(a) Interpretation: The average flow rate in the column is to be calculated. Concept introduction: The migration through a chromatographic column of a solute zone is made under the influence of the gas flow. On an average, all compound’s molecule spend the same amount of time in the mobile phase. During that time, the molecules move at the same speed along the column as the carrier gas.

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Answer 1

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Chapter 27, Problem 27.22QAP

Interpretation Introduction

(a)

Interpretation:

The average flow rate in the column is to be calculated.

Concept introduction:

The migration through a chromatographic column of a solute zone is made under the influence of the gas flow. On an average, all compound’s molecule spend the same amount of time in the mobile phase. During that time, the molecules move at the same speed along the column as the carrier gas.

Expert Solution

Answer to Problem 27.22QAP

The average flow rate in the column is 31.4 mL/min.

Explanation of Solution

The expression for the average flow rate is:

F=Fm((Tc+273) K(T+273) K)(P−PH2OP) ...... (I)

Here, the average flow rate in the column is F, the measured flow rate is Fm, the temperature of column is Tc, the temperature of flow meter is T, the pressure at the end of the column is P and the vapor pressure of the water is PH2O.

Substitute 25.3 mL/min for Fm, 102.0°C for Tc, 21.2°C for T, 748 torr for P and 18.88 torr for PH2O in Equation (I).

F=(25.3 mL/min)((102.0°C+273) K(21.2°C+273) K)((748 torr)−(18.88 torr)(748 torr))=(25.3 mL/min)(375 K294.2 K)(729.12 torr748 torr)=31.4 mL/min

Thus, the average flow rate in the column is 31.4 mL/min.

Interpretation Introduction

(b)

Interpretation:

The corrected retention volumes for air and the three esters is to be calculated.

Concept introduction:

Retention measurements could make in terms of chart or times, volumes as well as distances. If recorder speeds and flow are constant, the volumes are directly proportional to the charts and time distances.

Expert Solution

Answer to Problem 27.22QAP

The corrected retention volumes for air is 8.6 mL, for methyl acetate is 56.9 mL, for methyl propionate is 119.5 mL, and for n-butyrate is 227.6 mL.

Explanation of Solution

The expression for the corrected retention volume is:

VR°=jtRF ...... (II)

Here, the corrected retention volume is VR°, the compressibility factor is j and the retention time is tR.

The expression for the compressibility factor is:

j=3[(PiP)2−1]2[(PiP)3−1] ...... (III)

Here, the inlet pressure is Pi and the outlet pressure is P.

The expression for the inlet pressure is:

Pi=P+(26.1 psi) ...... (IV)

Substitute 748 torr for P in Equation (IV).

Pi=(748 torr)+(26.1 psi)=(748 torr)+(26.1 psi)(5.17 torr1 psi)=883 torr

Substitute 883 torr for Pi and 748 torr for P in Equation (III).

j=3[(883 torr748 torr)2−1]2[(883 torr748 torr)3−1]=3(1.394−1)2(1.645−1)=1.1821.29=0.915

Substitute (VM°)air for VR°, 0.915 for j, 18.0 s for tm and 31.4 mL/min for F in Equation (II).

(VM°)air=(0.915)(18.0 s)(31.4 mL/min)=(0.915)(18.0 s)(1 min60 s)(31.4 mL/min)=8.6 mL

Substitute (VR°)methyl acetate for VR°, 0.915 for j, 1.98 min for tm and 31.4 mL/min for F in Equation (II).

(VR°)methyl acetate=(0.915)(1.98 min)(31.4 mL/min)=(1.8117 min)(31.4 mL/min)=56.9 mL

Substitute (VR°)methyl propionate for VR°, 0.915 for j, 4.16 min for tm and 31.4 mL/min for F in Equation (II).

(VR°)methyl propionate=(0.915)(4.16 min)(31.4 mL/min)=(3.8064 min)(31.4 mL/min)=119.5 mL

Substitute (VR°)n-butyrate for VR°, 0.915 for j, 7.93 min for tm and 31.4 mL/min for F in Equation (II).

(VR°)n-butyrate=(0.915)(7.93 min)(31.4 mL/min)=(7.25 min)(31.4 mL/min)=227.6 mL

Thus, the corrected retention volumes for air is 8.6 mL, for methyl acetate is 56.9 mL, for methyl propionate is 119.5 mL, and for n-butyrate is 227.6 mL.

Interpretation Introduction

(c)

Interpretation:

The specific retention volumes for the three components.

Concept introduction:

The specific retention volume of a solute is defined as the corrected retention volume of the solute per unit mass of stationary phase.

Expert Solution

Answer to Problem 27.22QAP

The specific retention volume for methyl acetate is 25.1 mL/g, for methyl propionate is 57.7 mL/g, and for n-butyrate is 113.8 mL/g.

Explanation of Solution

The expression for the specific retention volume is:

VR=(VR°−VM°ms)(273 KT) ...... (V)

Here, the specific retention volume is VR, the volume that is not retained on the column is VM°, the retention volume of species retained is VR° and the mass of stationary phase is ms.

Substitute (VR)methyl acetate for VR, 56.9 mL for VR°, 8.6 mL for VM°, 1.4 g for ms and 375 K for T in Equation (V).

(VR)methyl acetate=((56.9 mL)−(8.6 mL)1.4 g)(273 K375 K)=(48.3 mL1.4 g)(273 K375 K)=25.1 mL/g

Substitute (VR)methyl propionate for VR, 119.5 mL for VR°, 8.6 mL for VM°, 1.4 g for ms and 375 K for T in Equation (V).

(VR)methyl propionate=((119.5 mL)−(8.6 mL)1.4 g)(273 K375 K)=(110.9 mL1.4 g)(273 K375 K)=57.7 mL/g

Substitute (VR)n-butyrate for VR, 227.6 mL for VR°, 8.6 mL for VM°, 1.4 g for ms and 375 K for T in Equation (V).

(VR)methyl propionate=((227.6 mL)−(8.6 mL)1.4 g)(273 K375 K)=(219.0 mL1.4 g)(273 K375 K)=113.8 mL/g

Thus, the specific retention volume for methyl acetate is 25.1 mL/g, for methyl propionate is 57.7 mL/g, and for n-butyrate is 113.8 mL/g.

Interpretation Introduction

(d)

Interpretation:

The distribution constants for each ester are to be calculated.

Concept introduction:

The distribution constant is defined as the equilibrium constant for the distribution of an analyte in two immiscible solvents.

Expert Solution

Answer to Problem 27.22QAP

The distribution constants for methyl acetate is 35.1, for methyl propionate is 80.8, and for methyl n-butyrate is 159.4.

Explanation of Solution

The expression for the distribution constant is:

K=Vsρs(Tc273 K) ...... (VI)

Here, the distribution constant is K, the stationary phase volume is Vs and the density of stationary phase is ρs.

Substitute Kmethyl acetate for K, 25.1 mL/g for Vs, 1.02 g/mL for ρs and 375 K for Tc in Equation (VI).

Kmethyl acetate=(25.1 mL/g)(1.02 g/mL)(375 K273 K)=(25.1 mL/g)(1.02 g/mL)(1.3736)=35.1

Substitute Kmethyl propionate for K, 57.7 mL/g for Vs, 1.02 g/mL for ρs and 375 K for Tc in Equation (VI).

Kmethyl propionate=(57.7 mL/g)(1.02 g/mL)(375 K273 K)=(57.7 mL/g)(1.02 g/mL)(1.3736)=80.8

Substitute Kmethyl n−butyrate for K, 113.8 mL/g for Vs, 1.02 g/mL for ρs and 375 K for Tc in Equation (VI).

Kmethyl n−butyrate=(113.8 mL/g)(1.02 g/mL)(375 K273 K)=(113.8 mL/g)(1.02 g/mL)(1.3736)=159.4

Thus, the distribution constants for methyl acetate is 35.1, for methyl propionate is 80.8, and for methyl n-butyrate is 159.4.

Interpretation Introduction

(e)

Interpretation:

The corrected retention volume and retention time for methyl n-hexanoate is to be determined.

Concept introduction:

Retention measurements could make in terms of chart or times, volumes as well as distances. If recorder speeds and flow are constant, the volumes are directly proportional to the charts and time distances.

Expert Solution

Answer to Problem 27.22QAP

The corrected retention volume and time are 1029.4 mL and 35.83 min.

Explanation of Solution

Consider, the value of slope is 0.3286 and line cut y axis at −0.7496.

The expression for the line is:

logtR'=(0.3286)(nc)−0.7496 ...... (VII)

Here, the slope is m, the number of carbon atoms is nc.

Substitute 7 for nc in Equation (VII).

logtR'=(0.3286)(7)−0.7496logtR'=1.55tR'=101.55tR'=35.53

Write the expression for the retention time.

tR=tR'+tm ...... (VIII)

Here, the retention time is tR.

Substitute 35.53 min for tR' and 0.30 min for tm in Equation (VIII).

tR=(35.53 min)+(0.30 min)=35.83 min

Substitute 0.915 for j, 35.83 min for tR and 31.4 mL/min for F in Equation (II).

VR°=(0.915)(35.83 min)(31.4 mL/min)=1029.4 mL

Thus, the corrected retention volume and time are 1029.4 mL and 35.83 min.

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