Retention times in a 1.25m chromatographic column: dead time, 0.395 min; isopropylamine, 4.59 min; and n-propylamine, 4.91 min.Peak widths: 0.365 min for isopropylamine and 0.382 min for n-propylamine. Calculate: (a) Value of the retention factor (k') for each amine (b) Value of the selectivity factor (alpha) between the two amines (c) Number of theoretical column trays (d) Column tray height (e) Resolution between the two amines

Retention times in a 1.25m chromatographic column: dead time, 0.395 min; isopropylamine, 4.59 min; and n-propylamine, 4.91 min.Peak widths: 0.365 min for isopropylamine and 0.382 min for n-propylamine. Calculate: (a) Value of the retention factor (k') for each amine (b) Value of the selectivity factor (alpha) between the two amines (c) Number of theoretical column trays (d) Column tray height (e) Resolution between the two amines

Principles of Instrumental Analysis

7th Edition

ISBN:9781305577213

Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch

Publisher:Douglas A. Skoog, F. James Holler, Stanley R. Crouch

Chapter27: Gas Chromatography

Section: Chapter Questions

Problem 27.24QAP

See similar textbooks

Similar questions

A 5.00-mL sample of blood was treated with trichloroacetic acid to precipitate proteins. After centrifugation, the resulting solution was brought to pH 3 and extracted with two 5-mL portions of methyl isobutyl ketone containing the lead-complexing agent APCD. The extract was aspirated directly into an air/acetylene flame and yielded an absorbance of 0.569 at 283.3 nm. Five-milliliter aliquots of standard solutions containing 0.400 and 0.600 ppm of lead were treated in the same way and yielded absorbances of 0.396 and 0.599. Please calculate the concentration of lead in the sample in ppm.
A 5.00-mL sample of blood was treated with trichloroacetic acid to precipitate proteins. After centrifugation, the resulting solution was brought to pH 3 and extracted with two 5-mL portions of methyl isobutyl ketone containing the lead-complexing agent APCD. The extract was aspirated directly into an air/acetylene flame and yielded an absorbance of 0.527 at 283.3 nm. Five-milliliter aliquots of standard solutions containing 0.400 and 0.600 ppm of lead were treated in the same way and yielded absorbances of 0.396 and 0.599. Find the concentration of lead in the sample in ppm assuming that Beer’s law is followed.
A juice concentrate was colorimetrically assayed using Nelson’s test. One milliliter (1.00 mL) of the sample solution and various concentrations of the standard glucose solution were treated with freshly prepared Nelson’s reagent and arsenomolybdate reagent and then diluted to 10.0 mL separately in properly labeled test tubes. Absorbances at 480 nm for distilled water, glucose standard, and for the sample are 0.052, 1.702, and 0.926, respectively. What is the reducing sugar concentration (mg/mL) in the juice concentrate? The equation of the line was plotted to be: y = 1.6656x - 0.0885 0.578 mg/mL 577.9 mg/mL 57.7 mg/mL 5.78 mg/mL
In a chromatographic analysis of low-molecular-weight acids, butyric acid elutes with a retention time of 7.63 min. The column's void time is 0.31 min. Calculate the retention factor for butyric acid
A 50.0 mL juice extract is colorimetrically assayed using Nelson’s test. One milliliter (1.00 mL) of the solution and the standard glucose solution (concentration: 1mg/mL) were treated with freshly prepared Nelson’s reagent and arsenomolybdate reagent and then diluted to 10.0 mL separately in properly labeled test tubes. Absorbance at 480 nm for the standard is 1.702 and for the sample, 0.926. A. What is the principle behind the assay? B. Calculate the concentration of the reducing sugar in the 50.0 mL sample in mg/mL. Apply Beer-Lambert’s law.
A 50.0 mL juice extract is colorimetrically assayed using Nelson’s test. One milliliter (1.00 mL) of the solution and the standard glucose solution (concentration: 1mg/mL) were treated with freshly prepared Nelson’s reagent and arsenomolybdate reagent and then diluted to 10.0 mL separately in properly labeled test tubes. Absorbance at 480 nm for the standard is 1.702 and for the sample, 0.926. A. What is the principle behind the assay? B. Calculate the concentration of the reducing sugar in the 50.0 mL sample in mg/mL. Apply Beer-Lambert’s law: A=bc
A 50.0 mL juice extract is colorimetrically assayed using Nelson’s test. One milliliter (1.00 mL) of the solution and the standard glucose solution (concentration: 1mg/mL) were treated with freshly prepared Nelson’s reagent and arsenomolybdate reagent and then diluted to 10.0 mL separately in properly labeled test tubes. Absorbance at 480 nm for the standard is 1.702 and for the sample, 0.926. A. What is the principle behind the assay?
A series of nonpolar aromatic compounds were separated by isocratic HPLC on a nonpolar stationary phase containing octadecyl groups [−(CH2)17CH3] covalently attached to silica particles. The mobile phase was 60/40 methanol/water. How would the retention times be affected if 90/10 methanol/water were used instead and why? What type of chromatography?
a) What would you see if you ran a Thin Layer Chromatography (TLC) on a mixture of the three solids shown below using silica gel as the stationary phase and ethyl acetate/hexanes as the mobile phase? Whichsolid would have the highest Rf and which would have the lowest Rf? Explain b) Would changing the mobile phase (from part a) and increasing thepolarity by introducing small amount of methanol to the mobile phase change the order ofRf of the above 3 compounds? Explain c) Another type of chromatography is Reverse Phase Chromatography. In this type ofchromatography, the stationary phase is non-polar and the mobile phase is polar. If astudent performed Reverse Phase TLC on the mixture of 3 solids (from part a),what would be the order of Rf’s of the compounds? Which would have the lowest Rf andwhich would have the highest Rf?
In a normal-phase liquid chromatographic column, a solute was found to have a retention time of 26.7 minutes, and an unretained component required 0.55 minutes for elution when the mobile phase was 40% (by volume) THF and 60 % acetonitrile. Calculate: a) Retention factor (k) b) the THF-acetonitrile composition that should bring k to a value of 4.
The separation of APAP-L,(12 = 0.28 min) from APAP-D (W12 = 0.35 min) on a chiral stationary phase was found to have retention times of 4.78 min and 5.26 min, respectively. With a unretained solute being, found to have a retention time of 1.1 min on the same column. Calculate the column resolution and selectivity factor
(i) Calculate the capacity factors (k') for both compounds A and B and the selectivity factor (a) between them, commenting on the values. (ii) Calculate the resolution between A and B and comment on it. (iii) Using the peak for A, calculate the number of theoretical plates for the GC column and comment on the quality of the column.