Asked by Viniana
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
Answers
Answered by
Viniana
Still haven't received my answer to my questions
Answered by
DrBob222
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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