Q. Determine the molecular formula of an organic compound containing 49.3% carbon 6.84% hydrogen and its vapour density is $73.$
NTA AbhyasNTA Abhyas 2020Organic Chemistry – Some Basic Principles and Techniques
Solution:
Element Atomic mass Simple ratio C 12 $\frac{49.3}{12} = 4.1$ $\frac{4.1}{2.7} = 1.5 ; \, 1.5 \times 2 = 3$ H 1 $\frac{6.84}{1} = 6.84$ $\frac{6.84}{2.7} = 2.5 \times 2.5$ O 16 $\frac{43.86}{16} = 2.7$ $\frac{2.7}{2.7} = 1 \times 2 = 2$
Empirical formula = $C_{3}H_{5}O_{2}$
Empirical Formula wt. $=12\times 3+1\times 5+16\times 2=73$
Molecular wt $=V.D.\times 2=73\times 2=146$
$n=\frac{M . w t}{E . F . w t}=\frac{146}{73}=2$
Molecular formula $=( E.F )_{ n }=\left( C _{3} H _{5} O _{2}\right)_{2}= C _{6} H _{10} O _{4}$.
| Element | Atomic mass | Simple ratio | |
| C | 12 | $\frac{49.3}{12} = 4.1$ | $\frac{4.1}{2.7} = 1.5 ; \, 1.5 \times 2 = 3$ |
| H | 1 | $\frac{6.84}{1} = 6.84$ | $\frac{6.84}{2.7} = 2.5 \times 2.5$ |
| O | 16 | $\frac{43.86}{16} = 2.7$ | $\frac{2.7}{2.7} = 1 \times 2 = 2$ |