Q.
Two organic compounds $A$ and $B$ both containing $C$ and $H$ yield on analysis, the same percentage composition by weight,
$C =\frac{12}{13} \times 100 ; H =\frac{1}{13} \times 100$
Which among the following will be the correct formula of the compounds ' $A$ ' and ' $B$ '? (' $A$ ' decolourises bromine water but ' $B$ ' does not.)
Organic Chemistry – Some Basic Principles and Techniques
Solution:
Element
Percentage
At. Mass
Relative Number of Atoms
Simplest Ratio
C
$\frac{12}{13} \times 100=92.30$
12
$\frac{92.30}{12}=7.69$
1
H
$\frac{1}{13} \times 100=7.69$
1
$\frac{7.69}{1}=7.69$
1
Thus, empirical formula $= CH$
The two compounds which satisfy this empirical formula are $C _{2} H _{2}$ and $C _{6} H _{6}$
As ' $A$ ' decolourises bromine water thus ' $A$ ' should be $C _{2} H _{2}$ and ' $B$ ' does not decolourise bromine water thus ' $B$ ' should be $C _{6} H _{6}$
| Element | Percentage | At. Mass | Relative Number of Atoms | Simplest Ratio |
|---|---|---|---|---|
| C | $\frac{12}{13} \times 100=92.30$ | 12 | $\frac{92.30}{12}=7.69$ | 1 |
| H | $\frac{1}{13} \times 100=7.69$ | 1 | $\frac{7.69}{1}=7.69$ | 1 |