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

Concentration=RF*Area (relationship between peak area and concentration.where RF=response factor is the is the proportionality constant for the analyte or component of mixture.Each component has its unique RF

RF for std mixture=(Area/concentration) 1/((Area/concentration)2=(314/1.85 mg/L)/(372/2.04mg/L)=0.93

RF for plant stream=(477/conc of methyl benzoate)/(409/2.43 mg/L)= conc of methyl benzoate *2.834

2.834/conc of methyl benzoate =0.93

conc of methyl benzoate =0.93*2.834=2.635 mg/L

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