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

1 Answer
Meave60
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.

#"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"#.

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