Q. In a compound $ C,\,\,H $ and $ N $ are present in $ 9:13.5 $ by weight. If molecular weight of the compound is $ 108 $ , then the molecular formula on the compound is
Rajasthan PMTRajasthan PMT 2007Organic Chemistry – Some Basic Principles and Techniques
Solution:
C
H
N
9
1
3.5
9/12 = 0.75
1/1=1
3.5/14 = 0.25
$\frac{0.75}{0.25} = 3$
$\frac{1}{0.25} = 4$
$\frac{0.25}{0.25} = 1$
So, empirical formulaa $= C_3H_4N$
$n = \frac{108}{54} = 2$
Molecular formula $ = (C_3H_4N)_2 = C_6H_8N_2$
C | H | N |
---|---|---|
9 | 1 | 3.5 |
9/12 = 0.75 | 1/1=1 | 3.5/14 = 0.25 |
$\frac{0.75}{0.25} = 3$ | $\frac{1}{0.25} = 4$ | $\frac{0.25}{0.25} = 1$ |