The weight % water in hydrous cupric sulfate is approximately 35%. Some of
you had some old cupric sulfate that had already begun to dehydrate due to the
extremely dry Nevada climate. Answers between 20 and 40 % are reasonable.
Calculating the formula extra credit.
1.) Recognize that the CuSO_{4} is many times heavier on a molecule
by molecule basis than the H_{2}O.
2.) We need to find a ratio of how many molecules of water to how many
molecules of CuSO_{4} are present.
3.) Recall that if you have a GFW then you have one mole
of molecules
4.) Calculate number of moles of H2O present.
1.8 g H_{2}0 
1 mole H_{2}0 

18.02 g H_{2}O 
= 0.1 mole H_{2}0
5.) Calculate the number of moles of dry CuSO_{4}
present (Subtract the mass water from the mass of hydrous CuSO_{4})
3.2 g CuSO_{4} 
1 mole CuSO_{4} 

160 g CuSO_{4} 
= 0.02 mole CuSO_{4 }
6.) Now we have a ratio, but the numbers are awkward. Lets
"normalize" them by dividing both by the smallest number.
0.1 mole/ 0.02 mole = 5 (the H_{2}O)
0.02 mole/ 0.02 mole = 1 ( the CuSO_{4} )
So for every 1 CuSO_{4} we have 5 H_{2}0 or we can write this
CuSO_{4} ^{.} 5H_{2}0 and its formal name is
copper (II) sulfate pentahydrate. 