Topic: Percentage of Cu in mineral? (Read 2640 times)

big · « **on:** November 30, 2011, 06:03:39 PM »

The amount of copper in a 2.00 g sample of the mineral cuprite was determined by dissolving the sample in nitric acid (HNO₃) to produce a copper nitrate [Cu(NO₃)₂] solution, and then adding an excess of iodide (I^-) solution. The iodine (I₂) liberated required 15.7 mL of a 0.200 molar sodium thiosulfate (Na₂S₂O₃) solution to reach an end point. What is the percentage of Cu in the mineral? The essential reactions are:

2 Cu²⁺ + 4I^-

2 CuI + I₂
I₂ + 2 S₂O₃^2-

S₄O₆^2- + 2 I^-

A) 5.0%
B) 10%
C) 20%
D) 40%

I found the amount of moles of thiosulfate ion first by doing .0157 L*0.200 M=0.00314 moles S₂O₃^2-. Then I did dimensional analysis:
0.00314 moles S₂O₃^2-*(1 mole I₂/2 moles S₂O₃^2-)*(2 moles Cu²⁺/1 mole I₂)*(63.55 g/mol Cu)=0.199547g Cu.
Finally, I used this to find the percentage of copper in the original 2.00 g sample: 0.199547/2.00*100=9.98%, and so I put down B, but the answer is A. I must have not divided by 2 somewhere, but where, and why?