Chapter 23: QBE (page 628)

Consider a chromatography experiment in which two components with retention factors $k_{1} = 4.00$ and $k_{2} = 5.00$ are injected into column with $N = 1.00 \times 10^{3}$ theoretical plates. The retention time for the less-retained component is $t r_{1} = 10.0 m i n$ .
(a)Calculate $t_{m}$ and $t_{12}$ .Find $w_{\frac{1}{2}}$ (width at half height) and w (at the base) for each peak.
(b)Using graph paper , sketch the chromatogram analogous to figure 23-7,supposing that two peaks have same amplitude(height). Draw the half widths accurately.
(c)Calculate the resolution of the two peaks and compare this value with those drawn in Figure 23-10.

Short Answer

Expert verified

(a) $t_{m} = 2.00 m i n, t_{r 2} = 12.00 m i n$

are 1.265 min and 1.518 min.

$w_{\frac{1}{2}}$ (for peak1 ) and $w_{\frac{1}{2}}$ (for peak2) are 0.745 min and 0.894 min

(b) Resolution =1.437

Step by step solution

Step1:

Finding the value of, $t_{m}, t_{r 2}$ w and $W_{\frac{1}{2}}$

We know that $k_{1} = \frac{t_{r 1} - t_{m}}{t_{m}}$ or, $t_{m} = \frac{t_{r 1}}{k_{1} + 1}$

Or, $t_{m} = \frac{10.00}{4 + 1} = 2.00 m i n$

We know that , $t_{r 2} = t_{m} (K_{2} + 1)$ Or, $t_{r 2} = 2.00 (5.00 + 1) = 12 m i n$

Now $N = \frac{5.55 \times t_{r}^{2}}{{w^{2}}_{\frac{1}{2}}}$ Or, $w_{\frac{1}{2}} = \sqrt{\frac{5.55 \times t_{r}^{2}}{N}}$

For $w_{\frac{1}{2}} = \sqrt{\frac{5.55 \times {(10)}^{2}}{1.00 \times 10^{3}}} = 0.745 m i n$

$w_{\frac{1}{2}} = \sqrt{\frac{5.55 \times {(12)}^{2}}{1.00 \times 10^{3}}} = 0.894 m i n$

$N = \frac{16 t_{r}^{2}}{w}$ Or, $w = \sqrt{\frac{16 \times t_{r}^{2}}{N}}$

$w_{1} = \sqrt{\frac{16 \times {(10)}^{2}}{1.00 \times 10^{3}}} = 1.265 m i n \ w_{2} = \sqrt{\frac{16 \times {(12)}^{2}}{1.00 \times 10^{3}}} = 1.518 m i n$

(b)Step 2 : Sketching the graph :

finding Resolution:

Or, Resolution= $\frac{(12 - 10)}{\frac{1}{2} (1.265 + 1.518)}$ =1.437

For quantitive analysis resolution >1.5 is highly desirable.

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Short Answer

Step by step solution

Step1:

(b)Step 2 : Sketching the graph :

finding Resolution:

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