Suppose that an HPLC column produces Gaussian peaks. The detector measures absorbance at 254 nm. A sample containing equal moles of compounds A and B was injected into the column. Compound A E254=2.26×104M-1cm-1has a height h=128mm and a half-width w1/2=10.1mm. Compound B E254=168×104M-1cm-1has w1/2=7.6mm. What is the height of peak B in millimeters?

Short Answer

Expert verified

The height of peak B in millimetershB=126.45mm

Step by step solution

01

Step 1:

Since the sample containing equal moles of compounds A and B, the peak Ares are proportional to molar absorptivity:

PeakApeakB=2.26.1041.68.104=1.064.hA.w1/21.064.hB.w1/2

02

Step 2:

The height of peak B is:

hB=1.064.hA.w1/2.1.68.1041.064.w1/2.2.26.104hB=1.064.128mm.10.1mm.1.68.1041.064.76mm.2.26.104hB=126.45mmHere,thefinalresultishB=126.45mm

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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).

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

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Question: Literature search problem: Human serum albumin (HSA) is an important protein ingredient in cryopreservation media used in procedures such as in vitro fertilization. Search the literature for a high-performance liquid chromatography method for the determination of human serum albumin and the stabilizer N-acetyl tryptophan in medical devices.

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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.]

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