Chemistry
posted by Graham on .
I had a 1.094g of mystery Alum. "AB(SO4)c. dH2O" (I assume that that the Alum is KAl(SO4)212H2O but we don't know A B c or d, we do know that mass percent of water is 51.9)
After processing the alum (Ba(NO3)2 and HNO3 through a gooch crucible) we end up with 1.739g of BaSO4.
We know that SO4 is 42.15% of BASO4 thus
SO4 has a mass of .7330 g
assuming we captured all of the SO4 of the original Alum.
so if we subtract .7330g from 1.094g = .361 g
therefor .361g is the the AB and the H2O of the Alum
but if water is 51 of the mass of the Alum. I did something wrong any ideas.

You're right. Something is wrong but it must have been in the experimental details. Your math is not quite right but it won't make that much difference.
I think the 42.15% SO4 in BaSO4 is not quite right. I think that should be (96/233.39)*100 = 41.13%. That won't make that much difference. Mass SO4 = 1.739 x (96/233.39) = 0.7153g
I don't think the H2O calculation is correct either. 1.094 x 0.519 = 0.5678 and 1.0940.5678 = 0.5262g for AB(SO4)c.
So you are faced with the situation of sulfate weighing more than the anhydrous salt.