compound formulas

Question

When 5.825 grams of a hydrocarbon, C_xH_y, were burned in a combustion analysis apparatus, 18.28 grams of CO₂ and 7.484 grams of H₂O were produced.

In a separate experiment, the molar mass of the compound was found to be 70.13 g/mol. Determine the empirical formula and the molecular formula of the hydrocarbon

Answer 1

moles C = 18.28 g CO2 x 1 mol CO2 / 44 g x 1 mol C / mol CO2 = 0.415 mol C

moles H = 7.484 g H2O x 1 mol H2O / 18 g x 2 mol H / mol H2O = 0.832 mol H

Divide both by 0.415 to get whole nubers:

mols C = 0.415 / 0.415 = 1 mol C

mols H = 0.832 / 0.415 = 2 mol H

Empirical formula = CH₂ (molar mass CH₂ = 14 g/mol)

Molecular formula

70.13 g/mol / 14 g/mol = 5

Molecular formula = C₅H₁₀