Analysis of Alum
Andy Glodowski
Ap Chem
Hour 6
Part 1.
This experiment was to find the MP of Alum. This was done by heating
the alum in a small capillary tube and then put in boiling water.
Published Value : 92.1 C
Data : Alum = .56 g
Mp = 92.9 C
% Error : 92.9 C – 92.1
---------------- = 0.00868 or 0.868 % Error
92.1C
Part 2.
This experiment was to find the ratio of moles of Anhydrous Alum to
water. This was done by heating the alum in a crucible to evaporate the
water.
Published Value : KAl(SO4)2 * 12 H2O
Data : Alum = 1.9935g
Crucible + Alum = 23.5882g
Crucible + Anhydrous Alum = 22.6736g
H2O = .9146 g
Calculations:
H2O = .9146g
Anhydrous = 1.0789g
1 mol H20 = 17.999g
1 mol Anhydrous = 258.19g
.9146g
1 mol
-------- * --------- = .05081 mol H2O
1
17.999g
1.0789g
1 mol
-------- * ---------- = .0041787 mol Alum
1
258.19g
Ratio : .0041787 mol : .05081 mol =
1 mol : 12.16mol
Equation : KAl(SO4)2 * 12 H2O
Part 3.
This experiment was to determine the % sulfate in Alum. This was done
by adding Ba(NO3)2 to form BaSO4 with the alum.
Data : Alum = .9740g
Filter Paper with BaSO4 = 3.5984g
Filter Paper = 2.5270g
BaSO4 = 1.0714
Calculations :
Amount of Ba(NO3)2 used:
1.000g alum
1 mole
2 Ba(NO3)2 1L 1000ml
-------------- * ------- * --------------- * --- * --------474.2g
1 Alum
.2 M
1L
= 21.1 ml * 2 = 42.2ml
1 mol SO4 = 96.07g
1 mol BaSO4 = 233.39g
96.07g / 233.39g = 41.15%
% sulfate in BaSO4 = 41.15 %
% of sulfate evaporated
1.0714g BaSO4 1 mole
1 SO4
96.07g
------------------- * --------- * ----------- * -------- = .4417 g SO4
233.39g 1 BaSO4 1 SO4
.4417 g SO4 / .9740g Alum = .4534 = 45.34% SO4
Difference:
45.34% - 41.15% = 4.19%
Explanation :
The experiment was more than 3% off because all of the water was
not evaporated from the filter paper, which caused it to weigh
more.