Compound Formulas

Question

When 7.370 grams of a hydrocarbon, C_xH_y, were burned in a combustion analysis apparatus, 21.57 grams of CO₂ and 13.25 grams of H₂O were produced.

In a separate experiment, the molar mass of the compound was found to be 30.07 g/mol. Determine the empirical formula and the molecular formula of the hydrocarbon.

Answer 1

moles C = 21.57 g CO2 x 1 mol CO2 / 44 g x 1 mol C / mol CO2 = 0.490 moles C

moles H = 13.25 g H2O x 1 mol H2O / 18 g x 2 mol H / mol H2O = 1.47 moles H

Divide both by 0.490 to get whole numbers:

moles C = 0.490 / 0.499 = 1 mol C

moles H = 1.47 / 0.490 = 3 mol H

EMPIRICAL FORMULA = CH₃ and molar mass of CH₃ = 15 g / mol

MOLECULAR FORMULA = 30.07 g/mol / 15 g/ mol = 2 = C₂H₆