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When 0.500 g of copper(II) sulfate pentahydrate is heated at a temperature of 300°C for twenty minutes it loses some water of hydration. If the resulting sample weights 0.427 g then what is the formula of the resulting compound?

What equation(s) can I use to solve this?

  • Chem -

    How much water was in CuSO4.5H2O? That is
    0.500g x (5mol H2O/1 mol CuSO4.5H2O) = xx g
    0.500 g - 0.427 = yy g = mass H2O lost.
    How much water still remains in the CuSO4.xH2O? That is xx g - yy g = zz g.

    How much of that 0.427 g is CuSO4 with no water. 0.427 - water remaining so
    0.427 - zz = w.
    So you now have w g CuSO4 and zz g water remaining. Determine the ratio of each and the empirical formula from that. I get CuSO4.3H2O but check my thinking. Check my work.

  • Chem -


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