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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?

Short Answer

Expert verified

a) The greatest and least retention of components in the mobile phase in reversed phase column chromatography, increasing the amount of organic solvent increases the mobile phase strength.

Acetonitrile at 40% yields the longest retention time. The retention time is the shortest at 60%.

Toluene and benzophenone are co-elutes at 51.6%acetonitrile.

b) The chromatogram:

c) 60%acetonitrile yield 2.4resolution.

d) The separation of peaks can be done by 60%acetonitrile and it has additional plus point of separation time less than six minutes.

Step by step solution

01

Concept used.

Resolution=Nα-1αk21+k2Where,N-numberoftheoreticalplatesα-relativeretentionofthetwopeaksk2-retentionfactorofthemoreretainedcomponentTheaboveequationshowsthatthevalueofresolutionisproportionaltoN

02

The greatest and least retention of components in mobile phase. 

a)

The greatest and least retention of components in the mobile phase in reversed phase column chromatography, increasing the amount of organic solvent increases the mobile phase strength.

Acetonitrile at 40%yields the longest retention time. The retention time is the shortest at 60%.

Toluene and benzophenone are co-elutes at51.6%acetonitrile

03

Step 3: Draw the chromatogram (a "stick diagram" representing each peak as a vertical line) of the separation at each mobile phase composition.

b)Eachcomposition'smobilephaseisseparated.logktoretentiontimeconversion.K=tf-tmtmtm×(1+K)tmcanbecalculatedbyEquation25-5~tm≈1AC22F≈15cm×(0.46cm2)2×1.0mil/min=1.6mintf=1.6min×(1+15.1)=25.8minThechromatogramisgivenbelowas

04

Step 4: Explain if 60%acetonitrile yield adequate resolution.

c)

Benzophenone and toluene, the chemicals with the closest peak, have the lowest resolution. These chemicals' retention factors are reported as

100.13=1.35100.21=1.62Theformulaforcalculatingrelativeretentionisasfollows:α=K2K1=1.611.35=1.20N=3000.1(cm)dp(1m)=3000×155=9000Resolution=N4α-1αk21+k2=900041.20-11.201.621+1.62=2.4

05

Step 5: Whether the separation at60%acetonitrile have the attributes of a good separation.

d)

The requirements for good peak separation are as follows:

0.5≤K≤20Resoluction≥2Operatingpressure≤15MPa0.9≤asymmetryfactor≤1.5

All of the following requirements for excellent peak separation are met by the 60%

acetonitrile. Thus,60%acetonitrile may be used to separate peaks, with the added benefit of a separation period of less than six minutes.

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Most popular questions from this chapter

Question: Explain how to use a gradient for the first run to decide whether isocratic or gradient elution would be more appropriate.

  1. According to equation 25-2if all conditions are constant, but particle size is reduced from3μ³¾to0.7μ³¾by what factor must pressure be increased to maintain constant linear velocity?
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