Encyclopedia of Lipidomics

Living Edition
| Editors: Markus R. Wenk

Liquid Extraction: Recovery Rate

  • Thusitha RupasingheEmail author
Living reference work entry
DOI: https://doi.org/10.1007/978-94-007-7864-1_92-1
The recovery rate of a compound is the amount of that compound present in the extract compared to the total amount of the compound found in both the extract and the raffinate expressed as a percentage (Wells 2003). As an example, when a compound A is extracted using liquid-liquid extraction shown in Fig. 1, the efficiency of extraction is calculated by the following formula:
Fig. 1

Schematic diagram of solute’s partitioning at the equilibrium in the liquid- liquid extraction system

$$ \mathrm{R}\% = \mathrm{A}\ {\left(\mathrm{extract}\right)}^{*}100 $$
$$ \mathrm{A}\ \left(\mathrm{extract} + \mathrm{raffinate}\right) $$
R% represents the percentage extraction recovery, A(extract) is the amount of analyte A in the extract and A(extract + raffinate) is the total amount of A in both the extract and raffinate. Selecting appropriate solvents based on the relative solubility of the analyte in the extract and raffinate will result in higher recovery rates.
The partition coefficient K, for analyte A between the extract and raffinate, is given by K = [A]e/[A]r, where [A]e and [A]r are the concentration of analyte A in the extract and raffinate at equilibrium. After the first extraction, if the fraction of analyte A in the extract is E and the fraction of analyte A in the raffinate is R, E and R are given by following equations:
$$ \mathrm{E}={\mathrm{m}}_{\mathrm{e}}/\left({\mathrm{m}}_{\mathrm{e}} + {\mathrm{m}}_{\mathrm{r}}\right)\ \mathrm{and}\ \mathrm{R} = {\mathrm{m}}_{\mathrm{r}}/\left({\mathrm{m}}_{\mathrm{e}} + {\mathrm{m}}_{\mathrm{r}}\right) $$
where m e and m r are the amount of A in the extract and raffinate and the sum of E and R is equal to one.
If the analyte A is initially present in the raffinate is Atotal, after the first extraction, Atotal is given as follows:
$$ {\left(\mathrm{moles}\ {\mathrm{A}}_{\mathrm{total}}\right)}_0 = {\left(\mathrm{moles}\ {\mathrm{A}}_{\mathrm{r}}\right)}_1 + {\left(\mathrm{moles}\ {\mathrm{A}}_{\mathrm{e}}\right)}_1 $$
where (moles Atotal)0 is the number of moles present in the system initially, before the extraction. The (moles Ar)1 and (moles Ae)1 represent the number of moles found in the extract and raffinate after the first extraction, respectively. The subscripts 0 and 1 represent the system has no extraction and after the first extraction:
$$ \mathrm{E} = {\mathrm{m}}_{\mathrm{e}}/\left({\mathrm{m}}_{\mathrm{e}}+{\mathrm{m}}_{\mathrm{r}}\right)\ \mathrm{and}\ \mathrm{R} = {\mathrm{m}}_{\mathrm{r}}/\left({\mathrm{m}}_{\mathrm{e}}+{\mathrm{m}}_{\mathrm{r}}\right) $$
After the first extraction, concentration of analyte A in the extract and raffinate, [Ae]1 and [Ar]1, is given by \( {\left({\mathrm{m}}_{\mathrm{e}}\right)}_1 = \Big[{\mathrm{A}}_{\mathrm{e}}{\Big]_1}^{*}{\mathrm{V}}_{\mathrm{e}} \) and \( {\left({\mathrm{m}}_{\mathrm{r}}\right)}_1={{\left[{\mathrm{A}}_{\mathrm{r}}\right]}_1}^{*}{\mathrm{V}}_{\mathrm{r}} \) where Ve and Vr are the volumes of extract and raffinate:
$$ \begin{array}{l}{\left[{\mathrm{A}}_{\mathrm{e}}\right]}_1 = \left\{{\left({\mathrm{moles}\ \mathrm{A}}_{\mathrm{total}}\right)}_0 - {\left({\mathrm{moles}\ \mathrm{A}}_{\mathrm{r}}\right)}_1\right\}/\ {\mathrm{V}}_{\mathrm{e}} \\ {}\mathrm{and} {\left[{\mathrm{A}}_{\mathrm{r}}\right]}_1 = {\left({\mathrm{moles}\ \mathrm{A}}_{\mathrm{r}}\right)}_1/\ {\mathrm{V}}_{\mathrm{r}}\end{array} $$
The partition coefficient is \( \mathrm{K} = {\left[{\mathrm{A}}_{\mathrm{e}}\right]}_1/{\left[{\mathrm{A}}_{\mathrm{r}}\right]}_1 \), and by substituting, \( \mathrm{K} = \left[\left({\left(\mathrm{moles}\ {\mathrm{A}}_{\mathrm{total}}\right)}_0 - {\left(\mathrm{moles}\ {\mathrm{A}}_{\mathrm{r}}\right)}_1\right)/\ {\mathrm{V}}_{\mathrm{e}}\right]/\ \left({\left(\mathrm{moles}\ {\mathrm{A}}_{\mathrm{r}}\right)}_1/\ {\mathrm{V}}_{\mathrm{r}}\right). \)

Therefore, \( \mathrm{K} = \left[{\left(\mathrm{moles}\ {\mathrm{A}}_{\mathrm{total}}\right)}_0 \times {\mathrm{V}}_{\mathrm{r}} - {\left(\mathrm{moles}\ {\mathrm{A}}_{\mathrm{r}}\right)}_1 \times {\mathrm{V}}_{\mathrm{r}}\right]/\ \left[{\left(\mathrm{moles}\ {\mathrm{A}}_{\mathrm{r}}\right)}_1 \times {\mathrm{V}}_{\mathrm{e}}\right]. \)

At the first liquid-liquid extraction, the fraction of analyte A that remains in the volume of raffinate Vr is R1, and if the amount of the analyte A transferred to the extract is E1 in the volume of extract Ve, then R1 and E1 are given as follows:
$$ \begin{array}{l}{\mathbf{R}}_{\mathbf{1}} = {\left(\mathrm{moles}\ {\mathrm{A}}_{\mathrm{r}}\right)}_1/{\left(\mathrm{moles}\ {\mathrm{A}}_{\mathrm{total}}\right)}_0 \mathrm{and} \\ {}{\mathbf{E}}_{\mathbf{1}} = {\left(\mathrm{moles}\ {\mathrm{A}}_{\mathrm{e}}\right)}_1/{\left(\mathrm{moles}\ {\mathrm{A}}_{\mathrm{total}}\right)}_0.\end{array} $$
By rearranging the above equations, the fractions of the analyte A in raffinate and extract are calculated as follows:
$$ \begin{array}{l}{\mathbf{R}}_{\mathbf{1}}={\mathrm{V}}_{\mathrm{r}}/\left({\mathrm{K}}^{*}{\mathrm{V}}_{\mathrm{e}} + {\mathrm{V}}_{\mathrm{r}}\right)\ \mathrm{and} \\ {}{\mathbf{E}}_{\mathbf{1}} = {\mathrm{K}}^{*}{\mathrm{V}}_{\mathrm{r}}/\left({\mathrm{K}}^{*}{\mathrm{V}}_{\mathrm{e}} + {\mathrm{V}}_{\mathrm{r}}\right).\end{array} $$
After the second consecutive liquid-liquid extraction, using the same volumes for the raffinate and the extract as the first extraction, the fraction of analyte A present in the raffinate, R2, and in the extract, E2, is given by the following equations:
$$ \begin{array}{l}{\mathbf{R}}_{\mathbf{2}} = {\left[{\mathrm{V}}_{\mathrm{r}}/\left({\mathrm{K}}^{*}{\mathrm{V}}_{\mathrm{e}} + {\mathrm{V}}_{\mathrm{r}}\right)\right]}^2\ \mathrm{and}\ \\ {}{\mathbf{E}}_{\mathbf{2}} = 1-1/{\left\{\left[1 + \left({\mathrm{K}}^{*}{\mathrm{V}}_{\mathrm{e}}/\ {\mathrm{V}}_{\mathrm{r}}\right)\right]\right\}}^2.\end{array} $$
After n number of continuous extractions are carried out, the fraction of analyte A remains in the raffinate, Rn, and the fraction of analyte A transferred into the extract, En, is given by the following equations:
$$ \begin{array}{l}{\mathbf{R}}_{\mathbf{n}} = {\left[{\mathrm{V}}_{\mathrm{r}}/\left({\mathrm{K}}^{*}{\mathrm{V}}_{\mathrm{e}} + {\mathrm{V}}_{\mathrm{r}}\right)\right]}^{\mathrm{n}} \mathrm{and} \\ {}{\mathbf{E}}_{\mathbf{n}} = 1 - + 1/{\left\{\left[1 + \left({\mathrm{K}}^{*}{\mathrm{V}}_{\mathrm{e}}/\ {\mathrm{V}}_{\mathrm{r}}\right)\right]\right\}}^{\mathrm{n}}.\end{array} $$
The above equations clearly demonstrate that, after several consecutive liquid-liquid extractions, the amount of analyte A that remains in the raffinate is negligible, and by using repeat extractions, a higher recovery rate can be achieved.

References

  1. Wells MJM. Principles of extraction, sample preparation techniques in analytical chemistry. John Wiley & Sons, New Jersey, 2003; 37–57Google Scholar

Copyright information

© Springer Science+Business Media Dordrecht 2016

Authors and Affiliations

  1. 1.The University of MelbourneParkvilleAustralia