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1. Experiment 1: Oxidation to Alkanal Method
This method gives good results for the experimental determination of ethanol content in
white wine.
The colour of red wine or beer will mask the end point of the titration.
The concentration of ethanol (ethyl alcohol, C2H5OH) is to be determined using a redox back
titration method.
Procedure:
1. Pipette 10.0 mL of wine into a 250.0 mL volumetric
flask.
2. Add distilled water to make up to 250.0 mL and mix
thoroughly.
3. Pipette 20.0 mL of diluted wine into a conical flask.
4. Add 20.0 mL aliquot of 0.04 mol L-1 potassium
dichromate solution.
5. Add 10 mL concentrated sulfuric acid (40%)
6. Place stopper in flask loosely and heat in a 45oC
water bath for 10 minutes then remove from heat.
7. Add 2 g potassium iodide.
8. Titrate solution with 0.010 mol L-1 sodium
thiosulfate solution under solution clears.
9. Add 1 mL starch solution and continue titrating until
blue colour disappears and solution is clear green.
10. Record the volume of sodium thiosulfate (the titre).
11. Repeat procedure on several samples of the diluted
wine until concordant titres are achieved.
Safety Considerations
 Sulfuric acid is
corrosive.
Wear safety glasses
(goggles)
 Avoid contact with
iodine.
Iodine stains skin,
clothing & benches.
 Potassium dichromate
is a carcinogen.
Avoid contact with
skin.
Calculations:
a. average titre (mL of Na2S2O3)
b. moles of S2O3
2- used in titration with dichromate ions.
c. moles of I2 present in the flask at the time of the titration.
d. moles of Cr2O7
2- in the flask at the time of titration.
e. moles of Cr2O7
2- added to the diluted wine sample at the start of the experiment.
f. moles of Cr2O7
2- that reacted with ethanol.
g. moles of C2H5OH in the diluted wine sample that reacted with Cr2O7
2-
h. concentration of C2H5OH in diluted wine sample.
i. concentration of C2H5OH in original, undiluted wine.
Calculating the Ethanol Content in Wine in mol L-1
10 mL of wine was placed in a 250 mL volumetric flask and water was added up to the
mark.
20.0 mL of the diluted wine was placed in a conical flask with 20.0 mL of 0.04 mol L-
1 K2Cr2O7(aq) and sulfuric acid to acidify the solution.
After gentle heating, 2 g of KI(s) was added and the solution titrated with 0.010 mol L-
1 Na2S2O3(aq).

2. The results of the experiment are shown below:
sample
volume of sodium thiosulfate solute
(titre)
A 31.92 mL
B 29.63 mL
C 29.61 mL
Calculation: concentration of ethanol in the wine.
a. Calculate the average titre:
Disregard the first titre as being a "rough" titration to help establish the approximate
end point.
average titre = (29.63 + 29.61) ÷ 2 = 29.62 mL of sodium thiosulfate solution.
b. Calculate the moles of sodium thiosulfate that reacted with the iodine:
concentration (mol L-1) = moles ÷ volume (L)
moles (Na2S2O3) = concentration (mol L-1) × volume (L)
concentration (Na2S2O3) = 0.01 mol L-1 (given in the steps above)
volume (Na2S2O3) = average titre = 29.62 mL = 29.62 mL ÷ 1000 mL/L = 0.02962
L
moles (Na2S2O3) = 0.01 × 0.02962 = 2.96 × 10-4 mol
c. Calculate the moles of iodine (I2) present in the flask that have reacted with the
thiosulfate solution (S2O3
2-):
Balanced redox equation:
I2(aq) + 2S2O3
2-(aq) → 2I-(aq) + S4O6
2-(aq)
1 mole of I2 reacts with 2 moles of S2O3
2-
therefore ½ mole I2 reacts with 1 mole S2O3
2-
so, 2.96 × 10-4 mol S2O3
2- reacted with ½ × 2.96 x 10-4 = 1.48 × 10-4 moles of I2
moles of I2 = 1.48 × 10-4 mol
d. Calculate the moles of dichromate that were in excess in the flask:
Balanced redox equation:
Cr2O7
2-(aq) + 14H+ + 6I-(aq) → 2Cr3+(aq) + 7H2O(l) + 3I2(aq)
1 mole dichromate (Cr2O7
2-) produces 3 moles I2
therefore 1 mole I2 is produced by 1/3 mole Cr2O7
2-
so, 1.48 × 10-4 mol I2 is produced by 1/3 × 1.48 x 10-4 = 4.94 × 10-5 mol of Cr2O7
2-
4.94 × 10-5 mol of Cr2O7
2- was present in the flask AFTER all the ethanol had been
oxidised.
e. Calculate how much dichromate was added to the sample intially:
concentration (mol L-1) = moles ÷ volume (L)
moles = concentration (mol L-1) × volume (L)

3. concentration (Cr2O7
2-) = 0.04 mol L-1 (give in the steps above)
volume (Cr2O7
2-) = 20.0 mL (given in the steps above) = 20.0 mL ÷ 1000 mL/L =
0.020 L
moles (Cr2O7
2-) = 0.04 × 0.020 = 8.0 × 10-4 mol
f. Calculate the moles of dichromate in excess in the flask (after all the ethanol has
been oxidised)
moles (Cr2O7
2-) added = 8.0 × 10-4 mol
moles (Cr2O7
2-) in excess (that is, moles that then reacted with I-) = 4.94 × 10-
5 mol
moles (Cr2O7
2-) that reacted with ethanol = moles added to flask - moles in excess =
8.0 × 10-4 - 4.94 × 10-5 = 7.51 × 10-4 mol
moles Cr2O7
2- that reacted with ethanol = 7.51 × 10-4 mol
g. Calculate the moles of ethanol that reacted with dichromate in the flask:
Balanced redox equation:
3C2H5OH(aq) + Cr2O7
2-(aq) + 8H+ → 3CH3CHO(aq) + 2Cr3+(aq) + 7H2O
3 moles ethanol (C2H5OH) reacts with 1 mole dichromate (Cr2O7
2-)
so, 7.51 × 10-4 moles of dichromate react with 3 × 7.51 × 10-4 mol = 2.25 × 10-
3 moles ethanol
Remember the first step in the experiment when a 20.0 mL aliquote of diluted wine
was placed in the conical flask?
This means that that 20.0 mL sample contained 2.25 × 10-3 moles ethanol.
h. Calculate the concentration of ethanol in the conical flask at the start of the
experiment:
concentration (mol L-1) = moles ÷ volume (L)
moles (C2H5OH) = 2.25 × 10-3 mol
volume (C2H5OH) = 20.0 mL = 20.0 mL ÷ 1000 mL/L = 0.020 L
concentration (C2H5OH) = 2.25 × 10-3 ÷ 0.020 = 0.11 mol L-1
i. Calculate the concentration of ethanol in the wine
Remember that the original wine was diluted, 10.0 mL of wine was placed in a 250.0
mL volumetric flask and water was added up to the mark.
Then a 20.0 mL aliquot of this diluted wine was used in the titration.
We have so far calculated the concentration of ethanol in that 20.0 mL sample and
found it to be 0.11 mol L-1
That means that the concentration of the diluted wine in the 250.0 mL volumetric
flask was 0.11 mol L-1.
Let's calculate how many moles of ethanol must have been in that 250.0 mL
volumetric flask:

4. concentration (mol L-1) = moles ÷ volume (L)
moles = concentration (mol L-1) × volume (L)
concentration (C2H5OH) = 0.11 mol L-1
volume = 250.0 mL = 250.0 ÷ 1000 mL/L = 0.250 L
moles (C2H5OH) = 0.11 × 0.250 = 0.028 mol
All 0.028 moles of ethanol must have come from the original 10.0 mL aliquot of wine
that was added to the 250.0 mL volumetric flask.
Let's calculate the concentration of ethanol in that 10.0 mL wine sample (before it is
diluted)
concentration (mol L-1) = moles ÷ volume (L)
moles (C2H5OH) = 0.028 mol
volume (wine) = 10.0 mL = 10.0 ÷ 1000 mL/L = 0.010 L
concentration (C2H5OH) = 0.028 ÷ 0.010 = 2.75 mol L-1
Since the 10.0 mL aliquot came directly from the wine to be tested, the
concentration of ethanol in the wine must be 2.75 mol L-1
(If you go back to the beginning of the experiment you will remember that we
said that the concentration of the wine will be 25 times the concentration of
the diluted sample we use in the titration ... well .... 25 × 0.11 mol L-1= 2.75
mol L-1!)
Calculation: Concentration of Ethanol in Wine as %(v/v)
Assume the density of ethanol is 0.79 g mL-1
1. concentration of ethanol in wine was found to be 2.75 mol L-1 (see above)
This means that 1 L, or 1000 mL, of wine contains 2.75 mol of C2H5OH
2. Calculate mass of ethanol in 1 L of wine:
moles = mass ÷ molar mass
mass (g) = moles × molar mass
moles (C2H5OH) = 2.75 mol
molar mass (C2H5OH) = (2 × 12.01) + (5 × 1.008) + 16.00 + 1.008 = 46.068 g
mol-1
mass (C2H5OH) = 2.75 × 46.068 = 126.69 g
3. Calculate the volume of 126.69 g of ethanol:
density = mass (g) ÷ volume (mL)
volume (mL) = mass (g) ÷ density (g mL-1)
density (C2H5OH) = 0.79 g mL-1
mass (C2H5OH) = 126.69 g
volume (C2H5OH) = 126.69 ÷ 0.79 = 160.37 mL
4. Calculate the concentration of ethanol in wine in %(v/v)

5. % (v/v) = (volume of ethanol ÷ total wine volume) × 100
volume (C2H5OH) = 160.37 mL
volume (wine solution) = 1 L = 1000 mL
% (v/v) = (160.37 ÷ 1000) × 100 = 16.0 %
Quicker Calculation: Concentration of Ethanol in Wine
Once you understand the process that is occurring in this analysis, it is possible to take a
couple of shortcuts in the calculations.
1. Since all the reactions are taking place in the same vessel and nothing is leaving the
system, we can add together all the iodine production equation and the titration
equation:
iodine
production:
Cr2O7
2-(aq) + 14H+ + 6I-(aq) → 2Cr3+(aq) + 7H2O(l) + 3I2(aq)
titration: 3I2(aq) + 6S2O3
2-(aq) → 6I(aq) + 3S4O6
2-(aq)
overall
equation:
Cr2O7
2-(aq) + 14H+ + 6S2O3
2-(aq) → 2Cr3+(aq) + 7H2O(l) + 3S4O6
2-(aq)
2. Therefore moles of Cr2O7
2- not reacted with ethanol = 1/6 × moles S2O3
2-
3. moles of Cr2O7
2- not reacted with ethanol = 1/6 × (concentration S2O3
2- ×
volume (L))
4. moles of Cr2O7
2- added to sample = concentration × volume (L)
So moles of Cr2O7
2- reacted with ethanol = (concentration Cr2O7
2- × volume Cr2O7
2-)
- 1/6 × moles S2O3
2-
5. moles of C2H5OH :
3C2H5OH(aq) + Cr2O7
2-(aq) + 8H+ → 3CH3CHO(aq) + 2Cr3+(aq) + 7H2O
moles of C2H5OH = 3 × moles Cr2O7
2- reacted
moles of C2H5OH = 3 × ((concentration Cr2O7
2- × volume Cr2O7
2-) - 1/6 ×
(concentration S2O3
2- × volume (L))
nC2H5OH = 3 × ((cCr2O7
2- × VCr2O7
2-) - 1/6 × (cS2O3
2- × VS2O3
2-)
concentration of ethanol in diluted sample = nC 2H5OH ÷ volumeC 2H5OH
concentration of ethanol in diluted sample = 3 × ((cC r2O7
2- × VC r2O7
2-) - 1/6 × (cS2O3
2- × VS2O3
2-
)) ÷ volumeC 2H5OH
concentration of ethanol in undiluted sample = dilution factor × (nC 2H5OH ÷ volumeC 2H5OH)
concentration of ethanol in undiluted sample = (final wine volume ÷ initial wine volume) ×
(3 × ((cC r2O7
2- × VC r2O7
2-) - 1/6 × (cS2O3
2- × VS2O3
2-)) ÷ volumeC 2H5OH)
cethanol in wine = (Vfinal wine volume ÷ Vinitial wine volume) × (3 × ((cCr2O7
2- × VCr2O7
2-) - 1/6 ×
(cS2O3
2- × VS2O3
2-)) ÷ VC2H5OH)

6. Substituting in the values from the experiment above:
cethanol in wine = (Vfinal wine volume ÷ Vinitial wine volume) × (3 × ((cC r2O7
2- × VC r2O7
2-) - 1/6 × (cS2O3
2- ×
VS2O3
2-)) ÷ VC 2H5OH)
cethanol in wine = (0.250 ÷ 0.010) x (3 x ((0.04 x 0.020) - 1/6 x (0.01 x 0.02962)) ÷ 0.020)
cethanol in wine = 25 x (3 x ((8.00 x 10-4 - 4.94 x 10-5) ÷ 0.020) = 2.81 mol L-1
Why isn't this answer EXACTLY the same as the one calculated above?
Because the step-by-step approach used above introduced more rounding errors into the
calculation!