Question

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

• A. $C_6{H}_8{N}_2$
• B. $C_7{H}_{10}N$
• C. $C_5{H}_6{N}_3$
• D. $C_4{H}_{18}{N}_3$

Answer 1

Answer: (A)

$C_6{H}_8{N}_2$

The molar mass is 108 g /mol.

The mass of C present in one mole (108 g) of compound $=\frac{108\times 9}{9+1+3.5}=72$ g.

The mass of H present in one mole (108 g) of compound $=\frac{108\times 1}{9+1+3.5}=8$ g.

The mass of N present in one mole (108 g) of compound $=\frac{108\times 3.5}{9+1+3.5}=28$ g.

The number of moles of C $=\frac{72g}{12g/mol}=6mol$.

The number of moles of H $=\frac{8g}{1g/mol}=8mol$.

The number of moles of N $=\frac{28g}{14g/mol}=2mol$.

The mole ratio $C:H:N=6:8:2$

Molecular formula can be $C_6{H}_8{N}_2$