[zerality]

A student dissolves 9.86g epsom salt, a MgSO4 hydrate into 50ml water. from this solution the student isolated 9.30g of anhydrous BaSO4 (s) which is formed in the following reaction. BaSO4 is very insoluble in water, and esentially all sulfate ions will precipitate as BaS04 (s). assume handling losses are small.

MgSO4(aq) + BaCl2 (aq) ---> MgCl2 (aq) + BaSO4 (s)


a)  determine the formula of the MgSO4 hydrate

not sure where to start. i know i can find the mass of SO4 (sup) -2 from the given mass of BaSO4.






b) calculate the % by mass of H20 in the hydrate.

[zerality]

a) Do I find the no mole of water: 50g/18.01   and the no moles of MgSO4 9.86/molar mass  and    then find the ratio of the two to give me the formula of the hydrate.

b) %mass of H20 = [ 50g/ (50g + 9.86g)] x 100


Am I anywhere close on this?

[Donaldson Tan]

Am I anywhere close on this?

Yes and no. You do not need to consider the mass of 50ml of water, but you need to take in account of sulphate precipitated.

Let the formula of epsom salt be MgSO₄.xH₂O

The solvation process can described as:
MgSO₄.xH₂O -> Mg²⁺ (aq) + SO₄^2- (aq) + x H₂O (l)

from this solution the student isolated 9.30g of anhydrous BaSO4 (s) which is formed in the following reaction.

From here, you can work the number of moles of sulphate and magnesiums ions present in 9.86g  of epsom salt given you know the required stoichiometric ratio.

Next, you can deduce how much of 9.86g Epsom salt is water by using mass balance.

The unknown quantity x can be found as the molar ratio of water molecules to sulphate ions in 9.86g of Epsom salt.

[zerality]

Ok,

I think I figured this out:  formula = MgSO4. 7H20

% by mass H20 in hydrate 51.32%

Is that correct