# What is the empirical chemical formula of a compound that is 69.9% of iron and 30.0% oxygen?

Apr 24, 2016

$F {e}_{2} {O}_{3}$

#### Explanation:

We assume there are $100 \cdot g$ of compound, and thus there are:

$\left(i\right)$ $\frac{69.9 \cdot g}{55.85 \cdot g \cdot m o {l}^{-} 1}$ $=$ $1.25 \cdot m o l \cdot F e$, and

$\left(i i\right)$ $\frac{30.0 \cdot g}{15.999 \cdot g \cdot m o {l}^{-} 1}$ $=$ $1.88 \cdot m o l \cdot O$.

If we divide thru by the lowest molar quantity we get a formula of $F e {O}_{1.5}$. Because, by definition the empirical formula is the simplest WHOLE number ratio, we double this result to give $F {e}_{2} {O}_{3}$.