Steam Distillation: Vapour Pressure Calculations
Steam distillation is a key separation technique tested regularly in Class 11 and 12 exams, JEE Main, NEET UG, and CUET UG. It purifies temperature-sensitive organic liquids that are immiscible with water and decompose near their normal boiling points, such as aniline and nitrobenzene. Passing steam through the mixture enables boiling well below 100 °C. Understanding these vapour pressure calculations allows students to determine distillate composition, required steam mass, and organic molar mass accurately.
Physicochemical Principles of Immiscible Liquids
Steam distillation is governed by immiscible liquids. Unlike miscible solutions obeying Raoult's law, each component in an immiscible mixture exerts its saturated vapour pressure independently. By Dalton's law of partial pressures, total vapour pressure (Ptotal) equals the sum of the saturated vapour pressures of pure water (P°water, aqueous tension) and the organic liquid (P°organic):
Ptotal = P°water + P°organic = Pext
Standard pressure: 760.0 mmHg = 760.0 torr = 101.325 kPa (or 1.000 bar = 100.0 kPa).
Mole Ratio: norganic / nwater = P°organic / P°water
Mass Ratio: worganic / wwater = (P°organic × Morganic) / (P°water × Mwater)
Units: masses w in g, molar masses M in g/mol, vapour pressures P in mmHg or torr.
Boiling occurs when Ptotal equals external pressure Pext (760 mmHg). Because both liquids contribute to total pressure, P°water is always less than 760 mmHg. Thus, boiling always occurs below 100 °C, preventing thermal decomposition.
Derivation of the Distillate Mass Ratio
Applying the ideal gas law (PV = nRT) to both vapours in volume V at temperature T gives:
norganic / nwater = P°organic / P°water
Substituting n = w / M gives (worganic / Morganic) / (wwater / Mwater) = P°organic / P°water.
Rearranging yields the mass ratio formula: worganic / wwater = (P°organic × Morganic) / (P°water × Mwater). The molar mass of water is 18.015 g/mol (commonly 18.0 g/mol in school textbooks).
Exam details
- Source: Practice problem written for this article (not a past-paper question)
- Relevant exams: CBSE Class 11, CBSE Class 12, JEE Main, NEET UG, CUET UG
- Topic: Steam Distillation and Immiscible Liquid Vapour Pressures
- Difficulty: Intermediate
Step-by-Step Worked Examples
Question: An immiscible mixture of water and nitrobenzene (molar mass = 123.11 g/mol) is steam distilled at 760.0 mmHg. The mixture boils at 99.0 °C, where water's vapour pressure is 733.0 mmHg. Calculate the mass of water that distils over with 100.0 g of nitrobenzene.
Step 1: Find nitrobenzene vapour pressure.
P°organic = Ptotal − P°water = 760.0 mmHg − 733.0 mmHg = 27.0 mmHg.
Step 2: Apply the mass ratio formula.
wwater = worganic × (P°water × Mwater) / (P°organic × Morganic)
Step 3: Calculate the numerical value.
wwater = 100.0 g × (733.0 × 18.015) / (27.0 × 123.11)
wwater = 100.0 × 13204.995 / 3323.97 = 397.3 g (using Mwater = 18.0 gives 397.3 g).
Second Route Verification:
Moles of nitrobenzene = 100.0 / 123.11 = 0.8123 mol.
Moles of water = 0.8123 × (733.0 / 27.0) = 22.052 mol.
Mass of water = 22.052 mol × 18.015 g/mol = 397.3 g.
Both calculation routes match.
Answer: 397.3 g
Question: An unknown organic liquid immiscible with water is distilled with steam at 760.0 torr. The mixture boils at 98.0 °C, where water vapour pressure is 712.0 torr. The distillate contains 2.00 g of the organic liquid and 3.00 g of water. Calculate the molar mass of the organic liquid.
Step 1: Find organic vapour pressure.
P°organic = 760.0 torr − 712.0 torr = 48.0 torr.
Step 2: Express molar mass from mass ratio.
Morganic = (worganic / wwater) × (P°water × Mwater / P°organic)
Step 3: Compute the molar mass.
Morganic = (2.00 / 3.00) × (712.0 × 18.015 / 48.0)
Morganic = 0.6667 × 267.22 = 178.1 g/mol (or 178.0 g/mol using Mwater = 18.0 g/mol).
Second Route Verification:
Moles of water = 3.00 / 18.015 = 0.1665 mol.
Moles of organic = 0.1665 × (48.0 / 712.0) = 0.01123 mol.
Molar mass = 2.00 g / 0.01123 mol = 178.1 g/mol.
Both methods agree exactly.
Answer: 178.1 g/mol
Question: Bromobenzene (molar mass = 157.01 g/mol) is steam distilled at 95.0 °C under 760.0 mmHg. The vapour pressure of water at 95.0 °C is 640.0 mmHg. Calculate the percentage by mass of bromobenzene in the distillate.
Step 1: Find partial vapour pressure.
P°organic = 760.0 mmHg − 640.0 mmHg = 120.0 mmHg.
Step 2: Calculate distillate mass ratio.
worganic / wwater = (120.0 × 157.01) / (640.0 × 18.015)
worganic / wwater = 18841.2 / 11529.6 = 1.6342.
Step 3: Calculate percentage by mass.
Percentage = [1.6342 / (1 + 1.6342)] × 100%
Percentage = (1.6342 / 2.6342) × 100% = 62.04% ≈ 62.0%.
Second Route Verification:
Assume 1.000 mol water (18.015 g). Organic moles = 1.000 × (120.0 / 640.0) = 0.1875 mol.
Organic mass = 0.1875 mol × 157.01 g/mol = 29.439 g.
Total mass = 18.015 + 29.439 = 47.454 g.
Percentage = (29.439 / 47.454) × 100% = 62.04%.
Both routes confirm the answer.
Answer: 62.0%
Specific Mistakes Students Make
Common Traps in Steam Distillation Problems:
- Applying Raoult's law to immiscible liquids: Raoult's law applies to miscible solutions. For immiscible liquids, each liquid exerts its full pure vapour pressure independent of liquid mole fractions.
- Inverting molar masses in the formula: Remember that mass equals moles multiplied by molar mass. Each partial pressure must be multiplied by its corresponding molar mass.
- Using water pressure as organic pressure: When given total pressure and aqueous tension, always subtract water's vapour pressure: P°organic = Ptotal − P°water.
- Assuming boiling point exceeds 100 °C: Since both liquids exert vapour pressure simultaneously, total pressure reaches atmospheric pressure below 100 °C.
Exam Relevance and Question Patterns
| Examination | How the Topic Is Tested | Key Concepts Emphasised |
|---|---|---|
| CBSE Class 11 & 12 | Descriptive questions on immiscibility criteria and numerical problems finding mass ratios. | Why boiling occurs below 100 °C, Dalton's law application, and distillate ratio formulas. |
| JEE Main | Numerical value questions determining required steam mass or molar mass of unknown liquids. | Dalton's partial pressure calculations, mole-to-mass conversions, and percentage purity. |
| NEET UG | Multiple-choice questions testing suitable organic compounds and vapour pressure principles. | Purification of aniline and nitrobenzene, aqueous tension subtraction, and mass relations. |
| CUET UG | Assertion-reason and match-the-column questions comparing purification techniques. | Steam distillation vs fractional distillation, immiscibility conditions, and distillate yields. |
Turn this concept into marks: solve previous-year CSIR-NET questions and take a timed quiz.
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