Solvent extraction and chromatography

Chemistry · Experimental techniques in chemistry. The idea in plain words, the rules to remember, the mistake to avoid and a worked example, written and checked by Mega Lecture.

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The idea

In solvent extraction a solute is moved from water into an organic solvent, such as ether, that does not mix with water. The two liquids are shaken in a separating funnel and the two layers are then run off separately. The solute shares itself between the layers in a fixed ratio, called the distribution coefficient.

Chromatography separates the components of a mixture because they distribute differently between a stationary phase and a mobile phase. In paper chromatography the stationary phase is the water held in the paper and the mobile phase is the solvent that rises up it. A component that is more soluble in the mobile phase travels further.

Rules to remember

Distribution law: at a constant temperature, a solute distributes itself between two immiscible liquids so that the ratio of its concentrations is constant
distribution coefficient K = concentration in organic solvent ÷ concentration in water
Several extractions with small portions of solvent remove more solute than one extraction with the same total volume
Adsorption chromatography: stationary phase is a solid. Partition chromatography: stationary phase is a liquid. Paper chromatography is partition chromatography
R_f = distance moved by the component ÷ distance moved by the solvent front, both measured from the base line
R_f has no unit and is always less than 1. It identifies a substance for a given solvent and temperature
Paper chromatography can be ascending, descending or radial (circular)

Common mistake

In distribution questions students use masses in the ratio when the two volumes are different. K is a ratio of concentrations, so divide each mass by its own volume first.

Worked example

0.60 g of a solute is dissolved in 100 cm³ of water. It is shaken once with 50 cm³ of ether. The distribution coefficient (ether/water) is 4. What mass of solute passes into the ether?

Let x g pass into the ether. Then (0.60 − x) g stays in the water.
K = (x ÷ 50) ÷ ((0.60 − x) ÷ 100) = 4, which gives 2x ÷ (0.60 − x) = 4.
2x = 2.40 − 4x, so 6x = 2.40 and x = 0.40 g.
Check: ether 0.40 ÷ 50 = 0.008 g cm⁻³; water 0.20 ÷ 100 = 0.002 g cm⁻³; ratio = 4.

Answer: 0.40 g passes into the ether and 0.20 g stays in the water.

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Chapter 2: Experimental Techniques in Chemistry

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