To solve this problem, we will use the partition coefficient, $k$, which describes the ratio of the concentrations of malononitrile in ether and water. The partition coefficient is given by:

$k = \frac{\text{concentration in ether}}{\text{concentration in water}} = \frac{x/100}{(30 - x)/300}$

where:

• $x$ is the amount of malononitrile extracted into the ether layer.
• The concentration of malononitrile in the ether layer is $\frac{x}{100}$.
• The concentration of malononitrile in the water layer is $\frac{30 - x}{300}$.
• $k = \frac{20.0 \, \text{g} / 100 \, \text{mL}}{13.3 \, \text{g} / 100 \, \text{mL}} = \frac{20}{13.3} \approx 1.504$.

Let's solve each part:

(a) Extraction with Three 100 mL Portions of Ether

Step 1: First Extraction

Let $x_1$ be the amount of malononitrile extracted into the first 100 mL portion of ether. Using the partition coefficient:

$\frac{x_1 / 100}{(30 - x_1) / 300} = 1.504$

Solving for $x_1$:

$x_1 = \frac{1.504 \times 100 \times (30 - x_1)}{300}$

Simplify:

$x_1 = \frac{150.4 \times (30 - x_1)}{300}$

$x_1 = 15.04 - 0.5013x_1$

$1.5013x_1 = 15.04$

$x_1 = \frac{15.04}{1.5013} \approx 10.02 \, \text{g}$

Step 2: Second Extraction

After the first extraction, $30 - 10.02 = 19.98 \, \text{g}$ of malononitrile remains in the water layer. Let $x_2$ be the amount extracted in the second 100 mL portion:

$\frac{x_2 / 100}{(19.98 - x_2) / 300} = 1.504$

Solving for $x_2$:

$x_2 = \frac{1.504 \times 100 \times (19.98 - x_2)}{300}$

$x_2 = \frac{150.4 \times (19.98 - x_2)}{300}$

$x_2 = 10.02 - 0.5013x_2$

$1.5013x_2 = 10.02$

$x_2 = \frac{10.02}{1.5013} \approx 6.67 \, \text{g}$

Step 3: Third Extraction

After the second extraction, $19.98 - 6.67 = 13.31 \, \text{g}$ remains in the water layer. Let $x_3$ be the amount extracted in the third 100 mL portion:

$\frac{x_3 / 100}{(13.31 - x_3) / 300} = 1.504$

Solving for $x_3$:

$x_3 = \frac{1.504 \times 100 \times (13.31 - x_3)}{300}$

$x_3 = \frac{150.4 \times (13.31 - x_3)}{300}$

$x_3 = 6.67 - 0.5013x_3$

$1.5013x_3 = 6.67$

$x_3 = \frac{6.67}{1.5013} \approx 4.44 \, \text{g}$

Total Amount Extracted:

$x_1 + x_2 + x_3 = 10.02 + 6.67 + 4.44 \approx 21.13 \, \text{g}$

(b) Extraction with One 300 mL Portion of Ether

Let $x$ be the amount of malononitrile extracted with a single 300 mL portion of ether:

$\frac{x / 300}{(30 - x) / 300} = 1.504$

Simplify:

$\frac{x}{30 - x} = 1.504$

$x = 1.504 \times (30 - x)$

$x = 45.12 - 1.504x$

$2.504x = 45.12$

$x = \frac{45.12}{2.504} \approx 18.02 \, \text{g}$

Summary of Results:

• (a) The weight of malononitrile recovered by extraction with three 100 mL portions of ether is approximately 21.13 g.
• (b) The weight of malononitrile recovered by extraction with one 300 mL portion of ether is approximately 18.02 g.

Would you like further explanation or details on any step?

Related Questions:

1. How does the partition coefficient affect the efficiency of the extraction process?
2. What happens to the amount recovered if more extraction steps are used?
3. How would the results change if the solubility of malononitrile in water were higher?
4. What are the advantages and disadvantages of using multiple extractions versus a single extraction?
5. How would you calculate the amount recovered if using different solvents?

Tip:

Always perform multiple extractions with smaller solvent volumes rather than a single extraction with a large volume to maximize the recovery of the desired compound.