[keishii]

A sample of copper ore of mass 0.4225g was dissolved in acid and the resulting solution of copper(II) reacted with the iodide ion.
2Cu ²⁺ + 4I ^- --> 2CuI + I₂
The I₂ produced required 29.96 cm³ of 0.02100 mol dm^-3 Na₂S₂O₃ solution for complete reaction according to the following equation.
2S₂O₃^2- + I₂ --> S₄O₆^2- + 2I^-
What was the % by weight of copper in the copper ore? 

[Borek]

Please read forum rules.

You have to do stoichiometry backwards - how many moles of thiosulfate were used? Which means how much iodine? For that amount of iodine - what was the amount of copper (II)?

[keishii]

Sorry I didn't read the forum rules...

So far for 2S₂0₃ ^2- I have moles = 29.96 x 0.021 = 0.62916/ 1000 = 6.2916 x 10^-4
And since the ratio is 2:1 .
6.2916 x 10^-4 / 2 = 3.1458 x 10 ^-4

Then I'm stumped! Would I need to work out the amount in g?

[Lozia]

 You x's 3.1458 x 10 -4 by 2 as 2Cu make 1 I₂

You then do Moles x Mr (of copper) = Actual amount in grams

Then just do (that/0.4225) x 100

That should give you a percentage.