Determine the formula of the MgSO4 hydrate & calculate the % by mass of H2O in the hydrate

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(Epson Salt)



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

9.30 g / 233.391 g/mol = 0.0398473 mol of BaSO4

BaSO4 and MgSO4 are in a 1:1 molar ratio, so this many moles of MgSO4 formed:

0.0398473 mol

grams of MgSO4 ---> 0.0398473 mol times 120.366 g/mol = 4.80 g (to three sig figs)

water in the hydrate ---> 9.86 g minus 4.80 g = 5.06 g

moles of water ---> 5.06 g / 18.015 g/mol = 0.280877 mol

We want to know a whole number ratio between MgSO4 and H2O where MgSO4 is set to 1.

MgSO4 ---> 0.0398473 / 0.0398473 = 1
H2O ---> 0.280877 / 0.0398473 = 7

MgSO4 · 7H2O

percent by mass of water ---> (5.06 g / 9.86 g)*100 = 51.3%

Nice problem.