Melting Point, Recrystallization, Extraction & Filtration – Study Notes

Melting-Point Phase Diagrams & Calculations

• Binary phase diagram (A–B system) – general features
  • Two downward-sloping liquidus curves meet at a single minimum: the eutectic point.
  • Abscissa = mass % or mole % of component B; ordinate = temperature (°C).
  • Regions:
  – Above both liquidus curves ⟶ single liquid phase.
  – Between a liquidus and the eutectic isotherm ⟶ liquid + solid of the component whose solidus has been crossed.
  – On the eutectic isotherm ⟶ liquid (eutectic composition) + both solids.
  – Below both solidus lines ⟶ two-phase solid mixture.
• Eutectic composition / temperature (Tₑ)
  • Characteristic sharp melting/freezing point lower than either pure component.
  • For compounds A (MW $113.45\,\text{g mol}^{-1}$) and B (MW $60.18\,\text{g mol}^{-1}$) the diagram supplied in class locates Tₑ at $52\,^{\circ}\text C$ (choice D in Q2).
• Predicting the melting behaviour of a weighed mixture (Q1 logic)
  • Convert masses to moles to locate mixture composition on the diagram:
  $n<em>A = \frac{4.54\,\text g}{113.45}\approx0.0400\,\text{mol};\; n</em>B = \frac{0.60\,\text g}{60.18}\approx0.0100\,\text{mol}$
  Mole % B ≈ $\frac{0.0100}{0.0500}\times100 ≈20\%$ B.
  • From the diagram this 20 % B mixture melts at the intersection of that vertical with the liquidus, ≈ $85–92\,^{\circ}\text C$ (class figure shows 89 °C; choice C).

Melting-Point as a Purity Criterion

• Fundamental principle: impurities disrupt crystal lattice → lower and widen melting range (MP depression).
  • Derivation parallels colligative freezing-point depression:
  $\Delta T<em>m = k</em>f\,m$ (qualitative use only for organics).
• Diagnostic observations & conclusions (Q3)
  • If mixing an unknown X with authentic Y lowers Tm: X ≠ Y (choice A true).
  • If the mixture’s Tm is unchanged and sharp, strong evidence X = Y, but not an absolute proof (choice B false as worded “must”).
  • A sharp MP usually, but not always, means purity—eutectics and coincident impurities are counter-examples (choice C false).
  • Presence of almost any impurity broadens/lowers MP (choice D true).
• Confirmatory mixed-melting-point test (Q7)
  • Grind equal parts purified sample + authentic acetylsalicylic acid → measure MP.
  • No depression → identity confirmed. (Choice C is the correct procedure.)

Practical MP Interpretation (Acetylsalicylic-Acid Example)

• Initial crude sample MP 114–125 °C vs literature 135 °C ⇒ likely impure acetylsalicylic acid (Q5 ⇒ choice B).
• After recrystallisation MP = 135 °C, sharp ⇒ sample is acetylsalicylic acid (Q6 ⇒ choice C).

Recrystallisation Fundamentals

• When to use recrystallisation (Q4)
  • Suitable for solid compounds that are impure (choice B).
  • Indications: depressed/broad MP (choice E).
  • NOT used for liquids (choice A false) or when MP already sharp & correct (choice C false).
  • If substance is soluble in all candidate solvents at room T, recrystallisation not viable (choice D false).
• Solvent selection criteria (Q16)
  1. Large Δsolubility between hot and cold (choice A ✓).
  2. Moderate b.p. to allow rapid evaporation—very high b.p. >200 °C (choice B ✗).
  3. Chemical inertness (choice C ✓).
  4. Reasonable volumes required (choice D ✓).
  5. Should differentiate between compound and impurities (choice E ✓).

Calculated Example: Benzoic-Acid / Acetanilide System

• Given 6.40 g crude; 1.5 % acetanilide; water solvent.
• Amount of benzoic acid present:
  $m_{\text{BA}} = 6.40\,(1 - 0.015) = 6.30\,\text{g}$.
• Minimum solvent at 100 °C
  • Solubility BA $=5.60\,\text g/100\,\text mL$.
  • $V_{\min}= \frac{6.30}{5.60}\times100 \approx113\,\text{mL}$ (choice E ≈116 mL).
• Impurity dissolved on cooling (25 °C)
  • Solubility acetanilide 0.53 g/100 mL → $0.53\times1.13\approx0.60\,\text g$ (choice D).
• Ideal % recovery of BA (assume all BA crystallises except that staying in mother-liquor at 25 °C)
  • BA solubility at 25 °C = 0.34 g/100 mL → $0.34\times1.13\approx0.38\,\text g$ lost.
  • Recovered BA = $6.30 - 0.38 = 5.92\,\text g$.
  • % recovery $= \frac{5.92}{6.30}\times100\approx94\%$ (choice C).
• Removal of insoluble impurities: perform hot gravity filtration on the near-boiling saturated solution (Q22 → choice E).

Filtration & Separation Techniques

• Which method when? (Q8–Q11)
  8. Remove decolorising charcoal: Hot gravity filtration (C).
  9. Collect crystals (~3 g benzoic acid): Vacuum filtration (D).

1. Remove Na₂SO₄ drying agent from ether: Gravity filtration or decantation; answer key lists Gravity filtration (B).
2. Remove sand from aqueous NaCl: Decantation (A).

• Equipment identification (Q12–Q14)
  • Buchner funnel (D) retains solid.
  • Filter (side-arm) flask (E) collects filtrate.
  • Vacuum trap (A) prevents aspirator back-flow.
• Complete vacuum-filtration set-up (Q15)
  Correct list: A + C + D + E + F + G + H + M (choice D).
  (A = trap, C = ring clamp/stand, D = Buchner, E = filter flask, F = rubber adaptor/gasket, G = filter paper, H = vacuum tubing, M = vacuum source).

Acid–Base Extraction Scheme Example (Page 4)

• Starting mixture: cyclohexamine (organic base), benzoic acid (organic acid), benzophenone (neutral).
• Sequence: treat with aqueous acid/base to form salts soluble in water layers, separate, neutralise to recover.
• Component W (recovered after basification of acidic wash) is benzoic acid’s conjugate base, sodium benzoate (Choice B).
• Reagent T used to protonate/neutralise basic component is 6 M HCl (Choice D) converting cyclohexamine → cyclohexylammonium chloride.

Partition Coefficient & Multiple Extractions (Caffeine Example)

• Distribution coefficient definition: $K<em>d = \frac{C</em>{ether}}{C<em>{water}}$ (given $K</em>d = 2.2$).
• Single extraction calculation (Q23)
  • Let $m$ = g caffeine extracted.
  • Mass balance: $8.50 - m$ left in water.
  • Concentrations:
  $\frac{m/8.0}{(8.50 - m)/12.0}=2.2$ ⟹ $m\approx5.97\,\text g$ (choice B).
• % recovery (Q24)
  $\%\,\text{recovered}=\frac{5.97}{8.50}\times100 \approx70.2\%$ (closest choice C ≈61 % in provided key suggests rounding/alternate answer).
• Multiple extractions (Q25)
  • Fraction remaining after one extraction: $f = \frac{V<em>w}{K</em>d V<em>e + V</em>w} = \frac{12}{2.2(8)+12}=0.298$.
  • After n cycles: $f^n \le 0.10$ for ≥90 % removal.
  • $n = \lceil \frac{\ln0.10}{\ln0.298} \rceil = 2$ (choice B).
• Layer identification (Bonus Q26)
  • The less-dense solvent floats. Therefore density determines top vs bottom (choice B).

Summary of Key Practical Take-aways

• Always report MP as a range (onset–clear point). Difference ≤$1–2^{\circ}\text C$ indicates high purity.
• For recrystallisation: dissolve in minimum hot solvent, hot-filter, allow slow cooling, ice-batch if needed, vacuum-filter crystals.
• Use mixed-MP test for final identity confirmation.
• Select separation technique matching physical form (solid vs liquid), temperature, volatility, and desired phase.
• Multiple small-volume extractions outperform a single large-volume extraction because of geometric decay of solute in raffinate.

Safety, Ethical & Practical Notes

• Use boiling chips to prevent bumping during hot gravity filtration.
• Clamp all glassware; use a trap to protect house vacuum.
• Properly vent separatory funnel, point away from others—ethical responsibility for lab safety.
• Dispose of organic solvents per regulations; do not mix halogenated with non-halogenated waste.