Question

In an organic compound of molar mass 108g${mol}^{-1}$ C, H and N atoms are present in 9 : 1 : 3.5 by wegith. Moleculer 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)

$C_6{H}_8{N}_2$

Mass of Carbon present in the compound$=\frac{9}{9+1+3.5}\times 108=72g$

Moles of $C=\frac{72}{12}=6$

Mass of Hydrogen present in the compound$=\frac{1}{9+1+3.5}=8g$

Moles of $H=\frac{8}{1}=8$

Mass of Nitrogen present in the compound$=\frac{3.5}{9+1+3.5}=28$

Moles of $N=\frac{28}{14}=2$

Hence, the enpirical formula of the compound is $C_6{H}_8{N}_2$