Question

A compound contains $38.8$% $C$, $16.0$% $H$ and $45.2$% $N$. The empirical formula of the compound would be :

• A. $C{H}_{3}N{H}_2$
• B. $C{H}_{3}CN$
• C. $C_2{H}_{5}CN$
• D. $C{H}_2{\left(NH\right)}_2$

Answer 1

Answer: (A)

$C{H}_{3}N{H}_2$

The ratio among $C:H:N$ is $38.8:16:45.2$.
The molar ratio is $\frac{38.8}{12}:\frac{16}{1}:\frac{45.2}{14}$ = $3.23:16:3.22$ = $1:5:1$.0.
Therefore, the empirical formula of the compound would be $CN{H}_5\to C{H}_{3}N{H}_2$.