Pause for thought

One of the quirks of the current syllabus is that sometimes it is not exactly clear what should be covered when teaching a topic. Much of this sub-topic is self-evident but one problem occurs in the ‘Applications and skills’ where it states:

“Description of the process of extraction and purification of an organic product”

It then goes on to state:

“Consider the use of fractional distillation, Raoult’s law, the properties on which extractions are based and explaining the relationship between organic structure and solubility.”

The use of solvent extraction relies on the relative solubility of the compound between two immiscible liquids, such as water and an organic solvent. The substance is extracted into the layer in which it is most soluble and the layers can then be separated using a separating funnel.

This describes the process but it is not clear from the guide whether students are also expected to understand and solve problems using the partition coefficient. Since ‘partition coefficient’ is not specifically mentioned and there is also no mention of ‘calculate’ or determine’ in the ‘Application and skills’ statement and no further reference under ‘Guidance’ one can only assume that it is not required, but at least one of the text books for the course does include it.

It may be worth explaining it and giving your students an example such as:

(a) Let mass of X extracted in ether be m, so [X(ether)] = 100m g dm^-3

Partition coefficient, K_pc = [X(ether)] ÷ [X(aq)]

[X(aq)] = (1 –m) g in 100 cm³ = (10 – 10m) g dm^-3

[X(ether)] = 40 x (10 – 10m) g dm^-3 = 100m g dm^-3

400 – 400m = 100m

so mass extracted using 10 cm³ of ether = 0.80 g

(b) First 5 cm³

Let mass of X extracted in 5 cm³ of ether be m₁ ,so [X(ether)] = 200m₁ g dm-³

[X(ether)] = 40 x (10 – 10m₁) g dm^-3 = 200m₁ g dm^-3

400 – 400m₁ = 200m₁

Mass of ether extracted, m₁ = 0.667 g

Second 5 cm³

Let mass of X extracted in 5 cm³ of ether be m₂ ,so [X(ether)] = 200m₂ g dm-³

Now only 0.333 g of X remains in the 100 cm³ aqueous layer

[X(aq)] = (0.333 –m₂) g in 100 cm³ = (3.33 – 10m₂) g dm^-3

[X(ether)] = 40 x (3.33–10m₂) g dm^-3 = 200m₂ g dm^-3

133.2 – 400m₂ = 200m₂ so m₂ = 0.22 g

So total amount extracted using two lots of 5 cm³ of ether = m₁ + m₂ = 0.67 + 0.22 = 0.89 g

Students can understand from this that it is better to carry out several extractions with smaller quantities than to use all the extracting solvent at once.