The concentration of ethanol in a wine sample was determined by gas chromatograp

Question

The concentration of ethanol in a wine sample was determined by gas chromatography (GC). A standard solution containing 2.82 mM ethanol and 3.65 mM internal standard (methanol) produced two GC peaks, and a ratio of their peak areas (Aethanol/Amethanol) was 2.60. Then a 5.00-mL solution of 20.0 mM methanol was added to 45.00 mL of a wine sample, producing 50.00 mL of a solution. This mixture solution gave a ratio of Aethanol/Amethanol = 3.80. What would be the concentration of ethanol in the wine sample?

Explanation / Answer

Solution:

Given: Concentration (Ethanol)= 2.82 mM

             Concentration (Methanol)= 3.65 mM

Also,

Aethanol/Amethanol = 2.60

We know, response factor can be given as follows:

Response factor, F = [Area(Ethanol)/Area(Methanol)] × [Conc. (Methanol)/Conc. (Ethanol)]

                               = 2.6 x 3.65/2.83                                

       = 3.35

Now, final concentration of methanol after 5.00-mL solution of 20.0 mM methanol was added to 45.00 mL of a wine sample, producing 50.00 mL of a solution can be calculated as follows:

5.00-mL solution = 20.0 mM methanol

50 mL solution (dilution to 10 times) = 20/10 = 2 mM

Now,

F = [Area(Ethanol)/Area(Methanol)] × [Conc. (Methanol)/Conc. (Ethanol)]

3.35 = 3.8 x 2/Conc. (Ethanol)

Or, Conc. (Ethanol) = 3.8 x 2/3.35

                                = 2.27 mM

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