(a). Sketch a graph of the van Deemnter equation (plate height versus linear flow rate).what would the curve look like if the multiple path term were 0? If the longtitundinal diffusinal diffusion term were 0?

(b). Explain why the van Deemter curve for 1.8μmparticles in figure 25-3is nearly flat at high flow rate.what can you say about each of the terms in the van Deemeter equation for 1.8μmparticles.

(c). Explain why the 2.7μmsuperficially porous particle enables separations similar to those achieved by 1.8μmtotally porous particles,but the superficially porous particle requires lower pressure.

Short Answer

Expert verified

a) Sketch a graph of the van Deemnter.

b) Comparing the van Deemter curve of the particles in Figure to the graph

c) Superficially porous particles have a diameter non porous silica core.

Step by step solution

01

Sketch a graph of the van Deemnter 

The van Deemter equation was graphed to help visualise each term in the equation. Figure 24-3 was then examined using the van Deemter equation and its implications.

The figure below shows the graph of the van Deemter equation when the multiple paths term ( A) is , when the longitudinal diffusion term ( B) is , when the equilibration time term (C ) is constant, and when all terms are .

02

Comparing the van Deemter

b) why the van Deemter curve

Comparing the van Deemter curve of the 1.8μmparticles in Figure 24-3to the graph in (a), we can see that the graph that it most resembles is the graph of c=0. Although not obvious from the graphs, the equation of the curve of 1.8μmparticles has a small multiple paths term (A ) relative to that of the other larger particles in the graph.

Based on the graph, the curve of the particles plateaus at the smallest plate height ( H) compared to the other graphs. The reduction in plate heights can be related to the decrease in the band spreading due to multiple flow paths.

Given this relationship, the decrease in the H plateau value of the particles means that there is a decrease in A as the particle size decreases.

c) superficially porous particle

Superficially porous particles have a 5μmdiameter non porous silica core. Compared to other fully porous particles with a larger radii, superficially porous particles can perform at the same level or better.

This is due to the structure of superficially porous particles, having a 0.25 thick porous silica layer on the outside of its 5μmdiameter non porous silica core

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Most popular questions from this chapter

The rate at which heat is generated inside a chromatographycolumn from friction of flowing liquid is power (watts, W=J/s)= volume flow raterole="math" localid="1656474760665" (m3/s)×pressure drop (pascals, role="math" localid="1656474828044" Pa=kg/[m?s2]).

(a) Explain the analogy between heat generated in a chromatography column and heat generated in an electric circuit (power = current×voltage).

(b) At what rate (watts = J/s) is heat generated for a flow of 1 mL/minwith a pressure difference of3500 bar between the inlet and outlet?

You will need to convert mL/min tom3/s.Also 1 bar=105Pa.

The antitumor drug gimatecan is available as nearly pure (S)-enantiomer. Neither pure (R)-enantiomer nor a racemic (equal) mixture of the two enantiomers is available. To measure small quantities of (R)-enantiomer in nearly pure (S)-gimatecan, a preparation was subjected to normal-phase chromatography on each of the enantiomers of a commercial, chiral stationary phase designated (S,S)- and (R,R)-DACH-DNB. Chromatography on the (R,R)-stationary phase gave a slightly asymmetric peak at tr 5 6.10 min with retention factor k 5 1.22. Chromatography on the (S,S)- stationary phase gave a slightly asymmetric peak at tr 5 6.96 min with k 5 1.50. With the (S,S) stationary phase, a small peak with 0.03% of the area of the main peak was observed at 6.10 min.

Chromatography of gimatecan on each enantiomer of a chiral stationary phase. Lower traces have enlarged vertical scale. [Data from E. Badaloni, W. Cabri, A. Ciogli, R. Deias, F. Gasparrini, F. Giorgi, A. Vigevani, and C. Villani, “Combination of HPLC ‘Inverted Chirality Columns Approach’ and MS/MS Detection for Extreme Enantiomeric Excess Determination Even in Absence of Reference Samples.” Anal. Chem. 2007, 79, 6013.]

(a) Explain the appearance of the upper chromatograms. Dashed lines are position markers, not part of the chromatogram. What Problems 709 would the chromatogram of pure (R)-gimatecan look like on the same two stationary phases?

(b) Explain the appearance of the two lower chromatograms and why it can be concluded that the gimatecan contained 0.03% of the (R)-enantiomer. Why is the (R)-enantiomer not observed with the (R,R)-stationary phase?

(c) Find the relative retention (a) for the two enantiomers on the (S,S)-stationary phase.

(d) The column provides N 5 6 800 plates. What would be the resolution between the two equal peaks in a racemic (equal) mixture of (R)- and (S)-gimatecan? If the peaks were symmetric, does this resolution provide baseline separation in which signal returns to baseline before the next peak begins?

Question:The graph shows retention data from aC8silica column with an acetonitrile/water mobile phase.


(a) What mobile phase composition provides greatest retention(k)for the components? Least retention? Coelution (equalk)of two components?

(b) Predict the retention time of each peak at 40%and60%acetonitrite. Draw a chromatogram (a "stick diagram" representing each peak as a vertical line) of the separation at each mobile phase composition.

(c) Would60%acetonitrile yield adequate resolution?

(d) Assuming Gaussian peaks, does the separation at60%acetonitrile have the attributes of a good separation?

A known mixture of compounds A and B gave the following HPLC results:

A solution was prepared by mixing 12.49mgof Bplus 10.00mLof unknown containing just and diluting to 25.00mL. Peak areas of 5.97and 6.38were observed for AandB, respectively. Find the concentration of A(mg/mL)in the unknown.

(a) Nonpolar aromatic compounds were separated by HPLC on an octadecyl(C18)bonded phase. The eluent was 65 vol% methanol in water. How would the retention times be affected if 90% methanol were used instead?

(b) Octanoic acid and 1-aminooctane were passed through the same column described in (a), using an eluent of 20% methanol/80% buffer (pH 3.0). State which compound is expected to be eluted first and why.

role="math" localid="1656416023291" CH3CH2CH2CH2CH2CH2CH2CO2HOctanoicacidCH3CH2CH2CH2CH2CH2CH2CH2NH21Aminooctane

(c) Polar solutes were separated by hydrophilic interaction chromatography (HILIC) with a strongly polar bonded phase. How would retention times be affected if eluent were changed from 80 vol% to 90 vol% acetonitrile in water?

(d) Polar solutes were separated by normal-phase chromatographyon bare silica using methyl t-butyl ether and 2-propanol solvent. How would retention times be affected if eluent were changed from 40 vol% to 60 vol% 2-propanol? (Hint: See Table 25-4.)

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