Chemistry Help Please

5.00g of a mixture of magnesium sulfate crystals, MgSO4.7H2O, and copper sulfate crystals, CuSO4.5H2O, was heated until both salts has lost all water of crystallisation. There was a decrease in mass of 2.00g. Find the percentage of each hydrated salt in the mixture.

Yes. Not an easy question, but it's more maths than chemistry.

If you let the mass of magnesium sulfate = m
Then the mass of copper sulfate = (5 - m)
Now work out the moles of each
mol MgSO4.7H2O = m/246
mol CuSO4.5H2O = (5-m)/249.5

Now calculate the moles of water lost by each when heated
mol water from MgSO4.7H2O = 7m/246
mol water from CuSO4.5H2O = 5(5-m)/249.5

Now calculate the mass of water lost by each
mass water from MgSO4.7H2O = 18 x 7m/246
mass water from CuSO4.5H2O = 18 x 5(5-m)/249.5

and you know that the sum of the water masses = 2.00g

Therefore
[18 x 7m/246] + [18 x 5(5-m)/249.5] = 2

Now solve for m

126m/246 + 450/249.5 - 90m/249.5 = 2

0.512m + 1.804 - 0.361m = 2

0.151m = 0.196

mass of magnesium sulfate heptahydrate = 1.3g
mass of copper(II) sulfate pentahydrate = 3.7g

Yes. Not an easy question, but it's more maths than chemistry.

If you let the mass of magnesium sulfate = m
Then the mass of copper sulfate = (5 - m)
Now work out the moles of each
mol MgSO4.7H2O = m/246
mol CuSO4.5H2O = (5-m)/249.5

Now calculate the moles of water lost by each when heated
mol water from MgSO4.7H2O = 7m/246
mol water from CuSO4.5H2O = 5(5-m)/249.5

Now calculate the mass of water lost by each
mass water from MgSO4.7H2O = 18 x 7m/246
mass water from CuSO4.5H2O = 18 x 5(5-m)/249.5

and you know that the sum of the water masses = 2.00g

Therefore
[18 x 7m/246] + [18 x 5(5-m)/249.5] = 2

Now solve for m

126m/246 + 450/249.5 - 90m/249.5 = 2

0.512m + 1.804 - 0.361m = 2

0.151m = 0.196

mass of magnesium sulfate heptahydrate = 1.3g
mass of copper(II) sulfate pentahydrate = 3.7g