Was your average experimental mass percent of oxygen in potassium chlorate higher or lower than the theoretical value (circle one)? Higher Lower Which 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 your raw data to obtain your experimental % of oxygen.

Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
icon
Related questions
Question

Experimental Value = 40.767%

Theoretical Value = 39.165%

Was your average experimental mass percent of oxygen in potassium chlorate higher or lower than the
theoretical value (circle one)?
Higher
Lower
Which 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 your
raw data to obtain your experimental % of oxygen.
Transcribed Image Text:Was your average experimental mass percent of oxygen in potassium chlorate higher or lower than the theoretical value (circle one)? Higher Lower Which 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 your raw data to obtain your experimental % of oxygen.
Mass of original KClO3 sample =
Mass of KCl residue =
Mass of Oxygen released =
Mass Percent of Oxygen in KClO3
Average Mass Percent Oxygen =
Sample 1
[Mass of crucible, lid + KCIO3] -
[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 KClO3 Average Mass Percent Oxygen = Sample 1 [Mass of crucible, lid + KCIO3] - [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 %
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 3 steps with 3 images

Blurred answer
Knowledge Booster
UV and Visible Spectroscopy
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Similar questions
  • SEE MORE QUESTIONS
Recommended textbooks for you
Chemistry
Chemistry
Chemistry
ISBN:
9781305957404
Author:
Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:
Cengage Learning
Chemistry
Chemistry
Chemistry
ISBN:
9781259911156
Author:
Raymond Chang Dr., Jason Overby Professor
Publisher:
McGraw-Hill Education
Principles of Instrumental Analysis
Principles of Instrumental Analysis
Chemistry
ISBN:
9781305577213
Author:
Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:
Cengage Learning
Organic Chemistry
Organic Chemistry
Chemistry
ISBN:
9780078021558
Author:
Janice Gorzynski Smith Dr.
Publisher:
McGraw-Hill Education
Chemistry: Principles and Reactions
Chemistry: Principles and Reactions
Chemistry
ISBN:
9781305079373
Author:
William L. Masterton, Cecile N. Hurley
Publisher:
Cengage Learning
Elementary Principles of Chemical Processes, Bind…
Elementary Principles of Chemical Processes, Bind…
Chemistry
ISBN:
9781118431221
Author:
Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:
WILEY