Chemistry - Eudiometry Question with Solution | TestHub

Let the formula of the hydrocarbon be ${\mathrm{C}}_{\mathrm{x}}{\mathrm{H}}_{\mathrm{y}}$.

Therefore, the combustion reaction is:

$${\mathrm{C}}_{\mathrm{x}}{\mathrm{H}}_{\mathrm{y}}(\mathrm{g})+\left(\mathrm{x}+\frac{\mathrm{y}}{4}\right)\cdot {\mathrm{O}}_2(\mathrm{g})\to \mathrm{x}\cdot {\mathrm{CO}}_2(\mathrm{g})+\frac{\mathrm{y}}{2}\cdot {\mathrm{H}}_2\mathrm{O}(\mathrm{g})$$

At $t=0$:

$\mathrm{a}mol$ of ${\mathrm{C}}_{\mathrm{x}}{\mathrm{H}}_{\mathrm{y}}$, $2\mathrm{a}\left(\mathrm{x}+\frac{\mathrm{y}}{4}\right)mol$ of ${\mathrm{O}}_2$, $0$ of ${\mathrm{CO}}_2$, $0$ of ${\mathrm{H}}_2\mathrm{O}$.

At $t=t_{\mathrm{f}}$:

$0$ of ${\mathrm{C}}_{\mathrm{x}}{\mathrm{H}}_{\mathrm{y}}$, $\mathrm{a}\left(\mathrm{x}+\frac{\mathrm{y}}{4}\right)mol$ of ${\mathrm{O}}_2$, $\mathrm{ax}mol$ of ${\mathrm{CO}}_2$, $\frac{\mathrm{ay}}{2}mol$ of ${\mathrm{H}}_2\mathrm{O}=\frac{0.162}{18}mol$.

So, $\frac{\mathrm{ay}}{2}=\frac{0.162}{18}=0.009$.

Therefore, $\mathrm{ay}=0.018\dots$ (i)

Also, the final number of moles $n_{final}$ is given by:

$$n_{final}=\frac{P_f\cdot V}{RT}=\frac{1.05\times 0.96}{0.08\times 400}\mathrm{mol}=0.0315\mathrm{mol}$$

This implies:

$$\mathrm{a}\left(\mathrm{x}+\frac{\mathrm{y}}{4}\right)+\mathrm{ax}+\frac{\mathrm{ay}}{2}=0.0315$$

$$2\mathrm{ax}+\frac{3\mathrm{ay}}{4}=0.0315$$

Substituting the value of $\mathrm{ay}$ from (i):

$$2\mathrm{ax}+\frac{3}{4}\times 0.018=0.0315$$

$$2\mathrm{ax}=0.018$$

Therefore, $\mathrm{ax}=0.009$.

Thus, the ratio $\mathrm{ax}:\mathrm{ay}=\mathrm{x}:\mathrm{y}=0.009:0.018=1:2$.

The empirical formula is ${\mathrm{CH}}_2$.