A 2.554-g sample of a certain hydrocarbon is burned in excess oxygen, producing 8.635 g CO2(g) and 1.768 g H2O(l). If the molecular weight of the hydrocarbon is found to be 78.11 u, what is its molecular formula? [m.w. CO2 = 44.01 u; m.w. H2O = 18.02 u]

Step 1: Determine the moles of CO2 and H2O produced

The moles of CO2 = mass of CO2 / molar mass of CO2 = 8.635 g / 44.01 g/mol = 0.196 mol The moles of H2O = mass of H2O / molar mass of H2O = 1.768 g / 18.02 g/mol = 0.098 mol

Step 2: Determine the moles of C and H in the hydrocarbon

The moles of C = moles of CO2 = 0.196 mol The moles of H = 2 * moles of H2O = 2 * 0.098 mol = 0.196 mol

Step 3: Determine the mass of C and H in the hydrocarbon

The mass of C = moles of C * atomic mass of C = 0.196 mol * 12.01 g/mol = 2.354 g The mass of H = moles of H * atomic mass of H = 0.196 mol * 1.008 g/mol = 0.198 g

Step 4: Determine the mass of O in the hydrocarbon

The mass of O = mass of hydrocarbon - mass of C - mass of H = 2.554 g - 2.354 g - 0.198 g = 0.002 g

Step 5: Determine the moles of O in the hydrocarbon

The moles of O = mass of O / atomic mass of O = 0.002 g / 16.00 g/mol = 0.000125 mol

Step 6: Determine the empirical formula of the hydrocarbon

The ratio of C:H:O = 0.196 : 0.196 : 0.000125 ≈ 1 : 1 : 0.0006 So, the empirical formula is CH

Step 7: Determine the molecular formula of the hydrocarbon

The molar mass of the empirical formula = 12.01 g/mol + 1.008 g/mol = 13.018 g/mol The number of empirical units in the molecular formula = molar mass of the molecular formula / molar mass of the empirical formula = 78.11 g/mol / 13.018 g/mol ≈ 6 So, the molecular formula is C6H6.