Column Efficiency Calculation , Analytical Wavelength

Methods for Describing Column Efficiency

There are two related terms used as quantitative measures of chromatographic efficiency:

  • Number of ‘theoretical plates’ N
  • Plate height H.

These parameters are used to measure a column’s resolving power (ability to differentiate between closely migrating peaks - sometimes called efficiency). The two parameters are related by the following equation:

N equals L divided by H

where L is the length of the column packing (usually in cm - this is a constant for a given column).

The term theoretical plates refers to an early explanation of the resolving power of a column where the column was viewed as a ‘stack’ of closely related but discrete layers (or plates). The more plates in the column the greater the resolving power of the column but as the column has a definite length, more plates equals thinner plates. In other words and H are inversely proportional. As N increases (for a given column) H decreases.

There are great differences in the plate height and numbers of theoretical plates between columns due to differences in the column type as well as the stationary and mobile phases used.

Thus variations of the following magnitudes exist:

Plate height:0.1 to 0.001 mm or smaller
Plate numbers:100 to 100 000

There are a number of complex methods for determining N but a simple approximation of N is given by:

N equals 5.54 times (tg divided by W over 2) squared.

where W/2 = half of the width of the peak at its base, which is also equal to the width of the peak at half its maximum height and tg = retention time (time taken for the peak to come off the column) of the analyte.

A peak on chromatogram showing a vertical line through the top of the peak and a horizontal line accross the peak at its base. The W/2 label on the chromatogram represents half of the base width.

Sample problem:
Calculate the number of theoretical plates N and the plate height H, when the retention time is 20.40 minutes, half of the base width (given in minutes) is 0.65 minutes and the column length is 30 cm.

Solution
N equals 5.54 times (tg divided by W over 2) squared, which equals 5.54 times (20.40 divided by 0.65) squared, which equals 5.54 times (31.38) squared, which equals 5.54 times 984.70 which equals 5455 plates.

N equals L divided by H, so H equals L divided by N, which equals 30 cm divided by 5455 which equals 0.05 mm.

Practical application of N
N is usually used to measure and verify the performance of a column.
Measurement of N can be used to:

  • measure the resolving power of a column
  • measure that the resolving power of a new column is to specification
  • monitor the resolving power of a column over time
  • determine if column replacement is required due to deterioration in performance
  • troubleshoot problems.

The number of theoretical plates n, is the number of discrete distillations that would have to be performed to obtain an equivalent separation.

Gas chromatography columns normally have 103 to 106 theoretical plates.

The number of theoretical plates is related to the retention time, tr, and the width of the peak containing the compound.

If the peaks are reasonably symmetric, it can be assumed that they have a Gaussian shape. Then

n=5.54(trw12)2

where w12 is the peak width at half-height

You find the peak width at half height by drawing a line vertically from the peak maximum to the baseline, measuring half-way up the peak, drawing a horizontal line, and measuring the length of the horizontal line.

You measure the retention time (designated avove as Ve for elution volume) at the point where the vertical line drawn through the maximum intersects the baseline.

Both tr and w12 must be measured in the same units.

other hand-----

EXAMPLE

A chromatogram from a certain column has a peak with a w12 of 12 mm and a tr of 650 mm as measured on the chart. What is the number of theoretical plates?

Solution

n=5.54(trw12)2=5.54(650mm12mm)2=16 250 theoretical plates


Analytical Wavelength

The analytical wavelength for quantitative analysis is normally lamdamax ie the wavelength corresponding to an absorption peak.

Absorption spectra with one peak

There are several reasons for this choice.

Absorbance per unit concentration of analyte is greatest at lambdamax - this provides maximum sensitivity. That is to say, the higher the absorbance, the lower the concentration that the analysis is able to reliably measure.

Consider the comparison (below) of using the analytical wavelength at lambdamax and at half lambdamax - the analyte concentration is some value called ‘x’. If absorbances cannot be read below 0.1 because of the potential for substantial error, you can see in this example that lambdamax enables the analyte concentration to be determined down to at least x/8 whereas this concentration falls below the 0.1 absorbance threshold at half lambdamax. The higher the absorbance per unit of concentration (highest at lambdamax), the more sensitive the analysis.

Analytical wavelength at lambda max and at half lambda max

The absorbance curve is normally flat (or plateaus out) around lambdamax and this circumstance leads to optimum precision and accuracy. There can be some spectrophotometer error in the setting of lambdamax but without there being significant change in the value of the absorbance. On the other hand, if the analytical wavelength is selected from a point on the side of a peak, the same error in wavelength setting can have a significant impact on absorbance. Error in the wavelength setting of a spectrophotometer translates to imprecision/inaccuracy in the calculated concentrations.

A peak on a spectrum showing that variation in wavelength setting leads to relatively small errors in absorbance at lambda max compared to much larger errors at wavelength settings on the shoulder of the peak.

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