Q. $ 64\,g $ of an organic compound contains $ 24\,g $ of carbon, $ 8 \,g $ of hydrogen and the rest oxygen. The empirical formula of the compound is
Chhattisgarh PMTChhattisgarh PMT 2007
Solution:
Weight of organic compound $=64\, g$
Weight of carbon $=24\, g$
Weight of liydrogen $=8 \,g$
Weight of oxygen $=64-[24+8] \,g$
$=32 \,g$ Percentage of carbon $=\frac{24}{64} \times 100$
$\therefore $ Percentage of carbon $=37.5$
Percentage of hydrogen $=\frac{8}{64} \times 100$
$\therefore $ Percentage of hydrogen $=12.5$
Percentage of oxygen $=\frac{32}{64} \times 100$
$\therefore $ Percentage of oxygen $=50$
$\%$ of element
$C$
$H$
$O$
$37.5$
$12.5$
$50$
Mole
$\frac{37.5}{12}$
$\frac{12.5}{1}$
$\frac{50}{16}$
Relative number of atoms
$3.125$
$12.5$
$3.125$
Simplest atomic ratio
$\frac{3.125}{3.125}$
$\frac{12.5}{3.125}$
$\frac{3.125}{3.125}$
$=1$
$=4$
$=1$
| $\%$ of element | $C$ | $H$ | $O$ |
|---|---|---|---|
| $37.5$ | $12.5$ | $50$ | |
| Mole | $\frac{37.5}{12}$ | $\frac{12.5}{1}$ | $\frac{50}{16}$ |
| Relative number of atoms | $3.125$ | $12.5$ | $3.125$ |
| Simplest atomic ratio | $\frac{3.125}{3.125}$ | $\frac{12.5}{3.125}$ | $\frac{3.125}{3.125}$ |
| $=1$ | $=4$ | $=1$ |