now, let's calculate the moles of O used in the reaction. Moles of O = (moles of CO + 2 * moles of CO2) + moles of H2O = (0.0167 + 2 * 0.0167) + 0.0250 = 0.0751 moles

Now, let's determine the empirical formula by finding the mole ratio of C and H in the hydrocarbon. Mole ratio of C and H = \(\frac{0.0334}{0.0500} = 0.668\) Since the ratio is close to 2:3, the empirical formula of the hydrocarbon is C2H3.

(a) The empirical formula of the compound is C2H3. (b) To calculate the mass of oxygen used, multiply the moles of oxygen used by the molar mass of oxygen: Mass of O2 used = 0.0751 moles * 32.00 g/mol = 2.40 g (c) To calculate the mass of oxygen required for complete combustion, we need to calculate the theoretical O2 requirement to completely oxidize the hydrocarbon. A completely oxidized hydrocarbon would produce CO2 and H2O as the only products: C2H3 + O2 -> 2CO2 + 3/2H2O Now, we calculate the required moles of O2: Moles of O2 required = moles of C * 2 + moles of H * 1/2 = (0.0334 * 2) + (0.0500 * 1/2) = 0.0668 + 0.0250 = 0.0918 moles Mass of O2 required = 0.0918 moles * 32.00 g/mol = 2.93 g So, 2.93 g of O2 would have been required for complete combustion.