Question

A 0.3999g sample of reagent grade Al2(SO4)3.xH2O was dissolved and analyzed for its aluminum content gravimetrically by precipitation as the aluminum oxinate complex, AL(C9H6NO)3, the precipitate weighing 0.4185g. Calculate the number of molecules of water of crystallization.

Answer 1

Ans. Moles of Aluminum oxinate, Al(C₉H₆NO)₃ produced = Mass / Molar mass

                                                            = 0.4185 g / (459.439879 g/ mol)

                                                            = 0.000910892 mol

Note that 1 mol aluminum oxinate contains 1 mol Al-atom whereas 1 mol aluminum sulfate, Al₂(SO₄)₃.xH₂O contains 2 mol Al-atom. That is, 2 mol aluminum sulfate produces 2 mol aluminum oxinate.

So,

            Moles of Al₂(SO₄)₃.xH₂O in sample = (1/2) x moles of Al(C₉H₆NO)₃ produced

                                                            = (1/2) x 0.000910892 mol

                                                            = 0.000455446 mol

1 mol Al₂(SO₄)₃.xH₂O consists of 1 mol Al₂(SO₄)₃.

So,

Mass of Al₂(SO₄)₃ in sample = Moles x Molar mass

                                                            = 0.000455446 mol x (342.153878 g/ mol)

                                                            = 0.155832574 g

Now,

            Mass of H₂O in sample = (Mass of Al₂(SO₄)₃.xH₂O – Mass of Al₂(SO₄)₃) in sample

                                                = 0.3999 g – 0.155832574 g

                                                = 0.244067426 g

            Moles of water in sample = Mass / Molar mass

                                                = 0.244067426 g / (18.01528 g /mol)

                                                = 0.013547801 mol

# Calculate stoichiometry:

Molar stoichiometry in Al₂(SO₄)₃.xH₂O = Moles of H₂O / Moles of Al₂(SO₄)₃.xH₂O

                                                = 0.013547801 mol / 0.000455446 mol

                                                = 29.746 : 1

                                                = 30 : 1                      (nearest whole number)

That is, there are 30 moles of H₂O every mol of Al₂(SO₄)₃.xH₂O

Hence,

            Molecules of water of crystallization, X = 30

            And, the compound is - Al₂(SO₄)₃.30H₂O