You wish to measure the concentration of methyl benzoate in a plant stream by ga

Question

You wish to measure the concentration of methyl benzoate in a plant stream by gas chromatography. You prepare a sample of butyl benzoate to use as an internal standard. A preliminary run, shown at the right involved a solution containing 1.21 mg/mL of methyl benzoate (peak A) and 1.33 mg/mL of butyl benzoate (peak B). The area of peak A was determined to be 327 and of peak B to be 386 (measured in arbitrary units by the computer). To measure the sample, 1.00 mL of a standard sample of butyl benzoate containing 2.71 mg/mL was mixed with 1.00 mL of the plant stream material, and analysis of the mixture gave a peak area of 477 for peak A and 425 for peak B. What is the concentration of the methyl benzoate in the plant stream?

Explanation / Answer

From the known measurement

Peak B : butyl benzoate 1.33 mg/ml gave an area 386

Peak A : methyl benzoate 1.21 mg/ml gave an area 327

Response factor = Peak B/Peak A = (386/1.33)/(327/1.21) = 1.074

From the unknown plant stream sample

Peak B : butyl benzoate 2.71 mg/ml gave an area 425

Peak A : methyl benzoate X mg/ml gave an area 477

So,

the concentration of methyl benzoate present in sample = 477/[(425/2.71)/1.074] = 3.266 mg/ml

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