Q. In a compound $ C=40\% $ and $ H=6.66\%, $ what is the empirical formula of the compound?
Haryana PMTHaryana PMT 2001
Solution:
Symbol
$\%$
Atomic mass
Relative no. of atoms
Simple Reaction
$C$
$40$
$12$
$\frac{40}{12}=3.33$
$\frac{3.33}{3.33}=1$
$H$
$6.66$
$1$
$\frac{6.66}{1}=6.66$
$\frac{6.66}{3.33}=2$
$O$
$53.34$
$16$
$\frac{53.34}{16} 3.33$
$\frac{3.33}{3.33}=1$
| Symbol | $\%$ | Atomic mass | Relative no. of atoms | Simple Reaction |
|---|---|---|---|---|
| $C$ | $40$ | $12$ | $\frac{40}{12}=3.33$ | $\frac{3.33}{3.33}=1$ |
| $H$ | $6.66$ | $1$ | $\frac{6.66}{1}=6.66$ | $\frac{6.66}{3.33}=2$ |
| $O$ | $53.34$ | $16$ | $\frac{53.34}{16} 3.33$ | $\frac{3.33}{3.33}=1$ |