Suppose a reaction mixture, when diluted with water, afforded 300 mL of an aqueous
solution of 30 g of the reaction product malononitrile [CH2(CN)2], which is to be isolated
by extraction with ether. The solubility of malononitrile in ether at room temperature is
20.0 g / 100 mL, and in water is 13.3 g / 100 mL.
(a) What weight of malononitrile would be recovered by extraction with three 100 mL
portions of ether?
(b) What weight of malononitrile would be recovered by extraction with one 300 mL
portion of ether?
Suggestion: For each extraction, let x equal the weight extracted into the ether layer. In
part (a), the concentration in the ether layer is x / 100 and in the water layer is (30 – x) /
300; the ratio of these quantities is equal to k = 20 / 13.3.

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.

Learn how to calculate the amount of malononitrile recovered by extraction using ether. This problem involves using the partition coefficient formula to determine extraction efficiency. Detailed step-by-step solution for extracting malononitrile from aqueous solution with ether. Suitable for college-level chemistry students.

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