Question #ba207

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

Let us assume that we started with 100 g of compound. On analysis it gives 25.24 % or 25.24 g of Sulfur, and 74.76% or 74.76g of fluorine.

#"25.24% S"# => #"25.24 g S"#
#"74.76% F"# => #"74.76 g 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"#.

#"25.24 g S"# x #"1 mol S"/"32.065g S"# = #"0.787 mol S"#

#"74.76 g F"# x #"1 mol F"/"19 g F"# = #"3.937 mol F"#

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

#"S"# =># "0.787mol"/"0.787mol"# = #"1"#

#"F"# => #"3.937mol"/"0.787mol"# = #"5"#

The empirical formula is #"SF"_5"#.

The empirical formula is #"SF"_5"# .The substance has empirical formula mass: One mole of S has mass 32.06 g /mol and 5 moles of F has mass 5x 19 g/mol = 95 g/mol

Empirical formula mass is 32.06 g /mol + 95 g /mol = 127.06 g/mol
Molecular formula mass is given as 254.1 g /mol.

Calculating n = Molecular formula mass / Empirical formula mass

n =254.1 g/mol / 127.06 g/mol

n = 2

Molecular formula is (#"SF"_5"#)2

which is #S_2# # F_10#