A mixture of cuso4?5h2o and mgso4?7h2o is heated until all the water is lost. if 5.020 g of the mixture gives 2.988 g of the anhydrous salts, what is the percent by mass of cuso4?5h2o in the mixture?

First, let's assign variables for the moles of CuSO₄·5H₂O as x and moles of MgSO₄·7H₂O as y. The molar mass of the substances are the following

CuSO₄·5H₂O: 249.685 g/mol

CuSO₄: 159.609 g/mol

MgSO₄·7H₂O: 246.49 g/mol

MgSO₄: 120.366 g/mol

H₂O: 18 g/mol

The solution is as follows:

2.988 = 249.685x + 246.49y - - > eqn 1

5.02 - 2.988 = 18 (5x + 7y) - - > eqn 2

Solving both equations simultaneously, the values of x and y are:

x = 0.0134 mol CuSO₄·5H₂O

y = 0.0257 mol MgSO₄·7H₂O

Thus, the percent CuSO₄·5H₂O is equal to

Mass Percentage = [ (0.0134 mol CuSO₄·5H₂O) * (249.685 g/mol) ]/{[ (0.0134 mol CuSO₄·5H₂O) * (249.685 g/mol) ] + [ (0.0257 mol MgSO₄·7H₂O) * (246.49 g/mol) ]} * 100

Mass percentage = 34.56%