Question

An oxide of nitrogen gave the following percentage composition,
$N=25.94\%$ and $=74.06\%$.
The empirical formula of the compound is:

• A. $N_2{O}_2$
• B. $N{O}_3$
• C. $N_2{O}_5$
• D. $N_2{O}_3$

Answer 1

Answer: (C)

$N_2{O}_5$

Convert the masses to moles:
$N=\frac{25.94}{14}=1.85$ mol

$O=\frac{74.06}{16}=4.62$ mol
Divide by the lowest, seeking the smallest whole-number ratio:
$N=\frac{1.85}{1.85}=1$

$O=\frac{4.62}{1.85}=\frac{5}{2}$
Hence, emperical formulae is $N_2{O}_5$.