Redox Titration of Ethanol in Wine and Beer

Key Concepts

• Ethanol (ethyl alcohol, C₂H₅OH) is present in alcoholic beverages such as wine, beer and spirits.
• Depending on the conditions of the experiment;

  Experiment 1: ethanol can be oxidised to ethanal (acetaldehyde, CH₃CHO) using a suitable oxidising agent such as acidified potassium dichromate solution (K₂Cr₂O₇)¹

  3C₂H₅OH(aq) + Cr₂O₇^2-(aq) + 8H⁺ → 3CH₃CHO(aq) + 2Cr³⁺(aq) + 7H₂O
  or
  Experiment 2: ethanol can be oxidised to ethanoic acid (acetic acid, CH₃COOH):
  3C₂H₅OH(aq) + 2Cr₂O₇^2-(aq) + 16H⁺ → 3CH₃COOH(aq) + 4Cr³⁺(aq) + 11H₂O

• In order to ensure that all the ethanol has reacted, an excess of the oxidising agent is used.
• Excess dichromate ions are reacted with iodide ions to produce coloured I₂:

  Cr₂O₇^2-(aq) + 14H⁺ + 6I^-(aq) → 2Cr³⁺(aq) + 7H₂O(l) + 3I₂(aq)

• The I₂, iodine, produced is then titrated with sodium thiosulfate (Na₂S₂O₃):

  I₂(aq) + 2S₂O₃^2-(aq) → 2I^-(aq) + S₄O₆^2-(aq)

  The volume of thiosulfate, S₂O₃^2-, is recorded.

• Calculations :

  The volume of thiosulfate used in the titration and its concentration is used to calculate the moles of I₂ present.

  Moles of I₂ is used to calculate the moles of dichromate in excess after the ethanol was oxidised.

  Moles of dichromate reacted is calculated using the volume and concentration of dichromate solution added to the solution originally minus the dichromate that was in excess.

  Moles of ethanol present in the prepared sample is calculated using the moles of dichromate that reacted.

  Concentration of ethanol can then be calculated for the prepared sample, and then for the original wine sample.

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Theory

Preparing the Wine Sample for Analysis

The ethanol content of wine can be as high 16%(v/v) which is quite a concentrated solution. So the wine is first prepared by diluting it.

1:25 dilution used in Experiment 1 (oxidation to ethanal)

1. Use a pipette to place a 10.0 mL aliquot of wine into a 250.0 mL volumetric flask and make up to the mark with water.
2. Use a pipette to transfer 20.0 mL of this diluted wine to a conical flask.
   This conical flask will be used in the next step in which the ethanol is oxidised.
3. It is important to remember that you have diluted the original wine.
   When you come to calculate the concentration of the ethanol in the wine later on you will need to remember that the ethanol concentration in the wine will be much greater than the concentration of ethanol in the sample you use in your titration.
   (250/10 = 25 times greater!)

1:50 dilution used in Experiment 2 (oxidation to acetic acid)

1. Use a pipette to place a 20.0 mL aliquot of wine into a 1.0 L volumetric flask and make up to the mark with water.
2. Use a pipette to transfer 1.0 mL of this diluted wine to a sample holder.
   This ethanol is to be oxidised in the next step.
3. It is important to remember that you have diluted the original wine.
   When you come to calculate the concentration of the ethanol in the wine later on you will need to remember that the ethanol concentration in the wine will be much greater than the concentration of ethanol in the sample you use in your titration.
   (1000/20 = 50 times greater!)

Oxidising the Ethanol

1. Potassium dichromate (K₂Cr₂O₇) is an oxidising agent, it causes something else (the ethanol) to be oxidised.
   In this reaction chromium will be reduced from an oxidation state of +6 in Cr₂O₇^2- to +3 in Cr³⁺.
2. The potassium dichromate solution must be acidified by adding concentrated sulfuric acid.
   A source of protons, H⁺, is required for this redox reaction. Concentrated sulfuric acid is providing the protons.
3. Ethanol is highly volatile, on heating the ethanol will evaporate.
   We need to ensure that ALL of the ethanol (C₂H₅OH) in the wine sample is oxidised to ethanal (CH₃CHO) in experiment 1, or to acetic acid (CH₃COOH) in experiment 2.
   In experiment 1, the flask is stoppered to prevent the ethanol being lost and the overall redox reaction is:
   

   3C2H5OH(aq)

   +

   Cr2O72-(aq)

   +

   8H+

   →

   3CH3CHO(aq)

   +

   2Cr3+(aq)

   +

   7H2O

   
   In experiment 2, a small wine sample is warmed overnight to evaporate all the ethanol and trap it in a sealed flask with the dichromate, and, the next morning the sides washed down with water to ensure all the ethanol in washed into the flask.
   The overall redox reaction equation is:
   

   3C2H5OH(aq)

   +

   2Cr2O72-(aq)

   +

   16H+

   →

   3CH3COOH(aq)

   +

   4Cr3+(aq)

   + 11H2O

   
   Note that you solution should NOT be green!
   Because we need to add an excess of the orange dichromate solution, the solution in our flask after all the ethanol has been oxidised should still be orange (or possibly a darker red-orange colour).
   The concentration of ethanol in the conical flask in either experiment 1 or 2 is now 0
   The conical flask contains excess dichromate ions (Cr₂O₇^2-), excess H⁺, K⁺ spectator ions, as well as the products of the reaction chromium(III) ions (Cr³⁺) and either ethanal (experiment 1) or acetic acid (experiment 2).

Reacting Excess Dichromate with Iodide

oxidation:     2I^- → I₂ + 2e^-

reduction:     Cr₂O₇^2- + 14H⁺ + 6e^- → 2Cr³⁺ + 7H₂O

The overall redox reaction equation:

By adding an excess of potassium iodide we ensure that all the excess dichromate ions will react with the iodide ions to form iodine.

The solution will look rather brown due to the presence of the I₂.

The colour of this solution can be used to indicate the end-point of the titration with sodium thiosulfate, which is the the next step.

Titration with Thiosulfate

1. Fill a burette with a standard sodium thiosulfate solution (Na₂S₂O₃).
2. Titrate the contents of the conical flask (from above) with the thiosulfate solution.

   Balanced redox equation:
   I₂(aq) + 2S₂O₃^2-(aq) → 2I^-(aq) + S₄O₆^2-(aq)
   Note that the presence of I₂ in the solution will make it appear brown.
   Thiosulfate ions act as a reducing agent, reducing I₂ to colourless I^- and thereby removing the brown colour from the solution.

3. Record the titre.
4. Repeat the experiment several times until concordant titres are produced.

   The average titre becomes the basis for calculating the concentration of ethanol in the alcholic beverage.

Experiment 1: Oxidation to Alkanal Method

The concentration of ethanol (ethyl alcohol, C₂H₅OH) is to be determined using a redox back titration method.

Calculations:

1. average titre (mL of Na₂S₂O₃)
2. moles of S₂O₃^2- used in titration with dichromate ions.
3. moles of I₂ present in the flask at the time of the titration.
4. moles of Cr₂O₇^2- in the flask at the time of titration.
5. moles of Cr₂O₇^2- added to the diluted wine sample at the start of the experiment.
6. moles of Cr₂O₇^2- that reacted with ethanol.
7. moles of C₂H₅OH in the diluted wine sample that reacted with Cr₂O₇^2-
8. concentration of C₂H₅OH in diluted wine sample.
9. 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).

The results of the experiment are shown below:

Calculation: concentration of ethanol in the wine.

1. 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.

2. 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

3. 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

4. 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.

5. Calculate how much dichromate was added to the sample intially:

   concentration (mol L^-1) = moles ÷ volume (L)
   moles = concentration (mol L^-1) × volume (L)
   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

6. 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

7. 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.

8. 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

9. 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:

   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)

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)

   % (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

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:	Cr2O72-(aq)	+	14H+	+	6I-(aq)	→	2Cr3+(aq)	+	7H2O(l)	+	3I2(aq)
   titration:			3I2(aq)	+	6S2O32-(aq)	→	6I(aq)	+	3S4O62-(aq)
   overall equation:	Cr2O72-(aq)	+	14H+	+	6S2O32-(aq)	→	2Cr3+(aq)	+	7H2O(l)	+	3S4O62-(aq)

   Therefore moles of Cr₂O₇^2- not reacted with ethanol = ¹/₆ × moles S₂O₃^2-

   moles of Cr₂O₇^2- not reacted with ethanol = ¹/₆ × (concentration S₂O₃^2- × volume (L))

2. 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-

3. 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))

n_C2H5OH = 3 × ((c_Cr2O7^2- × V_Cr2O7^2-) - ¹/₆ × (c_S2O3^2- × V_S2O3^2-)

concentration of ethanol in diluted sample = n_C2H5OH ÷ volume_C2H5OH

concentration of ethanol in diluted sample = 3 × ((c_Cr2O7^2- × V_Cr2O7^2-) - ¹/₆ × (c_S2O3^2- × V_S2O3^2-)) ÷ volume_C2H5OH

concentration of ethanol in undiluted sample = dilution factor × (n_C2H5OH ÷ volume_C2H5OH)

concentration of ethanol in undiluted sample = (final wine volume ÷ initial wine volume) × (3 × ((c_Cr2O7^2- × V_Cr2O7^2-) - ¹/₆ × (c_S2O3^2- × V_S2O3^2-)) ÷ volume_C2H5OH)

c_{ethanol in wine} = (V_{final wine volume} ÷ V_{initial wine volume}) × (3 × ((c_Cr2O7^2- × V_Cr2O7^2-) - ¹/₆ × (c_S2O3^2- × V_S2O3^2-)) ÷ V_C2H5OH)

Substituting in the values from the experiment above:

c_{ethanol in wine} = (V_{final wine volume} ÷ V_{initial wine volume}) × (3 × ((c_Cr2O7^2- × V_Cr2O7^2-) - ¹/₆ × (c_S2O3^2- × V_S2O3^2-)) ÷ V_C2H5OH)
c_{ethanol in wine} = (0.250 ÷ 0.010) x (3 x ((0.04 x 0.020) - ¹/₆ x (0.01 x 0.02962)) ÷ 0.020)
c_{ethanol 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!

Experiment 2: Oxidation to Ethanoic (Acetic) Acid Method

The concentration of ethanol (ethyl alcohol, C₂H₅OH) is to be determined using a redox back titration method.

Calculations:

1. average titre (mL of Na₂S₂O₃)
2. moles of S₂O₃^2- used in titration with dichromate ions.
3. moles of I₂ present in the flask at the time of the titration.
4. moles of Cr₂O₇^2- in the flask at the time of titration.
5. moles of Cr₂O₇^2- added to the diluted wine sample at the start of the experiment.
6. moles of Cr₂O₇^2- that reacted with ethanol.
7. moles of C₂H₅OH in the diluted wine sample that reacted with Cr₂O₇^2-
8. concentration of C₂H₅OH in diluted wine sample.
9. concentration of C₂H₅OH in original, undiluted wine.

Calculation: Concentration of Ethanol in Wine in mol L^-1

20.0 mL of wine was diluted to 1.0 L in a volumetric flask.
1 mL of diluted wine was suspended over 10.0 mL of acidified 0.010 mol L^-1 potassium dichromate solution and left over night in an incubator at 30°C.
After cooling, 100 mL of water and 1 mL of 1.2 mol L^-1 potassium iodide solution was added.
This solution was titrated with 0.030 mol L^-1 sodium thiosulfate solution.
The experiment was repeated several times until concordant titres were acheived.
The average titre was calculated to be 14.50 mL of sodium thiosulfate solution.
Calculate the concentration of ethanol in the wine in mol L^-1.

1. Calculate the moles of S₂O₃^2- used in the titration:

   concentration = 0.030 mol L^-1
   volume = 14.50 mL = 14.50 ÷ 1000 = 14.50 × 10^-3 L
   moles = concentration × volume = 0.030 × 14.50 x 10^-3 = 4.35 × 10^-4 moles

2. Calculate moles of I₂ that reacted with S₂O₃^2- in the titration:

   Balanced chemical equation:
   I₂(aq) + 2S₂O₃^2-(aq) → 2I^-(aq) + S₄O₆^2-(aq)
   2 moles S₂O₃^2- react with 1 mole I₂
   1 mole S₂O₃^2- react with ½ moles I₂
   So, 4.35 × 10^-4 moles S₂O₃^2- react with ½ × 4.35 × 10^-4 = 2.18 × 10^-4 moles I₂
   moles I₂ = 2.18 × 10^-4

3. Calculate moles of Cr₂O₇^2- in flask:

   Balanced redox equation:
   Cr₂O₇^2-(aq) + 14H⁺ + 6I^-(aq) → 2Cr³⁺(aq) + 7H₂O(l) + 3I₂(aq)
   1 mole Cr₂O₇^2- produces 3 moles I₂
   ¹/₃ mole Cr₂O₇^2- produces 1 mole I₂
   2.18 × 10^-4 moles I₂ is produced from ¹/₃ × 2.18 × 10^-4 = 7.25 × 10^-5 moles Cr₂O₇^2-.
   moles Cr₂O₇^2- = 7.25 × 10^-5 mol

4. Calculate moles of dichromate that reacted with ethanol.

   total moles Cr₂O₇^2- in flask = moles Cr₂O₇^2- reacted with ethanol + moles Cr₂O₇^2- in excess (reacted with I^- to produce I₂)
   total moles Cr₂O₇^2- = concentration × volume added to flask originally
   total moles Cr₂O₇^2- = 0.010 mol L^-1 × 10.0/1000 = 1.00 × 10^-4 mol
   moles Cr₂O₇^2- in excess = 7.25 × 10^-5 mol (see above)
   moles Cr₂O₇^2- reacted with ethanol = total moles - moles reacted with I^-
   moles Cr₂O₇^2- reacted with ethanol = 1.00 × 10^-4 - 7.25 × 10^-5 = 2.75 × 10^-5 mol

5. Calculate moles of ethanol that reacted with Cr₂O₇^2-

   Balanced redox equation :
   3C₂H₅OH(aq) + 2Cr₂O₇^2-(aq) + 16H⁺ → 3CH₃COOH(aq) + 4Cr³⁺(aq) + 11H₂O
   2 moles Cr₂O₇^2- react with 3 moles C₂H₅OH
   1 mole Cr₂O₇^2- react with ³/₂ moles C₂H₅OH
   So, 2.75 × 10^-5 moles of Cr₂O₇^2- react with ³/₂ × 2.75 × 10^-5 = 4.13 × 10^-5 moles C₂H₅OH
   moles C₂H₅OH = 4.13 × 10^-5 mol

6. Calculate concentration of ethanol in diluted wine sample:

   moles = 4.13 × 10^-5 mol
   volume = 1 mL = 1 ÷ 1000 = 1.00 × 10^-3 L
   concentration = moles ÷ volume = 4.13 × 10^-5 ÷ 1.00 × 10^-3 = 0.0413 mol L^-1
   concentration of ethanol in diluted wine sample = 0.0413 mol L^-1

7. Calculate concentration of ethanol in undiluted wine:

   ethanol concentration of diluted sample = 0.0413 mol L^-1 (same as in the 1 mL aliquot calculated above)
   volume of volumetric flask = 1.0 L
   moles ethanol in volumetric flask = concentration × volume = 0.0413 × 1 = 0.0413 mol
   All 0.0413 moles of ethanol must have been in the original 20.0 mL aliquot of wine added to the volumetric flask.
   concentration of ethanol in wine = moles ÷ volume = 0.0413 ÷ 20.0 × 10^-3 = 2.07 mol L^-1
   concentration of ethanol in wine = 2.07 mol L^-1

Calculation: %(v/v) Concentration of Ethanol in Wine

1. concentration of ethanol in wine was found to be 2.07 mol L^-1 (see above)

   This means that 1 L, or 1000 mL, of wine contains 2.07 mol of C₂H₅OH

2. Calculate mass of ethanol in 1 L of wine:

   moles = mass ÷ molar mass
   mass (g) = moles × molar mass
   moles (C₂H₅OH) = 2.07 mol
   molar mass (C₂H₅OH) = (2 × 12.01) + (5 × 1.008) + 16.00 + 1.008 = 46.068 g mol^-1
   mass (C₂H₅OH) = 2.07 × 46.068 = 95.36 g

3. Calculate the volume of 95.36 g of ethanol:

   volume (mL) = mass (g) ÷ density (g mL^-1)
   density (C₂H₅OH) = 0.79 g mL^-1
   mass (C₂H₅OH) = 95.36 g
   volume (C₂H₅OH) = 95.36 ÷ 0.79 = 120.7 mL

4. Calculate the concentration of ethanol in wine in %(v/v)

   % (v/v) = (volume of ethanol ÷ total wine volume) × 100
   volume (C₂H₅OH) = 120.7 mL
   volume (wine solution) = 1 L = 1000 mL
   % (v/v) = (120.7 ÷ 1000) × 100 = 12.1 %

Quicker Calculation: Concentration of Ethanol in Wine

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:	Cr2O72-(aq)	+	14H+	+	6I-(aq)	→	2Cr3+(aq)	+	7H2O(l)	+	3I2(aq)
   titration:			3I2(aq)	+	6S2O32-(aq)	→	6I(aq)	+	3S4O62-(aq)
   overall equation:	Cr2O72-(aq)	+	14H+	+	6S2O32-(aq)	→	2Cr3+(aq)	+	7H2O(l)	+	3S4O62-(aq)

   Therefore moles of Cr₂O₇^2- not reacted with ethanol = ¹/₆ x moles S₂O₃^2-
   moles of Cr₂O₇^2- not reacted with ethanol = ¹/₆ x (concentration S₂O₃^2- x volume (L))

2. 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-

3. moles of C₂H₅OH :

   3C₂H₅OH(aq) + 2Cr₂O₇^2-(aq) + 16H⁺ → 3CH₃COOH(aq) + 4Cr³⁺(aq) + 11H₂O
   moles of C₂H₅OH = ³/₂ × moles Cr₂O₇^2- reacted
   moles of C₂H₅OH = ³/₂ × ((concentration Cr₂O₇^2- × volume Cr₂O₇^2-) - ¹/₆ × (concentration S₂O₃^2- × volume (L))

n_C2H5OH = ³/₂ × ((c_Cr2O7^2- × V_Cr2O7^2-) - ¹/₆ × (c_S2O3^2- × V_S2O3^2-)

concentration of ethanol in diluted sample = n_C2H5OH ÷ volume_C2H5OH

concentration of ethanol in diluted sample = (³/₂ × ((c_Cr2O7^2- × V_Cr2O7^2-) - ¹/₆ × (c_S2O3^2- × V_S2O3^2-))) ÷ volume_C2H5OH

concentration of ethanol in undiluted sample = dilution factor × (n_C2H5OH ÷ volume_C2H5OH)

concentration of ethanol in undiluted sample = (final wine volume ÷ initial wine volume) × (³/₂ × ((c_Cr2O7^2- × V_Cr2O7^2-) - ¹/₆ × (c_S2O3^2- × V_S2O3^2-))) ÷ volume_C2H5OH)

c_{ethanol in wine} = (V_{final wine volume} ÷ V_{initial wine volume}) × (³/₂ × ((c_Cr2O7^2- × V_Cr2O7^2-) - ¹/₆ × (c_S2O3^2- × V_S2O3^2-))) ÷ V_C2H5OH)

Substituting in the values from the experiment above:

c_{ethanol in wine} = (V_{final wine volume} ÷ V_{initial wine volume}) × (³/₂ × ((c_Cr2O7^2- × V_Cr2O7^2-) - ¹/₆ × (c_S2O3^2- × V_S2O3^2-))) ÷ V_C2H5OH)

c_{ethanol in wine} = (V_{1 L} ÷ 0.020) × (³/₂ × ((0.010 × 0.010) - ¹/₆ × (0.030 × 0.0145))) ÷ 0.001)

c_{ethanol in wine} = 50 × (³/₂ × ((1.00 × 10^-4 - 7.25 × 10^-5) ÷ 0.001) = 2.06 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!