Question

When 3.793 grams of a hydrocarbon, C_xH_y, were burned in a combustion analysis apparatus, 12.34 grams of CO₂ and 3.791grams 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

Moles of CO2 = mass/molecular weight

= 12.34g / 44.00964g/mol

= 0.2803931 mol

Moles of C = moles of CO2 x (1 mol C/ 1 mol CO2)

= 0.2803931 mol x 1 / 1

= 0.2803931 mol

Moles of H2O = 3.791g / 18.01532g/mol

= 0.21043201 mol

Moles of H = moles of H2O x (2 mol H/ 1 mol H2O)

= 0.21043201 mol x 2/ 1

= 0.42086402 mol

Divide by the smaller number of moles:

(0.21043201 mol C) / 0.21043201 = 1.00

(0.42086402 mol H) / 0.21043201 = 2

Empirical formula = CH2

Molecular weight = 14.0266 g/mol

(54.09 g/mol) / (14.0266 g/mol) = 3.85

Round to nearest whole number = 4

Molecular formula = C4H8