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The chromatogram in Box 25-3 shows the supercritical fluid chromatography separation of seven steroids monitored by three detectors.

(a) In the middle chromatogram, ultraviolet detection provides near universal response for the steroids, whereas in the lower chromatogram the ultraviolet detector provides a selective response for a few of the steroids. How can ultraviolet detection act as either a selective or universal detector?

(b) Why is a sloping baseline observed at 210 nm, but the baseline is flat at 254 nm?

(c) Use the baseline disturbance early in the 254 nm chromatogram to measure tm. How does the measured value compare with that predicted using Equation 25-5 given that the column is 25 脳 0.46 cm and the flow rate is 2.0 mL/min.

Short Answer

Expert verified

(a)Many chromophores absorb ,210 nm which is in the middle range but only some absorb at 254 nm. So ultraviolet detector acts as universal detector below 210 nm. Above that wavelength it acts as a selective detector.

(b)The increasing gradient of CH3OH absorbs weakly at 210 nm. So,sloping baseline is observed at 210 nm, but the baseline is flat at 254 nm.

(c) The measured value of tm is 1.29 min, and the predicted value of tm is 1.32 min.

Step by step solution

01

Explanation regarding part (a)

In the middle chromatogram, ultraviolet detection provides near universal response for the steroids, whereas in the lower chromatogram the ultraviolet detector provides a selective response for a few of the steroids. An ultraviolet detector using a flow cell such is the most common HPLC detector because many solutes absorb ultraviolet light. Simple systems employing the intense 254-nm emission of a mercury vapor lamp were the backbone of early HPLC systems, but are little used today. More versatile, variable-wavelength detectors have broadband deuterium, xenon, or tungsten lamps and a monochromator, are used now a days. At wavelengths above 210 nm, detection is selective for compounds with an absorbing chromophore. Many compounds absorb wavelengths below 210nm and so ultraviolet detection below 210 nm is nearly universal whereas above that wavelength it acts as a selective detector.

02

Explanation regarding part (b)

The increasing gradient of CH3OH absorbs weakly at 210 nm. So, sloping baseline is observed at 210 nm, but the baseline is flat at 254 nm.

03

Explanation regarding part (c)

For calculating tm the formula need to be used is shown below

tm=Ldc22FL=Columnlengthdc=ColumndiameterF=Flowrate

Therefore, from the given data the following can be calculated

L=25cmdc=0.46cmF=2mL/mintm=250.46222min=1.29min

The measured value of tm is 1.29 min, and the predicted value of tm is 1.32 min.

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Most popular questions from this chapter

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, 鈥淐ombination of HPLC 鈥業nverted Chirality Columns Approach鈥 and MS/MS Detection for Extreme Enantiomeric Excess Determination Even in Absence of Reference Samples.鈥 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?

Use Figure 25-30 for the following questions:

(a) What pH would be best for the separation of benzoic acid, 4-nitrophenol, and 3-methylbenzoic acid?

(b) What pH would be best for the separation of benzoic acid, 3-methylbenzoic acid, and 4-methylaniline?

(c) What pH would be best for separation of 4-nitrophenol, 4-methylaniline, and codeine on a typical C18-silica column?

If along 15cmHPCL column has a place height of 5.0 what will be the half-width (in seconds) of a peak eluted at 10.0min? if plate height5渭尘,what will bew1/2?

25-2Why does the retention onder of peaks 2 and 3 change on the polar embedded column?

what are criteria for an adequate isocratic chromatographic separation?

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