Question

$0.27g$ of an organic compound gave on combustion $0.396g$ of $C{O}_2$ and $=0.216g$ of $H_{2}O$, $0.36g$ of the same substance gave $48.88mL$ of $N_2$ at $290K$ and $740mm$ pressure. Calculate the percentage composition of the compound.

Answer 1

of $C=\frac{14}{44}\times \frac{0.396}{0.27}\times 100$
of $C=40$
of $H=\frac{2}{18}\times \frac{0.216}{0.27}\times$
of $H=8.88$
$⟹PV=nRT$
$\frac{740}{760}\times 48.88\times {10}^{-3}=n\times 0.082\times 290$
$n_{N_2}=0.002$
Mass of $N_2=0.002\times 28=0.056$
of $N=\frac{14}{28}\times \frac{0.056}{0.36}\times 100$