Percentage Labelling of Oleum
Decode the SO₃ content of fuming sulphuric acid and calculate its H₂SO₄ yield
AI Generation Prompt
Ultra-wide cinematic banner (16:5 ratio). A dark industrial scene at night — large metallic storage tanks of oleum with faint orange safety lighting, thin white acidic fumes curling upward from valve fittings. In the foreground, a sealed glass ampoule holding a deep amber liquid with fine white mist rising off its surface. Conveys chemical potency and industrial scale. Photorealistic, dark cinematic atmosphere, orange accent lighting. No text.
Walk into a Contact Process sulphuric acid plant and the air itself stings. That sting is oleum — sulphuric acid so concentrated that it holds dissolved sulphur trioxide, which grabs moisture from the air instantly to form a white acidic mist.
Oleum is not a lab curiosity. It is the working intermediate in the industrial manufacture of almost all the world's , and it is potent enough that its containers must stay sealed at all times. Its other name — fuming sulphuric acid — is completely literal.
What oleum is. Oleum is a solution of free sulphur trioxide () dissolved in 100% sulphuric acid (). It is loosely written as (pyrosulphuric acid), but the real composition depends on how much was dissolved — it is not a fixed compound.
Oleum is made in the Contact Process by absorbing excess gas into concentrated . Dissolving straight into water is avoided — that reaction is so violently exothermic that it shatters glassware and throws up a dangerous acid mist. Oleum is the controlled route to ultra-concentrated .
The Dilution Chemistry
What happens when oleum is diluted. Add water to oleum and the free reacts with it to make more :
Treating the whole oleum as one unit:
Dilution is continued until every bit of free has converted to . Notice the mass bookkeeping: 80 g of takes in 18 g of to give 98 g of — heavier than the you started with. The total mass of the solution grows during dilution, and that growth is exactly what the % labelling number records.
AI Generation Prompt
Infographic-style diagram on a dark background. Left: a labelled flask "100 g Oleum (118%)" split into two coloured bands — 80 g SO₃ (orange, upper) and 20 g H₂SO₄ (blue, lower). A horizontal arrow labelled "+ 18 g H₂O" points right. Right: a labelled flask "118 g H₂SO₄" as a single unified blue band. Beneath the arrow, the equation SO₃ + H₂O → H₂SO₄ with masses 80 g + 18 g → 98 g. Dark background (#0a0a0a), orange and blue accent labels, clean technical illustration style.
What % Labelling Actually Means
The definition. The % labelling of oleum = the total grams of you obtain when exactly 100 g of the oleum is diluted with just enough water to convert all the free .
118% oleum — worked through. 100 g of it needs 18 g of for complete dilution. Moles of = 1, so moles of that reacted = 1 — meaning 80 g of free was present. The other 20 g was already . After dilution: 20 g (original) + 98 g (from ) = 118 g ✓
Patterns worth locking in:
- % labelling is always greater than 100 for oleum (you add water, so the mass overshoots 100 g)
- Pure 100% → exactly 100% labelling
- Water needed for full dilution (g)
- Free in a 100 g sample g
Composition of x% Oleum (per 100 g sample)
Both follow from SO₃ + H₂O → H₂SO₄ stoichiometry. Check for 118%: m(SO₃) = 40×18/9 = 80 g, m(H₂O) = 18 g.
Calculate the composition (mass of free and mass of ) of 109% labelled oleum.
0.5 g of fuming (oleum) is dissolved in water and neutralised completely by 26.7 mL of 0.4 N NaOH. Find the percentage of free in the sample.
Q1.A bottle is labelled "112% H₂SO₄". What does this mean?
AI Generation Prompt
Ultra-wide cinematic banner (16:5 ratio). A dark industrial scene at night — large metallic storage tanks of oleum with faint orange safety lighting, thin white acidic fumes curling upward from valve fittings. In the foreground, a sealed glass ampoule holding a deep amber liquid with fine white mist rising off its surface. Conveys chemical potency and industrial scale. Photorealistic, dark cinematic atmosphere, orange accent lighting. No text.
Walk into a Contact Process sulphuric acid plant and the air itself stings. That sting is oleum — sulphuric acid so concentrated that it holds dissolved sulphur trioxide, which grabs moisture from the air instantly to form a white acidic mist.
Oleum is not a lab curiosity. It is the working intermediate in the industrial manufacture of almost all the world's , and it is potent enough that its containers must stay sealed at all times. Its other name — fuming sulphuric acid — is completely literal.
What oleum is. Oleum is a solution of free sulphur trioxide () dissolved in 100% sulphuric acid (). It is loosely written as (pyrosulphuric acid), but the real composition depends on how much was dissolved — it is not a fixed compound.
Oleum is made in the Contact Process by absorbing excess gas into concentrated . Dissolving straight into water is avoided — that reaction is so violently exothermic that it shatters glassware and throws up a dangerous acid mist. Oleum is the controlled route to ultra-concentrated .
The Dilution Chemistry
What happens when oleum is diluted. Add water to oleum and the free reacts with it to make more :
Treating the whole oleum as one unit:
Dilution is continued until every bit of free has converted to . Notice the mass bookkeeping: 80 g of takes in 18 g of to give 98 g of — heavier than the you started with. The total mass of the solution grows during dilution, and that growth is exactly what the % labelling number records.
AI Generation Prompt
Infographic-style diagram on a dark background. Left: a labelled flask "100 g Oleum (118%)" split into two coloured bands — 80 g SO₃ (orange, upper) and 20 g H₂SO₄ (blue, lower). A horizontal arrow labelled "+ 18 g H₂O" points right. Right: a labelled flask "118 g H₂SO₄" as a single unified blue band. Beneath the arrow, the equation SO₃ + H₂O → H₂SO₄ with masses 80 g + 18 g → 98 g. Dark background (#0a0a0a), orange and blue accent labels, clean technical illustration style.
What % Labelling Actually Means
The definition. The % labelling of oleum = the total grams of you obtain when exactly 100 g of the oleum is diluted with just enough water to convert all the free .
118% oleum — worked through. 100 g of it needs 18 g of for complete dilution. Moles of = 1, so moles of that reacted = 1 — meaning 80 g of free was present. The other 20 g was already . After dilution: 20 g (original) + 98 g (from ) = 118 g ✓
Patterns worth locking in:
- % labelling is always greater than 100 for oleum (you add water, so the mass overshoots 100 g)
- Pure 100% → exactly 100% labelling
- Water needed for full dilution (g)
- Free in a 100 g sample g
Composition of x% Oleum (per 100 g sample)
Both follow from SO₃ + H₂O → H₂SO₄ stoichiometry. Check for 118%: m(SO₃) = 40×18/9 = 80 g, m(H₂O) = 18 g.
Calculate the composition (mass of free and mass of ) of 109% labelled oleum.
0.5 g of fuming (oleum) is dissolved in water and neutralised completely by 26.7 mL of 0.4 N NaOH. Find the percentage of free in the sample.
Q1.A bottle is labelled "112% H₂SO₄". What does this mean?