# A compound that contains only nitrogen and oxygen is 30.4% N by mass; the molar mass of the compound is 92 g/mol. What is the empirical formula and the molecular formula of the compound?

Nov 9, 2015

Empirical formula is $N {O}_{2}$, and the molecular formula is ${N}_{2} {O}_{4}$, dinitrogen tetroxide.

#### Explanation:

In $100 \cdot g$ of this compounds there are $30.4 \cdot g$ $N$, and the balance $O$. We divide thru by the atomic masses in order to approach the empirical formula: $\frac{30.4 \cdot g}{14.01 \cdot g \cdot m o {l}^{-} 1}$ $=$ $2.17 \cdot m o l \cdot N$;

$\frac{69.6 \cdot g}{16.0 \cdot g \cdot m o {l}^{-} 1}$ $=$ $4.35 \cdot m o l \cdot O$.

After we divide thru by 2.17, we get an empirical formula of $N {O}_{2}$. And as I (and many others) have said before, the empirical formula $\text{is the simplest whole number ratio defining constituent atoms}$
$\text{in a species.}$

Now the molecular formula is always a multiple of the empirical formula; i.e. $M F = {\left(E F\right)}_{n}$, or $92.4 \cdot g \cdot m o {l}^{-} 1$ $=$ $2 \times \left(14.01 + 2 \times 16.00\right) \left(g \cdot m o {l}^{-} 1\right)$. So $n$ $=$ $2$, and the molecular formula is ${N}_{2} {O}_{4}$, i.e. dinitrogen tetroxide. the dimer of $N {O}_{2}$ radical.