(a) use figure 25-28a select a tetrahydrofurn/water mobile phase strength of equivalent strength to 80%methanol.

(b)Describe how to prepare 1 L of this tetrahydrofurn mobile phase.

(c)what limitation would be imposed by the use tetrahydrofurn.

Short Answer

Expert verified

a) 53%

b)To prepare 1 L of this tetrahydrofuran mobile phase (53 %), we need to mix 530 mL of tetrahydrofuran with 470 mL of water.

c) The limitations that would be imposed by the use of tetrahydrofuran is that it absorbs UV radiation and attacks polyether, ether, ketone and plastic components.

Step by step solution

01

Definition of water mobile phase of equivalent strength

The weak solvent is commonly referred to as the A-solvent (percent A), and the strong solvent is referred to as the B-solvent ( percent B). The mobile phase is typically a mixture of A- and B-solvents, such as 40% buffer and 60% methanol (abbreviated 60 percent B).

a) Figure 25-28 depicts a nomograph of the volume percentage of solvents with the same eluent strength.

At the same eluent strength, a vertical line intersects each solvent line.

Tetrahydrofurn mobile phase

b) The figure shows that 53 percent of the tetrahydrofuran/water mobile phase strength is equivalent to a concentration of 80% methanol.

To prepare 1 L of this tetrahydrofuran mobile phase (53%), we need to mix 530 mL of tetrahydrofuran with 470 mL of water.

02

Limitations of tetrahydrofuran

c) The limitations that would be imposed by the use of tetrahydrofuran is that it absorbs UV radiation and attacks polyether, ether, ketone and plastic components.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

(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.)

Retention factors for three solutes separated on aC8non-polar stationary phase are listed in the table. Eluent was a 70 : 30 (vol/vol) mixture of 50 mM citrate buffer (adjusted to pH withNH3) plus methanol. Draw the dominant species of each compound at each pH in the table and explain the behavior of the retention factors.

Chromatography–mass spectrometry. HPLC separation of

enantiomers of the drug Ritalin on a chiral stationary phase was

shown in Problem 25-13.

(a) Detection is by atmospheric pressure chemical ionization with

selected reaction monitoring of the m/z 23484 transitions. Explain

how this detection works and propose structures for m/z 234 and m/z 84.

(b) For quantitative analysis, the internal standardH32-Ritalin with

a deuterated methyl group was added. Deuterated enantiomers have

the same retention times as unlabelled enantiomers. Which selected

reaction monitoring transition should be monitored to produce a

chromatogram of the internal standard in which unlabelled Ritalin

will be invisible?

Chromatography–mass spectrometry. Cocaine metabolism in rats can be studied by injecting the drug and periodically with drawing blood to measure levels of metabolites by HPLC–mass spectrometry. For quantitative analysis, isotopically labelled internal standards are mixed with the blood sample. Blood was analysed by reversed-phase chromatography with an acidic eluent and atmospheric pressure chemical ionization mass spectrometry for detection. The mass spectrum of the collisionally activated dissociation products from the m/z 304 positive ion is shown in the figure on the next page. Selected reaction monitoring (m/z 304 from mass filter Q1 and m/z 182 from Q3 in Figure 22-33) gave a single chromatographic peak at 9.22 min for cocaine. The internal standard H52-cocaine gave a single peak at 9.19 min for m/z 309 (Q1) 182(Q3).

(a) Draw the structure of the ion at m/z 304.

(b) Suggest a structure for the ion at m/z 182.

(c) The intense peaks at m/z 182 and 304 do not have C2isotopic partners at m/z 183 and 305. Explain why.

(d) Rat plasma is exceedingly complex. Why does the chromatogram show just one clean peak?

(e) Given that H52-cocaine has only two major mass spectral peaks at m/z 309 and 182, which atoms are labelled with deuterium?

(f) Explain how you would use H52-cocaine for measuring cocaine in blood.

Spectrum for Problem 25-25.

Left: Mass spectrum of collisionally activated dissociation products from m/z 304 positive ion from atmospheric pressure chemical ionization mass spectrum of cocaine.

Right: Chromatograms obtained by selected reaction monitoring. [Data from G. Singh, V. Arora, P. T. Fenn, B. Mets, and I. A. Blair, “Isotope Dilution Liquid Chromatography Tandem Mass Spectrometry Assay for Trace Analysis of Cocaine and Its Metabolites in Plasma,” Anal. Chem. 1999, 71, 2021.]

A bonded stationary phase for the separation of optical isomers has the structure

To resolve the enantiomers of amines, alcohols, or thiols, the compounds are first derivatized with a nitroaromatic group that increases their interaction with the bonded phase and makes them observable with a spectrophotometric detector.

When the mixture is eluted with 20 vol% 2-propanol in hexane, the (R)-enantiomer is eluted before the (S)-enantiomer, with the following chromatographic parameters:

Resolution=Δtrwav=7.7

Relative retention(α)=4.53

k for (R)-isomer =1.35

tm=1.00min

wherewavis the average width of the two Gaussian peaks at their base.

(a) Findt1,t2andwawith units of minutes.

(b) The width of a peak at half-height isrole="math" localid="1663582749865" w1/2(Figure 23-9). If the plate number for cach peak is the same, findw1/2for each peak.

(c) The area of a Gaussian peak is1.064×peakheight×w1/2. Given that the areas under the two bands should be equal, find the relative peak heights(heightR//heightS).

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free