Evaluation of an LC Method - - Chromatography Online
Evaluation of an LC Method

LCGC North America
Volume 30, Issue 11


Table I: System suitability test results
Aliquid chromatography (LC) method from the United States Pharmacopoeia (USP) for the assay of a drug product was set up in the laboratory of one of the authors (F.O.). Although all of the details of the method cannot be shared because of its proprietary nature, the key elements follow. A 250 mm 4.6 mm L1 column packed with 5-μm particles was used with a mobile phase of 50:50 acetonitrile–water and a flow rate of 0.6 mL/min. A refractive-index detector was used, with both the detector and column maintained at a temperature of 55 C. System suitability requirements included a minimum resolution between the active ingredient and a related compound, a relative standard deviation (RSD) of <2% for the peak area of replicate injections of the active ingredient, and a USP tailing factor (TF) of <2. When the method was run, the resolution requirements were met; TF = 1.13 was observed, and RSD = 0.2% was determined (see Table I). It can be seen that the system suitability requirements were passed, and because the method was operated according to the monograph specifications, no further testing was required before placing the method into routine use.

Other Factors




Although the method proved suitable for use, let's examine it a little more closely to see how it fits into our expectations for a "good" isocratic method. One system suitability parameter that often is included in methods such as this is a minimum column plate number requirement. The plate number can be calculated easily from a chromatogram, such as that of Figure 1. The plate number, N, is


Figure 1: Chromatogram for system suitability test injection 1.
where t R is the retention time and w 0.5 is the peak width at half its height, both in the same units; these values are included in the data system report generated for each chromatogram. For the last peak listed in Table I, t R = 6.243 min and w 0.5 = 0.1324 min, so N = 12,300 (as usual, numbers are rounded for simplicity, so if you try to repeat the calculations presented here, you may get slightly different values).




What does the calculated plate number tell us? As has been described in previous LC Troubleshooting discussions, for real samples operating under real conditions, we can estimate a reasonable plate number for any column as

where L is the column length (in millimeters) and d p is the particle diameter (in micrometers). The present column is a 250 mm 4.6 mm column packed with 5-μm particles, so the predicted value of N ≈ 15,000. If the measured value of N is within about 20% of this predicted value, the column is in good condition. So, although the column was new, the plate number is at the low end of its expected performance range. In the present case, we'll see in a moment that the retention factor is quite small, which makes the peaks more susceptible to extracolumn band broadening because of injection problems or extracolumn volume. We suspect that this is the reason for the lower-than-expected plate number in the present example.

Retention Factor




Another parameter that can be used to evaluate the quality of an LC method is the retention factor, k. As a general rule, for the best isocratic methods, we would like to have 2 < k < 10, and if that is not possible, 1 < k < 20 usually is acceptable (see the discussion associated with Figure 1 in reference 1 for more information on this). We calculate k as




where t 0 is the column dead time, usually identified as the first peak in the chromatogram or "solvent front." We can see in Figure 1 that the retention time of the first peak in the chromatogram is 4.0 min. It is a good idea to double-check that this is where the t 0 peak is expected. For 4.6-mm i.d. columns we can estimate t 0 from the column volume, V M:

where L is the length of the column (in millimeters) and V M is in milliliters. Therefore, in the present case V M ≈ 2.5 mL. Convert to t 0 by dividing by the flow rate: t 0 = 2.5 mL/0.6 mL/min = 4.2 min. The estimate should be within ˜≈ 10% of the observed value, so the observed t 0 = 4.0 min is reasonable.

Now we can calculate k = (6.2 – 4.0)/4.0 = 0.6. This is much less than even the minimum desired k = 1. Does this mean that the method is bad or that a bad method was approved and included in the USP? Not necessarily; let's look further.


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