R.A.M Anhydrous = 196.
R.A.M H20 = 18.
Grams of Anhydrous = 1.735g.
H3X ~ nH20 = 210g mol -1
Moles of H3X ~ nH20 = 0.00833 mol.
How do I work out the ratio of water molecules to H3X molecules using these?
Any help would be greatly appreciated
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water of crystallisation help :/ watch
- Thread Starter
- 21-03-2011 16:41
- 22-03-2011 08:16
Ok, imagine you had copper sulphate.
CuSO4 : ram = 159.6
CuSO4.5H2O : ram = 159.6 + (5x18) = 249.6
ram anhydrous (H3X) + n(ram H2O)=ram H3X~nH2O
196 + 18n=210
I guess, rounding up, you can say you have one water then. H3X.H2O
But I'd question your numbers because n should be an integerLast edited by Plato's Trousers; 22-03-2011 at 11:07.