I have included the entire lab information. I am having difficulties with g) and

Question

I have included the entire lab information.
I am having difficulties with g) and h) calculations on the last page. there is a similar chegg answer to this question that says there is information missing so i provided everything that is given. please help relieve this headache! My professor has not provided any help when asked.
thank you

Purpose The objective of this experiment is to determine the concentration of ethandol HCH OH, in a commercial beer sample. This is an illustration of volumetric redox titration often done in quantitative analysis. This experiment also teaches you techniques such as distillation and volumetric analysis Principle Primary alcohol such as ethanol (commonly referred to as drinking alcohol) can be oxidized to acetic acid (vinegar) in acid as follows 3C-HsOH(aq) + 2Cr3O72-(aq) + 16H+(aq) 3CH3CO2H(aq) + 4Cr3Taq) + 1 1 H2O (ethanol) (acetic acid) Equation (1) Note: When sulfuric acid is used as the source of H, the green complex ion CrSO4 formed rather than the blue Cr3+ion Unfortunately, one cannot do a simple minded one step titration of alcohol with dichromate as shown above. This is because complete oxidation of alcohol takes place only in the presence of excess dichromate and under continuous heating for a long time Another problem with the direct titration is that there ate substances that are oxidized besides alcohol in beer sample. In order to get around this problem, we just isolate alcohol from the beer sample by simple distillation (alcohol being more volatile than water will evaporate first upon heating). Then we do the indirect titration by adding excess dichromate ion to the alcohol and heat the mixture for a prolonged period of time The excess dichromate ion is back titrated with the standardized ferrous ion (Fe+) whose concentration is exactly known. This is another well known redox titration according to the reaction shown below: 6Fe"(aq) + Cr,072(aq) + 14H+(aq) 6Fe3+(aq) + 2Cr+(aq) + 7H20(1) Equation (2) Equation (2) will give you the amount of excess dichromate ion and this amount can be subtracted from the initial amount of dichromate added to the alcohol solution to obtain the net amount of dichromate which reacts in Equation (1) The end point in both of these redox titrations is obtained using a small amount of the redox indicator, sodium diphenylamine sulfonate. The indicator changes its color from gray-purple just before the end point to colorless as soon as the excess Fe ion has been added, and the solution turns green due to the presence of Cr (as the complex ion CrSOs). Phosphoric acid is added to mask the yellow color of Fe" as the colorless complex of FeHPO4 2+

Explanation / Answer

Answer:

a) Fe2+ added to titrate the sample = 1.6866 mmol

b) Cr2O72- in excess in titrated sample = 0.2811 mmol

c) Cr2O72- in excess in solution B = 2.811 mmol ( titrated sample is prepared by taking 25ml from Solution B= 250ml)

d) Cr2O72- added to solution B = 4.89 mmol (data given)

e) Cr2O72- reacted with C2H5OH in solution B = (Cr2O72- added to solution B) - (Cr2O72- in excess in solution B) = 2.079 mmol

f) C2H5OH in solution B = 3.1185 mmol

I hope you got the answers as mentioned above while calculating

Now let us go to

Question g:

C2H5OH in solution A:

Solution A is 250ml of the distilled alcohol solution.

Solution B was made by pipetting out 25ml of the sample A and diluting to 250ml.

We know the concentration of the alcohol in Solution B = 3.1185 mmol

= 25ml of the sample solution contains 3.1185 mmols of alcohol

so, 250ml of the sample solution = Solution A contains = 3.1185/25 x 250 mmol = 31.185 mmol

Answer: C2H5OH in solution A = 31.185 mmol

Question h:

mass % of C2H5OH in original unknown beer sample = mass % of C2H5OH in Solution A (whole of the alcohol in the beer would have transferred to solution A upon distillation)

No: of moles of C2H5OH in solution A = 31.185 x 10-3

mass of C2H5OH in solution A = 31.185 x 10-3 x molar mass of C2H5OH = 31.185 x 10-3  x 46.07 = 1.4367g

mass of sample taken = 50.532g

So, mass percentage of C2H5OH in solution A = 1.4367/50.532 x 100 = 2.8431%

Answer: mass % of C2H5OH in original unknown beer sample = 2.8431%

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