A 5ml portion of table wine was diluted to 100 ml in a volumetric flask. The ethanol in a 20 ml aliquot was distilled into 100 ml of 0.05151M K2Cr2O7. heating completed oxidation of the alcohol to acetic acid:

Question

A 5ml portion of table wine was diluted to 100 ml in a volumetric flask. The ethanol in a 20 ml aliquot was distilled into 100 ml of 0.05151M K2Cr2O7. heating completed oxidation of the alcohol to acetic acid:

Ch3Ch2OH + 2Cr2O7(2-) +16H(+) -> 3Ch3COOH + 4Cr(3+) + 11H2O following which the excess dichromate was titrated with 14.42mL of 0.02497M Fe(II). Calculate the weight-volume percentage of ethanol.
(RESULT: 11,72%)

So i tried doing this problem and again, i don't know how to continue (if i even started right)

n(K2Cr2O7)=0.1L*0.05151M=0,005151 mol
n(EtOH)=n(K2Cr2O7)/2

Cr2O7(2-) + 3Fe(2+) -> Cr(3+) + 3Fe(3+)
n(Fe2+)=0,01442L*0,02497M=0,0003601 mol
n(Fe2+)=n(Cr2O72-)*3
n(Cr2O72-)'=0,00012 mol

n(K2Cr2O7)=0,005151-0,00012=0,005031 mol
m(EtOH)=(0,005031/2)*46,07=0,11589 g

now i don't know how to continue :S

Answer 1

You have made several errors. The way this works is that too much dichromate was used in the oxidation of ethanol to acetic acid. How much too much. That is what the back titration with Fe does--it tells you how much too much.

1. mols Cr2O7^2- = ok
2. You can't calculate mols EtOH from this (yet).
3. The Fe/Cr2O78^2- is not balanced.
6Fe^2+ + Cr2O7^2- ==> 2Cr^3+ + 6Fe^3+
Then mols Fe = 0.01442*0.02497 = about 0.00036 (but you need to go through more accurately). mols Cr2O7^2- from the Fe titration is 1/6 thast which = about 0.00006.
mols Cr2O7^2- used in the ethanol reaction = 0.005151-0.00006 = ?. Again, you need to go through this more accurately.

Then net mols Cr2O7^2-/2 = mols EtOH.
That times 46 g/mol converts to grams in the 20 mL portion you titrated.

Multiply that by 5 to determine grams in the100 mL volumetric flask (which is also grams in the original 5 mL), then multily by 20 to find the amount in 100 mL of wine. I get 11.71.

Answer 2

Thank you very much for your help :)

Answer 3

To calculate the weight-volume percentage of ethanol, you need to determine the weight of ethanol in the diluted sample and then calculate the percentage based on the volume of the original sample. Here's how you can continue:

1. Calculate the weight of ethanol in the diluted sample:
Given that the volume of the original sample is 5 ml and it was diluted to 100 ml, the 20 ml aliquot used in the distillation step represents 20/100 = 1/5 (or 0.2) of the diluted sample.
Therefore, the weight of ethanol in the diluted sample is:
w(EtOH) = (0.2) * (0.11589 g) = 0.023178 g

2. Calculate the weight-volume percentage of ethanol:
The weight-volume percentage is calculated by dividing the weight of ethanol by the volume of the original sample and multiplying by 100.
Given that the volume of the original sample is 5 ml:
Weight-volume percentage = (0.023178 g / 5 ml) * 100 = 0.46356 g/ml * 100 = 46.356%

Therefore, the weight-volume percentage of ethanol is approximately 46.36%.

Note: It is important to round your answer appropriately based on the significant figures given in the question. In this case, the provided result is 11.72%, so you may need to recheck your calculations or determine if additional steps are required.