Q. A compound having mol., mass $=78 $ contains $ C=92.31% $ and $ H=7.69%. $ Its molecular formula is
Rajasthan PMTRajasthan PMT 2003Some Basic Concepts of Chemistry
Solution:
The empirical formula of compound can be calculated as
Element
Percentage
Percentage/At.wt.
Ratio
C
92.31
92.31/12=7.7
1
H
7.69
7.69/1 = 7.69
1
Thus the empirical formula is $CH$
The empirical formula wt $=12+1=13$
Value of ' $n'=\frac{\text { Molecular wt }}{\text { Empirical formula wt }}$
$=\frac{78}{13}=6$
Molecular formula $=$ (Empirical formula) $ \times n$
$=( CH ) \times 6= C _{6} H _{6}$
| Element | Percentage | Percentage/At.wt. | Ratio |
|---|---|---|---|
| C | 92.31 | 92.31/12=7.7 | 1 |
| H | 7.69 | 7.69/1 = 7.69 | 1 |