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chemistry

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If a 8.6195 g of a mixture which contains (20.0 % of CuSO4.XH2O and 80.0 % of sand) was heated in
a crucible then cooled in the desiccator the mass obtained was 7.9976 g.
a- Find X the number of water of hydration in the Hydrate (CuSO4.XH2O)
(MM of CuSO4 = 159.5 g/mol and MM of H2O = 18.00 g/mol).

  • chemistry -

    please help me ,becauce i have final exam tomorrow

  • chemistry -

    8.6195 = g CuSO4.xH2O + sand
    -7.9976 = g CuSO4 - H2O + sand
    ----------
    0.6219 = g H2O lost

    8.6195 = g CuSO4.xH2O + sand
    x 0.20 CuSO.xH2O

    = 1.7239 = g CuSO4.xH2O
    -0.6219 = mass xH2O
    --------------
    = 1.102 = g CuSO4

    mols CuSO4 = g/molar mass = 1.102/159.5 = estimated 0.0069
    mols H2O = g/molar mass = 0.6219/18 = estimated 0.0345
    Then find the ratio for mols H2O per 1 mol CuSO4. That's 0.0345/0.0069 = 5.0
    Then find ratio of

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