[Pineapple]

Hi,
I have attempted the calculations for an Alcohol Back Titration, however the result seems to be incorrect.
I need urgent help as this is for my last chemistry assignment, and I would like to get the correct values. 
Please find listed below the method for the calculations, known values and my working.
Any help would be greatly appreciated.
Thank you!

Method for Calculations:

1. Calculate the amount of dichromate ions, in mol in each 10mL aliquot of of potassium dichromate solution.
2. Calculate the average titre of standard sodium thiosulfate solution and the amount of thiosulfate ions in this titre.
3. Deduce the amount of iodine formed during the reaction with excess dichromate.
4. Use the equation between dichromate ions and I- ions to deduce the amount of dichromate ions in excess after the oxidation of ethanol.
5. By subtracting the amount of excess dichromate ions from the original amount of dichromate ions added to the diluted wine (Q1), calculate the amount of dichromate ions that reacted with ethanol.
6. Using the equation for the reaction between ethanol and dichromate ions, find the amount of ethanol in each 10mL aliquot of diluted wine.
7. Calculate the amount of ethanol in the 250mL volumetric flask.
8. Find the molar mass of ethanol and hence calculate the mass of ethanol in the original 10mL sample of wine.
9. Given the density of ethanol is 0.785g/mL-1, calculate the volume of ethanol in the 10mL sample of wine.
10. Determine the percentage of alcohol in the wine.

Know Values:

Wine sample: 10mL, then diluted to 250mL, after a 10mL sample taken
Titration result average: 15.55mL
Potassium Iodide: 1g
Potassium Dichromate:
Concentration 0.0400mol/L
10mL used
Sodium Thiosilfate:
Concentration 0.1M


My Calculations:
 1.  C=n/V therefore
n=C x V = 0.0400 x 0.01 = 0.0004 mol dichromate

2. n= C x V = 0.1 x 0.01555 = 1.555 x 10-3 mol of thiosulfate

3. I(2) + 2S(2)O(3)^2- -> 2I^- + S(4)O(6)^2-
Ratio: 2 thisoulfate : 1 iodide
Therefore, 1.555 x 10-3 mol/2 = 7.775 x 10-4 mol of I(2)

4. Cr(2)O(7)^2- + 14H^+ + 6I^- (Upper case i) -> 2Cr^3+ + 7H(2)O + 3I(2)
Ratio: 3 iodide: 1 dichromate
Therefore, 7.775 x 10-4 mol /3 = 2.592 x 10-4 mol of dichromate

5. Cant subtract, because i have less then what I started with?

6. 3C(2)H(5)OH + Cr(2)O(7)^2- + 8H^+ -> 3CH(3)CHO + 2Cr^3+ + 7H(2)O
Ratio: 1 Dichromate:  3 Ethanol
Therefore, 2.592 x 10-4 mol x 3 = 7.775 x 10-4 mol ethanol
C=n/V = 7.775 x 10-4 mol / 0.01 = 0.07775 mol/L-1

7. n=C.V =0.07775 x 0.25 = 0.0194373 mol ethanol

8. Molar mass of ethanol= 46.06844 g/mol
n=m/M, therefore m= n x M = 0.0194373 x 46.06844 = 0.895 g

9. D= 0.785 g
V=M/D = 0.895/0.785 = 1.141 mL

10. 1/100 = 1.141/ x
1x = 1.141 x 100
x = 114.1
This does not seem right

[Burner]

[...]
3. I(2) + 2S(2)O(3)^2- -> 2I^- + S(4)O(6)^2-
Ratio: 2 thisoulfate : 1 iodide
Therefore, 1.555 x 10-3 mol/2 = 7.775 x 10-4 mol of I(2)

In step 3, you need to calculate the no. of moles of iodide, not iodine.

4. Cr(2)O(7)^2- + 14H^+ + 6I^- (Upper case i) -> 2Cr^3+ + 7H(2)O + 3I(2)
Ratio: 3 iodide: 1 dichromate
Therefore, 7.775 x 10-4 mol /3 = 2.592 x 10-4 mol of dichromate

5. Cant subtract, because i have less then what I started with?

Your mole ratio is incorrect. Dirchromate ions refer to Cr₂O₇^2-, not Cr³⁺.

[Pineapple]

Sorry for the late reply. Thank you for your help.

Reading back over what I had written, I see that I made a mistake in question 3 as I was calculating the ratio  between thiosilfate and iodide, while the question is asking to
3. Deduce the amount of iodine formed during the reaction with excess dichromate.

So, would it be correct for me to say that the ratio for question 3 should be 1 dichromate: 3 iodine ?
Cr(2)O(7)^2- + 14H^+ + 6I^- (Upper case i) -> 2Cr^3+ + 7H(2)O + 3I(2)

[Burner]

Sorry for the late reply. Thank you for your help.

Reading back over what I had written, I see that I made a mistake in question 3 as I was calculating the ratio  between thiosilfate and iodide, while the question is asking to
3. Deduce the amount of iodine formed during the reaction with excess dichromate.

So, would it be correct for me to say that the ratio for question 3 should be 1 dichromate: 3 iodine ?
Cr(2)O(7)^2- + 14H^+ + 6I^- (Upper case i) -> 2Cr^3+ + 7H(2)O + 3I(2)

You should calculate the amount of iodine from step 2 using the number of moles of thiosulfate ions.

[...]
3. I(2) + 2S(2)O(3)^2- -> 2I^- + S(4)O(6)^2-
Ratio: 2 thisoulfate : 1 iodide
Therefore, 1.555 x 10-3 mol/2 = 7.775 x 10-4 mol of I(2)

In step 3, you need to calculate the no. of moles of iodide, not iodine.

As I re-read your previous steps, I think your step 3 should be correct, you just typed 'iodine' to 'iodide' wrongly.

[Pineapple]

Thank you so much! 😊