In a compound C, H and N atoms are present in 9 : 1 : 3.5 by weight .

Answer: \(\mathrm{C_6H_8N_2}\)

Divide each mass by the atomic mass to get the mole ratio:

• C: \(9/12 = 0.75\)
• H: \(1/1 = 1\)
• N: \(3.5/14 = 0.25\)

Dividing by the smallest (0.25): C : H : N = 3 : 4 : 1. So the empirical formula is \(\mathrm{C_3H_4N}\).

Empirical formula mass \(= 3(12) + 4(1) + 14 = 54\).

\[n = \frac{\text{molecular mass}}{\text{empirical formula mass}} = \frac{108}{54} = 2\]

Molecular formula \(= (\mathrm{C_3H_4N})_2 = \mathrm{C_6H_8N_2}\).