What is the percentage of KClO3 in a mixture after removing oxygen?

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Discussion Overview

The discussion revolves around calculating the percentage of potassium chlorate (KClO3) in a mixture after the removal of oxygen through heating. Participants explore the implications of mass changes in the mixture and the relationships between moles of reactants and products, focusing on the theoretical and practical aspects of the problem.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant presents a problem involving a 16.00g mixture of KClO3 and KCl, which after heating results in a 9.00g product of KCl and 3.00g of oxygen released.
  • Another participant suggests a method to find the moles of oxygen and subsequently the moles of KClO3, indicating a need for clarity on the teacher's solution which combines several steps.
  • Confusion arises regarding the mass of the original sample, with some participants questioning whether it is 16.00g or 12.00g.
  • Several participants express uncertainty about how to relate the moles of oxygen to the moles of KClO3, with one suggesting a multiplication approach that is questioned by others.
  • One participant calculates the mass of KCl from the moles of oxygen but finds discrepancies with the teacher's results, prompting further discussion on the correct relationships between the compounds.
  • Another participant correctly identifies that one mole of KClO3 corresponds to three moles of oxygen, leading to a calculation of moles of KClO3 from the given moles of oxygen.
  • Final calculations are presented by a participant who arrives at a percentage of KClO3 in the mixture, noting slight discrepancies from the teacher's solution.

Areas of Agreement / Disagreement

Participants express varying levels of understanding and confusion regarding the calculations involved, particularly in relating moles of oxygen to moles of KClO3. There is no consensus on the clarity of the teacher's solution, and multiple interpretations of the problem persist.

Contextual Notes

Some participants highlight missing intermediate steps in the teacher's solution, which may contribute to confusion. There are also discrepancies in the reported mass of the original mixture, leading to further uncertainty.

Who May Find This Useful

Students studying stoichiometry, particularly those dealing with chemical reactions involving the decomposition of compounds and the calculation of percent composition in mixtures.

brbrett
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Homework Statement


A 16.00g sample of a mixture of KClO3 and KCl is heated until all the oxgen has been removed from the sample. The product entirely KCl (no O), has a total mass of 9.00g. What is the percentage of KClO3 present in the original mixture?

16.00g mixture
9.00g mixture without oxygen
3.00g oxygen without mixture

Homework Equations


Percent Composition (percent = mass of element/mass of compound)
Various mole equations (eg. moles to mass)

The Attempt at a Solution


I received this question on a test awhile back and, while I have received the answer and the steps to reach it, I cannot understand why it is that it is done that way. From what I can see, we do not need to worry about percent composition of the mixture until the end. At the moment, we do not have KCl by itself, as O3 is bonded to it.

I think I need to find the mass of KCl in the original mixture, and so to do that, I chose to calculate the moles in KCl and O.

9.00gKCl/(75g/mol) = 0.12molKCl
3.00gO/(16g/mol)= 0.1875mol O
By this point, I had already figured that this was likely going nowhere, as I still did not know the amount of KCl in KClO3 by itself, but rather just the moles of KCLO3+KCL without the oxygen.

------------------------------
Next, is my teacher's solution.
gKClO3 = 3.00gO x (1.22gKClO3/48.9gO) = 7.6625
% = (7.6625/12.00) x 100% = 63.9%

I understand the percent part, but what I do not understand is why 1.22gKClO3 is divided by 48.0gO. If I could have insight onto that, then I would be most appreciative.

Thanks for reading my wall of text!
 
Last edited:
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The teacher's solution glosses over some intermediate steps, so it's probably not as clear as it should be. Here are the steps I would take:

1. Find the number of moles of oxygen atoms in 3g. (0.1875 moles as you calculated)
2. Find how many moles of KClO3 can be created from 0.1875 moles of oxygen atoms.
3. Find the mass of that number of moles of KClO3

If you compress all of these 3 steps into one step, you get your teacher's solution.
 
brbrett said:

Homework Statement


A 16.00g sample of a mixture of KClO3 and KCl is heated until all the oxgen has been removed from the sample. The product entirely KCl (no O), has a total mass of 9.00g. What is the percentage of KClO3 present in the original mixture?

12.00g mixture
9.00g mixture without oxygen
3.00g oxygen without mixture
I'm confused. Is the mass of the sample 16.00 g or 12.00 g?

Homework Equations


Percent Composition (percent = mass of element/mass of compound)
Various mole equations (eg. moles to mass)

The Attempt at a Solution


I received this question on a test awhile back and, while I have received the answer and the steps to reach it, I cannot understand why it is that it is done that way. From what I can see, we do not need to worry about percent composition of the mixture until the end. At the moment, we do not have KCl by itself, as O3 is bonded to it.

I think I need to find the mass of KCl in the original mixture, and so to do that, I chose to calculate the moles in KCl and O.

9.00gKCl/(75g/mol) = 0.12molKCl
3.00gO/(16g/mol)= 0.1875mol O
By this point, I had already figured that this was likely going nowhere, as I still did not know the amount of KCl in KClO3 by itself, but rather just the moles of KCLO3+KCL without the oxygen.

------------------------------
Next, is my teacher's solution.
gKClO3 = 3.00gO x (1.22gKClO3/48.9gO) = 7.6625
% = (7.6625/12.00) x 100% = 63.9%

I understand the percent part, but what I do not understand is why 122.6gKClO3 is divided by 48.0gO. If I could have insight onto that, then I would be most appreciative.

I'm confused again. You write 1.22 g on one line, and then you mention 122.6 g on the next line.

You should proofread your posts carefully before submitting them.
 
Could I multiply the mass of KClO3 by 0.1875 molesO to find the number of moles of KClO3? I'm not entirely sure
SteamKing said:
I'm confused. Is the mass of the sample 16.00 g or 12.00 g?
I'm confused again. You write 1.22 g on one line, and then you mention 122.6 g on the next line.

You should proofread your posts carefully before submitting them.
You will have to forgive me. The information in the original question is the correct information. Anything else is my own error.
 
Ygggdrasil said:
2. Find how many moles of KClO3 can be created from 0.1875 moles of oxygen atoms.
I'm not entirely certain of how to find the number of moles KClO3 has using the number of moles of Oxygen atoms. Would multiplying 0.1875molO by the total mass of the mixture give me the moles of KClO? (Though I do not believe that to be correct...)
 
brbrett said:
I'm not entirely certain of how to find the number of moles KClO3 has using the number of moles of Oxygen atoms. Would multiplying 0.1875molO by the total mass of the mixture give me the moles of KClO? (Though I do not believe that to be correct...)
If you have 3 moles of oxygen atoms, how many moles of KClO3 do you have? Always look at the chemical formula.
 
I have one mole of KClO3 then, with 1 mole of both K and Cl?
(39+35.5)*0.1875 = 13.96875gKCl

However, this number is more than double the number my teacher had gotten.
 
brbrett said:
I have one mole of KClO3 then, with 1 mole of both K and Cl?
(39+35.5)*0.1875 = 13.96875gKCl

However, this number is more than double the number my teacher had gotten.
If you have one mole of KClO3, how many moles of oxygen do you have?
 
SteamKing said:
If you have one mole of KClO3, how many moles of oxygen do you have?
3
 
  • #10
brbrett said:
3
Now, work backwards. You have 0.1875 moles of oxygen. How many moles of KClO3 would 0.1875 moles of oxygen make?
 
  • #11
0.0625 moles?
 
  • #12
brbrett said:
0.0625 moles?
Yes, this seems reasonable. What is the mass of this number of moles of KClO3 ?
 
  • #13
0.0625*122.55 = 7.659375g
I see what I needed to do now. I guess I had forgotten that a mole is just a representative unit.
% = (7.659375/12.00) x 100% = 63.828125 = 63.8%
The number is slightly off, but that is just error.
Thanks for your help!
 

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