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?
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.