Q. An organic compound contains $C =40 \%, H =13.33 \%$ and $N =46.67 \%$. Its emperical formula is
Solution:
Element
Percent by mass
Atomic mass
Relative number of moles of elements
Simplest molar ratio
C
40
12
$\frac{40}{12}=3.33$
$\frac{3.33}{3.33}=1$
H
13.33
1
$\frac{13.33}{1}=13.33$
$\frac{13.33}{3.33}=4$
N
46.67
14
$\frac{46.67}{14}=3.33$
$\frac{3.33}{3.33}=1$
Hence, emperical ratio is $CH_4N$
| Element | Percent by mass | Atomic mass | Relative number of moles of elements | Simplest molar ratio |
|---|---|---|---|---|
| C | 40 | 12 | $\frac{40}{12}=3.33$ | $\frac{3.33}{3.33}=1$ |
| H | 13.33 | 1 | $\frac{13.33}{1}=13.33$ | $\frac{13.33}{3.33}=4$ |
| N | 46.67 | 14 | $\frac{46.67}{14}=3.33$ | $\frac{3.33}{3.33}=1$ |