The eluted compounds are transported by the mobile phase to the detector and recorded as Gaussian (bell-shaped) curves. The signals are known as peaks (Fig. 2.8) and the whole entity is the chromatogram.
The peaks give qualitative and quantitative information on the mixture in question:
(a) Qualitative: the retention time of a component is always constant under identical chromatographic conditions. The retention time is the period that
Fig. 2.8 Shape of a peak.
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1J. J. van Deemter, F. J. Zuiderweg and A. Klinkenberg,Chem. Engng. Sci.,5, 271 (1956).
elapses between sample injection and the recording of the signal maximum.
The column dimensions, type of stationary phase, mobile phase composi- tion and flow velocity, sample size and temperature provide the chromato- graphic conditions. Hence, a peak can be identified by injecting the relevant substance and then comparing retention times.
(b) Quantitative: both the area and height of a peak are proportional to the amount of a compound injected. A calibration graph can be derived from peak areas or heights obtained for various solutions of precisely known concentration and a peak-size comparison can then be used to determine the concentration of an unknown sample.
The chromatogram can be used to provide information on separation efficiency (Fig. 2.9). Herew is the peak width at the baseline,2t0 is thedead time or retention time of an unretained solute, i.e. the time required by the mobile phase to pass through the column (also called thebreakthrough time).
Hence the linear flow velocity,u, can be calculated as uẳ L
t0
Fig. 2.9 The chromatogram and its characteristic features.
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2wẳ4s, wheresis the standard deviation of a Gaussian peak.
where L is the column length. A non-retained compound, i.e. one that is not retained by the stationary phase, appears at the end of the column att0.tR is theretention time;3this is the period between sample injection and recording of the peak maximum. Two compounds can be separated if they have different retention times.t0Ris thenet retention timeoradjusted retention time. Figure 2.7 shows thattRẳt0ỵt0R.t0 is identical for all eluted substances and represents the mobile-phase residence time.t0Ris the stationary phase residence time and is different for each separated compound. The longer a compound remains in the stationary phase, the later it becomes eluted.
Retention time is a function of mobile phase flow velocity and column length.
If the mobile phase is flowing slowly or if the column is long, thent0 is large and hence so istR; tR is therefore not suitable for characterizing a compound.
Therefore the retention factor or k value (formerly known as the capacity factor,k0) is preferred:
kẳt0R
t0 ẳtRt0
t0
kis independent of the column length and mobile phase flow-rate and represents the molar ratio of the compound in the stationary and the mobile phase, as mentioned earlier (Section 2.1).
Problem 1
Calculate thekvalues of compounds 1 and 2 in Fig. 2.9.
Solution
t0ẳ12:5 mm; tR1ẳ33:1 mm; tR2ẳ70:5 mm.
k1 ẳtR1t0 t0
ẳ33:112:5 12:5 ẳ1:6 k2 ẳtR2t0
t0 ẳ70:512:5 12:5 ẳ4:6
Retention factors between 1 and 10 are preferred. If thekvalues are too low, then the degree of separation may be inadequate (if the compounds pass too rapidly through the column, no stationary phase interaction occurs and hence no chromatography). Highkvalues are accompanied by long analysis times.
Thekvalue is connected with the distribution coefficient described in Section 2.1 in the following way:4
kẳK VS VM
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3Retention volumeVRẳFtR(Fẳvolume flow-rate in ml min1). Void volumeV0ẳFt0.
4This is only valid within the so-called linear range wherekis independent of sample load but no longer under conditions of mass overload.
whereVSis the volume of stationary phase andVMthe volume of mobile phase in the column.
The retention factor is directly proportional to the volume occupied by the stationary phase and more especially to its specific area (m2g1) in the case of adsorbents. A column packed with porous-layer beads produces lowerkvalues and hence shorter analysis times than a column containing completely porous particles if the other conditions remain constant. Silica with narrow pores produces largerk values than a wide-pore material.
Two components in a mixture cannot be separated unless they have different kvalues, the means of assessment being provided by the separation factor,a, formerly known as therelative retention
aẳk2 k1
ẳtR2t0 tR1t0
ẳK2 K1
withk2>k1. Ifaẳ1, then no separation takes place as the retention times are identical. The separation factor is a measure of the chromatographic system’s potential for separating two compounds, i.e. its selectivity. Selection of the stationary and mobile phases can affect the value ofa.
Problem 2
Calculate theavalue of compounds 1 and 2.
Solution
Problem 1 showed thatk1 ẳ1:6 andk2 ẳ4:6; hence aẳ4.6=1.6ẳ2.9.
The resolution, R, of two neighbouring peaks is defined by the ratio of the distance between the two peak maxima, i.e. the distance between the two retention times,tR, and the arithmetic mean of the two peak widths, w:5
Rẳ2tR2tR1 w1þw2
ẳ1:18 tR2tR1 w1=21þw1=22 wherew1=2 is the peak width at half-height.
The peaks are not completely separated with a resolution of 1, but two peaks can be seen. The inflection tangents touch each other at the baseline. For
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5These equations are less suited for peak pairs of highly unequal area and for asymmetric peaks.
In such cases the Peak Separation Index is a better criterion: PSIẳ1ba.
quantitative analysis a resolution of 1.0 is too low in most cases. It is necessary to obtain baseline resolution, e.g. Rẳ1.5. If one of the peaks is markedly smaller than its neighbour even higher resolution is needed. See paragraph 19.5.
Problem 3
Calculate the resolution of compounds 1 and 2.
Solution
tR1ẳ33:1 mm,tR2ẳ70:5 mm; w1ẳ17 mm,w2ẳ29 mm.
Rẳ2ð70:533:1ị 17ỵ29 ẳ1:6 Manual Determination of Peak Width at the Baseline:
Inflection point tangents6 are drawn to each side of the Gaussian curve; this generally causes few problems. However, the recorder line width must be considered and this is best done by adding it to the signal width on one side but not the other (Fig. 2.10). The base width is the distance along the baseline between the two inflection tangents.
Fig. 2.10 Construction of inflection tangents.
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6Inflection pointẳposition at which the curvature or slope changes sign; ẳpositive curvature;
ẳnegative curvature.
Finally, the chromatogram can be used to calculate thenumber of theoretical plates,N, in the column:
N ẳ16 tR w
2
N ẳ5:54 tR w1=2
2
wherew1=2 is the peak width at half-height;
N ẳ2 hPtR A
2
wherehP is the peak height andAthe peak area.
All three equations yield correct results only if the peak has a Gaussian shape. This is hardly ever the case with real-life chromatograms.7 Correct values for asymmetric peaks are obtained by the momentum method.8 Approximately correct values are obtained by the equation9
N ẳ41:7ðtR=w0:1ị2 Tþ1:25
wherew0:1is the peak width at 10% of the peak height andTis tailingb0:1=a0:1
(Fig. 2.25).
The plate number calculated from a non-retained peak is a measure of the column packing efficiency, whereas in the case of peaks eluted later, mass- transfer processes also contribute to the plate number. As a general rule,N is higher for retained compounds because their relative band broadening by extra- column volumes (see Section 2.6) is lower than for the early eluted peaks.
Problem 4
How many theoretical plates emerge from calculations based on the last peak in Fig. 2.9?
Solution
tR2ẳ70:5 mm; w2 ẳ29 mm:
N ẳ16 70:5 29
2
ẳ94
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7B. A. Bidlingmeyer and F. V. Warren,Anal. Chem.,56, 1583A (1984).
8Definition:mnẳéx
0tnfðtịdt. The second moment withnẳ2 corresponds to the variancesof the peak. Withwẳ4s it is possible to calculateN. Literature: N. Dyson, Chromatographic Integration Methods, Royal Society of Chemistry, London, 2nd ed., 1998, p. 23.
9J. P. Foley and J. G. Dorsey,Anal. Chem.,55, 730 (1983).
Theheight of a theoretical plate,H, is readily calculated provided the length of the column is known:
Hẳ L N
where H is the distance over which chromatographic equilibrium is achieved (see Fig. 2.1) and is referred to as the height equivalent to a theoretical plate (HETP).
Fig. 2.11 Chromatogram of red test dyes.
Problem 5
Fat Red 7B, 1-[(p-butylphenyl)azo]-2-naphthol, 1-[(p-methoxyphenyl)azo]-2- naphthol, 1-[(m-methoxyphenyl)azo]-2-naphthol and 1-[o-methoxyphenyl)- azo]-2-naphthol red test dyes were chromatographed on a Merck low-pressure column. Silica gel 60 was the stationary phase (40–63mm) and 50% water- saturated dichloromethane was used as the mobile phase at a flow-rate of 1 ml min1.
Using the chromatogram shown in Fig. 2.11, calculate:
(a) retention factors for peaks 1–5;
(b) separation factors for the best and worst resolved pair of peaks;
(c) resolution of these two pairs of peaks;
(d) plate number for each of peaks 1–5.
Solution
(a) k1ẳ0:51;k2ẳ1:15;k3ẳ1:51;k4 ẳ1:88;k5 ẳ2:50.
(b) a12ẳ2:2; a34ẳ1:2.
(c) R12ẳ2:4;R34 ẳ0:79.
(d) N1 ẳ720;N2 ẳ850;N3 ẳ760; N4ẳ770;N5ẳ850.