Separating a Damp Four-Component Mixture Without Losing the Salt
A laboratory assistant drops a tray holding 40 g of a dry mixture of iron filings, common salt, chalk powder and ammonium chloride. The powders are now thoroughly mixed and slightly damp from a spilled beaker. Available equipment: a bar magnet, filter paper and funnel, two beakers, an evaporating dish, a water bath, a tripod and burner, and a glass rod. The teacher's conditions are: the ammonium chloride must be recovered as a solid, the salt must be recovered dry, the iron must contain no chalk, and no component may be discarded. One student suggests adding 100 mL water and stirring vigorously first, then filtering. Another suggests heating the entire mixture strongly in a test tube and collecting everything that comes off. A third says the order hardly matters because each component has a different property. The assistant has 45 minutes and one class period.
1. Give a workable separation sequence and name the property that makes each step selective. State why the order you choose matters.
Read the worked answer
Sequence and property used
- Magnet over the dry mixture - iron is ferromagnetic; chalk, salt and ammonium chloride are not. Remove the iron first so it does not tear the filter paper later.
- Gentle heating in an evaporating dish or water bath - ammonium chloride sublimes (solid to vapour and back to solid on a cool surface); salt and chalk do not. Collect the sublimate on a cool watch glass. Do this before adding water, because ammonium chloride is soluble and would otherwise contaminate the salt.
- Add water, stir, filter - salt dissolves; chalk is insoluble. The residue is chalk; the filtrate contains salt.
- Evaporate the filtrate to dryness - salt is recovered; avoid strong heating, which can decrepitate and spit salt.
Why order matters: the properties overlap in water. Both salt and ammonium chloride dissolve, so if water is added first, filtration cannot separate them and evaporation gives a mixed solid. Removing the ammonium chloride by sublimation while the mixture is dry keeps the two soluble solids apart. Removing iron first protects the filter and keeps the iron clean.
2. Evaluate the water-first proposal and the heat-everything proposal. Which one fails the teacher's conditions, and on what specific ground?
Read the worked answer
Water-first proposal: it fails. Salt and ammonium chloride both dissolve, so filtration separates only chalk. Evaporating the filtrate leaves a mixture of salt and ammonium chloride, not dry pure salt. Sublimation is no longer practical from solution, so the ammonium chloride cannot be recovered as a separate solid by the stated method. The proposal therefore violates two conditions at once.
Heat-everything proposal: it also fails. Ammonium chloride would sublime, but the sublimate would be mixed with water vapour from the damp mixture and possibly with fine chalk dust carried by convection. Strong heating may also decompose chalk, oxidise the iron (so it is no longer clean iron), and crack a closed tube. The collected solid would not be pure ammonium chloride, and the iron condition would be lost.
Third student: wrong. The order matters because the separation depends on which property is used while the components are still in a favourable state - dry for sublimation and magnetic removal, wet for filtration.
3. The recovered dry salt weighs 9.6 g and the mixture is known to contain 12 g salt. Calculate the percentage recovery and give two defensible reasons for the shortfall that are consistent with your scheme.
Read the worked answer
Percentage recovery = (9.6 g / 12 g) × 100 = 80%.
Two consistent reasons:
- Transfer losses: some salt solution remains in the beaker, on the glass rod and in the filter paper after filtration; rinsing would recover part of it.
- Spray loss during evaporation: if the filtrate was heated too strongly, fine droplets of salt solution were thrown out of the dish.
A further acceptable reason is that some salt remained trapped in the chalk residue if the mixture was not washed. An 80% recovery is plausible for school apparatus but is on the low side; a second extraction of the chalk residue with a little water would improve it. The calculation does not by itself prove contamination - it only measures loss, so purity would need a separate check.