Question

You need 4.00g of copper (II) surface, but only have copper surface pentahydrate, determine the mass of copper sulfate pentahhydrate required in order to have 8.00g of copper (II) sulfate. Carefully measure the calculated mass and transfer it into the 250ml beaker

Answer 1

Solution:

Supposing your auto-correct got away from you and "surface" means "sulfate":

You also can't make up your mind about how much copper sulfate you need, so I will choose 4.00 g to be made:

$\dfrac{(4g\;of\;CuSO_4)}{159.6086\;\frac{g}{mole}}\;*\;\dfrac{1\;mole\;CuSO_4\cdot H_2O}{1\;mole\;CuSO_4}\;*\;(249.685\;\tfrac{g}{mole})=6.25g\;CuSo_4\cdot H_2O$

[If you really need 8.00 grams of CuSO₄, then double the previous answer:

6.2574 g x 2 = 12.51 g CuSO₄·5H₂O ]