Question

A gas contains $14.3$% hydrogen and $85.7$% carbon by mass has a density of $2.5$ gL${}^{-1}$ at STP. What is the molecular formula of the gas?

• A. $C{H}_2$
• B. $C_2{H}_4$
• C. $C_4{H}_8$
• D. $C_6{H}_{12}$

Answer 1

The correct option is C $C_4{H}_8$

$100$ g of a sample contains $14.3$ g of $H$ and $85.7$ g of $C$.

The number of moles of $H$ present is $\frac{14.3}{1}=14.3$ moles.

The number of moles of $C$ present is $\frac{85.7}{12}=7.14$ moles.

The molar ratio of $C:H$ is $7.14:14.3=1:2$

Therefore, the empirical formula is $C{H}_2$.
The empirical formula mass is $12+2=14$ g.

At STP, the density is $2.5$ g/L.

Thus, $1$ L of gas weighs $2.5$ g.

Hence, $22.4$ L ($1$ mol) of gas will weigh $22.4\times 2.5=56g$ which is the molecular weight.

The ratio of the molecular weight to the empirical formula mass ($n$) is $\frac{56}{14}=4$.

$\text{Molecular formula}=n\times \text{empirical formula}=4\times C{H}_2=C_4{H}_8$

Thus, the molecular formula of the gas is $C_4{H}_8$.

Hence, option $C$ is correct.