Question

What is the empirical formula for a compound containing 63.8% $N$ and 36.2% $O$?

• 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 E $N_{2}O$

The compound contains $63.8$% $N$ and $36.2$% $O$.
100 g of compound will contain $63.8gN$ and $36.2gO$.
The atomic masses of $N$ and $O$ are $14$ g/mol and $16$ g/mol respectively.
The mass is divided with atomic mass to get number of moles.

The number of moles of $N$ $=\frac{63.8}{14}=4.6$ moles.

The number of moles of $O$ $=\frac{36.2}{16}=2.3$ moles.

The mole ratio $N:O$ is $4.6:2.3$ or $2:1$.
Hence, the empirical formula is $N_{2}O$.