Q. An organic compound on analysis was found to contain $10.06 \%$ carbon, $0.84 \%$ hydrogen and $89.10 \%$ chlorine. What will be the empirical formula of the substance?
Some Basic Concepts of Chemistry
Solution:
Element
$\%$
Atomic mass
Molar ratio
Simpler molar ratio
$C$
$10.06\%$
$12$
$\frac{10.06}{12}=0.84$
$\frac{0.84}{0.84}=1$
$H$
$0.84\%$
$1$
$\frac{0.84}{1}=0.84$
$\frac{0.84}{0.84}=1$
$Cl$
$89.10\%$
$35.5$
$\frac{89.10}{35.5}=2.5$
$\frac{2.5}{0.84}=3$
Thus, the empirical formula of the substance is $CHCl _{3}$.
| Element | $\%$ | Atomic mass | Molar ratio | Simpler molar ratio |
|---|---|---|---|---|
| $C$ | $10.06\%$ | $12$ | $\frac{10.06}{12}=0.84$ | $\frac{0.84}{0.84}=1$ |
| $H$ | $0.84\%$ | $1$ | $\frac{0.84}{1}=0.84$ | $\frac{0.84}{0.84}=1$ |
| $Cl$ | $89.10\%$ | $35.5$ | $\frac{89.10}{35.5}=2.5$ | $\frac{2.5}{0.84}=3$ |