Hi, I am stuck on this question for my lab assignment. There are 2 different tri

Question

Hi, I am stuck on this question for my lab assignment. There are 2 different trials. Trial 1 Mass of Bottle+Sample--31.2418 Mass of Bottle-Sample--30.9285 Mass of Sample--.3137 Crucible Number- Mass of empty Crucible:29.2615 Mass of crucible +BaSO4--29.6810 Mass of dry BaSO4--0.4195 Trial 2: Mass of Bottle+Sample--30.9285 Mass of bottle-Sample--30.5795 Mass of Sample--0.3490 Mass of empty Crucible:28.4582 Mass of Crucible+BaSO4--28.9251 Mass of dry BaSO4--0.4688(g) Calculation: Calculate the mass percent sulfate for each of the 2 trials. Then calculate the average(mean)mass% SO4 2-

Explanation / Answer

Analysis of sulfate in unknown samples

Molecular weight of BaSO4 is 233g

Molecular weight of sulfate (SO42-) = 96g

Percentage of sulfate present in BaSO4 = (96/233 ) x 100 = 41.2%

Trial 1

Mass of BaSO4 = 0.4195g

Mass of sulfate present = (41.2 x 0.4195)/ 100

Mass of sulfate present = 0.172g

Mass of sulfate in precipitate BaSO4 = mass of sulfate present in sample

Sample = 0.3137g

Sulfate present in the sample = 0.172g

Mass percent of sulfate in the sample = (0.172/ 0.3137) x 100 = 54.83%

Trial 2

Mass of BaSO4 = 0.4688g

Mass of sulfate present = (41.1 x 0.4688)/ 100 = 0.193g

sulfate present in BaSO4 = sulfate present in sample

Sample = 0.3490g

Sulfate present in the sample = 0.193g

Mass percentage of sulfur in the sample = (0.193/ 0.3490) x 100 = 55.30%

Average mass % of sulfate = (54.83 + 55.3)/ 2 = 55.07%

Average (mean) mass% of sulfate (SO42-) = 55.07%

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