A mixture of CuSO4 * 5H2O and MgSO4 * 7H2O weighing 3 g after dehydration loses 50% of the mass

A mixture of CuSO4 * 5H2O and MgSO4 * 7H2O weighing 3 g after dehydration loses 50% of the mass. Find the mass fraction of MgSO4 * 7H2O in the mixture.

Given:
m mixture1 = 3 g
Δm mixture1 = 50%

To find:
ω (MgSO4 * 7H2O) -?

Solution:
1) Let n (CuSO4 * 5H2O) = (x) mol, and n (MgSO4 * 7H2O) = (y) mol;
2) m (CuSO4 * 5H2O) = n * M = (x * 250) g;
3) m (MgSO4 * 7H2O) = n * M = (y * 246) g;
4) m mixture1 = m (CuSO4 * 5H2O) + m (MgSO4 * 7H2O);
5) n (CuSO4) = (x) mol;
6) m (CuSO4) = n * M = (x * 160) g;
7) n (MgSO4) = (y) mol;
8) m (MgSO4) = n * M = (y * 120) g;
9) m mixture2 = Δm mixture1 * m mixture1 / 100% = 50% * 3/100% = 1.5 g;
10) m mixture2 = m (CuSO4) + m (MgSO4);
3 = 250x + 246y;
1.5 = 160x + 120y;
x = 0.012 – 0.984y;
1.5 = 160 * (0.012 – 0.984y) + 120y;
y = 0.011;
11) m (MgSO4 * 7H2O) = y * 120 = 0.011 * 246 = 2.706 g;
12) ω (MgSO4 * 7H2O) = m (MgSO4 * 7H2O) * 100% / m mixture1 = 2.706 * 100% / 3 = 90.2%.

Answer: The mass fraction of MgSO4 * 7H2O is 90.2%.