(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

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?

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.

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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.

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(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).

Literature search problem: The purity of cocaine bought on the street varies dramatically. Cutting agents include levamisole, a compound normally used to kill parasites. Search the literature for a reversed-phase liquid chromatography method with diode array detection for the determination of the purity of street cocaine.

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(b) What alternative methods could be used for analysis of adulterated street cocaine?

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(e) How was the method validated?

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