Molar mass of Al = 27.0 g/mol
Molar mass of S = 32.1 g/mol
Molar mass of Aluminium sulfide = 27.0 x 2 + 32.1 x 3 = 150.3 g/mol
% of Al (there are 2 Al atoms) in Aluminium sulfide = \(\frac{54.0}{150.3} * 100 = 35.9%\)
% of S (there are 3 S atoms) in Aluminium sulfide = \(\frac{96.3}{150.3} * 100 = 64.1%\)
Assume the percentages to be masses, so:
In 100.0 g of Aluminium sulfide, n (Al) = \(\frac{35.9}{27.0} = 1.33 mol\)
In 100.0 g of Aluminium sulfide, n (S) = \(\frac{64.1}{32.1} = 2.00 mol\)
Convert values to a molar ratio,
1.33 : 2.00
Divide by smallest ratio value,
1 : 1.50
Multiply each ratio value by a factor of 2 to eliminate the decimal,
2 : 3
Hence why the empirical formula of Aluminium sulfide is \(Al_{2} S_{3}\)