The Relationship between Resolution and Percent Band Overlap In discussing the chromatographic separation of two components, many analytical texts point out that the so-called "separation factor" (ratio of the retention volumes or retention times) does not give an accurate picture of an analytical separation because it fails to take intoaccount the width of the t 1" I. elutinn hands. "Resolution", defined as twice the retention time difference Sussian bands "parated by The portion divided by the sum of the band widths, is usually offered ass better measure AthatOveriaps shaded. of separation. Unfortunately, there is no simple way to look a t a resolution Of number and determine the degree of band overlap; a calculation must be made. On surveying several texts on instrumental analysisand separationscience, I found many that related resolution to band overlap in a table (e.g., R = 0.5 and 1.0correspond to band overlaps of 15.8%and 2.3%, respectively), but most did not include a discussion of how to make this calculation. Littlewood2 gives a method for calculating the impurity fraction in terms of the two retention volumes (or times) and the number of theoretical plates in the column, based on the original work of Gluckauf.Vhe following discussion of the relationship between resolution and band overlap for two gaussian peaks of equal areas is a simplified version of this mrl hurl. .a t:tmiliar paramrtrr J n ~ n u s t i a nrurvr :ithe .tandnrd dmmrwn, n . Area? under l l w r u r w h r r w r n rhr mem and any numl~rr8 ~ i n - mttahulaltd t~ n i ' ? " I rrn ht. fimnd in rnl,les in mlrt 5tati41.. tt.\l*and 3150 in d d e r ~ d n ~ m the s d "Hnndlmnk .d('hem~,trvand I'hwic 5". ' F8.r cxam~>lr. under := I.. w u will find Il.l%~,.l,%othe n r m hrtuwn rhr mcrn and I yo is 15.51 nf the tota~Breaunder the curve. ~ h figure d shows two identical gaussian bands whose maxima are separated by 20. For purposes of calculating resolution, band widths are definedGas 40; therefore, the resolution is: R = 2(20)/(40 40) = 0.5. To determine the overlap between the two bands, note that the area between the retention time of band A(ta) and 1 . 0 ~ is0.3413 of the total peak area, while the area between ta and the limit of the curve is0.5000 of the total badd area. Therefore theshaded overlap area in the figure is0.5000 = 0.3413 = 0.1587. A resolution of0.5 corresponds to an overlapof about 15.9%. The general application of this reasoning can be illustrated by the following two problems:
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1) What is the degree of band overlap for two peaks havinga resolution = 0.8? Solution: R = 2(tn - ta)l(4o 40) = 0.8; t" - tn = 3.2c, and the distance between ta and the band intersection = 1.60. The area between t n and 1.60 = 0.4452 of the total area under the curve, thus the overlap area = 0.5000 - 0.4452 = 0.0648. A resolution of 0.8 corresponds to a band overlap of about 5.5%. 2) What resolution corresponds t o a 10%hand overlap? Solution: The area between ta and the band intersection = 0.5000 - 0.1000 = 0.4000 of the total area under the curve, and the corresponding value of r = 1.28. Thus t" - tn = 2(1.28) = 2.560, and the resolution, R = 2(2.560)/(40 40) = 0.64.
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Contribution No. 241 of the Belle W. Baruch Institute for Coastal and Marine Sciences. "ittlewood, A. B., "Gas Chromatography", 2nd ed., Academic Press, New York, 1970, p. 156-162. Wluckauf, E., Trans. Foradmy Soc 51,34 (1955). Weast. Robert C.. "Handbook of Chemistrv and Phvsics", 45th ed., Chemical Rubber Co., Cleveland, OH, 1964, A107. W a g e r , Barry L., Snyder, Lloyd R., and Horvath, Cnaba, "An Introduction to Separation Science", John Wiley & Sons, Inc., New Ynrk. 1973.p. 147-150.
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University of South Carolina Columbia, 29208
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T e r r y F. Ridleman
Volume 56, Number 5. May 1979 / 293
