How to calculate the remaining mass in the mother solution after multiple extractions? During an extraction, a solute ($\ce{S}$) will partition itself between the two solvent phases with a partition coefficient ($K$) defined as:

$$\ce{S(solvent 1) &lt;=&gt; S(solvent 2)}$$

$$K = \frac{[\ce{S}]_2}{[\ce{S}]_1}$$

Solute $\ce{S}$ is found to have a partition coefficient of 15 between solvent 1 and solvent 2; i.e., with 15 times as much in solvent 2. Suppose $\pu{150 mL}$ of a $\pu{0.02 M}$ solution of $\ce{S}$ in solvent 1 is extracted with \ce{\pu{50 mL}} of solvent 2. What fraction of $\ce{S}$ remains in solvent 1? If three $\pu{50 mL}$ extractions are performed, what fraction of $\ce{S}$ remains in solvent 1?
In old lecture notes of mine I found this formula

$ g_i = g\left(\frac{V_1}{V_1 + V_2K_2}\right)^n$

where $g_i$ is the remnant mass of a product initially dissolved in volume $V_1$, to be $n$-times extracted by volume $V_2$ with a given Nernst dissipation coefficient $K = \frac{a_2}{a_1}$.  Here, $g$ refers to the mass of the product initially dissolved in $V_1$.

Applied to your case, assuming a molar mass of 180 g/mol, this would read like 0.54 g initially dissolved in you 150 mL solution 1 (3 mmol).  After a single extraction with 50 mL of 2, only 90 mg were remaining (0.5 mmol) in $V_1$, after the third time 2.5 mg (about 14 umol).

I speculate a textbook like P. W. Atkins' Physical Chemistry may contain an example to confirm such a calculation.