MCQ
An organic compound containing $C, H$ and $N$ gave following analysis : $C = 40\%$, $H = 13.33\%$ and $N = 46.67\%$. Its empirical formula would be
- A${C_2}{H_7}{N_2}$
- B$C{H_5}N$
- ✓$C{H_4}N$
- D${C_2}{H_7}N$
|
$C = 40\%$ |
$40/12$ |
$3.33$ |
$1$ |
|
$H = 13.33\%$ |
$13.33/1$ |
$13.33$ |
$4$ |
|
$N = 46.67\%$ |
$46.67/14$ |
$3.33$ |
$1$ |
Thus formula $C{H_4}N$.
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