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The average experimental mass percent of oxygen in potassium chlorate is higher than the theoretical value. This is because some of the potassium chloride product splattered out of the crucible during the heating process.  Explain this statement using an analysis of the image. TV = 39.165% EV = 40.767%

The average experimental mass percent of oxygen in potassium chlorate is higher than the theoretical value. This is because some of the potassium chloride product splattered out of the crucible during the heating process.  Explain this statement using an analysis of the image. TV = 39.165% EV = 40.767%

Chemistry: Principles and Reactions
Chemistry: Principles and Reactions
8th Edition
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:William L. Masterton, Cecile N. Hurley
Chapter1: Matter And Measurements
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The average experimental mass percent of oxygen in potassium chlorate is higher than the theoretical value. This is because some of the potassium chloride product splattered out of the crucible during the heating process. 

Explain this statement using an analysis of the image.

TV = 39.165%

EV = 40.767%

Mass of original KClO3 sample= Mass of KCl residue = Mass of Oxygen released = Mass Percent of Oxygen in KC103 Average Mass Percent Oxygen = -KCIO3] - Sample 1 [Mass of crucible, lid + [Mass of crucible + lid] = 33.745 g - 32.721 g = 1.024 g [Mass of crucible, lid + residue after 2nd heating] - [Mass of crucible + lid] = 33.314 g - 32.721 g = 0.593 g Mass of original KCIO3 sample - Mass of KCI residue = 1.024 g - 0.593 g = 0.431 g 42.090 +39.444 2 Mass Percent of Oxygen (experim-Mass Percent of Oxygen (experim- ental) = Mass of Oxygen released /ental) = Mass of Oxygen released Mass of Potassium chlorate used x Mass of Potassium chlorate used x 100 = 0.431 g / 1.024 g x 100 = 100 = 0.383 g / 0.971 g x 100 = 42.090 % 39.444% 81.534 Sample 2 [Mass of crucible, lid + KCIO3] - [Mass of crucible + lid] = 33.692 g - 32.721 g = 0.971 g 2 [Mass of crucible, lid + residue after 2nd heating] - [Mass of crucible + lid] = 33.309 g - 32.721 g = 0.588 g Mass of original KCIO3 sample - Mass of KCI residue= 0.971 g - 0.588 g = 0.383 g = 40.767 %
Transcribed Image Text:Mass of original KClO3 sample= Mass of KCl residue = Mass of Oxygen released = Mass Percent of Oxygen in KC103 Average Mass Percent Oxygen = -KCIO3] - Sample 1 [Mass of crucible, lid + [Mass of crucible + lid] = 33.745 g - 32.721 g = 1.024 g [Mass of crucible, lid + residue after 2nd heating] - [Mass of crucible + lid] = 33.314 g - 32.721 g = 0.593 g Mass of original KCIO3 sample - Mass of KCI residue = 1.024 g - 0.593 g = 0.431 g 42.090 +39.444 2 Mass Percent of Oxygen (experim-Mass Percent of Oxygen (experim- ental) = Mass of Oxygen released /ental) = Mass of Oxygen released Mass of Potassium chlorate used x Mass of Potassium chlorate used x 100 = 0.431 g / 1.024 g x 100 = 100 = 0.383 g / 0.971 g x 100 = 42.090 % 39.444% 81.534 Sample 2 [Mass of crucible, lid + KCIO3] - [Mass of crucible + lid] = 33.692 g - 32.721 g = 0.971 g 2 [Mass of crucible, lid + residue after 2nd heating] - [Mass of crucible + lid] = 33.309 g - 32.721 g = 0.588 g Mass of original KCIO3 sample - Mass of KCI residue= 0.971 g - 0.588 g = 0.383 g = 40.767 %
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