A solution containing 3.45 mM of p-xylene (analyte) and 6.86 mM of nonane (stand

Question

A solution containing 3.45 mM of p-xylene (analyte) and 6.86 mM of nonane (standard) gave peak areas of 13587 and 14769 respectively, in a chromatographic analysis. then 1.5 mL of 6.86mM nonane solution was added to 5 mL of an unknown sample containing p-xylene. This solution gave peak areas of11875 and 10765 for p-xylene and nonane.
(A)calculate the response factor for the analyte.
(B)find the concentration of the standard (mM) in the mixed solution
(c) find the concentration of p-xylene in the mixed solution
(D) find the concentration of p-xylene in the original unknown sample

Explanation / Answer

I am not sure what the "(Mr 150) and (Mr 225)" means. Molar masses? But anyway, we can calculate response factors on the basis of molar masses, masses, volumes or concentrations. In this example, I will use the concentrations. The concentration of analyte C is 0,65 mg dissolved in a final (total) volume of 50 ml. This is equal to c(C) = 13 mg/L The concentration of analyte D is 0,9 mg dissolved in a final (total) volume of 50 ml. This is equal to c(D) = 18 mg/L The peak area of analyte C is A(C) = 1 The peak area of analyte D is A(D) = 1/85 The correspondent response factors are: RF(C) = A(C)/c(C) = 1/13 [L/mg] RF(D) = A(D)/c(D) = (1/85)/18 [L/mg] If we consider analyte C to be the internal standard, we set the response factor of C to 1 and calculate the relative response factor of D: RF(D rel.C) = {A(D)/c(D)}/{A(C)/c(C)} = {(1/85)/18}/{1/13} RF(D rel.C) = 0,00850 That's the way I do quantitative gas chromatography. And I have done it for decades!

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