Question

(a) For the asymmetric chromatogram in Figure 23-14, calculate the asymmetry factor, $\frac{\mathbf{B}}{\mathbf{A}}$.(b) The asymmetric chromatogram in Figure 23-14 has a retention time equal to 15.0 min and a ${\mathbf{w}}_{\mathbf{0}\mathbf{.}\mathbf{1}}$of 44s . Find the number of theoretical plates. (c) The width of a Gaussian peak at a height equal to$\frac{\mathbf{1}}{\mathbf{10}}$ of the peak height is $\mathbf{4}\mathbf{.}\mathbf{297}\mathbf{\sigma }$. Suppose that the peak in (b) is symmetric with $\mathbf{A}\mathbf{=}\mathbf{B}\mathbf{=}\mathbf{22}\mathbf{s}$. Use Equations 23-30 and 23-32 to find the plate number.

Short answer

The solution is

$$\begin{gathered} \mathrm{a})\mathrm{Asymmetry}\mathrm{factor}=2 \\ \mathrm{b})\mathrm{N}=5368 \\ \mathrm{c})\mathrm{\rho }=10.24\mathrm{s} \\ \mathrm{d})\mathrm{N}=7.72\times {10}^3 \end{gathered}$$

Step-by-step solution

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

asymmetry factor (B/A) .

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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 $=\frac{1/1}{1/2}=2$

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

$$\begin{gathered} {\mathrm{t}}_{\mathrm{r}}=15\min =900\mathrm{s}\mathrm{and}\mathrm{a}{\mathrm{w}}_{0.1}=44\mathrm{s}: \\ \mathrm{N}=\frac{41.7{({\mathrm{t}}_{\mathrm{r}}/{\mathrm{w}}_{0.1})}^2}{(\mathrm{B}/\mathrm{A})+1.25} \end{gathered}$$

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

$\mathrm{N}=\frac{41.7{(900\mathrm{s}/44\mathrm{s})}^2}{\left(2\right)+1.25}$

$\mathrm{N}=5368$

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c) A Gaussian peak with a width of 4.297$\mathrm{\rho }$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.

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

$\mathrm{\rho }=\frac{44\mathrm{s}}{4.297}=10.24\mathrm{s}$

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

$$\begin{gathered} \mathrm{N}={\left(\frac{{\mathrm{t}}_{\mathrm{r}}}{\mathrm{\rho }}\right)}^2 \\ \mathrm{N}={\left(\frac{900\mathrm{s}}{10.24\mathrm{s}}\right)}^2 \\ \mathrm{N}=7.72\times {10}^3 \end{gathered}$$