Extracting Copper II Sulphate

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Homework Statement



For a final task, I need to determine a procedure to extract pure copper (Cu(s)) from a mixture of CuSO4 pentahydrate and NaCl.

The Attempt at a Solution



Well, I honestly don't know where to begin. I think I would need to dissovle the micture in water to seperate the hydrate from the copper II sulphate. Then I'm stukc. Can anyone offer any guidance?
 

Answers and Replies

  • #2
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Is this single displacement reaction possible?

Cu(OH)2(aq) + Mg(s) [or Fe(s)] --> Cu(s) + Mg(OH)2(aq) [or Fe(OH)2(aq)]

this way, you can get solid copper from an aqueous solution of copper hydroxide?
 
  • #3
Borek
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Hydroxide? What for?
 
  • #4
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Ok so what I came up with is to dissolve the copper sulphate pentahydride and the NaCl in water. Then I mixed in sodium hydroxide:

NaCl + NaOH --> NR
CuSO4(aq) + NaOH(aq) --> Cu(OH)2(s) + Na2SO4(aq)

So now that copper hydroxide is a precipitate, i can take it out using filter paper or something, then do a single displacement reaction with Mg (higher than copper in activity series):

Cu(OH)2 (aq) + Mg (s) --> Cu(s) + Mg(OH)2(s)

i just wanted to know if the last reaction is possible, i tried googling this reaction and nothing comes up.
 
  • #5
Borek
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That's not the bad idea, although you won't be able to reduce hydroxide - you will need to convert it into something soluble.

But it can be done much simpler, without filtering hydroxide (which is a thing you don't want to do - it is a colloidal precipitate, that clogs filtering paper). You can do it "in one pot".
 
  • #6
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If you wanted to do it in the same container, then wouldn't you need to first remove the NaCl ?

And what do you mean by "you won't be able to reduce hydroxide - you will need to convert it into something soluble" ?
 
  • #7
Borek
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If you wanted to do it in the same container, then wouldn't you need to first remove the NaCl ?
No - you just have to be sure its presence will not interfere. Once you have solid copper, you can filtrate it and NaCl will be gone with the solution.

And what do you mean by "you won't be able to reduce hydroxide - you will need to convert it into something soluble" ?
How are you going to mix solid Mg and solid copper hydroxide? Even if you do, they will never react to the end.
 
  • #8
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Ok I see now. So I just have to make sure that whatever reactions I'm doing to the copper sulphate, it cannot react with the NaCl.

OMG! I just realized that! So I'm guessing that after I get the Cu(OH)2, then I would need to react it further to make it soluble. But then, how will I get it back to solid copper? :|
 
  • #9
Borek
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You don't have to reduce solid hydroxide to get solid copper. You can reduce dissolved copper - it will precipitate as solid.
 
  • #10
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Are you suggesting that after the CuSO4 and NaCl has been dissolved in water, we can add in Mg, which, in the reactivity series, is higher than copper and lower that sodium so that only the copper II sulphate will react with it, and produce solid copper? And from there, we can filter that precipitate out?
 
  • #11
Borek
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Yes, although you will need one more operation. When you add Mg you have to add it in excess, to be sure all copper was reduced. That means after reduction you are left with mixture of solid copper and leftover Mg. However, removing Mg is pretty simple.
 
  • #12
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So I need to do stoichiometry to find out how much Mg I need. Another question, for Copper II sulphate pentahydrate, the chemical forumula is CuSO4 * 5H20. When calculating the molar mass, since it is dissolved in water, do I need to consider 5H20?
 
  • #13
Borek
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Depends on what do you need molar mass for.

Using stoichiometrical amount of Mg is not bad, but using small excess is even better. Besides, you can't use stoichiometric amount if your are not given copper sulfate concentration. Best approach then is to add Mg in small batches till it obvious that the last batch didn't react. However, that means after reaction you have a mixture of solid magnesium and copper.

What methods of metal dissolving do you know?
 
  • #14
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Actually, its a mixture of CuSO4 * 5H20 (s) and NaCl(s) and that will be dissolved in water. Once I put magnesium strips in it, it'll react overnight, then the next day, I will need to filter it, so I will have the pure copper and some excess magnesium strip on the filter paper. That is my next problem. How do I get the copper alone?

I was thinking that I should use HCl(aq) on it, because it will not react with the copper, but with the magnesium to produce an aqueous MgCl2 and hydrogen gas. But the question is, how much HCl do I need? Do I use an arbitrary amount, or can it be calculated by stoichiometry?

I think what I should do is find the stoichiometrical amount of Cu produced, and subtract it from the weight of the filter paper with the magnesium strips and copper on it. Then use that to find out how much HCl to use. Does this sound good?
 
  • #15
Borek
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If you know how much copper to expect, that's OK. I would go for much simpler procedure of using a reasonable excess - when Mg bubbles, it dissolves, when it stops to bubble - add more acid. If it doesn't bubble after adding acid - there is no more magnesium.
 
  • #16
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OK, so the amount of HCl used isn't really necessary to any calculations anyway (purpose is to find percent compostiion of copper in the ore sample), it is just to separate the magnesium from the copper. So I think it would be a good idea to use Mg in excess and then just add some HCl after the precipitate and Mg is filtered out
 
  • #17
Borek
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Possible, but not necessary - you can dissolve excess Mg before filtration.
 
  • #18
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That's not a bad idea. I did the chemical equations and it doesn't affect anything other than the Mg in the solution. That's great.

So let's say all of the copper is now filtered out and I have water, NaCl and other aqueous substances in the mixture. Would it be possible to just heat the mixture, and separate the "liquids" from the salt in it?
 
  • #19
Borek
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It is possible to evaporate water and recrystallize salts, with proper techniques it will be even possible to separate them - but what for? Your task was to separate copper, if it is already filtrated you can safely throw away everything else. Just wash your copper with distilled water and let it dry.
 
  • #20
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"Explain how you can hypothetically extract the salt from the leftovers of the reaction."

So with the Mg and the Cu all gone, there is now MgSO4(aq), MgCl2(aq) and NaCl(aq) in the dissolved water. So then, lets say you take an ordinary Bunsen burner to heat up and evaporate the water. Do the MgSO4(aq) and MgCl2(aq) go away with in the air and leave behind NaCl?
 
  • #21
Borek
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If you have followed the procedure with excess Mg and excess HCl, it won't be easy. If you want to separate NaCl other approaches will be better.

First that I can think of requires using iron filings and precipitating iron hydroxide with strong base. Try to design the procedure (order of operations) by yourself. Note that when copper is reduced by iron Fe2+ is produced - it won't hurt to oxidize it to Fe3+ before precipitation. It is not that difficult to oxidize iron - even atmospheric oxygen will do, it just takes time.
 
  • #22
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so I'd have to redesign the entire experiment from the beginning (i.e instead of using Mg use Fe)?
 
  • #23
Borek
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If you want to separate NaCl as well - then yes.

You initially stated you want to extract copper using magnesium and you have not stated anything about reclaiming NaCl. As Mg is a reasonable idea I followed this path, but it is not the only one.
 
  • #24
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Okay thank you very much!

And one last question: is it possible to calculate the percent composition of copper in the mixture if I know the mass of one of the products?

ie. Na2CO3(s) + CuSO4(aq) --> Na2SO4(aq) + CuCO3(s)

Cu = 63.546g/mol C = 12.011g/mol O = 15.999g/mol
Assume1 mole of CuCO3,
mass of CuCO3 = 123.554g
%Cu in compound = m(Cu) / m(CuCO3) x 100% = 63.546g / 123.554g x 100% = 51%
∴ in a mass of 0.33g of CuCO3, there is 0.51 x 0.33g = 0.1683g of Cu
%Cu = mass(Cu) / mass(Ore sample) x 100% = 0.1683g / 1.00g x 100% = 0.1683 x 100% = 16.83%

would this be correct?
 
  • #25
Borek
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Your calculation can be right, but whether it is right depends on the input data - and you have not told us what is the question you are solving!
 

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