Reasons for error in percentage of oxygen in KClO3 lab

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SUMMARY

The discussion focuses on the percentage of oxygen in potassium chlorate (KClO3) and the factors contributing to a significant percent error of approximately 30%. Key reasons identified for this error include rounding off masses during calculations and incomplete combustion of oxygen. Additionally, external factors such as open windows in the lab, potentially affecting the reaction environment, were also considered. The experimental percentage of oxygen calculated was 27.83%, while the theoretical percentage was determined to be 39.34%, leading to a percentage error of 29.26%.

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GiantSheeps
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Homework Statement



I did this lab on percentage of oxygen in potassium chlorate and I got a percent error of about 30%. I now need to find three reasons for error, and human error does not count. I think one could be that the masses used in calculating the theoretical percentage of oxygen were all rounded off. Another could be that we did not allow all of the oxygen to burn away. Do those two make sense? Also, I need help with a third. My teacher said something about the fact that our lab was on the fourth floor of a building and the windows were all open, could that somehow lead to error? If so, how? Any help would be greatly appreciated. Thanks so much!

Procedure

1. Weigh a clean dry crucible and cover or Pyrex test tube.

2. Obtain a sample of potassium chlorate, KClO3 , as directed by your instructor. Place about

1.5 g of KClO3 in the weighed crucible, and reweigh. Use the same balance for all weighings

of the same sample

3. Place the crucible with KClO3 on a clay triangle and set the cover slightly ajar.

4. Heat gently for 8 minutes, then strongly for 10 minutes. Be sure the inner-blue cone of the

flame is just below the crucible bottom while you are heating strongly, and that the crucible bottom and/or clay triangle are heated to redness. Allow the crucible to cool to room temperature, which takes at least 10 minutes, and then weigh the crucible and residue.

5. Heat, strongly, a second time for about 5 minutes, cool and weigh. If the weighings after the first and the second heatings are not the same, i.e. they differ by more than 0.02 g; repeat the process of heating and cooling and weighing as needed, until 2 successive weights agree (within +/- 0.02 g).

6. Calculate the experimental percentage of oxygen in KClO3 based on the final weight of container’s contents.
 
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Wouldn't hurt to see your numbers. 30% plus or minus?

Was the original compound dry, or could it be wet?
 
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Borek said:
Wouldn't hurt to see your numbers. 30% plus or minus?

Was the original compound dry, or could it be wet?

It was dry, I believe. Here are my numbers:
  1. A. Mass of clean dry crucible 22.130
  2. B. Mass of crucible and KClO3 23.887
  3. C. Mass of crucible and contents after the first heating 23.441
  4. D. Mass of crucible and contents after the second heating 23.407
  5. E. Mass of crucible and contents after the third heating if needed 23.398
1) Calculate the mass of KClO3 used. (From Data B-A)
B - A
23.887 - 22.130
1.757g

2) Calculate the mass of oxygen lost by heating (from data B- final heating)
B -E
23.887 - 23.398
0.489g

3) Calculate the experimental percentage of oxygen in KClO3
__mass lost__ x 100 = calc 2/calc 1 x 100
mass KClO3
0.489/1.757 x 100 =
27.83%

4) Calculate the theoretical percentage of oxygen in KClO3 (from the chemical formula like we do in class).
(O + O + O)/(K + Cl + O + O + O)
(16 + 16 + 16)/(39 + 35 + 16 + 16 +16)
39.34%

5) Calculate the percentage error.
Percentage Error = experimental value– accepted value x 100
accepted value
((27.83 - 39.37)/39.34) x 100 =
29.26%Thank you so much for you help!
 
What effect would contaminants have?
 

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