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 1.17 mg/mL of methyl benzoate (peak A) and 1.29 mg/mL of butyl benzoate (peak B). The area of peak A was determined to be 304 and of peak B to be 353 (measured in arbitrary units by the computer). To measure the sample, 1.00 mL of a standard sample of butyl benzoate containing 2.59 mg/mL was mixed with 1.00 mL of the plant stream material, and analysis of the mixture gave a peak area of 487 for peak A and 427 for peak B. What is the concentration of the methyl benzoate in the plant stream?

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 intenal standard. A preliminary run, shown at the right involved a solution containing 1.17 mg/mL of methyl benzoate peak A) and 1.29 mg/mL of butyl benzoate (peak B). The area of peak A was determined to be 304 and of peak B to be 353 (measured in arbitrary units by the computer). To measure the sample, 1.00 mL of a standard sample of bu benzoate containing 2.59 mg/mL was mixed with 1.00 mL of the plant stream material, and analysis of the mixture gave a peak area of 487 for peak A and 427 for peak B. What is the concentration of the methyl benzoate in the plant stream? 5 15 Time (min) Number concmethyl benzoate mg/mL

Explanation / Answer

Given :

butyl benzoate to use as an internal standard

for 1.17 mg/mL of methyl benzoate ,peak area is 304

for 1.29 mg/mL of butyl benzoate , peak area is 353

1.00 mL of a standard sample of butyl benzoate containing 2.59 mg/mL was mixed with 1.00 mL of the plant stream material, and analysis of the mixture gave a peak area of 487 for peak A and 427 for peak B.

I have worked out the concentrations of the target compounds using the formula:

conc of target compound = conc. of standard (mg/ml) / peak area of standard x Peak area of sample

                                             = 2.59 (mg/ml) / 427 x 487

                                             = 0.0060 x 487

                                            = 2.95 mg/ml

concentration of the methyl benzoate in the plant stream is 2.95 mg/ml

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