Gas Chromatography Directions: Show all work for this problem. You may want to w
ID: 850927 • Letter: G
Question
Gas Chromatography Directions: Show all work for this problem. You may want to write your answers on a piece of notebook paper. The chromatogram in the figure below has a peak for isooctane at 13.81 min., heptanes at 14.56 min, and p-difluorobenzene at 14.77 min. Calculate or determine the following information: (a) retention times of each peak, (b) the widths of the peaks, (c) the resolution between isooctane and heptanes; (d) the resolution between isooctane and heptane; (e) the number of theoretical plates for each peak; (f) the area of each peak; and (g) the area percent for each component. Reprinted from Exploring Chemical Analysis (2007), W.H. Freeman & Company Publishers.Explanation / Answer
1. (a) The retention time, tR, is the time it takes for a compound to travel from the injection port to the detector; it is reported in minutes on the GC plot. So here the RT are: 13.81, 14.56 and 14.77 min. It is more accurate to report adjusted retention time, which is retention time minus unretained peak time (tM). But your chromatogram does not show a peak for tM at the beginning.
(b) To calculate peak width, you will need the peak heights (mmax) at its apex, which you can measure with a ruler (do this on your own!). Peak width is meaured at half-height (wh) and is given by: wh = 0.94 A /mmax where A is the area under the peak.
(c) The distance between two GC peaks is known as their resolution. So resolution between isooctane and heptane is: 14.56 - 13.81 = 0.75 min
(d) Same question as before!
(e) Number of theoretical plates for each peak is given by: N = 5.545 (tR / wh)2
(f) Area of peak can be calculated using geometrical approximation. It will be equal to mmax * wh
(g) For area percent of each peak, you have to divide the peak area by the total area of all peaks and then multiply with 100.
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