The efficiency of a separating system is best demonstrated by its peak capacity.
This shows how many components can be separated in theory within a certaink range as peaks of resolution 1. The number of theoretical plates, N, of the column used must be known, as the peak capacity, n, is proportional to its square root:18
nẳ1ỵ ffiffiffiffiN p
4 lnð1ỵkmaxị wherekmaxẳmaximumkvalue.
Problem 10
A column with 10 000 theoretical plates has a breakthrough time of 1 min.
Calculate the number of peaks of resolution 1 that can possibly be separated over a 5 min period.
Solution
t0ẳ1 min;tR maxẳ5 min.
kmaxẳ51 1 ẳ4 nẳ1ỵ
ffiffiffiffiffiffiffiffiffiffiffiffiffi 10 000 p
4 lnð1ỵ4ị ẳ1ỵ ð251:61ị nẳ41
The above-mentioned equation for n in fact is only valid for isocratic separations and if the peaks are symmetric; the peak capacity is larger with gradient separations. Tailing decreases the peak capacity of a column. In real separations the theoretical plate number is not constant over the fullk range.
However, it is even more important to realize that a hypothetical parameter is discussed here. It is necessary to deal with peaks that arestatisticallydistributed over the accessible time range. The theory of probabilities allows us to proceed from ideal to near-real separations. Unfortunately, the results are discouraging.
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17V. R. Meyer and Th. Welsch,LC GC Int.,9, 670 (1996); A. Felinger and M. C. Pietrogrande, Anal. Chem.,73, 619 A (2001).
18E. Grushka,Anal. Chem.,42, 1142 (1970).
What is the probability that a certain component of the sample mixture will be eluted as a single peak and not overlapped by other components?
Pe2m=n where
Pẳprobability related to a single component mẳnumber of components in the sample mixture
Note that m is never known in real samples. One always has to presume the presence of unknown and unwanted compounds which will be eluted at any time, perhaps even together with the peak of interest. Nevertheless, it is possible to calculate the following problem with the equation given above.
Problem 11
The sample mixture consists of ten compounds. Calculate the probability that a given peak will be eluted as a single peak on the column with peak capacity 41.
Solution
mẳ10 nẳ41
Pe210=41ẳ0:61
The probability of sufficient resolution to the neighbouring peaks is only little more than 60% for each peak. We can expect that six of the compounds present will be resolved; the other four will be eluted with inadequate or missing resolution. Quantitative analysis can be impeded or almost impossible and fractions obtained by preparative chromatography can be impure, perhaps without any sign of warning to the user. The situation is even worse if the peaks are of unequal size, which in fact is the rule.
It is possible to calculate the necessary peak capacity for a 95% probability by converting the equation. For the ten-component mixture this is 390, which corresponds to a plate number of almost 1 million (kmaxẳ4) or to a retention factor of 5.7 millions (Nẳ10 000).
What is the probability that all components of a mixture will be separated?
P0ẳ 1m1 n1
m2
P0ẳprobability related to all components
Problem 12
Calculate the probability that all components of the ten-compound mixture will be separated on the column with a peak capacity of 41.
Solution
mẳ10 nẳ41 P0ẳ 1 9
40
8
ẳ0:13
P0is only 13% and is markedly lower thanPbecause now, in contrast to the above-discussed problem,allpeaks need to be resolved. To reach this goal with a 95% probability it would be necessary to perform the separation on a column with a peak capacity of 1400, i.e. with 12 million theoretical platesðkmaxẳ4)!
Computer generated chromatograms, obtained with random numbers, are shown in Fig. 2.26. They represent four possible separation patterns in a case withmẳ10 andnẳ41, i.e.Nẳ10 000 and a retention time window between 1 and 5 min. In one chromatogram all ten peaks are visible. In the others, only nine peaks can be seen, sometimes with poor resolution. In two cases it is not
Fig. 2.26 Simulated chromatograms, generated with random numbers, representing a separation at peak capacity 41 and a sample with 10 compounds.
known where the tenth peak is and in one separation its position can be supposed. These graphs look like real chromatograms, and real chromatograms often follow the same rules of random elution patterns.
These considerations do not mean that the separation methods available today are inadequate. Often the situation is better than presented here because the peaks are not statistically distributed but are eluted as given by their physical properties. Homologues will be separated by increasing hydrocarbon chain length on a reversed phase, as an example. Nevertheless, even the highest separation performance does not help against troublesome surprises in qualitative or quantitative analysis. The knowledge of the equations presented here is an aid in the realistic judgement of separations obtained from complex mixtures.
Remedy: highly specific detection, derivatization (see Section 19.10), coupling with spectroscopic methods (Section 6.10), optimized gradient elution (Sections 18.2 and 18.5) or column switching (multidimensional separation, Section 18.3).