Some CuSO4.5H20 was heated at 120C with the following results: Mass of crucible=10.00g Mass of crucible + CuSO4.5H20=14.98g Mass of crucible + recidue=13.54g. How many molecules of water of crystalization were lost? (H=1Cu=63.5O=16S=32)

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    I’m assuming you mean ##weight of## instead of ##water of## right? Here’s how I think the data actually looks like:
    The weight of the crucible is ##10.0 g##.
    The weight of the crucible + the weight of the copper (II) sulfate pentahydrate is ##14.98 g##. Automatically you know how much copper (II) sulfate pentahydrate you have
    ##m_(hydrate) = 14.98 g – 10.00 g = 4.980 g##
    The weight of the crucible + the weight of the residue is ##13.54 g##. The residue actually represents the anhydrated ##CuSO_4## which means that you now know how much water has been evaporated
    ##m_(water) = m_(hydrate) – (13.54 g – 10.00 g) = 4.980 g – 3.540 g##
    ##m_(water) = 1.440 g##
    You know that water has a molar mass of ##18.0 g/mol##. This will help you find the number of you have
    ##1.440 g * (1 mole)/(18.0 g) = 0.0800## ##moles##
    The number of molecules is
    ##0.0800 moles * (6.022 * 10^(23) molecules)/(1 mole) = 4.82 * 10^(22)## ##molecules##

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