A 15.0 mL sample of whiskey was diluted to 500.0 mL. A 3.00 mL aliquot of the diluted sample was removed and the ethanol, C2H5OH, was distilled into 50.00 mL of 0.02400 M K2Cr2O7 and oxidized to acetic acid.

The excess Cr2O72– was then back titrated with 19.0 mL of 0.1255 M Fe2 , producing Cr3 and Fe3 . Calculate the weight per volume percent, %w/v, of ethanol in the original whiskey sample.

3C2H5OH + Cr2O7^2- ==> 2Cr^3 + 3CH3COOH

Again, I've not worried about the rest of the stuff here.

In the back titration part.
6Fe^2+ + Cr2O7^2- ==> 6Fe^3+ + 2Cr^3+

How many moles did you have for dichromate initially. That's M x L = ?
How many mol Fe^2+ used in the back titration? That's M x L = ?. Convert that to mols Cr2O7^2- (that's 1/6 the iron(II) and subtract from the initial dichromate to find how much dichromate was used by the oxidation of the ethanol to acetic acid. Convert from mols dichromate to mols ethanol with the equation above and that's mols ethanol in 3 mL. Convert that to mols in the 15 mL sample and change to %w/v. Post your work if you get stuck.

No. But large chunks are ok. Here is your work.

moles for dichromate initially. That's 0.024M x 0.05L = 0.0012
mol Fe^2+ used in the back titration 0.1255M x 0.019L = 0.0023845.
Convert that to mols Cr2O7^2- (that's 1/6 the iron(II)
So 3.974 x 10^-4
and subtract from the initial dichromate to find how much dichromate was used by the oxidation of the ethanol to acetic acid.
So 0.0012-3.974 x 10^-4 = 8.0258x10^-4

Convert from mols dichromate to mols ethanol with the equation above and that's mols ethanol in 3 mL.
So then 8.0258x10^-4 / 3 = 2.6753 x 10^-4
Down to here appears to be ok. What you have with the 2.6753E-4 is mol in the 3 mL. That was a 3 mL aliquot from 500 mL total(that was initially 15 ml whiskey) so the 500 mL contained 2.6753E-4 mols x (500/3). That value is the amount ethanol in mols in 15 mL of the whiskey sample. Then you must convert that to grams (it appears you kept it in mols) ethanol in 15 mL and convert that to % w/v. What follows is part of your post and can be disregarded.
Convert that to mols in the 15 mL sample
then 2.6753 x 10^-4 / 15 = 1.784x10^-5
and change to %w/v. Post your work if you get stuck.
1.784x10^-5 x 100% = 0.00178

I've gone over this many times and found nothing UNTIL I checked the formula. I found two errors. The first is mols Cr2O7^2- x 3 (not divided by 3) = mols ethanol (and you should have caught that too). When I first responded I balanced the equation with dichromate + C2H5OH --> 2Cr^3+ + C2H4O (acetaldehyde) which is the first step of the oxidation. That equation is 3 mols ethanol to 1 mol Cr2O7^2- and I just ASSUMED that the next step was the same. It isn't. If we do it right let's balance by the half reaction method. Here is Cr2O7^2- ( and we really still don't need H^+ and H2O but I'll put them in)

Cr2O7^2- + 6e + 14H^+ ==> 2Cr^3+ + 7H2O
C2H6O + H2O ==> C2H4O2 + 4H^+ + 4e
----------------------------------
So 12 is the LCD. Multiply dichromate by 2 and ethanol by 3 so the equation is
2Cr2O7^2- + 3C2H5OH ==> CH3COOH + 2Cr^3+ and changes the ratio thing. I was in a hurry and I shouldn't have assumed that. By the way, I know the Fe/Cr2O7^2- is ok because I balanced that before I typed it in.
So we are ok to 0.000803 mols Cr2O7^2- used. You should confirm that.
Then 0.000803 x 3/2 = mols ethanol.
That x 500/3 is mols ethanol in the 15 mL
That divided by 15 is fraction ethanol
That x 100 to convert to %.
If I didn't goof that should be close to 60% ethanol. Watch that you report only 3 s.f. Sorry about my bad assumption.

Check all of this.

moles for dichromate initially. That's 0.024M x 0.05L = 0.0012

mol Fe^2+ used in the back titration 0.1255M x 0.019L = 0.0023845.
Convert that to mols Cr2O7^2- (that's 1/6 the iron(II)
So 3.974 x 10^-4
and subtract from the initial dichromate to find how much dichromate was used by the oxidation of the ethanol to acetic acid.
So 0.0012-3.974 x 10^-4 = 8.0258x10^-4

Convert from mols dichromate to mols ethanol with the equation above and that's mols ethanol in 3 mL.
So then 8.0258x10^-4 / 3 = 2.6753 x 10^-4
Convert that to mols in the 15 mL sample
then 2.6753 x 10^-4 / 15 = 1.784x10^-5
and change to %w/v. Post your work if you get stuck.
1.784x10^-5 x 100% = 0.00178

Is this correct?

Well I ended up finding the moles amount to be 2.054114959 g

Then to change to w%/v do you take the gram amount from above and divide by 15?

Because I got 13.6% and that is incorrect.

To calculate the weight per volume percent (%w/v) of ethanol in the original whiskey sample, we need to follow a series of steps.

Step 1: Calculate the number of moles of ethanol oxidized.

Given that the diluted sample was 500.0 mL and the 3.00 mL aliquot was taken, we can find the dilution factor as follows:

Dilution Factor = Total volume after dilution / Volume taken
Dilution Factor = 500.0 mL / 3.00 mL = 166.67

Since a 15.0 mL sample was diluted, the diluted sample amount is:

Diluted Sample Amount = Volume taken * Dilution Factor
Diluted Sample Amount = 15.0 mL * 166.67 = 2499.99 mL ≈ 2500.0 mL

Now we can determine the concentration of ethanol in the diluted sample:

Ethanol Concentration = (Volume of ethanol in aliquot / Diluted Sample Amount) * Concentration of K2Cr2O7
Ethanol Concentration = (3.00 mL / 2500.0 mL) * 0.02400 M ≈ 2.88 x 10^-5 M

From the balanced chemical equation of the oxidation reaction (C2H5OH + 2Cr2O72– → 2Cr3+ + 2CO2 + 3H2O), we see that 1 mole of ethanol is oxidized to form 1 mole of acetic acid, which is then titrated with Fe2+.

Step 2: Calculate the number of moles of K2Cr2O7 reacted.

Using the equation Fe2+ + Cr2O72– → Cr3+ + Fe3+, we can see that the mole ratio of Fe2+ to Cr2O72– is 1:1.

Moles of K2Cr2O7 reacted = Moles of Fe2+ titrated
Moles of K2Cr2O7 = (Volume of Fe2+ solution used * Molarity of Fe2+) / 1000
Moles of K2Cr2O7 = (19.0 mL * 0.1255 M) / 1000 = 0.0023885 mol

Since there is a 1:1 stoichiometric ratio between moles of K2Cr2O7 and moles of ethanol, the number of moles of ethanol oxidized is also 0.0023885 mol.

Step 3: Calculate the weight of ethanol.

The molecular weight of ethanol (C2H5OH) is:

2(12.01 g/mol) + 6(1.01 g/mol) + 16.00 g/mol = 46.07 g/mol

Weight of ethanol = Moles of ethanol * Molecular weight of ethanol
Weight of ethanol = 0.0023885 mol * 46.07 g/mol ≈ 0.11 g

Step 4: Calculate the weight per volume percent.

Weight per volume percent (%w/v) = (Weight of ethanol / Original sample volume) * 100%
Weight per volume percent = (0.11 g / 15.0 mL) *100% ≈ 0.73%w/v

Therefore, the weight per volume percent of ethanol in the original whiskey sample is approximately 0.73%w/v.