Alcohol titration calculations help (1 Viewer)

[DBC]

Hi.
I am doing a titration to see how much alcohol is in wine when left exposed to air at different times. I'm really confused about my calculations.

Method:
1. Dilute wine samples 1:50
2. Transfer 10mL of acid dichromate solution to a 250mL flask
3. Pipette 1.0 of diluted wine into a sample holder
4. Suspend sample holder over dichromate with rubber stopper and leave overnight
5. Next day remove sample holder carefully. Add 100mL of water and 1mL of potassium iodide to dichromate
6. Fill a burette with sodium thiosulfate
7. Pipette 50mL of solution in flask into another flask and titrate the flask against the sodium thiosulfate. When the brown colour fades to yellow add starch solution and keep titrating until the blue colour disappears.
<o> </o>
So I've done this and got a result of 6.6mL average.
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Calculations
I calculated my concentration of my thiosulfate to be 0.047058823molL-1
I also calculated my dichromate concentration to be 0.010197144molL-1
(
I have tried calculating n(C2H5OH) in two ways:
(rounded calcuations to 3sf)
1st way
n(S2O32-)= cxv
= 0.0471x(6.6/1000)
= 0.00031086mol
n(I2)= n(S2O32-)/2
= 0.000311/2
= 0.0001555mol
n(Cr2O72-) {unreacted}= n(I2)/3
= 0.000156/3
= 0.000052mol
total n(Cr2072-)= cxv
= 0.0102x(5/1000)
= 0.000051mol
Then this is where I am stuck
n(Cr2O72-)= total n (Cr2O72-)-n(Cr2O72-)
= -.........
2nd way
n(S2O32-)= cxv
= 0.0471x(6.6/1000)
= 0.00031086mol
n(Cr2O72-) remaining= 1/6n(S2O32-)
= 0.00005181mol
n(Cr2O72-) original= cxv
= 0.0102x(5/1000)
= 0.000051mol
n(Cr2O72-) used up= original-remaining
=-.........
then do I go:
n(C2H5OH)= 3/2n(Cr2O72-)?
<o> </o>
What calculations should I use?
What am I doing wrong?
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These are the equations given
2Cr2O72- + 16H+ + 3C2H5OH ----> 4Cr3+ + 11H2O + 3CH3COOH
Cr2O72- + 14H+ + 6I------> 2Cr3+ + 3I2 + 7H2O
2S2O32- + I2---> S4O6 2- + 2I
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Much appreciated