The compund consists of 35.98% aluminium and 64.02% sulfur, so in a 100 gram sample there's 35.98 grams of aluminium and 64.02 grams sulfur.
From those, the amount of substances are:
n(Al)= 35.98 g/ 26.98 g/mol ≈ 1.33 mol and n(S) = 64.02 g / 32.06 g/mol ≈ 2.00 mol
Then we divide these by the smallest amount (1.33):
1.33 / 1.33 = 1
2.00 / 1.33 ≈ 1.5
Empirical compund is therefore Al2S3.
