Question

What is the empirical formula for a compound containing $36.7$ $N$ and $63.3$ $O$?

• A. $N_2{O}_5$
• B. $N_2{O}_3$
• C. ${NO}_2$
• D. $NO$
• E. $N_{2}O$

Answer 1

The correct option is B $N_2{O}_3$

The compound contains 36.7% $N$ and 63.3% $O$.
100 g of compound will contain $36.7g$ $N$ and $63.3g$ $O$. The atomic masses of $N$ and $O$ are 14 g/mol and 16 g.mol respectively.
The number of moles of $N$ $=\frac{36.7}{14}=2.62$

The number of moles of $O$ $=\frac{63.3}{16}=3.98$

The mole ratio $N:O=2.62:3.98=1:1.5=2:3$
Hence, the empirical formula is $N_2{O}_3$