Answer to Question #324863 in General Chemistry for Alfie

Question

On analysis a 0.500 g sample of aluminium chloride is found to contain 0.101 g of aluminium. Determine the emperical formula of aluminium chloride

Answer 1

m (Cl) = 0.500 - 0.101 = 0.399 g

n = m / M

M (Al) = 27 g/mol

M (Cl) = 35.5 g/mol

n (Al) = 0.101 / 0.27 = 0.004 mol

n (Cl) = 0.399 / 35.5 = 0.011 mol

Cl : Al = 0.011 / 0.004 = 3

The emperical formula of aluminium chloride is AlCl₃.