Q. An organic compound containing carbon, hydrogen and oxygen contains $52.2 \%$ carbon and $13.04 \%$ hydrogen. Vapour density of the compound is $23 .$ Its molecular formula will be
Organic Chemistry – Some Basic Principles and Techniques
Solution:
Element
C
H
O
Percentage
52.2
13.04
$100-(52.2+13.04)$ $=34.76$
Atomic ratio
$\frac{52.2}{12}=4.35$
$\frac{13.4}{1}=13.04$
$\frac{34.76}{16}=2.17$
Simplest ratio
$\frac{4.35}{2.17}=2$
$\frac{13.04}{2.17}=6$
$\frac{2.17}{2.17}=1$
Simple whole no. ratio
2
6
1
Hence, empirical formula is $C _{2} H _{6} O$.
Empirical formula mass $=2 \times 12+6 \times 1+16=46$
Molecular mass $=2 \times 23=46$
$n=\frac{46}{46}=1$
Molecular formula $=(\text { Empirical formula })_{n}=\left( C _{2} H _{6} O \right)_{1}= C _{2} H _{6} O$
| Element | C | H | O |
|---|---|---|---|
| Percentage | 52.2 | 13.04 | $100-(52.2+13.04)$ $=34.76$ |
| Atomic ratio | $\frac{52.2}{12}=4.35$ | $\frac{13.4}{1}=13.04$ | $\frac{34.76}{16}=2.17$ |
| Simplest ratio | $\frac{4.35}{2.17}=2$ | $\frac{13.04}{2.17}=6$ | $\frac{2.17}{2.17}=1$ |
| Simple whole no. ratio | 2 | 6 | 1 |