Q. An organic compound contains carbon, hydrogen and oxygen. Its elemental analysis gave C, 38.71% and H, 9.67%. The empirical formula of the compound would be
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Solution:
Element % Atomic mass mole ratio simple ratio $C$ 38.71 $12$ $\frac{38.71}{12}=3.22$ $\frac{3.22}{3.22}=1$ $H$ 9.67 $1$ $\frac{9.67}{1}=9.67$ $\frac{9.67}{3.22}=3$ $O$ 51.62 $16$ $\frac{51.62}{16}=3.22$ $\frac{3.22}{3.22}=1$
Hence empirical formula of the compound would be $CH_{3}O$ .
| Element | % | Atomic mass | mole ratio | simple ratio |
|---|---|---|---|---|
| $C$ | 38.71 | $12$ | $\frac{38.71}{12}=3.22$ | $\frac{3.22}{3.22}=1$ |
| $H$ | 9.67 | $1$ | $\frac{9.67}{1}=9.67$ | $\frac{9.67}{3.22}=3$ |
| $O$ | 51.62 | $16$ | $\frac{51.62}{16}=3.22$ | $\frac{3.22}{3.22}=1$ |