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(a) When you try separating an unknown mixture byreversed-phase chromatography with 48%acetonitrile 50%water,the peaks are too close together and are eluted in the range k = 2- 6Should you use a higher or lower concentration of acetonitrile in thenext run?

(b) When you try separating an unknown mixture by normal-phasechromatography with 50%hexane50% methyl t-butyl ether, thepeaks are too close together and are eluted in the range k = 2 - 6Should you use a higher or lower concentration of hexane in thenext run?

Short Answer

Expert verified

(a) Lower concentration of acetonitrile should be used because lower solvent strength increases the difference in retention between compounds.

(b) Solvent that would lower the solvent strength is recommended for the use. In this case it is hexane.

Step by step solution

01

Define Acetonitrile:

Acetonitrile, often abbreviated MeCN (methyl cyanide), is the chemical compound with the formula CH3CN. This colourless liquid is the simplest organic nitrile (hydrogen cyanide is a simpler nitrile, but the cyanide anion is not classed as organic). It is produced mainly as abyproduct of acrylonitrile manufacture.

02

Concentration of Acetonitrile:

We should use lower concentration of acetonitrile because lower solvent strength increases the difference in retention between compounds.

01

DefineChromatography

Chromatography is a group of laboratory techniques used to separate the components of a mixture by passing the mixture through a stationary phase.

02

The Concentration of Hexane:

We should use a higher concentration of hexane in the next run because in normal-phase chromatography, solvent strength increases as the solvent becomes more polar (in this case methyl t-butyl ether is more polar). So, to improve resolution and increase the retention time,such solvent should be used that would lower the solvent strength. In this case it is hexane.

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Most popular questions from this chapter

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 ร— 0.46 cm and the flow rate is 2.0 mL/min.

In hydrophilic interaction chromatography (HILIC), why is eluent strength increased by increasing the fraction of water in the mobile phase?

Chromatographyโ€“mass spectrometry. Cocaine metabolism in rats can be studied by injecting the drug and periodically with drawing blood to measure levels of metabolites by HPLCโ€“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, โ€œIsotope Dilution Liquid Chromatography Tandem Mass Spectrometry Assay for Trace Analysis of Cocaine and Its Metabolites in Plasma,โ€ Anal. Chem. 1999, 71, 2021.]

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.

Chromatographyโ€“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?

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