Question

In an organic compound with molar mass 108 g, C, H, and N are present in the ratio 9: 1: 3.5 by mass. Molecular formula of the compound is:

• A. $C_6{H}_8{N}_2$
• B. $C_7{H}_{10}N$
• C. $C_5{H}_6{N}_3$
• D. $C_4{H}_5{N}_3$

Answer 1

Answer: (A)

$C_6{H}_8{N}_2$

Given,

Molar mass of compound=$108g$

$C:H:N=9:1:3.5$

$\Rightarrow$ Mass of $1$ mole of compound=$108g$

Now, Let in $1$ mole

Mass of $C$ be $9x$

Mass of $H$ be $1x$; Mass of $N$ be $3.5x$

$\Rightarrow 9x+x+3.5x=108$

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

$\Rightarrow$ Mass of $C$ in $1$ mole compound=$9x=72g$

Mass of $H$ is $x=8g$

Mass of $N$ is $3.5x=28g$

$\therefore$ We know Mass of $1$ mole $C$ atom=$12g$

$\therefore$ No. of moles of $C$ atom in $72g=6$

$\Rightarrow 1$ mole $H$ atom=$1g$

$\Rightarrow 8g=8$ mole $H$ atom

$\Rightarrow 1$ mole $N$ atom=$14g$

Moles of $N$ atom in $28=2$

$\therefore$ Molecular Formula is $C_6{H}_8{N}_2$.