A scientist wishes to measure the concentration of methyl benzoate in a plant stream by gas chromatography. He prepares a sample of butyl benzoate to use as an internal standard. The results of a preliminary run, which used a solution known to contain 1.35 mg/mL of methyl B benzoate (peak A) and 1.49 mg/mL of butyl benzoate (peak B), are shown. The area of peak A is determined to be 319 and the area of peak B is determined to be 381 measured in arbitrary units by the computer. To measure the sample, 1.00 mL of a standard sample of butyl benzoate containing 2.27 mg/mL is mixed with 1.00 mL of the plant stream material. Analysis of the mixture gave a peak area of 10 Time (min) 15 465 for peak A and 397 for pcak B. What is the concentration of methyl benzoate in the plant stream? methyl benzoate concentration: mg/ Detector Response

A scientist wishes to measure the concentration of methyl benzoate in a plant stream by gas chromatography. He prepares a sample of butyl benzoate to use as an internal standard. The results of a preliminary run, which used a solution known to contain 1.35 mg/mL of methyl B benzoate (peak A) and 1.49 mg/mL of butyl benzoate (peak B), are shown. The area of peak A is determined to be 319 and the area of peak B is determined to be 381 measured in arbitrary units by the computer. To measure the sample, 1.00 mL of a standard sample of butyl benzoate containing 2.27 mg/mL is mixed with 1.00 mL of the plant stream material. Analysis of the mixture gave a peak area of 10 Time (min) 15 465 for peak A and 397 for pcak B. What is the concentration of methyl benzoate in the plant stream? methyl benzoate concentration: mg/ Detector Response

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter31: Introduction To Analytical Separations

Section: Chapter Questions

Problem 31.19QAP

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A scientist wishes to measure the concentration of methyl benzoate in a plant stream by gas chromatography. He prepares a sample of butyl benzoate to use as an internal standard. The results of a preliminary run, which used a solution known to contain 1.37 mg/mL of methyl benzoate (peak A) and 1.51 mg/mL of butyl benzoate (peak B), are shown. The area of peak A is determined to be 3.40×102 and the area of peak B is determined to be 398 measured in arbitrary units by the computer. To measure the sample, 1.00 mL of a standard sample of butyl benzoate containing 2.49 mg/mL is mixed with 1.00 mL of the plant stream material. Analysis of the mixture gave a peak area of 437 for peak A and 415 for peak B. What is the concentration of methyl benzoate in the plant stream?
The standard addition method is used to analyze a sample of a river water for mercury. Solution A is made by pipetting 5.00 mL of undiluted sample in to a 10 mL volumetric flask and filling to the mark with DI water. Solution B is made by pipetting 5.00 mL of undiluted sample and 3.00 mL of 15.0 ppb of Hg standard into same 10.0 mL volumetric flask and filling to the mark with DI. Solution A and B are analyzed using atomic absorption spectroscopy and give a percent transmittance values of 56 % and 33 % respectively (not blank corrected). A blank has a transmittance of 96%. What is the corrected absorbance of both solution A and B? A. Solution A: 0.123 Solution B: 0.463 B. Solution A: 0.463 Solution B: 0.234 C. Solution A: 0.123 Solution B: 0.234 D. Solution A: 0.234 Solution B: 0.463
25 ml of pickled brine was taken from an apple pickled with vinegar and diluted to 500 ml, 25 ml of this was taken and acidity was determined with 0.2 N (Factor: 1.05) NaOH. In the main trials, 6.3 ml - 6.5 ml - 6.9 ml, respectively, 0.1 ml base was used in the analysis without using the sample.For example, how many ppm is its acidity?
Suppose that you try separating an unknown mixture by normal-phase chromatography with the solvent mixture 50% hexane-50% methyl t-butyl ether (which is more polar than hexane). The peaks are too close together and are eluted rapidly. Should you use a higher or lower percentage of hexane in the next run?
5) Retention factors for three solutes separated on a C8 nonpolar stationary phase at pH = 3 are given. Eluant was a 70:30 vol/vol mixture of 50 mM citrate buffer and methanol. a)Draw the dominant chemical structure of each of the substances at pH = 3. b)The retention factor for acetophenone is 4.21, for salicylic acid is 2.97, and for nicotine is zero. Explain the different retention factors.
What is the minimum distribution constant that permits removal of 99% of a solute from 50.0 mL of water with two 25.0-mL extractions with toluene? five 10.0-mL extractions with toluene?
If an instrument gives a response of 1240 for a standard containing 8 ppm of a substance, how much if this substance is in a sample that gives a response of 1705? Are any assumptions needed?
A brewed tea brand sample was subjected to solvent extraction to extract caffeine. The cup of tea measuring 200 mL was extracted with petroleum ether. With KD = 20.0 (caffeine in pet ether-water), the % caffeine (194.10) extracted after 3 extractions with a total of 75.0 mL pet ether used is: A. 97.7% B. 28.6% C. 71.4% D. 2.33%
3. Assume that in extraction from water into toulene, analyte A has distribution ratio of 10. A 20 mL portion of an aqueous solution of A is extracted with toulene. Which of thw following procedures will result in most efficient removal of A from the aqueous phase into toulene? (Please show the solution of the correct answer) A. 1 extraction with 20 mL of touleneB. 1 extraction with 40 mL of touleneC. 2 extractions with 20 mL of toulene eachD. 4 extractions with 10 mL of toulene each
Oral rehydration salts are stated to contain the following components: Sodium Chloride 3.5g Potassium Chloride 1.5g Sodium Citrate 2.9g Anhydrous Glucose 20.0g 8.342 g of oral rehydration salts are dissolved in 500 ml of water. 5 ml of the solution is diluted to 100 ml and then 5 ml is taken from the diluted sample and is diluted to 100 ml. The sodium content of the sample is then determined by flame photometry. The sodium salts used to prepare the mixture were: Trisodium citrate hydrate (C6H5Na3O7, 2H2O) MW 294.1 and sodium chloride (NaCl) NW 58.5. Atomic weight of Na = 23. The content of Na in the diluted sample was determined to be 0.3210 mg/100 ml. Determine the % of stated content of Na in the sample. The stated should be 104.5, how??
Oral rehydration salts are stated to contain the following components: Sodium Chloride 3.5g Potassium Chloride 1.5g Sodium Citrate 2.9g Anhydrous Glucose 20.0g 8.342 g of oral rehydration salts are dissolved in 500 ml of water. 5 ml of the solution is diluted to 100 ml and then 5 ml is taken from the diluted sample and is diluted to 100 ml. The sodium content of the sample is then determined by flame photometry. The sodium salts used to prepare the mixture were: Trisodium citrate hydrate (C6H5Na3O7, 2H2O) MW 294.1 and sodium chloride (NaCl) NW 58.5. Atomic weight of Na = 23. The content of Na in the diluted sample was determined to be 0.3210 mg/100 ml. Determine the % of stated content of Na in the sample.
in HPLC, what does it mean to equilibrate the system. Let us assume that the system is currently running 100%B. You have told them to run the first standard of caffeine using Reverse Phase chromatography with a C18 column, 30% A (Water/0.1%TFA) and 70% B (MeOH/0.1%TFA).What should I equilibrate the system to and for how long?