Q. If a compound, on analysis was found to contain $C =18.5 \%, H =1.55 \%, Cl =55.04 \%$ and $O =$ $24.81 \%$, then its empirical formula is
Solution:
Elements % At. mass Relative no. of atom Simplest ratio
of atoms C 18.5 12 $\frac{18.5}{12}=1.542$ 1 H 1.55 1 $\frac{1.55}{1}=1.55$ 1 Cl 55.04 35.5 $\frac{55.04}{35.5}=1.55$ 1 O 24.81 16 $\frac{24.16}{16}=1.538$ 1
Therefore, empirical formula of the compound is CHClO.
| Elements | % | At. mass | Relative no. of atom | Simplest ratio of atoms |
|---|---|---|---|---|
| C | 18.5 | 12 | $\frac{18.5}{12}=1.542$ | 1 |
| H | 1.55 | 1 | $\frac{1.55}{1}=1.55$ | 1 |
| Cl | 55.04 | 35.5 | $\frac{55.04}{35.5}=1.55$ | 1 |
| O | 24.81 | 16 | $\frac{24.16}{16}=1.538$ | 1 |