What is the percentage of Fool's Gold in the original sample?

[thejosh]

Homework Statement

Joshua has been gold panning and extracts a mixture of silicate stone and shiny gold-like particles. However Nkosi says these are Fool's Gold i.e Chalcopyrities (Cu Fe S2). They decide to find out how much iron is in the sample by:

-Grinding a 30.00g sample to a fine powder
-Dissolving the powder in 100ml of 0.1mol/dm3 Nitric acid
-Filtering the mixture to separate the silica and gold(?) from the copper and iron as the former compound do not react with nitric acid.
-Titrating 25ml of the filtrate with potassium permanganate (KMnO4) until the purple colour disappears.

In this titration Fe(ii) ion are being oxidised and manganese (vii) ions reduced to manganese (ii) ions.

Nkosi weighed the dried residue and obtained a mass of 24.48g.

What was the percentage of fool's gold in the original sample?

Homework Equations

volume * concentration = mol
mass/Molecular formula = mol

The Attempt at a Solution

I am at an almost complete loss whether I should find the number of moles of iron dissolved or write a chemical equation between the iron and the nitric acid, any help would be greatly appreciated and I know this may be annoyingly easy to most of you but we have to start somewhere right? Thanks for the help in advance.

 

[Borek]

The only reaction that matters is the one between Fe(II) and permanganate. Assuming all iron has been dissolved, number of moles of iron doesn't depend on the dissolution reaction.

 

[thejosh]

So I should write the chemical equation for the reaction between Fe(ii) and permanganate then find the number of moles of iron using it and the titration volume? but what is the weighed amount all about ? Is it the Copper and Iron without the silicon?

 

[Borek]

Does silica and gold (if any present) dissolve in the nitric acid?

Or, to put it differently: what was dissolved?

 

[thejosh]

I guess the copper and iron was dissolved but copper isn't mentioned in the reaction with potassium permanganate. To get this straight copper and iron were dissolved but gold and silicon wasn't, the gold and silicon is then filtered and the filtrate( which consists of copper and iron) is reacted with potassium permanganate(25ml). So you're supposed to calculate the number of moles of iron from the residue mass and the reacted potassium permanganate then you use this to calculate the total mass of fool's gold then you use this as the percentage of the whole thing, am I at least in the right direction?

 

[Borek]

Sorry, on the second read (I just skimmed at first) I realized there is either a trap here, or you have not posted the whole question.

If the volume of the titrant is not given (which is the part I missed), can you calculate amount of the iron?

But: if you assume as I suggested, that ONLY chalcopiryte got dissolved, don't you have enough information to calculate its mass?

 

[thejosh]

I feel so embarrased,:
Peter repeated the titration three times and obtained the following titre volumes using 3.8g /dm3 potassium permanganate solution: 32, 32.05, 31.95 ml
I didn't read this part as it was sandwiched at the bottom of the paper near other questions, my apologies

 

[Borek]

Now I am getting lost as well, as the mass of the fools gold I get from the titration is different than the mass difference of the sample before and after the dissolution. That suggests something else what dissolved as well, but I have no idea what.

Two things you can do easily:
a. calculate mass of what was dissolved in the nitric acid
b. calculate amount of iron present (and use it to calculate mass of the chalcopyrite - you don't need mass of copper nor sulfur for that, just think how many moles of CuFeS₂ contain calculated number of moles of iron).

If the chalcopyrite were the only thing dissolved, these numbers should be identical. As far as I can tell they are not.

 

[thejosh]

I'll look into it and see what I come up with.

 

[thejosh]

I still have not found a plausible solution

 

[Borek]

I have a feeling something is wrong with the problem.