Question

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

Answer 1

`"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 `"100-g"` sample of this unknown compound. This sample with contain, as given by the known 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 `2`. You will have

`"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)|)))`