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Q.
In an organic compound of molar mass 108 g $ mo{{l}^{-1}},C,H $ and N atoms are present in 9 : 1 : 3.5 by weight. Molecular formula can be
Rajasthan PETRajasthan PET 2012
Solution:
Molar mass $ =108\text{ }g\text{ }mo{{l}^{-1}} $
Total part by weight $ =9+1+3.5=13.5
$ Weight of carbon $ =\frac{9}{13.5}\times 108=72\,g $
Number of carbon atoms $ =\frac{72}{12}=6 $
Weight of hydrogen $ =\frac{1}{13.5}\times 108=8g $
Number of hydrogen atoms $ =\frac{8}{1}=8 $
Weight of nitrogen $ =\frac{3.5}{13.5}\times 108=28g $
Number of nitrogen atom
$ =\frac{28}{14}=2 $
Hence, molecular formula $ ={{C}_{6}}{{H}_{8}}{{N}_{2}} $