(a)
Interpretation: The component of the sample that elutes faster needs to be determined. The reported boiling points of these substances need to be determined.
Concept Introduction:
Gas chromatography is used to separate the components of a sample mixture. There are two phases that is stationary and mobile. The mobile phase flows with the stationary phase which is the phase on which compounds of the mixture that needs to be separated are selectively absorbed. The mobile phase carries these compounds.
(b)
Interpretation: The GLC experiment that should be performed to identified the peaks at 9.56 and 16.23 minutes needs to be determined.
Concept Introduction:
Gas chromatography is used to separate the components of a sample mixture. There are two phases that is stationary and mobile. The mobile phase flows with the stationary phase which is the phase on which compounds of the mixture that needs to be separated are selectively absorbed. The mobile phase carries these compounds.
(c)
Interpretation: The organic compound in bear responsible for the peak needs to be determined and explained.
Concept Introduction:
Gas chromatography is used to separate the components of a sample mixture. There are two phases that is stationary and mobile. The mobile phase flows with the stationary phase which is the phase on which compounds of the mixture that needs to be separated are selectively absorbed. The mobile phase carries these compounds.
(d)
Interpretation: The two ways that should be suggested to reduce the retention time for all the components of the sample needs to be explained.
Concept Introduction:
Gas chromatography is used to separate the components of a sample mixture. There are two phases that is stationary and mobile. The mobile phase flows with the stationary phase which is the phase on which compounds of the mixture that needs to be separated are selectively absorbed. The mobile phase carries these compounds.
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Chapter 6 Solutions
EBK EXPERIMENTAL ORGANIC CHEMISTRY: A M
- A student researcher performed a chromatographic separation of caffeine and aspartame. The retention time for caffeine, tc, was found to be 209.6 s with a baseline peak width, w., of 14.4 s. The retention time for aspartame, ta, was 260.0 s with a baseline peak width, wa, of 21.2 s. The retention time for the unretained solvent methanol was 49.2 s. Calculate the average plate height, H, in micrometers for this separation, given that it was performed on a 22.1 cm long column. H = um %3D Calculate the resolution, R, for this separation using the widths of the peaks. R = Calculate the resolution if the number of theoretical plates were to increase by a factor of 2. R2N =arrow_forwardA student researcher performed a chromatographic separation of caffeine and aspartame. The retention time for caffeine, te, was found to be 200.8 s with a baseline peak width, we, of 16.1 s. The retention time for aspartame, ta, was 258.7 s with a baseline peak width, wa, of 20.6 s. The retention time for the unretained solvent methanol was 44.2 s. Calculate the average plate height, H, in micrometers for this separation, given that it was performed on a 22.1 cm long column. H = Calculate the resolution, R, for this separation using the widths of the peaks. R = 88.2 R₂5N = 3.16 Calculate the resolution if the number of theoretical plates were to increase by a factor of 2.5. 3.533 Incorrect μmarrow_forward6. A certain mixture contains the following compounds: propionic acid, 2-hexanol, isoamyl acetate, and 3,4-dimethylheptane. Rank the compounds in the order of decreasing rate of elution using normal phase liquid chromatography. Explain your answer.arrow_forward
- ▪ In a blood alcohol analysis by GC-FID, methanol elutes at 1 min, ethanol at 2 min, IPA at 3 min, and IS at 4 min. In the patient sample, the analyte elutes t 2 min and the IS at 8 min. What is the relative retention time and the analyte in the patient? A) methanol B) Ethanol chu C) IPAarrow_forwardPleasearrow_forwardWhat is the best Chromatography instrument for analysis of Soluble Protein molecules CG-FID GC-MS LC-MS IC-MS O IC-ECarrow_forward
- A student developed a new dye called Dye C. It was analyzed against other dyes using paper chromatography. A solution of 1:1 ethyl acetate:hexane was used as the eluent. The resulting chromatogram is shown below. What other dye/s are present in dye C? Which dye contains the least number of pigments?arrow_forwardBased on the given chromatograph, answer the following questions:1) Which of the diet sodas have the highest concentration of caffeine? Which standard best reflects this concentration?2) Which of the diet sodas have the lowest concentration of aspartame? Which standard best reflects this concentration?3) What is the approximate retention time of benzoate?arrow_forwardFrom the six analytes listed below, n-octanal, n-octane, 1-chlorooctane, n-octanol, n-heptane, and n-octanoic acid predict retention order (shortest to longest retention times) if the SP is: 100% polyethylene glycol or 100% dimethylpolysiloxane.arrow_forward
- An unretained solute is eluted from a GC column in 1.80 min. Decane (C10H22) is eluted in 15.63 min and undecan (C11H24) is eluted in 17.22 min. What is the retention time (report to 2 decimal places, in minutes) of a compound whose retention index is 1050?arrow_forward2-pentanone has a retention index of 987 on a poly(ethylene glycol) column (also called Carbowax). (a) Between which two straight-chain hydrocarbons is 2-pentanone eluted? (b) An unretained solute is eluted from a certain column in 1.80 min. Decane (C10H22) is eluted in 15.63 min and undecane (C11H24) is eluted in 17.22 min. What is the retention time of a compound whose retention index is 1 050?arrow_forwardUsing triangulation, calculate the percentage of each component in a mixture composed of two substances, A and B. The chromatogram is shown.arrow_forward
Principles of Instrumental AnalysisChemistryISBN:9781305577213Author:Douglas A. Skoog, F. James Holler, Stanley R. CrouchPublisher:Cengage Learning