In an organic compound of molar mass 108g ${m o l}^{- 1}$ C, H and N atoms are present in 9 : 1 : 3.5 by wegith. Moleculer formula can be:

C_{6} H_{8} N_{2}

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C_{7} H_{10} N

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C_{5} H_{6} N_{3}

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C_{4} H_{18} N_{3}

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Solution

The correct option is B $C_{6} H_{8} N_{2}$
Mass of Carbon present in the compound $= \frac{9}{9 + 1 + 3.5} \times 108 = 72 g$
Moles of $C = \frac{72}{12} = 6$

Mass of Hydrogen present in the compound $= \frac{1}{9 + 1 + 3.5} = 8 g$
Moles of $H = \frac{8}{1} = 8$

Mass of Nitrogen present in the compound $= \frac{3.5}{9 + 1 + 3.5} = 28$
Moles of $N = \frac{28}{14} = 2$

Hence, the enpirical formula of the compound is $C_{6} H_{8} N_{2}$

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