In a compound C, H and N are present in 9: 1: 3. 5 by weight. If...

$∴$ Empirical formula $= {C}_{3}{H}_{4}N$

The empirical formula mass $= 3 \times 12 + 4 \times 1 + 14 = 54g$

Given, molecular mass $= 108g$

Thus, $n = \dfrac{108}{54} = 2$

Hence, molecular formula = (Empirical formula) $\times 2$

\left({C}_{3}{H}_{4}N\right) \times 2 = {C}_{6}{H}_{8}{N}_{2}