Question #b6ba3
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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 moles 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"#
