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Offline Trueolive

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Redox titration question
« on: February 11, 2022, 11:21:14 AM »
Question:
A piece of impure copper of mass 0.90 g was reacted with concentrated nitric acid. The copper (II) solution made was neutralised and reacted with excess iodide ions to produce iodine.
The iodine liberated in the reaction was titrated with 0.50 mol dm-3 Na2S2O3. It was found that the average titre was 23.70 cm3.
Calculate the percentage of copper in the sample. Give your answer to 3 sig fig.

Working:
I wrote out the two equations I would need,
2Cu2+ + 4I-  :rarrow: 2CuI + I2 (equation 1)

I2 + 2S2O32-  :rarrow: 2I- + S4O62- (equation 2)

I then calculated the moles of I2 from 2S2O32-
(0.5 mol dm-3 X 0.0237 dm3) / 2 = 0.005925 = n of I2

I used this number to calculate the moles of Cu2+
(0.005925 X 2) = 0.01185 moles

I calculated the moles of Cu2+ by doing mass/Mr (0.90g / 63.5) = 0.01417323 moles

It is the next bit I am confused with,
To calculate the percentage of copper in the sample I did (0.01185 moles / 0.01417323 moles) X 100 = 83.608324%

My answer:

83.6%

Obviously, it's not a hard question but I feel like I've made a stupid mistake. Any corrections or feedback would be greatly appreciated.
 
« Last Edit: February 11, 2022, 02:24:55 PM by Trueolive »

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Re: Redox titration question
« Reply #1 on: February 11, 2022, 11:55:15 AM »
I calculated the moles of Cu2+ by doing mass/Mr (0.90g / 63.5) = 0.01417323 moles

No, that's not true - sample was not just copper.

You know how many moles of copper were in the sample - can you calculate its mass?
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Offline Trueolive

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Re: Redox titration question
« Reply #2 on: February 11, 2022, 12:19:20 PM »
I calculated the moles of Cu2+ by doing mass/Mr (0.90g / 63.5) = 0.01417323 moles

No, that's not true - sample was not just copper.

You know how many moles of copper were in the sample - can you calculate its mass?

I am confused, I am trying to calculate what percentage of copper is in the sample. So I did the moles of a pure sample of 0.9g Copper (0.01417323 moles) and have "compared" it to the impure 0.9g Copper compound (0.01185 moles) because if it were 100% pure it would be (0.01417323 / 0.01417323) X 100 = 100% copper

Please could you explain what you mean by calculating the mass and the relevance to the question?

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Re: Redox titration question
« Reply #3 on: February 11, 2022, 12:59:14 PM »
You can express the purity only as mass percentage of the copper in the sample.

Say, you have 10 g piece of impure copper, you check that it contains 9 g of copper - so the copper is 90% of the sample.

Apply the same logic to your case.
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Offline Trueolive

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Re: Redox titration question
« Reply #4 on: February 11, 2022, 01:24:16 PM »
You can express the purity only as mass percentage of the copper in the sample.

Say, you have 10 g piece of impure copper, you check that it contains 9 g of copper - so the copper is 90% of the sample.

Apply the same logic to your case.

With the information I have been provided within the question surely I can only do the reverse? Use the percentage to calculate the mass of the impure substance (which would be 0.7525g of Copper and 0.1475g of the impure substance)

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Re: Redox titration question
« Reply #5 on: February 11, 2022, 02:36:27 PM »
With the information I have been provided within the question surely I can only do the reverse? Use the percentage to calculate the mass of the impure substance (which would be 0.7525g of Copper and 0.1475g of the impure substance)

I feel like you are trolling. If there is 0.7525 g of copper in the 0.9 g sample, what is the percentage of copper in that very sample?
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Offline Trueolive

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Re: Redox titration question
« Reply #6 on: February 11, 2022, 02:51:45 PM »
With the information I have been provided within the question surely I can only do the reverse? Use the percentage to calculate the mass of the impure substance (which would be 0.7525g of Copper and 0.1475g of the impure substance)

I feel like you are trolling. If there is 0.7525 g of copper in the 0.9 g sample, what is the percentage of copper in that very sample?

Well, that would be the answer I gave in my original post (83.6%). I am not trolling I just presented a question I wasn't too sure about and gave my answer asking for feedback or corrections I need to make. Perhaps I didn't understand your original point when you first replied to the post.

Offline mjc123

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Re: Redox titration question
« Reply #7 on: February 11, 2022, 03:12:22 PM »
Your approach gets the right numerical answer, but is problematic because it appears to assume there are 14.17 mmol of "stuff" in the impure material, which is not true if the molar mass of the impurity is different from that of copper. Actually you weren't doing that, you were comparing the moles of Cu in the impure sample with the moles of Cu in the same mass of a pure sample, but this may not be clear on first reading. It looks as if you are calculating a mole fraction (wrongly) rather than a mass fraction. If you want to calculate a mass fraction, do it by comparing masses, not moles - it worked out in this case, but in more complicated situations it might lead you astray.

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Re: Redox titration question
« Reply #8 on: February 11, 2022, 03:16:04 PM »
Well, that would be the answer I gave in my original post (83.6%).

But you calculated it wrong: you converted 0.9 g to moles of copper, even if there is only 0.75 g of copper. The number you got was accidentally correct, but the logic behind your calculations was wrong.

You have 0.01185 moles of Cu in 0.9 g sample. Copper percentage is

[tex]\frac{0.01185\times63.5}{0.9}\times100\%[/tex]

you did

[tex]\frac{0.01185}{\frac{0.9}{63.5}}\times100\%[/tex]

These numbers happen to be mathematically equivalent, but the first one makes sense, the latter doesn't - you can't use mass of a mixture to calculate number of moles of one of the mixture components.

edit: mjc123 posted while was typing, so we more or less say the same thing
« Last Edit: February 11, 2022, 04:11:04 PM by Borek »
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Offline Trueolive

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Re: Redox titration question
« Reply #9 on: February 11, 2022, 05:00:41 PM »
 :)
Well, that would be the answer I gave in my original post (83.6%).

But you calculated it wrong: you converted 0.9 g to moles of copper, even if there is only 0.75 g of copper. The number you got was accidentally correct, but the logic behind your calculations was wrong.

You have 0.01185 moles of Cu in 0.9 g sample. Copper percentage is

[tex]\frac{0.01185\times63.5}{0.9}\times100\%[/tex]

you did

[tex]\frac{0.01185}{\frac{0.9}{63.5}}\times100\%[/tex]

These numbers happen to be mathematically equivalent, but the first one makes sense, the latter doesn't - you can't use mass of a mixture to calculate number of moles of one of the mixture components.

edit: mjc123 posted while was typing, so we more or less say the same thing

I see it. I can't believe I made such a stupid error, thanks for opening my eyes. It would have been a lot easier to realise my mistake if it didn't lead to the same answer, funny how that works.

Offline Trueolive

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Re: Redox titration question
« Reply #10 on: February 11, 2022, 05:04:39 PM »
Your approach gets the right numerical answer, but is problematic because it appears to assume there are 14.17 mmol of "stuff" in the impure material, which is not true if the molar mass of the impurity is different from that of copper. Actually you weren't doing that, you were comparing the moles of Cu in the impure sample with the moles of Cu in the same mass of a pure sample, but this may not be clear on first reading. It looks as if you are calculating a mole fraction (wrongly) rather than a mass fraction. If you want to calculate a mass fraction, do it by comparing masses, not moles - it worked out in this case, but in more complicated situations it might lead you astray.

Thank you for the explanation.

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