Q. An organic compound contains $49.3\%$ carbon, $6.84\%$ hydrogen and its vapour density is $73$. Molecular formula of the compound is
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Solution:
Element
$\%$
Relative number of atoms
Simplest ratio
C
49.3
$\frac{49.3}{12}=4.10$
$\frac{4.10}{2.74} =1.5 \times 2=3$
H
6.84
$\frac{6.84}{1}=6.84$
$\frac{6.84}{2.74} =2.5 \times 2=5$
O
43.86
$\frac{43.86}{16}=2.74$
$\frac{2.74}{2.74} = 1 \times2=2$
$\therefore $ The empirical formula is $C _{3} H _{5} O _{2}$
Empirical formula weight
$=12 \times 3+1 \times 5+16 \times 2 $
$=73$
Molecular weight of the compound
$=2 \times$ vapour density
$=2 \times 73=146 $
. $n=\frac{\text { molecular weight }}{\text { empirical formula weight }} $
$=\frac{146}{73}=2$
Molecular formula $=$ empirical formula $\times 2$
$=2\left( C _{3} H _{5} O _{2}\right)= C _{6} H _{10} O _{4}$
| Element | $\%$ | Relative number of atoms | Simplest ratio |
|---|---|---|---|
| C | 49.3 | $\frac{49.3}{12}=4.10$ | $\frac{4.10}{2.74} =1.5 \times 2=3$ |
| H | 6.84 | $\frac{6.84}{1}=6.84$ | $\frac{6.84}{2.74} =2.5 \times 2=5$ |
| O | 43.86 | $\frac{43.86}{16}=2.74$ | $\frac{2.74}{2.74} = 1 \times2=2$ |