In an environmental case you are given a mixed sample and asked to plan or interpret a sequence of separations. Each step needs a named method and a reason that refers to a difference between the substances.
This lesson belongs to chemical reasoning in mixed tasks. The data below is invented for teaching and is labelled as such. Only the reasoning is taught, and hazards are discussed conceptually.
Which method matches which difference?
- Filtration separates an insoluble solid from a liquid, because the solid particles are too large to pass through the filter paper.
- Evaporation or crystallisation separates a dissolved solid from its solvent, because the solvent leaves as vapour and the solid stays.
- Simple distillation separates a pure solvent from a solution, because the solvent boils off and is condensed and collected.
- Using a separating funnel separates two liquids that do not mix, because one floats on the other.
Worked example (invented data)
A 500 cm³ sample is taken from a coastal drain. It contains sand and mud, a thin film of oil on the surface and dissolved salt.
Question: Suggest a sequence to obtain pure water and give the reason for each step. Then find the concentration of the sand and mud, and of the salt.
Step 1, remove the oil. The oil does not mix with water and floats, so a separating funnel (or careful skimming) takes it off first.
Step 2, filter. The sand and mud are insoluble. Filtration traps them on the paper while the salt solution passes through.
Step 3, distil the filtrate. Water boils at a lower temperature than the salt, which does not boil off. The water vapour is condensed and collected as pure water, and the salt stays behind.
Step 4, concentration of solids. The dried sand and mud have mass 4.0 g from 500 cm³. 500 cm³ = 0.500 dm³, so concentration = 4.0 ÷ 0.500 = 8.0 g/dm³.
Step 5, concentration of salt. A 50.0 cm³ portion of the filtrate is evaporated, leaving 1.75 g of salt. 50.0 cm³ = 0.0500 dm³, so concentration = 1.75 ÷ 0.0500 = 35.0 g/dm³.
Check: 8.0 × 0.500 = 4.0 g and 35.0 × 0.0500 = 1.75 g, so both match the data.
The mistake to watch for
Mistaken answer: “Filter the sample to remove the salt, then evaporate to remove the sand.”
The student matched the methods to the wrong substances. Sand is insoluble and is trapped by filtration. Salt is dissolved and passes through the filter.
The correction is to ask “is it dissolved or not?” before choosing a method. A second slip is giving a concentration in g/cm³ when the question wants g/dm³.
Check yourself
1. A 250 cm³ sample gives 2.0 g of dried sand. What is the concentration of sand in g/dm³?
Show answer
250 cm³ = 0.250 dm³. Concentration = 2.0 ÷ 0.250 = 8.0 g/dm³.
2. Evaporating 20.0 cm³ of filtrate leaves 0.70 g of salt. What is the concentration of salt in g/dm³?
Show answer
20.0 cm³ = 0.0200 dm³. Concentration = 0.70 ÷ 0.0200 = 35 g/dm³.
3. Why is distillation, not filtration, used to obtain pure water from the filtrate?
Show answer
The salt is dissolved, so its particles pass through filter paper along with the water. Distillation separates the two because water boils and is collected as a condensed liquid, while the salt does not boil off.
Where this leads next
Next, link acidity evidence with a proposed substance class to practise drawing a conclusion from a measurement. You can revisit connecting an equation with a supplied mass change for the unit conversion habits, and the mixed practice set includes a sequence question.
If your answers name the right methods but miss the reasons, our teachers can practise that wording with you in online one-to-one Combined Science tuition.