Master Oleum Labeling & % Free SO3
Learn the exact meaning of Oleum percentage labeling, stoichiometric calculations, and how to instantly find the % free SO3 with Abhishek Sengar.
Video Chapters
Quick Theory & Cheat Sheet
What is Oleum?
Oleum (H2S2O7) is essentially a mixture of Sulfuric Acid (H2SO4) and free Sulfur Trioxide (SO3). It is formed during the Contact Process by dissolving SO3 into H2SO4.
The Core Reaction
SO3 + H2O → H2SO4
- 1 mole of SO3 (80g) reacts with 1 mole of H2O (18g).
- Mass ratio: 80g of SO3 requires 18g of H2O.
Decoding the Label (e.g., 104.5%)
A label of 104.5% means that if you take 100g of this Oleum sample and add water, you will get exactly 104.5g of pure H2SO4.
The extra 4.5g is the mass of the Water (H2O) added, which reacted entirely with the free SO3 present in the 100g sample.
Calculation Formula
Mass of H2O added = (Label % - 100)g
Mass of free SO3 = Mass of H2O × (80 / 18)
% Free SO3 = (Mass of free SO3 / 100g) × 100
Video Challenge Quiz
Test what you learned from the end of the video!
What will be the % Free SO3 in a sample of Oleum labeled as 113.5%?
