3.05g of hydrated copper sulfate produces 1.94g of anhydrous copper sulfate.Assuming complete removal of all water of crystallisation,determine the formula of the hydrated copper sulfate
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CuSO4.xH2O --> CuSO4 + xH2O
3.05 g.....................1.94 g.......y
g H2O = 3.05 - 1.94 = 1.11 g
mols CuSO4 = 1.94/159.5 = 0.0122
mols H2O = 1.11/18.01 = 0.0616
Convert these numbers to mols H2O for 1 mol CuSO4. That's
mols CuSO4 = 0.0122/0.0122 = 1.000
mols H2O = 0.0616/0.0122 = 5.04 which rounds to 5 in whole numbers so the formula is CuSO4.5H2O
3.05 g.....................1.94 g.......y
g H2O = 3.05 - 1.94 = 1.11 g
mols CuSO4 = 1.94/159.5 = 0.0122
mols H2O = 1.11/18.01 = 0.0616
Convert these numbers to mols H2O for 1 mol CuSO4. That's
mols CuSO4 = 0.0122/0.0122 = 1.000
mols H2O = 0.0616/0.0122 = 5.04 which rounds to 5 in whole numbers so the formula is CuSO4.5H2O
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