Is the explination correct? Was your average experimental mass percent of oxygen in potassium chlorate higher or lower than thetheoretical value (circle one)?HigherLowerWhich of the following sources of error could be used to explain this discrepancy (circle one)?A. The potassium chlorate sample was not heated strongly or long enough.B. Some of the potassium chloride product splattered out of the crucible during the heating process.Explain your choice. Your response should include an analysis of the calculations you performed with yourraw data to obtain your experimental % of oxygen.The mass of oxygen released becomes higher than the actual value only when the mass of KCI residueis smaller than the actual mass of KCI because of the equation. The higher experimental mass percentof oxygen is possible when the mass of oxygen released is higher because of the equation. If some ofthe KCI product is splattered out of the crucible during the heating process then the mass of the KCIresidue weighed is less than the actual KCI residue. Mass of oxygen released = (Mass of original KCIO3sample) - (Mass of KCI residue). As the KCI residue's mass becomes smaller, the value for the "mass ofoxygen released" increases. Now, look at the equation. As the "mass of oxygen released" increases,the value of the mass percent of oxygen in KCIO3becomes higher. Mass of original KClO3 sample=Mass of KCl residue =Mass of Oxygen released =Mass Percent of Oxygen in KC103Average 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 + residueafter 2nd heating] -[Mass of crucible + lid] =33.314 g - 32.721 g = 0.593 gMass of original KCIO3 sample -Mass of KCI residue =1.024 g - 0.593 g = 0.431 g42.090 +39.4442Mass Percent of Oxygen (experim-Mass Percent of Oxygen (experim-ental) = Mass of Oxygen released /ental) = Mass of Oxygen releasedMass of Potassium chlorate used x Mass of Potassium chlorate used x100 = 0.431 g / 1.024 g x 100 = 100 = 0.383 g / 0.971 g x 100 =42.090 %39.444%81.534Sample 2[Mass of crucible, lid + KCIO3] -[Mass of crucible + lid] =33.692 g - 32.721 g = 0.971 g2[Mass of crucible, lid + residueafter 2nd heating] -[Mass of crucible + lid] =33.309 g - 32.721 g = 0.588 gMass of original KCIO3 sample -Mass of KCI residue=0.971 g - 0.588 g = 0.383 g= 40.767 %

Given : Mass percent of Oxygen Experimental yield = 40.767% To find : why experimental yield is…

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 %

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 %

Organic Chemistry: A Guided Inquiry

2nd Edition

ISBN:9780618974122

Author:Andrei Straumanis

Publisher:Andrei Straumanis

ChapterL2: Mass Spectrometry

Section: Chapter Questions

Problem 5CTQ

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