Percentage labeling of Oleum ( H2S2O7)

Labeling of oleum explained: definition, types, conversion into equivalent H₂SO₄, formulas, and solved examples for JEE, NEET & Class 12 Chemistry.

Labeling of Oleum — Definition, Conversion & Solved Examples

Oleum (fuming sulfuric acid) is a mixture of H₂SO₄ and free SO₃. It is often written as H2SO4 · x SO3, where x denotes the amount of free SO₃ present per formula unit. Understanding how oleum is labelled is essential for industrial chemistry and for solving numerical problems in JEE/NEET/Class 12 chemistry.

Inorganic Chemistry Notes Physical Chemistry Practice Tests (JEE/NEET)

What is Oleum?

Oleum (also called fuming sulfuric acid) is essentially concentrated sulfuric acid that contains free sulfur trioxide (SO₃) dissolved in it. Represented as:

H₂SO₄ · x SO₃

Here x is the percentage (by mass) of free SO₃ in the oleum formula when expressed per 100 g of oleum (or as grams per 100 g depending on convention used in problems).

Why Label Oleum?

Oleum contains both H₂SO₄ and free SO₃, so its strength is not directly given by a single % H₂SO₄ value. Free SO₃ reacts with water to form additional H₂SO₄:

SO₃ + H₂O → H₂SO₄

Therefore oleum is commonly labelled either:

• by its percentage of free SO₃ (e.g., “oleum 20%” means 20% free SO₃), or
• by its equivalent percentage of H₂SO₄ after hydration (calculated from % SO₃).

Key Conversion — SO₃ → H₂SO₄

When SO₃ reacts with water it forms H₂SO₄. The molar masses are:

• M(SO₃) = 80 g·mol⁻¹
• M(H₂SO₄) = 98 g·mol⁻¹

So 80 g SO₃ produces 98 g H₂SO₄. Use the ratio 98 / 80 in conversions.

Practical Conversion Formula

If oleum contains %SO₃ as free SO₃ (by mass), the equivalent extra H₂SO₄ formed after hydration is:

Extra H₂SO₄ (%) = %SO₃ × (98 / 80)

If the oleum already contains some H₂SO₄ (for a 100 g sample), total equivalent % H₂SO₄ can be computed by adding initial acid portion and the converted portion.

Solved Example 1 — 20% Oleum to Equivalent % H₂SO₄

1. Given: 20% SO₃ (i.e., 20 g SO₃ per 100 g oleum)
2. Extra H₂SO₄ formed = 20 × (98 / 80) = 20 × 1.225 = 24.5 g
3. Assume remaining part (100 − 20 = 80 g) is H₂SO₄ already present, so total H₂SO₄ = 80 + 24.5 = 104.5 g per 100 g oleum
4. Hence 20% oleum ≡ 104.5% H₂SO₄ (useful for exam numericals)

Solved Example 2 — Water Required to Convert x% Oleum to 98% Acid

Let oleum be labelled x% free SO₃. For 100 g oleum:

• Free SO₃ = x g
• SO₃ reacts with water to give H₂SO₄: x g SO₃ requires (x × 18 / 80) g water (because 80 g SO₃ + 18 g H₂O → 98 g H₂SO₄)
• After reaction, total H₂SO₄ mass = (100 − x) + x × (98/80) grams

Set required H₂SO₄ mass (for 98% strength) = 98 g per 100 g final solution and solve for x or required water as per the problem statement — this is a common type of JEE/NEET numerical.

Important Notes & Exam Tips

• Always use the ratio 98/80 to convert SO₃ → H₂SO₄.
• When a question gives oleum labelled as a % of SO₃, treat the rest of the mass as H₂SO₄ unless otherwise stated.
• Carefully track mass basis (100 g samples) to avoid mistakes in percents.
• Watch for traps: some problems ask for % strength after dilution — always calculate masses first, then percentages.

Industrial Relevance

Oleum is widely used in the sulphuric acid industry (contact process) and as a sulfonating/halogenating agent in organic synthesis. Labelling ensures correct stoichiometry in downstream processes and safe handling during transportation.

Related Resources

• Inorganic Chemistry – Notes & PYQs
• Practice Tests – JEE/NEET Chemistry
• Revision Notes – Class 11 & 12

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Conclusion: Labeling of oleum is simply a way of describing the amount of free SO₃ present and converting it to equivalent sulfuric acid uses the 98/80 ratio. Practise such conversions with 100 g basis calculations to master JEE/NEET numericals.

Happy studying — Chemca, Chemistry Made Easy.