Question

The formula mass (in gm) of ammonia from the following data is
(i) Volume of ammonia $=$ 25 mL; (ii) Volume on addition of $O_2$ after explosion $=$ 71.2 mL; (iii) Volume after explosion and reaction with $O_2$ on cooling v 14.95 mL; (iv) Volume after being absorbed by alkaline pyrogallol $=$ 12.5 mL.

Answer 1

Let the formula of ammonia be $N_a{H}_b$.
I step

$N_a{H}_b+O_2\stackrel{explosiononly}{\to }\left(a/2\right)N_2+\left(b/2\right)H_2$
Volume before $25$ $V$ $0$ $0$
explosion (in mL):
Volume after $0$ $V$ $\frac{25a}{2}$ $\frac{25b}{2}$
explosion (in mL):
II step
$N_2+H_2+\frac{1}{2}{O}_2\to H_2{O}_{\left(l\right)}$
Volume before reaction: $\frac{25a}{2}$ $\frac{25b}{2}$ $V$ $0$
Volume after reaction: $\frac{25a}{2}$ $0$ $[V-(\frac{25b}{4}\left)\right]$ -
Given in II step,
$\frac{25a}{2}+\frac{25b}{2}+V=71.2...\left(i\right)$
Given in III step
$\frac{25a}{2}+V-\frac{25b}{4}=14.95...\left(ii\right)$
Given in IV step
$\frac{25a}{2}=12.5$ ($\because$ pyrogallol absorbs the oxygen)
From Eq. (iii), a =1
From Eq. (i) and (ii), b= 3
Formula of ammonia is $N{H}_3$.
Formula mass of ammonia is $17$ g/mol.