Question

In a compound $C$, $H$ and $N$ are present in $9:1:3.5$ by weight. If the molecular mass of the compound is 108, what is the molecular formula?

• A. $C_6{H}_8{N}_2$
• B. $C_6{H}_{7}N$
• C. $C_2{H}_8{N}_2$
• D. $C_6{H}_5{N}_2$

Answer 1

The correct option is A $C_6{H}_8{N}_2$

By weight $C,H$ and $N$ ratio is $9:1:3.5$

Simple ratio is $18:2:7$

$C\to \frac{18}{12}=\frac{3}{2}=1.5$

$H\to \frac{2}{1}=2$

$N\to \frac{7}{14}=0.5$

Dividing these numbers with small number

$C\to \frac{1.5}{0.5}=3$

$H\to \frac{2}{0.5}=4$

$N\to \frac{0.5}{0.5}=1$

Emperical Formula is $(C_3{H}_{4}N{)}_n$

Its weight is $54n$

But the given weight =108

So, $54n=108$

$⟹n=2$

Molecular formula $=(C_3{H}_{4}N{)}_2=C_6{H}_8{N}_2$