The Student Room Group
Can you link moles to mass with an equation?

Posted from TSR Mobile
Original post by Year11guy
0.0185mol of hydrated magnesium sulfate MgSO4.xH2O has a mass of 4.56g. Work out the value of x, number of moles of water of crystallisation in the hydrated salt.
Normally, the questions like these give the mass of the water even if you have to work it out from experimental data. I don't recognize what to do here.


1 mole of MgSO4.xH2O contains 1 mole of MgSO4 and x moles of H2O

Therefore 0.0185 moles of MgSO4.xH2O will contain 0.0185 moles of MgSO4 and 0.0185x moles of H2O.
Reply 3
Original post by addylad
Can you link moles to mass with an equation?

Posted from TSR Mobile



moles=mass/Ar or Mr
Original post by Year11guy
moles=mass/Ar or Mr


Rearrange to make Mr the subject, and then substitute in the values.

You can work out the molecular weight of anhydrous (no water) MgSO4. The difference between these molecular weights accounts for the water. Then you can use this equation again to find the number of moles of water, i.e. x.
Reply 5
Original post by Year11guy
0.0185mol of hydrated magnesium sulfate MgSO4.xH2O has a mass of 4.56g. Work out the value of x, number of moles of water of crystallisation in the hydrated salt.
Normally, the questions like these give the mass of the water even if you have to work it out from experimental data. I don't recognize what to do here.


Molar mass of hydrated salt: mass/moles = 4.56g/0.0185 mol = gmol–1

MgSO4 contributes [24.3 + 32.1 + (4 × 16.0)] = 120.4 gmol–1; therefore, 246 120.4 = 125.6 gmol–1 is due to x moles of water

x(2.0 + 16.0) = 125.6

So x = 125.6/18.0

x = 6.98

The number of molecules of water of crystallisation must be a whole number, so hydrated magnesium sulfate contains seven molecules of water of crystallisation.

So x = 7
MgSO4.7H2O
Reply 6
Original post by Ir_xa
MgSO4.7H2O


Before you go to all of the trouble answering a Q, make sure that it isn't several years old.
OP has since changed his name to "year14bloke"
I'm sure the OP doesn't need any help any more 3 years later!
If you want to ask a similar question, you can create a new thread in the chemistry forum :smile:
Thread closed.

Latest