Question

The aluminum in a 1.258-g sample of impure ammonium aluminum sulfate was precipitated with aqueous ammonia as the hydrous Al2O3 . XH2O The precipitate was filtered and ignited at 1000°C to give anhydrous Al2O3, which weighed 0.1060 g. Express the result of this analysis in terms of:

% NH4 Al(SO4)2 = %

% Al2O3= %

% Al = %

Answer 1

molar mass of NH₄(AlSO₄)₂ = 237.1452g/mol

molar mass of Al₂O₃ =101.96g/mol

molar mass of Al =26.9815g/mol

first calculate gram of each

gram of Al₂O₃ = 0.1060 gm (given)

gram of NH₄(AlSO₄)₂ = According to molar masses 101.96 gm of Al₂O₃ = 237.1452 gm of NH₄(AlSO₄)₂then 0.1060 gram of Al₂O₃ = 0.1060237.1452/101.96 = 0.24654 gm

gram of NH₄(AlSO₄)₂ = 0.24654 gm

gram of Al = According to molar masses 101.96 gm of Al₂O₃ = 26.9615 gm of Al then 0.1060 gram of Al₂O₃=

0.106026.9815/101.96 = 0.02805 gm

gram of Al = 0.02805 gm

% calculation

% of NH₄(AlSO₄)₂ =

1.258 gm = 100% then 0.24654 gm = 0.24654100/1.258 = 19.5977 %

% of NH₄(AlSO₄)₂ = 19.5977 %

% of Al₂O₃ =

1.258 gm = 100% then 0.1060 gm = 0.1060100/1.258 = 8.426 %

% of Al₂O₃ = = 8.426 %

% of Al =

1.258 gm = 100% then 0.02805 gm = 0.02805100/1.258 = 2.229 %

% of Al = 2.229 %