Amixture 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%

Temperature and pressure