# Analysis of a compound shows that it contains 7.0 g nitrogen and 16.0 g oxygen. What is its empirical formula?

Oct 23, 2016

$N {O}_{2}$

#### Explanation:

The empirical formula is the simplest whole number that defines constituent atoms in a species.

Using the given masses, and the atomic masses of each constituent, we define an empirical formula this way:

$\text{Moles of nitrogen}$ $=$ $\frac{7.0 \cdot g}{7.0 \cdot g + 16 \cdot g} \times \frac{1}{14.01 \cdot g \cdot m o {l}^{-} 1} = 0.0217 \cdot m o l$.

$\text{Moles of oxygen}$ $=$ $\frac{16.0 \cdot g}{7.0 \cdot g + 16 \cdot g} \times \frac{1}{15.999 \cdot g \cdot m o {l}^{-} 1} = 0.0435 \cdot m o l$.

And thus we have the molar quantities of nitrogen, and oxygen respectively. To access the empirical formula, we simply divide thru by the LOWEST molar quantity, that of nitrogen:

$N : \frac{0.0217 \cdot m o l}{0.0217 \cdot m o l} = 1$

$O : \frac{0.0435 \cdot m o l}{0.0217 \cdot m o l} = 2$

And thus the empirical formula is $N {O}_{2}$.

Note that to find the molecular formula we need an estimate of molecular mass.