# 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%.