16) 5.00g of a mixture of MgSO4 . 7H2O and CuSO4 . 5H2O was heated at 120 degrees Celsius until a mixture of the anhydrous salts was formed, which weighed 3.00g. Calculate the percentage by mass of MgSO4 . 7H2O in the mixture.
Ok, as promised, i will try to get this question right. Let the initial mass of the hydrated MgSO4 be x, therefore the initial mass of the hydrated Na2SO4 = (5-x) g
Mass of water lost = 5 - 3 = 2g
mole of water lost = 2/18 = 1/9 mole
However, if you construct the decomposition eqn for both the hydrated Mg and Na sulphate, you would get 7 moles of H2O and 5 respectively.
Hence, there are 12 moles of water in total, mole of H20 produced by hydrated MgSO4 = 7/12 x 1/9
Mole of hydrated MgSO4 = 1/7 x mole of water produced
= 1/7 x 7/12 x 1/9 = 1/108 = mass/RMM of the hydrated MgSO4
Therefore mass of hydrated MgSO4 = 1/108 x 246 = 2.278 g
So mass of hydrated NaSO4 = 5 - 2.278 = 2.722 g
Hence the percentage by mass of the Mg... to Na... is like 45.6:54.4.
I think this should be the correct method. Anyone who finds this wrong please quote me, i'd like to know where i'd gone wrong anyway.