Principles of Instrumental Analysis
Principles of Instrumental Analysis
7th Edition
ISBN: 9781305577213
Author: Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher: Cengage Learning
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Chapter 26, Problem 26.23QAP
Interpretation Introduction

(a)

Interpretation:

The time spends by the component A and B in the stationary phase is to be stated.

Concept introduction:

The chromatography is the separation method for the substances of the mixture. The mobile phase consist the dissolved mixture in the fluid. This mixture is carried through the stationary phase of the material. Due the different speed of the particles the separation of the substance from the mixture takes place.

Expert Solution
Check Mark

Answer to Problem 26.23QAP

The time spends by the component A and B in the stationary phase are 25min and 45min respectively.

Explanation of Solution

Draw the diagram for the chromatogram of a two component mixture with the time period.

Principles of Instrumental Analysis, Chapter 26, Problem 26.23QAP , additional homework tip  1

Figure-(1)

The Figure (1) shows that the time spend by the component A in the stationary phase is 25min and the time spend by the component in the stationary phase is 45min.

Thus, the time spends by the component A and B in the stationary phase are 25min and 45min respectively.

Interpretation Introduction

(b)

Interpretation:

The retention time for A and B are to be stated.

Concept introduction:

The chromatography is the separation method for the substances of the mixture. The mobile phase consist the dissolved mixture in the fluid. This mixture is carried through the stationary phase of the material. Due the different speed of the particles the separation of the substance from the mixture takes place.

Expert Solution
Check Mark

Answer to Problem 26.23QAP

The retention time for the component A and B are 30min and 50min respectively.

Explanation of Solution

Write the expression for the retention time of component A.

tR=tSA+tM   ...... (I)

Here, the time spend by the component A in stationary phase is tSA and the time spend in the mobile phase is tM.

Write the expression for the retention time of component B.

tR=tSB+tM   ...... (II)

Here, the time spend by the component B in stationary phase is tSB

Draw the diagram for the mixture of two components to show the time spent by the components in the stationary phase and the mobile phase.

Principles of Instrumental Analysis, Chapter 26, Problem 26.23QAP , additional homework tip  2

Figure-(2)

The Figure (2) shows that the time spend by the component A in the stationary phase is 25min, the time spend by the component in the stationary phase is 45min and the time spent in the mobile phase is 5min.

Substitute 25min for tSA and 5min for tM in Equation (I).

tR=25min+5min=30min

Thus, the retention time of the component A is 30min.

Substitute 45min for tSB and 5min for tM in Equation (II).

tR=45min+5min=50min

Thus, the retention time of the component B is 50min.

Interpretation Introduction

(c)

Interpretation:

The retention factors for the two components are to be stated.

Concept introduction:

The chromatography is the separation method for the substances of the mixture. The mobile phase consist the dissolved mixture in the fluid. This mixture is carried through the stationary phase of the material. Due the different speed of the particles the separation of the substance from the mixture takes place.

Expert Solution
Check Mark

Answer to Problem 26.23QAP

The retention factors for the component A and B are 5 and 9.

Explanation of Solution

Write the expression for the retention factor for the component A.

kA=tSAtM   ...... (III)

Write the expression for the retention factor for the component B.

kB=tSBtM   ...... (IV)

Substitute 25min for tSA and 5min for tM in Equation (III).

kA=25min5min=5

Thus, the retention factor for the component A is 5.

Substitute 45min for tSB and 5min for tM in Equation (IV).

kB=45min5min=9

Thus, the retention factor for the component B is 9.

Interpretation Introduction

(d)

Interpretation:

The full width of the each peak and full width at half maximum values are to be stated.

Concept introduction:

The chromatography is the separation method for the substances of the mixture. The mobile phase consist the dissolved mixture in the fluid. This mixture is carried through the stationary phase of the material. Due the different speed of the particles the separation of the substance from the mixture takes place.

Expert Solution
Check Mark

Answer to Problem 26.23QAP

The full width for the component A and B are 18min and 22min respectively.

The half width for the component A and B are 9min and 11min respectively.

Explanation of Solution

Draw the diagram to show the full width and half width of both the components.

Principles of Instrumental Analysis, Chapter 26, Problem 26.23QAP , additional homework tip  3

Figure-(3)

The Figure-(3) shows that the full width of the component A is WA=18min and the full width for the component B is WB=22min.

Thus, the full width for the component A and B are 18min and 22min respectively.

Write the expression for the half width of the component A.

WA1/2=WA2   ...... (V)

Write the expression for the half width of the component B.

WB1/2=WB2   ...... (VI)

Substitute 18min for WA in Equation (V).

WA1/2=18min2=9min

Thus, the half width for the component A is 9min.

Substitute 22min for WB in Equation (VI).

WB1/2=22min2=11min

Thus, the half width for the component B is 11min.

Interpretation Introduction

(e)

Interpretation:

The resolution of two peaks is to be stated.

Concept introduction:

The chromatography is the separation method for the substances of the mixture. The mobile phase consist the dissolved mixture in the fluid. This mixture is carried through the stationary phase of the material. Due the different speed of the particles the separation of the substance from the mixture takes place.

Expert Solution
Check Mark

Answer to Problem 26.23QAP

The resolution of two peaks is 1.

Explanation of Solution

Write the expression for the resolution of two peaks.

RS=2(tRBtRA)WA+WB   ...... (VII)

Substitute 18min for WA, 22min for WB, 30min for tRA and 50min for tRB in Equation (VII).

RS=2(50min30min)18min+22min=2(20min)40min=4040=1

Thus, the resolution of two peaks is 1.

Interpretation Introduction

(f)

Interpretation:

The average number of plates is to be stated.

Concept introduction:

The chromatography is the separation method for the substances of the mixture. The mobile phase consist the dissolved mixture in the fluid. This mixture is carried through the stationary phase of the material. Due the different speed of the particles the separation of the substance from the mixture takes place.

Expert Solution
Check Mark

Answer to Problem 26.23QAP

The average number of plates is 64.

Explanation of Solution

Write the expression for the number of the plate for the component A.

NA=16(tRAWA)2   ...... (VIII)

Write the expression for the number of the plate for the component B.

NB=16(tRBWB)2   ...... (IX)

Write the expression for the average number of plates.

Navg=NA+NB2   ...... (X)

Substitute 30min for tRA and 18min for WA in Equation (VIII).

NA=16(30min18min)2=16(900324)=16(2.777)=44.44

Substitute 50min for tRB and 22min for WB in Equation (IX).

NB=16(50min22min)2=16(2500484)=16(5.165)=82.64

Substitute 82.64 for NB and 44.44 for NA in Equation (X).

Navg=44.44+82.642=127.082=63.5464

Thus, the average number of plates is 64.

Interpretation Introduction

(g)

Interpretation:

The average height of the plate is to be stated.

Concept introduction:

The chromatography is the separation method for the substances of the mixture. The mobile phase consist the dissolved mixture in the fluid. This mixture is carried through the stationary phase of the material. Due the different speed of the particles the separation of the substance from the mixture takes place.

Expert Solution
Check Mark

Answer to Problem 26.23QAP

The average height of plate is 0.40cm.

Explanation of Solution

Write the expression for the average height of the plate for the column.

Havg=LNavg   ...... (XI)

Here, the length of the column is L.

Substitute 25cm for L and 64 for Navg in Equation (XI).

Havg=25cm64=0.3906cm0.40cm

Thus, the average height of plate is 0.40cm.

Interpretation Introduction

(h)

Interpretation:

The length of the column to achieve resolution of 1.75 is to be stated.

Concept introduction:

The chromatography is the separation method for the substances of the mixture. The mobile phase consist the dissolved mixture in the fluid. This mixture is carried through the stationary phase of the material. Due the different speed of the particles the separation of the substance from the mixture takes place.

Expert Solution
Check Mark

Answer to Problem 26.23QAP

The length of the column to achieve resolution of 1.75 is 78.4cm.

Explanation of Solution

Write the expression for the average number of plate at 1.75 resolution.

(RS)1(RS)2=N1N2N2=(N1(RS)2(RS)1)2N2=N1((RS)2(RS)1)2   ...... (XII)

Here, the average number of plate at resolution 1 is N1, the average number of plate at resolution 1.75 is N2, the initial resolution is (RS)1 and the final given resolution is (RS)2.

Write the expression for the column length to achieve resolution of 1.75.

L=N2Havg   ...... (XIII)

Substitute 1 for (RS)1, 1.75 for (RS)2 and 64 for N1 in Equation (XII).

N2=64(1.751)2=64(3.0625)=196

Substitute 196 for N2 and 0.40cm for Havg in Equation (XIII).

L=196(0.40cm)=78.4cm

Thus, the length of the column to achieve resolution of 1.75 is 78.4cm.

Interpretation Introduction

(i)

Interpretation:

The time required to achieve resolution of 1.75 is to be stated.

Concept introduction:

The chromatography is the separation method for the substances of the mixture. The mobile phase consist the dissolved mixture in the fluid. This mixture is carried through the stationary phase of the material. Due the different speed of the particles the separation of the substance from the mixture takes place.

Expert Solution
Check Mark

Answer to Problem 26.23QAP

The time required to achieve resolution of 1.75 is 153min.

Explanation of Solution

Write the expression for the time required at 1.75 resolution.

(RS)1(RS)2=(tR)1(tR)2   ...... (XIV)

Here, the time required to achieve the resolution 1 is (tR)1, the time required to achieve resolution 1.75 is (tR)2, the initial resolution is (RS)1 and the final given resolution is (RS)2.

Substitute 1 for (RS)1, 1.75 for (RS)2 and 50min for (tR)1 in Equation (XIV).

(11.75)2=50min(tR)2(tR)2=50min0.3264(tR)2=153.14min(tR)2153min

Thus, the time required to achieve resolution of 1.75 is 153min.

Interpretation Introduction

(j)

Interpretation:

The measures required to increase the resolution to achieve the baseline separation are to be stated.

Concept introduction:

The chromatography is the separation method for the substances of the mixture. The mobile phase consist the dissolved mixture in the fluid. This mixture is carried through the stationary phase of the material. Due the different speed of the particles the separation of the substance from the mixture takes place.

Expert Solution
Check Mark

Answer to Problem 26.23QAP

The resolution of the peak is increased by increasing the retention time of the species.

Explanation of Solution

Write the expression for the resolution of the peak.

RS=[(tr)B(tr)A]WA+WB   ...... (XV)

Here, the retention time for the species A is (tr)A, the retention time for the species B is (tr)B, the width of the peak A is WA and the width of the peak B is WB.

The Equation (XV) shows that the retention time is directly proportional to the resolution of the peak. Hence, the retention time have to increase to increase the resolution of the peak.

Thus, the resolution of the peak is increased by increasing the retention time of the species.

Interpretation Introduction

(k)

Interpretation:

The measures to achieve a better separation in a shorter time with the same column as in part j is to be stated.

Concept introduction:

The chromatography is the separation method for the substances of the mixture. The mobile phase consist the dissolved mixture in the fluid. This mixture is carried through the stationary phase of the material. Due the different speed of the particles the separation of the substance from the mixture takes place.

Expert Solution
Check Mark

Answer to Problem 26.23QAP

The on increasing the resolution keeping the length of the column constant the better separation in a shorter time with the same column as in part j is achieved.

Explanation of Solution

Write the expression for the average number of the plates.

Navg=16RS2(αα1)2(1+KBKB)   ...... (XVI)

Here, the resolution of the peaks is RS, the average number of the plates is Navg, the selectively factor is α and the retention factor for the solute B is kB.

The Equation (XVI) shows that the average number of the plates is directly proportional to the square of the resolution. Hence, the average number of the plates increases on increasing the resolution keeping the length of the column constant. Due to this the better separation in a shorter time with the same column as in part j may be achieved.

Thus, the on increasing the resolution keeping the length of the column constant the better separation in a shorter time with the same column as in part j is achieved.

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Students have asked these similar questions
(a) An analyst has conducted a separation of a mixture of substance X and Y using liquid chromatography technique. The column used is 30-cm long and the flow rate was 0.65 mL/min. The chromatogram obtained is shown in Figure Q1.1. i) Determine the adjusted retention times for substance x and y. ii) Calculate the retention factors for substance x and y. iii) Calculate the resolution of the two peaks. Then, describe the quality of separation between the two adjacent peaks. iv) Calculate the plate height value of the column.
A 20-cm long packed column was used to perform the separation of A and B by LC. TRA=16.28 min, TR,B=17.50 min, WA=1.11 min, WB=1.19 min, tм=1.30 min. (a) Calculate the retention factor (k') for A and B. (b) Calculate the average number of theoretical plates and plate height.
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