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Calculate the mass of the anhydrous alum and the mass of the water that was given off.?
I was also asked to find the moles of alum and water. and to calculate the ratio of moles of water to the moles of alum and give the calculated formula of KAl(SO4)2 * X2O where X is the ratio of moles of water to moles of alum.
in my lab
the mass of the crushes alum crystal was 0.490g
the melting point of alum is 93 degrees Celsius
the melting range i observed was 91-93 degrees Celsius
the mass of the cruicible and lib was 21.700g
the mass of the crystal was 1.999g
the mass of the cruicible and alum was 22.746
therefore the mass of alum was 22.746-21.700=1.046g
using this information can you please help me answer my question :]
1 Answer
- Steve OLv 71 decade agoFavorite Answer
the info you provided shows:
the mass of the crystal was 1.999g
the mass of anhydrous alum was 1.046g (ANSWER #1)
this implies the residue 1.046 grams of KAl(SO4)2 was left over, after 0.953 grams of water, (ANSWER #2), was heated off
-------------------
1.046 grams of KAl(SO4)2 @ 258grams / mole =
0.00405 moles of KAl(SO4)2
ratios with
0.953 grams of water @18 grams / mole =
0.0529 moles of water
--------------------
0.00405 moles of KAl(SO4)2 / 0.00405 = 1 moles
0.0529 moles of water / 0.00405 = 13 moles water
----------------
this implies that the original formula was
KAl(SO4)2·13(H2O)
which is close to the right answer of KAl(SO4)2·12(H2O)
