A gaseous hydrocarbon contains 82.76% of carbon. Given that its vapour density is 29, find its molecular formula.

Element	% composition	At. wt.	Relative no. of atoms	Simplest ratio
Carbon	82.76	12	$\frac{82.76}{12}$ = 6.89	$\frac{6.89}{6.89}$ = 1
Hydrogen	17.24	1	$\frac{17.24}{1}$ = 17.24	$\frac{17.24}{6.89}$ = 2.5

Simplest ratio of whole numbers = C : H = 1 : 2.5 = 2 : 5

Hence, empirical formula is C₂H₅

Empirical formula weight = 2(12) + 5(1) = 29

V.D. = 29

Molecular weight = 2 x V.D. = 2 x 29

$$\begin{aligned}\text{n} = \frac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \frac{2 \times 29}{29} =2\end{aligned}$$

∴ Molecular formula = n[E.F.] = 2[C₂H₅] = C₄H₁₀