Question

Percentage of $C,H$ and $N$ are given as follows $C=40$%, $H=13.33$%, $N=46.67$%. The empirical formula will be:

Answer 1

Given percentages are

$C\to 40$ %

$H\to 13.33$ %

$N\to 46.67$ %

$\Rightarrow \frac{\text{Percentages of elements}}{\text{Atomic weight of element}}\to C=\frac{40}{12}=3.33\to \left(1\right)$

$H=\frac{13.33}{1}=13.33\to \left(2\right)$

$N=\frac{46.67}{14}=3.33\to \left(3\right)$

Simple ratio of $C,H$ and $N$ obtained by $1,2$ and $3$ is

$\Rightarrow C:H:N=3.33:13.33:3.33$

$=1:4:1$

For one molecule, $C$ has $1$ atom

$H$ has $4$ atoms

$N$ has $1$ atom.

So, the Empirical formula is $C{H}_{4}N$ .