Question

At STP, one litre of a gas weighs $1.25$ g. The gas contains $85.71$% of carbon and $14.29$% of hydrogen. The formula of the compound is:

• A. $C{H}_4$
• B. $C_2{H}_6$
• C. $C_3{H}_8$
• D. $C_2{H}_4$

Answer 1

The correct option is D $C_2{H}_4$

Given, $1$ L of gas weighs $1.25$ g.

$\therefore 22.4$ L weighs $1.25\times 22.4$ g $=28$ g $=$ Molecular weight.

The ratio of number of moles of $C$ and $H$ is,

$C:H=\frac{85.71}{12}:\frac{14.29}{1}=1:2$

$\therefore$ The empirical formula is $C{H}_2$ and empirical weight $=$ $14$.

$\text{Molecular Formula}=(\text{Empirical Formula}{)}_x$

$x=\frac{\text{Molecular weight}}{\text{Empirical weight}}=\frac{28}{14}=2$

$\therefore \text{Molecular Formula}=(C{H}_2{)}_2=C_2{H}_4$