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QUESTION

# 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=1,Cu=63.5,O=16,S=32)

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"