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

Nov 23, 2014

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.

$\text{21.96% S}$ => $\text{21.96g S}$
$\text{78.04% F}$ => $\text{78.04g F}$

Convert the mass of each element to moles using the molar mass of each element. The molar mass of sulfur = $\text{32.065g/mol}$. The molar mass of fluorine = $\text{18.9984032g/mol}$.

$\text{21.96g S}$ x $\text{1 mol S"/"32.065g S}$ = $\text{0.6849 mol S}$

$\text{78.04% F}$ x $\text{1 mol F"/"18.9984032g F}$ = $\text{4.108 mol F}$

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

$\text{S}$ =>$\text{0.6849mol"/"0.6849mol}$ = $\text{1}$

$\text{F}$ => $\text{4.108mol"/"0.6849mol}$ = $\text{6}$

The empirical formula is $\text{SF"_6}$.