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 CuSO4 is many times heavier on a molecule by molecule basis than the H2O.
2.) We need to find a ratio of how many molecules of water to how many molecules of CuSO4 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 H20 | 1 mole H20 |
| 18.02 g H2O |
= 0.1 mole H20
5.) Calculate the number of moles of dry CuSO4 present (Subtract the mass water from the mass of hydrous CuSO4)
| 3.2 g CuSO4 | 1 mole CuSO4 |
| 160 g CuSO4 |
= 0.02 mole CuSO4
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 H2O)
0.02 mole/ 0.02 mole = 1 ( the CuSO4 )
So for every 1 CuSO4 we have 5 H20 or we can write this CuSO4 . 5H20 and its formal name is copper (II) sulfate pentahydrate.