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Chapter 23: Q45P (page 631)

(a) For the asymmetric chromatogram in Figure 23-14, calculate the asymmetry factor, BA.(b) The asymmetric chromatogram in Figure 23-14 has a retention time equal to 15.0 min and a w0.1of 44s . Find the number of theoretical plates. (c) The width of a Gaussian peak at a height equal to110 of the peak height is 4.297蟽. Suppose that the peak in (b) is symmetric with A=B=22s. Use Equations 23-30 and 23-32 to find the plate number.

Short Answer

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The solution is

a) Asymmetryfactor=2b) N=5368c)蚁=10.24sd) N=7.72脳103

Step by step solution

01

of 6

a) For the asymmetric chromatogram in Figure 23-14, we shall calculate the

asymmetry factor (B/A) .

02

of 6

A horizontal line was drawn across the peak at of the height, as shown in Figure

23-14:

- The quantities A and B are then determined by measuring the left half A and right half B of the peak width at 10% of the peak height - start with the left half and find that

the width is around 1 / 2(A).

- Next, look at the right-hand half; you can see that the width is roughly 1 / 1 ( B ).

We can calculate the asymmetry factor ( B / A ) from these findings, which is around:

Asymmetry factor =1/11/2=2

03

of 6

b) If the chromatogram is asymmetric, this is where we'll find the number of theoretical plates.



tr=15min=900sandaw0.1=44s:N=41.7(tr/w0.1)2(B/A)+1.25


Use the asymmetry factor from section a) but make sure to stick to the same time values (seconds).

N=41.7(900s/44s)2(2)+1.25

N=5368

04

of 6

c) A Gaussian peak with a width of 4.297蚁has a height of 1/10 of the peak height.

We'll apply Equations 23-30 and 23-32 to find the plate number, assuming the

peak in (b) is symmetric with A = B = 22 s.

05

of 6

If we know that the width at 1 / 10 is 22 s + 22 s = 44 s, we'll start by calculating the standard deviation.

蚁=44s4.297=10.24s

06

of 6

We'll then figure out how many plates there are:

N=tr蚁2N=900s10.24s2N=7.72脳103

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Most popular questions from this chapter

A chromatogram with ideal Gaussian bands hastr=9minand
w1/2=2min.

(a) How many theoretical plates are present?

(b) Find the plate height if the column is 10cm long.

An open tubular column is 30.01 mlong and has an inner diameter of 0.530mm. It is coated on the inside wall with a layer of stationary phase that is3.1渭尘thick. Unretained solute passes through in 2.16min, whereas a particular solute has a retention time of 17.32min. (a) Find the linear velocity and volume flow rate. (b) Find the retention factor for the solute and the fraction of time spent in the stationary phase.(c) Find the partition coefficient, K 5 cs/cm, for this solute.

23-42. A separation of 2.5mgof an unknown mixture has been optimized on a column of length L and diameter d .

(a) Explain why you might not achieve the same resolution if the 2.5mgof unknown mixture were injected in twice the injection volume.

(b) Explain why you might not achieve the same resolution if5.0mg of unknown mixture were injected in the original injection volume.

The theoretical limit for extracting solute Sfrom Phase1(volumeV1)into phase2 (volumeV2)is attained by dividing V2into an infinite number of infinitesimally small portions and conducting an infinite number of extractions. With a partition coefficient, K=[S]2/[S]1the limiting fraction of solute remaining in phase 1is2qlimit=e-(V2/V1)K(V1=V2=50mLandK=2). Let volume V2be divided intoequal portions to conduct extractions. Find the fraction of S extracted into phase 2 for n = 1,2, and 10extractions. How many portions are required to attain 95%of the theoretical limit?

Why is longitudinal diffusion a more serious problem in gas chromatography than in liquid chromatography?
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