In: Chemistry
Consider two compounds for which tm = 1 min, t1 = 14.3 min and t2 = 15.0 min. The peak widths at half heights are 13.0 sec and 14.9 sec, respectively. What are the peak widths? (This requires that you think about the mathematics of a Gaussian curve.) Calculate the resolution of this separation. How many theoretical plates does the column have for these analytes? What is the value of the selectivity factor, α?
tm = Void time
t1 = Retention time of 1st component = 14.3 min
t2 = Retention time of 2nd component = 15 min
For finding resolution and Theoretical Plates, it is necessary to know the width of the peak. So, lets discuss about the peak width, peak width at half height. Generally, the peaks obtained are called as Gaussian Peak. The width-at-half-height (w0.5) is the distance from the front slope of the peak to the back slope of the peak measured at 50% of the maximum peak height. The peak width can be calculated from peak width at half height by using the following equation - w = 1.7 w0.5 (only if the peak is gaussain shape). Here 1.7 is standard deviation () units expressed for width, when Gaussain peak is observed by plotting Amplitude - Y-axis Vs Width ()-X-axis.
so, w = 1.7 X 13 = 22.1 sec ; w = 1.7 X 14.9 = 25.33 sec
Therefore, the peak widths are 22.1 and 25.33 sec respectively for 1st and 2nd component.
Resolution can be expressed by the following equation -
= 15-14.3/0.5(22.1 + 25.33) = 0.0295
So the resolution is 0.0295
Number of theoretical plates (N) = 16 tr2/w2 ; where tr = retention time of each component, w = width of peak;
Hence,
Number of theoretical plates for 1st component (N1) = 16 (14.3)2/(22.1)2 = 6.69
Number of theoretical plates for 2nd component (N2) = 16 (15)2/(25.33)2 = 5.61
Average no. of theoretical plates = 6.69 + 5.61 / 2 = 6.15
Selectivity factor () = t2 - tm / t1 - tm = 15-1 / 14.3-1 = 14/13.3 = 1.05, This value indicates both the components are separated well.