Question

In a compound C, H and N atoms are present in 9 : 1 : 3.5 weight. Molecular weight of compound is 108. Molecular formula of the compound is

• A. $C_2{H}_6{N}_2$
• B. $C_3{H}_{4}N$
• C. $C_6{H}_8{N}_2$
• D. $C_9{H}_{12}{N}_3$

Answer 1

Answer: (C)

$\begin{array}{ccccc}C& & H& & N\\ 9& :& 1& :& 3.5\\ \frac{9}{12}& :& \frac{1}{1}& :& \frac{3.5}{14}\\ \frac{3}{4}& :& \frac{1}{1}& :& \frac{1}{4}\\ 3& :& 4& :& 1\end{array}$
Empirical formula $=C_3{H}_{4}N$
${\left(C_3{H}_{4}N\right)}_n=108$
${\left(12\times 3+1\times 4+14\right)}_n=108$
$54n=108\Rightarrow n=108/54=2$
Molecular formula $=C_6{H}_8{N}_2$