The mass fraction of water in a mixture of crystalline hydrates consisting of CuSO4 * 5H2O and FeSO4 * 7H2O

The mass fraction of water in a mixture of crystalline hydrates consisting of CuSO4 * 5H2O and FeSO4 * 7H2O is 40%. Calculate the mass fraction of copper cupros in the mixture.

Given:
mixture (CuSO4 * 5H2O and FeSO4 * 7H2O)
ω (H2O) = 40%

To find:
ω (H2O) -?

Decision:
1) Find the amount of substance CuSO4 * 5H2O and FeSO4 * 7H2O:
Let n (CuSO4 * 5H2O) = x mol;
n (CuSO4 * 5H2O) = y mol;

2) Find the mass of CuSO4 * 5H2O and FeSO4 * 7H2O:
m (CuSO4 * 5H2O) = n (CuSO4 * 5H2O) * Mr (CuSO4 * 5H2O) = 250x g;
m (FeSO4 * 7H2O) = n (FeSO4 * 7H2O) * Mr (FeSO4 * 7H2O) = 278y g;
m mixture = 250x + 278y;

3) Find the amount of substance Н2О in CuSO4 * 5H2O and FeSO4 * 7H2O:
n (H2O in CuSO4 * 5H2O) = n (CuSO4 * 5H2O) * 5 = 5x mol;
n (H2O in FeSO4 * 7H2O) = n (FeSO4 * 7H2O) * 7 = 7y mol;
n (H2O) = 5x + 7y mol;

4) Find the mass of H2O in CuSO4 * 5H2O and FeSO4 * 7H2O:
m (H2O) = n (H2O) * Mr (H2O) = (5x + 7y) * 18 = 90x + 126y g;

5) Make and solve the equation according to the formula for finding the mass fraction of H2O in the mixture:
ω (H2O) = m (H2O) * 100% / m mixture;
40% = (90x + 126y) * 100% / (250x + 278y);
90x + 126y = 0.4 * (250x + 278y);
90x + 126y = 100x + 111.2y;
x = 1.48y;

6) Find the mass fraction of CuSO4 * 5H2O in the mixture (taking into account the solution of the previous equation):
ω (CuSO4 * 5H2O) = m (CuSO4 * 5H2O) * 100% / m (mixture) = 250x * 100 / (250x + 278y);
Because x = 1.48y, then we substitute this value into the formula:
ω (CuSO4 * 5H2O) = 250 * 1.48y * 100 / (250 * 1.48y + 278y) = 37000y / (370y + 278y) = 57.1%.

Answer: Mass fraction of CuSO4 * 5H2O in the mixture is 57.1%.