# What is the empirical formula of a compound that contains 53.73% Fe and 46.27% of S?

Sep 5, 2016

${\text{Fe"_2"S}}_{3}$

#### Explanation:

Your strategy here will be to

• pick a sample of this ionic compound
• use the molar masses of iron and of sulfur to find how many moles of each it contains
• find the smallest whole number ratio that exists between iron and sulfur in the compound

To make the calculations easier, pick a $\text{100-g}$ sample of this unknown compound. This sample with contain, as given by the known percent composition

• $\text{53.73 g Fe}$
• $\text{46.27 g S}$

Convert these masses to moles

"For Fe: " 53.73 color(red)(cancel(color(black)("g"))) * "1 mole Fe"/(55.845color(red)(cancel(color(black)("g")))) = 0.962

"For S: " 46.27 color(red)(cancel(color(black)("g"))) * "1 mole S"/(32.066color(red)(cancel(color(black)("g")))) = 1.44

Next, divide both values by the smallest one

"For Fe: " (0.95 color(red)(cancel(color(black)("moles"))))/(0.95color(red)(cancel(color(black)("moles")))) = 1

"For S: " (1.44 color(red)(cancel(color(black)("moles"))))/(0.95color(red)(cancel(color(black)("moles")))) = 1.52

Now, to find the smallest whole number ratio that exists between the two elements in the compound, multiply both values by $2$. You will have

$\text{For Fe: } 2 \times 1 = 2$

$\text{For S: } 2 \times 1.52 = 3.04 \approx 3$

The empirical formula of the unknown compound will thus be

$\textcolor{g r e e n}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{\text{Fe"_2"S}}_{3}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$