Question

A gaseous hydrocarbon with a vapour density of 14 contains 85.2% carbon. Calculate its molecular formula.

Answer 1

Step 1: Molar mass = 2 ✕ vapour density

= 2 ✕ 14 = 28 g mol^-1

Given mass % of Carbon = 85.2 %

Hydrocarbon is made up of hydrogen and carbon. It means the rest mass % will be of Hydrogen

Mass percentage of hydrogen = (100 – 85.2)%

= 14.8%

Number of moles of carbon = Given mass / Molar mass of C

= 85.2 / 12 = 7.1 mol

Number of moles of hydrogen = Given mass / Molar mass of H

= 14.8/ 1 = 14.8 mol

Mole ratio of carbon = 7.1/7.1 = 1

Mole ratio of hydrogen =14.8/7.1 = 2.08 approximately equal to 2.
Step 2: Therefore, empirical formula of the given compound = CH₂

Empirical formula mass = 12 + 1 ✕ 2

= 14 g

The factor n can be obtained by dividing the molar mass of the compound by the Empirical formula mass of the compound.

n = molar mass/ empirical formula mass

= 28/14 = 2

Step 3: The Molecular formula of a compound can be obtained by multiplying the Empirical formula by factor n. Molecular formula = n (empirical formula)

= 2 (CH₂)

= C₂H₄ ( Ethene)

Final answer:The molecular formula of a gaseous hydrocarbon with a vapor density of 14 contains 85.2% carbon is C₂H₄ ( Ethene).