Question

A gaseous hydrocarbon of vapour density 14 con-
tains 85.2% of carbon. Calculate its molecular
formula.​

Answer 1

Answer

$\sf C_2H_4$

Explanation

Given that vapour density = 14

⇒ molar mass = 2 × vapour density = 2 × 14 = 28

This means that molar mass of the given hydrocarbon = 28 g

The hydrocarbon contains 85.2% carbon

⇒ The hydrocarbon contains 85.2% carbon by mass

Let the weight of carbon in the hydrocarbon be x

⇒ 85.2% of 28 = x

⇒ 85.2/100 × 28 = x

⇒ x = 23.856 g

⇒ x ≈ 24 g

This means that rest of mass is of hydrogen

Mass of hydrogen = 28 - 24 = 4 g

Now,

Molar mass of carbon = 12 g

Molar mass of hydrogen = 1 g

Let the number of molecules of carbon in hydrocarbon be n and that of hydrogen be m, so molecular formula becomes $\sf C_nH_m$

Molar mass of carbon = 12 g

Mass of carbon in compound = 12n = 24 g

⇒ n = 24/12 = 2

molar mass of hydrogen = 1 g

mass of hydrogen in compound = 1 × m = 4 g

⇒ m = 4

So, molecular formula of compound = $\sf C_nH_m = C_2H_4$