Q. An organic compound containing $C$ and $H$ gave the following analysis $C =40 \%, H =6.7 \%$. Its empirical formula would be
Some Basic Concepts of Chemistry
Solution:
$\%$ age
Atomic mass
Moles
Simple ratio
$C$
$40\%$
$12$
$\frac{40}{12}= 3.33$
$1$
$H$
$6.7\%$
$1$
$\frac{6.7}{1}= 6.7$
$2$
$O$
$53.3\%$
$16$
$\frac{53.3}{16}= 3.33$
$1$
$\therefore EF = CH _{2} O$
| $\%$ age | Atomic mass | Moles | Simple ratio | |
|---|---|---|---|---|
| $C$ | $40\%$ | $12$ | $\frac{40}{12}= 3.33$ | $1$ |
| $H$ | $6.7\%$ | $1$ | $\frac{6.7}{1}= 6.7$ | $2$ |
| $O$ | $53.3\%$ | $16$ | $\frac{53.3}{16}= 3.33$ | $1$ |