# What is the empirical formula of a compound that contains 25.92 percent nitrogen and 74.07 percent oxygen?

Dec 19, 2016

${N}_{2} {O}_{5}$

#### Explanation:

As with all these problems we assume a $100 \cdot g$ mass of compound. And then we divide the individual elemental masses by the ATOMIC masses of each constituent.

$\text{Moles of nitrogen}$ $=$ $\frac{25.92 \cdot g}{14.01 \cdot g \cdot m o l} = 1.85 \cdot m o l$.

$\text{Moles of oxygen}$ $=$ $\frac{74.07 \cdot g}{15.999 \cdot g \cdot m o l} = 4.62 \cdot m o l$.

And we divide thru the individual molar quantities by the smallest such quantity, that of nitrogen.

N:(1.85*mol)/(1.85*mol)=1; O:(4.62*mol)/(1.85*mol)=2.50

Because, by definition, empirical formulae are the simplest WHOLE number ratio definining constituent atoms in a species, we double this ratio to get:

${N}_{2} {O}_{5}$, $\text{dinitrogen pentoxide}$. Of course, this is an empirical formula; we would need a molecular weight determination before we declared that ${N}_{2} {O}_{5}$ was the mollykule.