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Revision notes on theory and practice of steam distillation and uses

Doc Brown's Chemistry Theoretical–Physical Advanced Level Chemistry – Equilibria

 Chemical Equilibrium Revision Notes PART 8.5

 8.5 The Theory and Practice of Steam Distillation

What is steam distillation?

Why is steam distillation useful for thermally unstable compounds? What apparatus do you need?

How do you do a steam distillation? A steam distillation calculation to theoretically predict the composition of the distillate are fully explained.

(c) doc b KS4 Science GCSE/IGCSE Notes on reversible reactions and chemical equilibrium

Part 8 sub–index: 8.1 Vapour pressure origin and examples * 8.2.1 Introduction to Intermolecular Forces * 8.2.2 Detailed comparative discussion of boiling points of 8 organic molecules * 8.3 Boiling point plots for six organic homologous series * 8.4 Other case studies of boiling points related to intermolecular forces * 8.5 Steam distillation – theory and practice * 8.6 Evidence and theory for hydrogen bonding in simple covalent hydrides * 8.7 Solubility of covalent compounds, miscible and immiscible liquids

Advanced Equilibrium Chemistry Notes Part 1. Equilibrium, Le Chatelier's Principle–rules * Part 2. Kc and Kp equilibrium expressions and calculations * Part 3. Equilibrium and industrial processes * Part 4. Partition, solubility product and ion–exchange * Part 5. pH, weak–strong acid–base theory and calculations * Part 6. Salt hydrolysis, Acid–base titrations–indicators, pH curves and buffers * Part 7. Redox equilibria, half–cell electrode potentials, electrolysis and electrochemical series



8.5 Extraction using steam distillation of immiscible liquids

(NOT fractional distillation of miscible liquids involving water e.g. fractionally distilling fermented sugar solution to extract ethanol)

  • The technique of steam distillation is a very useful method for extracting molecules with a high boiling point, which under normal distillation conditions might thermally decompose.
    • i.e. the kinetic energies of the molecules at the boiling point may be sufficient to overcome the activation energies of possible reactions such as decomposition into smaller molecules or transformation into another molecule of similar size.
  • As has been stated earlier, when a liquid is heated, the vapour pressure rises, and when it equals the ambient pressure the liquid boils.
  • If a mixture of two immiscible liquids (or solutions) is heated, BOTH molecules can contribute to the vapour pressure and BOTH will express their full saturated vapour pressure because being immiscible, the two liquids act independently..
    • so at a given temperature
    • Ptot = pA + pB, and is irrespective of the actual ratio of the volumes of liquids.
    • or Ptot = pH2O + product, where Pproduct is the vapour pressure of the material extracted
    • If the larger proportion of Ptot is from water vapour, you can then distil over substances at ~100oC which might normally only distil over at much higher temperatures of say 150–300oC.
    • This means you can predict two things from vapour pressure tables data:
      • (1) the temperature at which the mixture will distil over i.e. at normal pressure in the laboratory
        • i.e. when Ptot = pH2O + pproduct = 760 mmHg
        • In the calculation procedure below to theoretically calculate the percentage by mass of water and desired product by mass, p = partial pressure, n = mol, m = mass in g, M = relative molecular mass
      • (2) the composition of the distilled mixture if you know the distillation temperature
        • you do this calculation from the vapour pressure ratio as follows
        • pproduct/pH2O = nproduct/nH2O where n = number of moles of each component
          • this assumes mole ratio and vapour pressure ratio are identical, which is strictly speaking only true for ideal gases, but accurate enough in this context.
        • So, bringing in relative molecular masses and mols n = mass m/Mr
        • Pproduct/PH2O = (mproduct/Mproduct) / (mH2O/18)
        • mproduct/mH2O = (Pproduct x Mproduct) / (PH2O x 18)
        • to give the mass ratio in the distillate
        • using the mass ratio numbers gives.
        • % mext = (mproduct x 100) / (mproduct + mH2O)
  • In practice the vapour pressure due to water often far exceeds that of the extracted material BUT the important thing to realise is that a thermally unstable material can then be distilled over at a temperature much lower than its decomposition or transformation temperature.
  • Below is shown the experimental technique for extracting phenylamine from a nitrobenzene reduction in section 8.5 Organic Redox reactions
  • orgPD4
  • The technique is widely used in the extraction of molecules from natural products e.g.
    • In the perfumery industry, perfume molecules like those from lavender can be extracted from natural plant materials using steam distillation – an unforgettable 'odour' experienced from the rural distilleries of Provence whilst driving in France! – not that the 'kids' where interested back in those days!
  • Vacuum or reduced pressure distillation is a superior method BUT is not always as convenient? or practical? or cheap? in the context of a rural industry!

 


Doc Brown's Advanced Level Chemistry Revision Notes

WHAT NEXT?

Part 8 sub–index: 8.1 Vapour pressure origin and examples * 8.2.1 Introduction to Intermolecular Forces * 8.2.2 Detailed comparative discussion of boiling points of 8 organic molecules * 8.3 Boiling point plots for six organic homologous series * 8.4 Other case studies of boiling points related to intermolecular forces * 8.5 Steam distillation – theory and practice * 8.6 Evidence and theory for hydrogen bonding in simple covalent hydrides * 8.7 Solubility of covalent compounds, miscible and immiscible liquids

Advanced Equilibrium Chemistry Notes Part 1. Equilibrium, Le Chatelier's Principle–rules * Part 2. Kc and Kp


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