# A compound is found to contain 53.70% iron and 46.30% sulfur. What is its empirical formula?

Feb 13, 2016

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

#### Explanation:

In 100 g of the compound, there are 53.7 g iron, and 46.3 g of sulfur.

We divide thru by the ATOMIC masses of each constituent:

$F e :$$\frac{53.7 \cdot \cancel{g}}{55.85 \cdot \cancel{g} \cdot m o {l}^{-} 1}$ $=$ $0.961$ $m o l$.

$S :$$\frac{46.3 \cdot \cancel{g}}{32.06 \cdot \cancel{g} \cdot m o {l}^{-} 1}$ $=$ $1.44$ $m o l$.

Now we divide thru by the LOWEST molar quantity to give...

$F e : S$ $=$ $1 : 1.50$.

Now, the empirical formula is, by definition, the simplest $\text{WHOLE number ratio}$ that defines constituent atoms in a species. To get whole numbers, clearly we have to multiply the given ratio by 2, to give (FINALLY):

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