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 can be:

• A. $C_6{H}_8{N}_2$
• B. $C_7{H}_{11}N$
• C. $C_5{H}_6{N}_3$
• D. $C_4{H}_{18}{N}_4$

Answer 1

The correct option is A $C_6{H}_8{N}_2$

Let the common multiple variable presents in each is $x$

The number of atoms of C (carbon), H(Hydrogen) and N(Nitrogen) is $9x,x,03.5x.$

$9x+x+3.5x=108$ (molar mass of compound)

$13.5x=108$

$x=8gm$

Mass of carbon(C)$=9x=\left(9\right)\left(8\right)=72$

mass of Hydrogen (H)$=1x=8$

Mass of nitrogen (N)$=3.5x=28$

Number of carbon $=\frac{totalmass}{molarmass}=\frac{72}{12}=6$

Number of Hydrogen $=\frac{totalmass}{molarmass}=\frac{8}{1}=8$

Number of Nitrogen $=\frac{totalmass}{molarmass}=\frac{28}{14}=2$

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

Hence, the correct option is $\text{A}$