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Analytical chemistry - help

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 i don't know how to continue (if i even started right)
Can someone please help me

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
Original post by findingneverland
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 i don't know how to continue (if i even started right)
Can someone please help me

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


i found my mistake
glad that i could help
Original post by findingneverland
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


This equation is not balanced correctly. There are only two carbon atoms on the left hand side and six on the right hand side...




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 i don't know how to continue (if i even started right)
Can someone please help me

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



Let's go through it one step at a time:

Moles of iron(II) = 0.01442 * 0.02497 = 0.0003601 mol

Fe2+ --> Fe3+ + 1e
Cr2O72- + 14H+ + 6e --> 2Cr3+ + 7H2O
------------------------------------------------------------------------------------------------------------------
6Fe2+ + Cr2O72- + 14H+ --> 6Fe3+ + 2Cr3+ + 7H2O

and as 6 mole Fe(II) reacts with 1 mol of dichromate, then moles of dichromate excess = 0.0003601/6 = 0.00006 mol

Total initial moles of dichromate = 0.05151* 0.1 = 0.005151 mol

Therefore moles of dichromate reacted with the ethanol = 0.005151 - 0.00006 = 0.005091 mol

The dichromate and ethanol half equations are as follows:

C2H5OH + H2O --> CH3COOH + 4H+ + 4e
Cr2O72- + 14H+ + 6e --> 2Cr3+ + 7H2O

Hence 3 moles of ethanol reacts with 2 moles of dichromate.

Hence 0.005091 mol of dichromate reacted with 0.007636 mol ethanol.

This came from a 20 ml aliquot taken from a 100ml volumetric flask, so the total ethanol in the 100ml flask = 5 * 0.007636 mol = 0.03818 mol

.. and this was originally in a 5ml portion of table wine.

relative mass of ethanol = 46

hence mass to volume ratio = 1.756 : 5

As a percentage = 35.13%

I don't agree with your answer...

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