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

[brbrett]

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!

 

[Ygggdrasil]

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.

 

[SteamKing]

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.

 

[brbrett]

Could I multiply the mass of KClO₃ by 0.1875 molesO to find the number of moles of KClO₃? I'm not entirely sure

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.

 

[brbrett]

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...)

 

[SteamKing]

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.

 

[brbrett]

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.

 

[SteamKing]

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?

 

[brbrett]

If you have one mole of KClO₃, how many moles of oxygen do you have?

3

 

[SteamKing]

3

Now, work backwards. You have 0.1875 moles of oxygen. How many moles of KClO₃ would 0.1875 moles of oxygen make?

 

[brbrett]

0.0625 moles?

 

[SteamKing]

0.0625 moles?

Yes, this seems reasonable. What is the mass of this number of moles of KClO₃ ?

 

[brbrett]

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!