Question

A gaseous saturated hydrocarbon (aliphatic) was mixed with excess of oxygen. When an electric spark was passed through the mixture, it suffered a contraction in volume amounting to $20 mL$. This residual volume on being shaken with concentrated $KOH$ solution suffered a further contraction of $10 mL$. Name the hydrocarbon.

Answer 1

The general formula of alkane $=C_{n} H _{2 n}+2$

Vol. of $CO _{2}$ formed $=10 mL$

COMBUSTION REACTION :

$C_{n} H_{2 n+2}(g)+\left(\frac{3 n+1}{2}\right) O_{2}(g)=n CO _{2}(g)+(n+1) H _{2} O (l)$

1 vol. $\frac{3 n+1}{2}$ vol. $n$ vol. ov.

$\because n$ vol. of $CO _{2}$ from 1 vol. of $C _{n} H _{2 n+2}$

$\therefore 10 mL$ of $CO _{2}$ from $\frac{10}{n} mL$ of $C_{n} H_{2 n+2}$

$\quad\therefore \quad$ Vol. of alkane $=\frac{10}{n} mL$

Vol. of oxygen reacted $=\frac{10}{n}\left(\frac{3 n+1}{2}\right) mL$ and vol. of $CO _{2}=10 mL$.

Contraction in volume $=\left(\right.$ vol. of hydrocarbon $+$ vol. of $\left.O _{2}\right)-$ vol. of $CO _{2}$

$=\frac{10}{n}+\frac{10}{n}\left(\frac{3 n+1}{2}\right)-10=\frac{10}{n}+15+\frac{5}{n}-10=\frac{15}{n}+5$

Contraction in volume $=20 mL$.

$\therefore \quad \frac{15}{n}+5=20 ; \frac{15}{n}=15 ; \quad \therefore \quad n=1 .$

$\therefore$ Formula of hydrocarbon $= CH _{4}$, Methane.