Answer:
Remember that there are 5 moles of water for every 1 mole of hydrate
159.61g mol−1(159.61+5×18.015)g mol−1×100=63.92% CuSO4
This tells you that for every 100 g of copper(II) sulfate pentahydrate, you get 63.92 g of anhydrous copper(II) sulfate. Use this as a conversion factor to get
1.596g CuSO4⋅100 g CuSO4⋅5H2O63.92g CuSO4=2.497 g CuSO4⋅5H2O
Rounded to two sig figs, the answer will be
