Question

A sample contains `"21.96% S"` and `"78.04% F"`. What is its empirical formula?

Answer 1

The empirical formula represents the lowest whole-number ratio of elements in a compound.

Since the percentages add up to 100%, you can change the percentages to mass in grams.

`"21.96% S"` => `"21.96g S"`
`"78.04% F"` => `"78.04g F"`

Convert the mass of each element to moles using the molar mass of each element. The molar mass of sulfur = `"32.065g/mol"`. The molar mass of fluorine = `"18.9984032g/mol"`.

`"21.96g S"` x `"1 mol S"/"32.065g S"` = `"0.6849 mol S"`

`"78.04% F"` x `"1 mol F"/"18.9984032g F"` = `"4.108 mol F"`

Divide the number of moles by the lowest number of moles to find the lowest whole-number ratio.

`"S"` =>`"0.6849mol"/"0.6849mol"` = `"1"`

`"F"` => `"4.108mol"/"0.6849mol"` = `"6"`

The empirical formula is `"SF"_6"`.