Question

A 2.50 gram sample of bronze was dissolved in sulfuric acid. The copper in the sample reacted with sulfuric acid to form copper sulfate.

Cu (s) +2 H2SO4 (aq) -> CuSO4 (aq) +2 H2O (l)
The copper sulfate formed in this reaction was mixed with I to form copper iodide:
CuSO4 (aq) + 5 I- (aq) -> 2CuI (s) + I3- (aq) + 2 SO4^2- (aq)

The I3- (aq) formed in this reaction was then titrated with S2O3^2- (aq):
I3- (aq) + S2O3^2- (aq) -> 3 I- (aq) + S4O6^2- (aq)

Calculate the mass percent of copper in the original sample if 31.5 mL of 1.00 M S2O3^2- were consumed in this titration.

I haven't taken Gen. Chem. in 2 years and can't remember how to do titration questions. I'm trying to help a freshman that has an exam on Wed., but she doesn't have any notes to help me with this problem.

Answer 1

moles S2O3^-2 = M x L = 0.0315*1M = 0.0315.
Now you want to convert this to moles Cu, using the coefficients in the balanced equation. I have shortened the above sequence into the following
1 mole Cu ==>1 mole CuSO4 ==> 1 mole I3^- ==> 2 moles S2O3^-2; therefore, 1/2 mole Cu = 1 mole S2O3^-2.
Since mole S2O3^-2 = 0.0315, then moles Cu must be 1/2 that.
g Cu = moles Cu x molar mass Cu.
mass %Cu = (mass Cu/mass sample)*100 = ??

Answer 2

80%