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.
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.
5 answers
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?
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?
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
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
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.
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.
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.
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.
0 answers