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 1.65 mg/mL of methyl benzoate (peak A) and 1.82 mg/mL of butyl benzoate (peak B). The area of peak A was determined to be 325 and of peak B to be 394 (measured in arbitrary units by the computer). To measure the sample, 1.00 mL of a standard sample of butyl benzoate containing 2.19 mg/mL was mixed with 1.00 mL of the plant stream material, and analysis of the mixture gave a peak area of 463 for peak A and 411 for peak B. What is the concentration of the methyl benzoate in the plant stream?

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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 1.65 mg/mL of methyl benzoate (peak A) and 1.82 mg/mL of butyl benzoate (peak B). The area of peak A was determined to be 325 and of peak B to be 394 (measured in arbitrary units by the computer) To measure the sample, 1.00 mL of a standard sample of butyl benzoate containing 2.19 mg/mL was mixed with 1.00 mL of the plant stream material, and analysis of the mixture gave a peak area of 463 for peak A and 411 for peak B What is the concentration of the methyl benzoate in the plant stream? 10 Time (min) 15 Number conc. mg/mL methyl benzoate Previous Give Up & View Solution e Check Answer e Next Exit Hint First use the preliminary run to determine the relative response of the chromatograph detector to methyl benzoate and butyl benzoate. Determine F, the response factor, from the following equation area peak A area peak B concb conc

Explanation / Answer

I'm assuming (and it is very likely) that the peaks for each experiment are of the same respective compounds So to calculate the concentration of the second, you have to know that the chromatography equations can be set to be equal and then manipulated for the desired variable. So here is the equation I used:

{[A(bb1)/ M(bb1)]/ [A(mb1)/ M(mb1)]}

={ [A(bb2)/ M(bb2)]/ [A(mb2)/M(mb2)]}

Where the A is the area under the peaks, and M is concentration. All of the examples I looked at used molarity, but I'm assuming the equation works the same since it deals with ratios. This eqaution can be simplified to:

M(mb2)= [A(bb1) × M(mb1) × A(mb2) × M (bb2)]/ [ A(mb1) × M(bb1) × A(bb2)]

And when you plug in all of the variables,

M(bb2)= (394 × 1.65 × 463 × 2.19)/ (325 × 1.82 × 411) = 2.711mg/ml

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