Answer to Question #75596 in General Chemistry for Anaeis Alexander

Question

0.8750 gram of amuminum was heated in a stream of chlorine gas to give 4.3250 grams of aluminum cloride. find the empirical formula of aluminum chloride

Answer 1

0.8750 g Al / 26.98 g/mol = 0.0324 mol Al atoms
4.3250 g – 0.8750 g = 3.45 g Cl atoms
3.45 g / 35.453 g/mol = 0.0973 mol Cl atoms
Divide both by the smallest.
Al: 0.0324 / 0.0324 = 1
Cl: 0.0973 / 0.0324 = 3
AlCl3 - the empirical formula of aluminum chloride.