Question

In an organic compound of molar mass, $108gmo{l}^{-1}$, C, H, and N atoms are present in 9: 1: 3.5 by weight. The molecular formula of the compound 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)

$C_6{H}_8{N}_2$

Let the ratio be $x$.
$\therefore$ Molar mass $=9x+1x+3.5x=13.5x=108$

$\therefore x=\frac{108}{13.5}=8$

$\therefore$ Weight of carbon $=9\times 8=72$

Weight of H $=1\times 8=8$

Weight of $N=3.5\times 8=28$

No. of atoms of Carbon $=\frac{72}{12}=6$

No. of atoms of Hydrogen $=\frac{8}{1}=8$

No. of atoms of Nitrogen $=\frac{28}{14}=2$

$\therefore$ Molecular formula $=C_6{H}_8{N}_2$