Question

An organic compound (Mol. wt. $=$ 60) is found to contain $C=20\%,H=6.67\%,N=46.67\%$ and remaining is oxygen. On heating, it gives $N{H}_3$ along with a solid residue. The solid residue gives violet colour with alkaline $CuS{O}_4$. The compound is:

• A. $C{H}_{3}NCO$
• B. $C{H}_{3}CON{H}_2$
• C. $N{H}_{2}CON{H}_2$
• D. $C{H}_{3}C{H}_{2}CON{H}_2$

Answer 1

The correct option is C $N{H}_{2}CON{H}_2$

Let the total weight of the compound is 100 grams.

Then, $\%C=20=>$ moles of C $=\frac{20}{12}=1.666$.
Moles of H $=\frac{6.67}{1}=6.67$.
Moles of N $=\frac{46.67}{14}=3.333$.
Moles of O $=\frac{26.66}{16}=1.666$.
$⟹$ Ratio of moles of C:H:N:O $=1.66:6.67:3.33:1.66$

$⟹1:4:2:1$

Hence, molecular formula is $C{H}_4{N}_{2}O$.

$\therefore$ Compound is $N{H}_{2}CON{H}_2$.