Question

An organic compound on analysis gave $C=42.8$%, $H=7.20$%, and $N=50$%. Volume of $1$ g of the compound was found to be $200$ mL at STP. Molecular formula of the compound is :

• A. $C_4{H}_8{N}_4$
• B. $C_{16}{H}_{32}{N}_{16}$
• C. $C_{12}{H}_{24}{N}_{12}$
• D. $C_2{H}_4{N}_2$

Answer 1

The correct option is A $C_4{H}_8{N}_4$

The ratio of the number of moles of atoms in the compound is given below:
$\begin{array}{ccccc}C& :& H& :& N\\ \frac{42.8}{12}& :& \frac{7.2}{1}& :& \frac{50}{14}\\ 3.56& :& 7.2& :& 3.57\\ 1& :& 2& :& 1\end{array}$
Therefore, the empirical formula is $C{H}_{2}N$.
$200$ ml = $1$ g.
$22400$ ml $=\frac{22400}{200}=112g$
Molecular weight = $112$ g
Empirical Formula weight = $C{H}_{2}N=12+2+14=28$ g
$n=\frac{\text{Molecular weight}}{\text{Empirical weight}}=\frac{112}{28}=4$
The molecular formula is $C_4{H}_8{N}_4$.