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

##### 1 Answer

#### Answer:

#### Explanation:

Your strategy here will be to

pick a sample of this ionic compounduse themolar massesof iron and of sulfur to find how manymolesof each it containsfind thesmallest whole number ratiothat exists between iron and sulfur in the compound

To make the calculations easier, pick a **percent composition**

#"53.73 g Fe"# #"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

#"For Fe: " 2 xx 1 = 2#

#"For S: " 2 xx 1.52 = 3.04 ~~ 3#

The **empirical formula** of the unknown compound will thus be

#color(green)(bar(ul(|color(white)(a/a)color(black)("Fe"_2"S"_3)color(white)(a/a)|)))#