Chapter 26, Problem 26.23P

(a)

Interpretation Introduction

Interpretation:

The elution volume of excluded molecules should be calculated.

Concept introduction:

Molecular exclusion chromatography:

The molecules are separated by its size in chromatographic separation is known as molecular exclusion chromatography.

In this chromatography, the large molecules are eluted first then small molecules are separated.

The elution volume is directly proportional to the molar mass of the molecule so it is called as gel filtration or permeation chromatography.

Hydrophobic interaction chromatography:

The substance, which is repels the water is known as hydrophobic substance. In Hydrophobic interaction chromatography, the stationary phase is hydrophobic substance.

In the chromatography, the molecule is interact with stationary phase of the column and the solubility of molecule, which is going to separate is decreasing salt concentration of reagent ti increasing solubility molecule.

Explanation of Solution

To calculate the elution volume of excluded molecules.

Given,

Column diameter is $\text{7}\text{.8 mm}$

Column length is $\text{30}\text{cm}$

Occupation of gel particles is $\text{20}\text{\%}$

Occupation of pores is $\text{40\%}$

The volume between particles occupies is $\text{40\%}$

In then affinity column chromatography, only the bonded molecule is remains in the column rest of other are eluted so the eluted volume is equal to the volume used in the elution of excluded molecules.

Total volume of the column is,

$\begin{array}{l}\text{Volume}\text{=}{\text{πR}}^{\text{2}}\text{l}\\ \text{=}\text{3}\text{.14}\text{×}{\text{(0}}^{\text{.39)}}\text{×(30)}\\ \text{=}\text{14}\text{.3}\text{mL}\end{array}$

The volume of elution of excluded molecules is equal to the $\text{40\%}$ void volume.

$\text{40\%}\text{of}\text{14}\text{.3}\text{mL}\text{=}\text{5}\text{.7}\text{mL}$

Therefore, the volume of elution volume of excluded molecules is $\text{5}\text{.7}\text{mL}$.

Conclusion

The elution volume of excluded molecules was calculated.

(b)

Interpretation Introduction

Interpretation:

The elution volume of small molecules should be calculated.

Concept introduction:

Molecular exclusion chromatography:

The molecules are separated by its size in chromatographic separation is known as molecular exclusion chromatography.

In this chromatography, the large molecules are eluted first then small molecules are separated.

The elution volume is directly proportional to the molar mass of the molecule so it is called as gel filtration or permeation chromatography.

Hydrophobic interaction chromatography:

The substance, which is repels the water is known as hydrophobic substance. In Hydrophobic interaction chromatography, the stationary phase is hydrophobic substance.

In the chromatography, the molecule is interact with stationary phase of the column and the solubility of molecule, which is going to separate is decreasing salt concentration of reagent ti increasing solubility molecule.

Explanation of Solution

To calculate the elution volume of excluded molecules.

Given,

Column diameter is $\text{7}\text{.8 mm}$

Column length is $\text{30}\text{cm}$

Occupation of gel particles is $\text{20}\text{\%}$

Occupation of pores is $\text{40\%}$

The volume between particles occupies is $\text{40\%}$

In then affinity column chromatography, only the bonded molecule is remains in the column rest of other are eluted so the eluted volume is equal to the volume used in the elution of excluded molecules.

Total volume of the column is,

$\begin{array}{l}\text{Volume}\text{=}{\text{πR}}^{\text{2}}\text{l}\\ \text{=}\text{3}\text{.14}\text{×}{\text{(0}}^{\text{.39)}}\text{×(30)}\\ \text{=}\text{14}\text{.3}\text{mL}\end{array}$

The volume of small of excluded molecules is equal to the $\text{80\%}$ void volume.

$\text{80\%}\text{of}\text{14}\text{.3}\text{mL}\text{=}11.5\text{mL}$

Therefore, the volume of elution volume of small molecules is $\text{11}\text{.5}\text{mL}$.

Conclusion

The elution volume of small molecules was calculated.

(c)

Interpretation Introduction

Interpretation:

Elution of polyethylene glycols from polystyrene resin in HPLC should be explained.

Concept introduction:

Molecular exclusion chromatography:

The molecules are separated by its size in chromatographic separation is known as molecular exclusion chromatography.

In this chromatography, the large molecules are eluted first then small molecules are separated.

The elution volume is directly proportional to the molar mass of the molecule so it is called as gel filtration or permeation chromatography.

Hydrophobic interaction chromatography:

The substance, which is repels the water is known as hydrophobic substance. In Hydrophobic interaction chromatography, the stationary phase is hydrophobic substance.

In the chromatography, the molecule is interact with stationary phase of the column and the solubility of molecule, which is going to separate is decreasing salt concentration of reagent ti increasing solubility molecule.

Explanation of Solution

To calculate the elution volume of excluded molecules.

Given,

Column diameter is $\text{7}\text{.8 mm}$

Column length is $\text{30}\text{cm}$

Occupation of gel particles is $\text{20}\text{\%}$

Occupation of pores is $\text{40\%}$

The volume between particles occupies is $\text{40\%}$

In then affinity column chromatography, only the bonded molecule is remains in the column rest of other are eluted so the eluted volume is equal to the volume used in the elution of excluded molecules.

Total volume of the column is,

$\begin{array}{l}\text{Volume}\text{=}{\text{πR}}^{\text{2}}\text{l}\\ \text{=}\text{3}\text{.14}\text{×}{\text{(0}}^{\text{.39)}}\text{×(30)}\\ \text{=}\text{14}\text{.3}\text{mL}\end{array}$

The volume of elution of excluded molecules is equal to the $\text{40\%}$ void volume.

$\text{40\%}\text{of}\text{14}\text{.3}\text{mL}\text{=}\text{5}\text{.7}\text{mL}$

Therefore, the volume of elution volume of excluded molecules is $\text{5}\text{.7}\text{mL}$.

The volume of small of excluded molecules is equal to the $\text{80\%}$ void volume.

$\text{80\%}\text{of}\text{14}\text{.3}\text{mL}\text{=}11.5\text{mL}$

Therefore, the volume of elution volume of small molecules is $\text{11}\text{.5}\text{mL}$.

Hence, the solute is absorbed on the polystyrene resin or otherwise they are eluted between $\text{5}\text{.7}\text{mL}$ and $\text{11}\text{.5}\text{mL}$.

The elution volume of polyethylene glycols are $\text{23}\text{mL to 27}\text{mL}$.

From the above calculations the given column size and elution volume of column, the molecules are eluted between $\text{5}\text{.7}\text{mL}$.

Therefore, polyethylene glycols are must adsorbed on the polystyrene resin and they are eluted between $\text{5}\text{.7}\text{mL}$ and $\text{11}\text{.5}\text{mL}$.

Conclusion

Elution of polyethylene glycols from polystyrene resin in HPLC was explained.