(a)
Interpretation:
Retention factors for each component should be calculated.
Concept introduction:
The retention factor can be calculated as follows:
(b)
Interpretation:
Selectivity factor for methyl acetate and methyl propionate and methyl propionate and n-butyrate should be calculated.
Concept introduction:
Selectivity factor can be calculated as follows:
Methyl acetate and methyl propionate.
Methyl propionate and n-butyrate.
(c)
Interpretation:
Average plate number and plate height should be calculated.
Concept introduction:
Number of theoretical plates can be calculated as follows:
Plate height can be calculated using following equation.
(d)
Interpretation:
Resolution formethyl acetate and methyl propionate and methyl propionate and n-butyrate should be calculated.
Concept introduction:
Resolution can be calculated as follow:
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Principles of Instrumental Analysis
- From the data in Problem 26-14, calculate for species C and D) (a) the resolution. (b) the length of column necessary to separate the two species with a resolution of 2.5.arrow_forwardA peak with a retention time of 10 minutes has a width at half-height of 7.6 s neighboring peak is eluted 15 s later with W½ = 7.5 s. (4) Find out the resolution for these two components. (a) (b) Is this resolution value good for quantitative analysis, why? (c) What difference in retention times is required for an adequate resolution of 1.5?arrow_forwardIn a HPLC separation of two compounds, A and B, the retention times observed are 3.5 min. for A and 4.5 min. for B on a 15 cm column. The retention time for the mobile phase is 1.0 min. The peak widths (at the base) for A and B are 0.5 and 0.7 min respectively. Calculate the following: (i) The capacity factor of A and B. (ii) The peak resolution. (iii) The average number of theoretical plates in the columnarrow_forward
- Substances A and B have retention times of 16.40 and 17.63 min, respectively,on a 30.0-cm column. An unretained species passes through the column in 1.30 min. The peakwidths (at base) for A and B are 1.11 and 1.21 min, respectively. Calculate(a) the column resolution,(b) the average number of plates in the column,(c) the plate height,(d) the length of column required to achieve a resolution of 1.5.arrow_forwardA spectrophotometric method for the quantitative determination of the concentration ofPb2+ in blood yields an Ssamp of 0.133 for a 1 mL sample of blood that has been diluted to 6 mL. A second sample is spiked with 1 µL of a 1467 ppb Pb2+ standard and diluted to 6 mL, yielding an Sspike of 0.491. Determine the concentration of Pb2+ in the original sample of blood.arrow_forward3arrow_forward
- A 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_forwardSubstances A and B have retention times of 12.75 and 14.28 min, respectively, on a 30.0-cm column. An unretained species passes through the column in 1.10 min. The peak widths (at base) for A and B are 1.11 and 1.27 min, respectively. What is the resolution?arrow_forwardA solute with a retention time of 407 seconds has a width at the base of 13.0 seconds on a column 12.2 meters long. Find the number of plates and plate height. The half-width of the same peak is 7.6 seconds. Find the plate height.arrow_forward
- Two components in an HPLC separation have retention times thatdiffer by 22 s. The first peak elutes in 10.5 min and the peakwidths are approximately equal. Use a spreadsheet to find theminimum number of theoretical plates needed to achieve thefollowing resolution, Rs, values: 0.50, 0.75, 0.90, 1.0, 1.10, 1.25,1.50, 1.75, 2.0, and 2.5. How would the results change if peak 2were twice as broad as peak 1?arrow_forwardConsider the separation of components A and B on a GC column. The retention time for component A is 1.4 min and for component B is 2.1 min. The width at the base for the peak due toA is 0.38 min and for the peak due to B is 0.53 min. Calculate the resolution, R, and tell whetherthe two peaks are considered to be resolved and why.arrow_forwardA 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_forward
- Principles of Instrumental AnalysisChemistryISBN:9781305577213Author:Douglas A. Skoog, F. James Holler, Stanley R. CrouchPublisher:Cengage Learning