Question

When 4.979 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 16.20 grams of CO2 and 4.976 grams of H2O were produced.
In a separate experiment, the molar mass of the compound was found to be 54.09 g/mol. Determine the empirical formula and the molecular formula of the hydrocarbon.

Answer 1

For Hydrocarbon combustion:

CₓHₐ + O₂ → xCO₂ + a/2 H₂O

Moles of CO₂ = 16.2 / 44 = 0.37
Moles of Carbon = 0.37
Moles of H₂O = 4.976 / 18 = 0.28
Moles of Hydrogen = 0.28 x 2 = 0.56

Molar ratio = C : H = 1 : 1.5
= 2 : 3
C₂H₃

Mass of empirical unit = 12 x 2 + 1 x 3
= 27
Mr = 54.09
Number of empirical units repeated: 54.09 / 27
= 2

Molecular formula = C₄H₆