Question

the formula of aluminium sulphide is Al₂S₃. Explain why the formula has a ratio of two aluminium ions for every three sulphide ions.

Answer 1

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 =
`(54.0)/(150.3) * 100 = 35.9%`
% of S (there are 3 S atoms) in Aluminium sulfide =
`(96.3)/(150.3) * 100 = 64.1%`
Assume the percentages to be masses, so:
In 100.0 g of Aluminium sulfide, n (Al) =
`(35.9)/(27.0) = 1.33 mol`
In 100.0 g of Aluminium sulfide, n (S) =
`(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)`