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 1.85 mg/mL of methyl benzoate (peak A) and 2.04 mg/mL of butyl benzoate (peak B). The area of peak A was determined to be 314 and of peak B to be 372 (measured in arbitrary units by the computer). To measure the sample, 1.00 mL of a standard sample of butyl benzoate containing 2.43 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 409 for peak B. What is the concentration of the methyl benzoate in the plant stream?

Explanation / Answer

[Analyte]sample = ([Analyte]standard / Peak area of standard )*Peak area of sample

= ( 2.43 / 409 )* 477

[Analyte]sample = 2.83mg/mL

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