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 1.094-0.5678 = 0.5262g for AB(SO4)c.
So you are faced with the situation of sulfate weighing more than the anhydrous salt.
I had a 1.094g of mystery Alum. "AB(SO4)c. dH2O" (I assume that that the Alum is KAl(SO4)2-12H2O 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.
1 answer
1 answer