1. Sulfate (S042-) is typically determined by dissolving in H2O followed by prec
ID: 884609 • Letter: 1
Question
1. Sulfate (S042-) is typically determined by dissolving in H2O followed by precipitation as BaSO4 via addition of BaCl2. Calculate the percent (w/w) Na2SO4 (FW = 142.04) in an unknown sample based on the data below? What percent SO3 (FW = 80.06) is in the unknown sample? Constant weight of crucible: 25.6390 g Weight of vial with solid unknown sample: 13.1131 g Weight of vial minus some sample: 12.4122 g Weight of crucible + BaSO4 after 1st heating (ignition): 25.9365 g Weight of crucible + BaSO4 after 2nd heating (ignition): 25.9348 g Weight of crucible + BaSO4 after 3rd heating (ignition): 25.9347 gExplanation / Answer
Weight of unknown sample
Weight of vial with solid unknown sample =13.1131 g
Weight of vial minus sample = 12.4122 g
= 13.1121 – 12.4122
= 0.7009 g
No of moles = Wt / MW
= .7009/ 142.04
= 0.00493 moles
Moleecular Weigt of BaSO4 = 233.43
Weight of BaSO4 = .00493 * 233.43
= 1.1508 g
Weight of crucible = 25.6390 g
Weight of crucible + BaSO4 1st heating = 25.9365 g
Weight of sample = 25.9365 – 25.6390
= 0.2975 g
= (1.1508/233.43)*(80.06/0.2975)*100
= 132 %
Weight of crucible + BaSO4 2nd heating = 25.9348 g
Weight of sample = 25.9348 – 25.6390
= 0.2958 g
= (1.1508/233.43)*(80.06/0.2958)*100
= 133 %
Weight of crucible + BaSO4 3rd heating = 25.9347 g
Weight of sample = 25.9347 – 25.6390
= 0.2957 g
= (1.1508/233.43)*(80.06/0.2957)*100
= 133 %
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