Question

The molecular formula of a compound having a molecular weight of $92$ and containing $30.5$% of nitrogen and $69.5$% of oxygen by weight is:

• A. $N_2{O}_4$
• B. $N{O}_2$
• C. $N_2{O}_3$
• D. $N_{2}O$

Answer 1

Answer: (A)

$N_2{O}_4$

Since compound contains 30.5 % of nitrogen and 69.5% of oxygen by weight,

$N:O=\frac{30.5}{14}:\frac{69.5}{16}=1:2$

Empirical Formula is $N{O}_2$.

Molecular weight is $92$.

$X=\frac{92}{46}=2$

$\text{Molecular formula}=(\text{Empirical formula}{)}_X$

$\text{Molecular formula}=N_2{O}_4$