Question:

A piece of impure copper of mass 0.90 g was reacted with concentrated nitric acid. The copper (II) solution made was neutralised and reacted with excess iodide ions to produce iodine.

The iodine liberated in the reaction was titrated with 0.50 mol dm

^{-3} Na

_{2}S

_{2}O

_{3}. It was found that the average titre was 23.70 cm

^{3}.

Calculate the percentage of copper in the sample. Give your answer to 3 sig fig.

Working:

I wrote out the two equations I would need,

2Cu

^{2+} + 4I

^{-} 2CuI + I

_{2} (equation 1)

I

_{2} + 2S

_{2}O

_{3}^{2-} 2I

^{-} + S

_{4}O

_{6}^{2-} (equation 2)

I then calculated the moles of I

_{2} from 2S

_{2}O

_{3}^{2-}(0.5 mol dm

^{-3} X 0.0237 dm

^{3}) / 2 = 0.005925 = n of I

_{2}I used this number to calculate the moles of Cu

^{2+}(0.005925 X 2) = 0.01185 moles

I calculated the moles of Cu

^{2+} by doing mass/Mr (0.90g / 63.5) = 0.01417323 moles

It is the next bit I am confused with,

To calculate the percentage of copper in the sample I did (0.01185 moles / 0.01417323 moles) X 100 = 83.608324%

My answer:

83.6%

Obviously, it's not a hard question but I feel like I've made a stupid mistake. Any corrections or feedback would be greatly appreciated.