Exercises — Separation techniques — filtration, distillation, chromatography, centrifugation, sublimation

Technique: filtration. The clay is insoluble (visible grains), so the components differ in particle size — the exact property filtration exploits.

• Pour through filter paper. Clay grains ($>10\mu \text{m}$) cannot fit through the pores.
• The clay trapped on the paper is the residue.
• The clear water passing through is the filtrate.

Why not distillation? You could boil it, but that wastes enormous energy to separate something a $2$-cent filter handles by size alone. Match the method to the property difference — here, size.

Technique: sublimation. Iodine sublimes — it goes solid → gas without a liquid stage — at gentle heat, while salt stays solid. The property difference is sublimation temperature.

• Warm the mixture in a dish covered by a cool surface.
• Iodine vapour rises, hits the cool lid, and re-solidifies (deposition) there.
• Salt is left behind in the dish.

The rule from the parent note: a particle is trapped when its diameter exceeds the pore diameter. $\text{diameter}>8\mu \text{m}⟹\text{trapped}$

• Sand: $400>8$ → trapped (residue).
• Pollen: $30>8$ → trapped (residue).
• Sugar: $0.001<8$ → passes (stays in filtrate).

So filtration removes sand and pollen but leaves the sugar dissolved — you'd still need evaporation to recover the sugar.

The retention factor is the fraction of the solvent's journey the component completes: $R_f=\frac{\text{distance moved by component}}{\text{distance moved by solvent front}}=\frac{6.5}{10.0}=0.65$ A higher $R_f$ means the component travelled farther, so it spent more time in the mobile phase and stuck less to the polar stationary phase — meaning it is less polar. Therefore the $R_f=0.65$ dye is less polar than the $R_f=0.20$ dye (which clung to the polar paper and barely moved).

See the figure below for the two spots side by side.

Why filtration fails: Salt dissolves into ${\text{Na}}^{+}$ and ${\text{Cl}}^{-}$ ions, each roughly $0.0003\mu \text{m}$ across. Filter pores are about $8\mu \text{m}$ — over $25,000\times$ larger: $\frac{8}{0.0003}\approx 2.7\times 10^4$ The ions slip through the pores as easily as water does. Filtration separates only by size, and dissolved ions are not size-distinguishable from the solvent.

What succeeds — distillation: Water and salt differ enormously in boiling point / volatility. Water boils at $100^{\circ }\text{C}$; salt does not vaporise until it decomposes near $1465^{\circ }\text{C}$. Heat to $100^{\circ }\text{C}$: only water leaves as vapour, condenses to pure water, salt stays behind. We switched the exploited property from size to boiling point.

At $80^{\circ }\text{C}$ both liquids have appreciable vapour pressure — water is below its boiling point but still evaporating. By Raoult's Law the total vapour is a blend: $P_{\text{total}}=P_{\text{eth}}^0{\chi }_{\text{eth}}+P_{\text{water}}^0{\chi }_{\text{water}}$ Because ethanol is the more volatile (lower-BP) component, its $P_{\text{eth}}^0$ is larger, so ethanol dominates the vapour — but water's non-zero term keeps the vapour a mixture. Hence the first distillate is ethanol-enriched (≈$40\text{–}60\%$), never $100\%$. To push higher you re-distil (fractional distillation), but you hit the ethanol–water azeotrope at $95.6\%$, where vapour and liquid have the same composition and simple distillation can go no further.

Attack one property difference at a time. Order matters.

Step 1 — Filtration (property: particle size). Sand is insoluble; salt and iodine's soluble/vaporisable parts stay with the water. Filter: sand is the residue (recovered #1); filtrate = salt + a little iodine in water. (Iodine is only slightly soluble; assume it passes as a trace in solution/suspension — we clean it up next.)

Step 2 — Distillation (property: boiling point). Heat the filtrate to $100^{\circ }\text{C}$. Water vaporises and condenses in the receiver → pure water (recovered #2). Salt (non-volatile) stays in the flask; any iodine escaping is caught below.

Step 3 — What's left in the flask: solid salt + solid iodine. Dry it, then use sublimation (property: sublimation temperature). Gently warm: iodine sublimes to purple vapour, deposits as pure crystals on a cool lid → iodine (recovered #3); salt cannot sublime and stays in the dish → salt (recovered #4).

Summary chain: size → boiling point → sublimation temperature. Each step removes exactly one component by a different physical handle.

(a) Salt mass. Total mass $=1025\text{kg}$ (one cubic metre at that density). Salt fraction $=3.5\%=0.035$: $m_{\text{salt}}=0.035\times 1025=35.875\text{kg}$ (b) Energy cost. Energy $=50\text{MJ}$. Cost =50\ \text{MJ}\times\dfrac{\0.10}{10\ \text{MJ}}: $$\text{cost}=\frac{50}{10}\times0.10=\0.50$$ Interpretation: recovering pure water from a whole cubic metre of seawater costs roughly half a dollar in raw energy and leaves ~$36\text{kg}$ of salt — a vivid reason desalination is called "energy-intensive."

Ethanol present: $40.0\%$ of $100.0\text{g}=40.0\text{g}$. This ethanol is the limiting ingredient of the $95.6\%$ azeotrope. If the collected distillate has mass $D$ and is $95.6\%$ ethanol: $0.956D=40.0\text{g}⟹D=\frac{40.0}{0.956}=41.8\text{g}$ So at most $\approx 41.8\text{g}$ of $95.6\%$ distillate. You physically cannot distil all $100\text{g}$ into pure ethanol — the azeotrope caps purity, and mass balance on ethanol caps yield. (See Colligative Properties for why real mixtures deviate from ideal.)

Filtration separates by ::: particle size Distillation separates by ::: boiling point (volatility) Chromatography separates by ::: polarity / affinity for the stationary phase Sublimation separates by ::: whether a component turns straight from solid to gas A higher $R_f$ means the component is ::: less attracted to the (polar) stationary phase, i.e. less polar Simple distillation cannot pass the ethanol–water ::: azeotrope at 95.6%