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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