[mrorganiclover]

Good day ladies and gentlemen

To set the scene:

I performed a synthesis of copper oxalate complex, which I had to analyse by titration against [KMnO4] = 0.02024M to determine the percent by mass of oxalate in my sample.

My calculation thus far:

2MnO4- + 5C2O42- +16H+ 2Mn2+ + 10CO2 + 8H2O

avg. titre g-1 = 102.15cm^3 g^-1

(n) KMnO4 = 102.15cm^3 g^-1 / 1000 = 0.1022 dm^3 g^-1

(n) 0.02024 mol dm^-3 * 0.1022dm^3 g^-1 = 2.069*10^-3 mol g^-1

(n) KMnO4 = 2.069*10^-3 mol g^-1

(n) oxalate = 2/5 * (2.069*10^-3 mol g^-1) = 8.276*10^-4 mol

Mass of oxalate = 8.276*10^-4 mol * 88.02g mol^-1 =0.07285 g

% by mass = 0.07285 g * 100 = 7.285%


My value for % by mass seems rather low to me.

All help greatly appreciated.

Thanks

[mjc123]

Moles of oxalate = 5/2 * moles of KMnO₄, not 2/5.

[mrorganiclover]

Many thanks

With your correction I have % by mass as 45.53% which seems better.

Would you say this is correct?

[mjc123]

Looks OK to me.

[mrorganiclover]

Thanks for your above help.

I have a further question:

To determine Cu2+ a titration was performed against thiosulfate + KI

I2 + S2O32- = 2I- + S4O62- (1:1)

[S2O32-] = 0.1388M

avg titre g-1 = 100.49 cm3 g-1 = 0.1005 dm3 g-1

(n) = 0.1388 mol dm-3 * 0.1005 dm3 g-1 = 0.01395 mol g-1

(n) thiosulfate = 0.01395 mol g-1

Therefore (n) liberated I2 = 0.01395 mol g-1 (1:1)

Cu2+ + 2I- = CuI + 1/2 I2

And (n) Cu2+ = (0.01395 mol g-1)/2 = 6.975*10-3 mol

63.55 g mol-1 * 6.975*10-3 mol = 0.4433g * 100 = 44.33%

All help appreciated.

[Borek]

What does "g-1" mean?

[mrorganiclover]

Per gram

[mjc123]

You've got the stoichiometric calculation wrong again. Do you understand it at all?

Cu2+ + 2I- = CuI + 1/2 I2

1 mol Cu²⁺ reacts to give 1/2 mol I₂. Therefore the amount of Cu²⁺ is twice (not half) the amount of I₂.
The problem with this is that it gives an answer of more than 100%. Check your working. Have you copied down a number wrongly? Should the thiosulfate concentration be 0.01388 M? If so, then the answer comes out right for what I think your compound is.

[mrorganiclover]

For my calculation I used titre per gram which is why I think the answer was incorrect.

How does this look now:

[S2O3^2-] = 0.1388M

avg. titre = 27.615 cm^3

27.615 cm^3/1000 = 0.02762 dm^3

(n) = 0.02762 dm^3 * 0.1388 mol dm^3 = 3.834 * 10^-3 mol

therefore moles I2 = 3.834 * 10^-3 mol

moles Cu^2+ = (3.834 * 10^-3 mol) * 2 = 7.668*10^-3 mol

mass Cu^2+ = (7.668*10^-3 mol) * 63.55 g mol^-1 = 0.4873g

percent by mass = 0.4873g * 100 = 48.73%

[mjc123]

No, this is wrong. You need titre per gram in order to do the last step, which is assuming that 0.4873g Cu is 48.73%. You can only do this if your sample mass is 1g. Your sample mass for the average titre is not given, so you can't solve it this way. If your previous value of titre per gram of 100.49 cm³ is correct, that implies your sample mass was 0.2748g - so again, the Cu is over 100%. There must be something wrong in your data. Did you double check the thiosulfate concentration? Sample mass? Everything else?