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.

Answer to Problem 27.22QAP

The average flow rate in the column is $31.4\text{mL}/\text{min}$.

Explanation of Solution

The expression for the average flow rate is:

$F=F_m\left(\frac{\left(T_c+273\right)\text{K}}{\left(T+273\right)\text{K}}\right)\left(\frac{P-P_{H_{2}O}}{P}\right)$   ...... (I)

Here, the average flow rate in the column is $F$, the measured flow rate is $F_m$, the temperature of column is $T_c$, 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 $P_{H_{2}O}$.

Substitute $25.3\text{mL}/\text{min}$ for $F_m$, $102.0°\text{C}$ for $T_c$, $21.2°\text{C}$ for $T$, $748\text{torr}$ for $P$ and $18.88\text{torr}$ for $P_{H_{2}O}$ in Equation (I).

$\begin{array}{c}F=\left(25.3\text{mL}/\text{min}\right)\left(\frac{\left(102.0°\text{C+273}\right)\text{K}}{\left(21.2°\text{C+273}\right)\text{K}}\right)\left(\frac{\left(748\text{torr}\right)-\left(18.88\text{torr}\right)}{\left(748\text{torr}\right)}\right)\\ =\left(25.3\text{mL}/\text{min}\right)\left(\frac{375\text{K}}{294.2\text{K}}\right)\left(\frac{729.12\text{torr}}{748\text{torr}}\right)\\ =31.4\text{mL}/\text{min}\end{array}$

Thus, the average flow rate in the column is $31.4\text{mL}/\text{min}$.

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.

Answer to Problem 27.22QAP

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

Explanation of Solution

The expression for the corrected retention volume is:

$V_R^{°}=j{t}_{R}F$   ...... (II)

Here, the corrected retention volume is $V_R^{°}$, the compressibility factor is $j$ and the retention time is $t_R$.

The expression for the compressibility factor is:

$j=\frac{3\left[{\left(\frac{P_i}{P}\right)}^2-1\right]}{2\left[{\left(\frac{P_i}{P}\right)}^3-1\right]}$   ...... (III)

Here, the inlet pressure is $P_i$ and the outlet pressure is $P$.

The expression for the inlet pressure is:

$P_i=P+\left(26.1\text{psi}\right)$   ...... (IV)

Substitute $748\text{torr}$ for $P$ in Equation (IV).

$\begin{array}{c}{P}_i=\left(748\text{torr}\right)+\left(26.1\text{psi}\right)\\ =\left(748\text{torr}\right)+\left(26.1\text{psi}\right)\left(\frac{5.17\text{torr}}{1\text{psi}}\right)\\ =883\text{torr}\end{array}$

Substitute $883\text{torr}$ for $P_i$ and $748\text{torr}$ for $P$ in Equation (III).

$\begin{array}{c}j=\frac{3\left[{\left(\frac{883\text{torr}}{748\text{torr}}\right)}^2-1\right]}{2\left[{\left(\frac{883\text{torr}}{748\text{torr}}\right)}^3-1\right]}\\ =\frac{3\left(1.394-1\right)}{2\left(1.645-1\right)}\\ =\frac{1.182}{1.29}\\ =0.915\end{array}$

Substitute ${\left(V_M^{°}\right)}_{\text{air}}$ for $V_R^{°}$, $0.915$ for $j$, $18.0\text{s}$ for $t_m$ and $31.4\text{mL}/\text{min}$ for $F$ in Equation (II).

$\begin{array}{c}{\left(V_M^{°}\right)}_{\text{air}}=\left(0.915\right)\left(18.0\text{s}\right)\left(31.4\text{mL}/\text{min}\right)\\ =\left(0.915\right)\left(18.0\text{s}\right)\left(\frac{1\min }{60\text{s}}\right)\left(31.4\text{mL}/\text{min}\right)\\ =8.6\text{mL}\end{array}$

Substitute ${\left(V_R^{°}\right)}_{\text{methyl}\text{acetate}}$ for $V_R^{°}$, $0.915$ for $j$, $1.98\min$ for $t_m$ and $31.4\text{mL}/\text{min}$ for $F$ in Equation (II).

$\begin{array}{c}{\left(V_R^{°}\right)}_{\text{methyl}\text{acetate}}=\left(0.915\right)\left(1.98\min \right)\left(31.4\text{mL}/\text{min}\right)\\ =\left(1.8117\min \right)\left(31.4\text{mL}/\text{min}\right)\\ =56.9\text{mL}\end{array}$

Substitute ${\left(V_R^{°}\right)}_{\text{methyl}\text{propionate}}$ for $V_R^{°}$, $0.915$ for $j$, $4.16\min$ for $t_m$ and $31.4\text{mL}/\text{min}$ for $F$ in Equation (II).

$\begin{array}{c}{\left(V_R^{°}\right)}_{\text{methyl}\text{propionate}}=\left(0.915\right)\left(4.16\min \right)\left(31.4\text{mL}/\text{min}\right)\\ =\left(3.8064\min \right)\left(31.4\text{mL}/\text{min}\right)\\ =119.5\text{mL}\end{array}$

Substitute ${\left(V_R^{°}\right)}_{\text{n-butyrate}}$ for $V_R^{°}$, $0.915$ for $j$, $7.93\min$ for $t_m$ and $31.4\text{mL}/\text{min}$ for $F$ in Equation (II).

$\begin{array}{c}{\left(V_R^{°}\right)}_{\text{n-butyrate}}=\left(0.915\right)\left(7.93\min \right)\left(31.4\text{mL}/\text{min}\right)\\ =\left(7.25\min \right)\left(31.4\text{mL}/\text{min}\right)\\ =227.6\text{mL}\end{array}$

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

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.

Answer to Problem 27.22QAP

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

Explanation of Solution

The expression for the specific retention volume is:

$V_R=\left(\frac{V_R^{°}-V_M^{°}}{m_s}\right)\left(\frac{273\text{K}}{T}\right)$   ...... (V)

Here, the specific retention volume is $V_R$, the volume that is not retained on the column is $V_M^{°}$, the retention volume of species retained is $V_R^{°}$ and the mass of stationary phase is $m_s$.

Substitute ${\left(V_R\right)}_{\text{methyl}\text{acetate}}$ for $V_R$, $56.9\text{mL}$ for $V_R^{°}$, $8.6\text{mL}$ for $V_M^{°}$, $1.4\text{g}$ for $m_s$ and $375\text{K}$ for $T$ in Equation (V).

$\begin{array}{c}{\left(V_R\right)}_{\text{methyl}\text{acetate}}=\left(\frac{\left(56.9\text{mL}\right)-\left(8.6\text{mL}\right)}{1.4\text{g}}\right)\left(\frac{273\text{K}}{375\text{K}}\right)\\ =\left(\frac{48.3\text{mL}}{1.4\text{g}}\right)\left(\frac{273\text{K}}{375\text{K}}\right)\\ =25.1\text{mL}/\text{g}\end{array}$

Substitute ${\left(V_R\right)}_{\text{methyl}\text{propionate}}$ for $V_R$, $119.5\text{mL}$ for $V_R^{°}$, $8.6\text{mL}$ for $V_M^{°}$, $1.4\text{g}$ for $m_s$ and $375\text{K}$ for $T$ in Equation (V).

$\begin{array}{c}{\left(V_R\right)}_{\text{methyl}\text{propionate}}=\left(\frac{\left(119.5\text{mL}\right)-\left(8.6\text{mL}\right)}{1.4\text{g}}\right)\left(\frac{273\text{K}}{375\text{K}}\right)\\ =\left(\frac{110.9\text{mL}}{1.4\text{g}}\right)\left(\frac{273\text{K}}{375\text{K}}\right)\\ =57.7\text{mL}/\text{g}\end{array}$

Substitute ${\left(V_R\right)}_{\text{n-butyrate}}$ for $V_R$, $227.6\text{mL}$ for $V_R^{°}$, $8.6\text{mL}$ for $V_M^{°}$, $1.4\text{g}$ for $m_s$ and $375\text{K}$ for $T$ in Equation (V).

$\begin{array}{c}{\left(V_R\right)}_{\text{methyl}\text{propionate}}=\left(\frac{\left(227.6\text{mL}\right)-\left(8.6\text{mL}\right)}{1.4\text{g}}\right)\left(\frac{273\text{K}}{375\text{K}}\right)\\ =\left(\frac{219.0\text{mL}}{1.4\text{g}}\right)\left(\frac{273\text{K}}{375\text{K}}\right)\\ =113.8\text{mL}/\text{g}\end{array}$

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

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.

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:

$K=V_s{\rho }_s\left(\frac{T_c}{273\text{K}}\right)$   ...... (VI)

Here, the distribution constant is $K$, the stationary phase volume is $V_s$ and the density of stationary phase is ${\rho }_s$.

Substitute $K_{\text{methyl}\text{acetate}}$ for $K$, $25.1\text{mL}/\text{g}$ for $V_s$, $1.02\text{g}/\text{mL}$ for ${\rho }_s$ and $375\text{K}$ for $T_c$ in Equation (VI).

$\begin{array}{c}{K}_{\text{methyl}\text{acetate}}=\left(25.1\text{mL}/\text{g}\right)\left(1.02\text{g}/\text{mL}\right)\left(\frac{375\text{K}}{273\text{K}}\right)\\ =\left(25.1\text{mL}/\text{g}\right)\left(1.02\text{g}/\text{mL}\right)\left(1.3736\right)\\ =35.1\end{array}$

Substitute $K_{\text{methyl}\text{propionate}}$ for $K$, $57.7\text{mL}/\text{g}$ for $V_s$, $1.02\text{g}/\text{mL}$ for ${\rho }_s$ and $375\text{K}$ for $T_c$ in Equation (VI).

$\begin{array}{c}{K}_{\text{methyl}\text{propionate}}=\left(57.7\text{mL}/\text{g}\right)\left(1.02\text{g}/\text{mL}\right)\left(\frac{375\text{K}}{273\text{K}}\right)\\ =\left(57.7\text{mL}/\text{g}\right)\left(1.02\text{g}/\text{mL}\right)\left(1.3736\right)\\ =80.8\end{array}$

Substitute $K_{\text{methyl}n-\text{butyrate}}$ for $K$, $113.8\text{mL}/\text{g}$ for $V_s$, $1.02\text{g}/\text{mL}$ for ${\rho }_s$ and $375\text{K}$ for $T_c$ in Equation (VI).

$\begin{array}{c}{K}_{\text{methyl}n-\text{butyrate}}=\left(113.8\text{mL}/\text{g}\right)\left(1.02\text{g}/\text{mL}\right)\left(\frac{375\text{K}}{273\text{K}}\right)\\ =\left(113.8\text{mL}/\text{g}\right)\left(1.02\text{g}/\text{mL}\right)\left(1.3736\right)\\ =159.4\end{array}$

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.

Answer to Problem 27.22QAP

The corrected retention volume and time are $1029.4\text{mL}$ 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:

$\log t_R^{\text{'}}=\left(0.3286\right)\left(n_c\right)-0.7496$   ...... (VII)

Here, the slope is $m$, the number of carbon atoms is $n_c$.

Substitute $7$ for $n_c$ in Equation (VII).

$\begin{array}{l}\log t_R^{\text{'}}=\left(0.3286\right)\left(7\right)-0.7496\\ \log t_R^{\text{'}}=1.55\\ t_R^{\text{'}}={10}^{1.55}\\ t_R^{\text{'}}=35.53\end{array}$

Write the expression for the retention time.

$t_R=t_R^{\text{'}}+t_m$   ...... (VIII)

Here, the retention time is $t_R$.

Substitute $35.53\min$ for $t_R^{\text{'}}$ and $0.30\min$ for $t_m$ in Equation (VIII).

$\begin{array}{c}{t}_R=\left(35.53\min \right)+\left(0.30\min \right)\\ =35.83\min \end{array}$

Substitute $0.915$ for $j$, $35.83\min$ for $t_R$ and $31.4\text{mL}/\text{min}$ for $F$ in Equation (II).

$\begin{array}{c}{V}_R^{°}=\left(0.915\right)\left(35.83\min \right)\left(31.4\text{mL}/\text{min}\right)\\ =1029.4\text{mL}\end{array}$

Thus, the corrected retention volume and time are $1029.4\text{mL}$ and $35.83\min$.