Oleum and its % labelling

Oleum, otherwise known and referred to as fuming sulfuric acid, is a concentrated solution of sulfur trioxide, SO₃, in sulfuric acid, H₂SO₄. The chemical compound retains a very key position in various industrial processes because of its strongly acidic properties and application as a dehydrating agent. Oleum is formed upon dissolving SO₃ in concentrated sulfuric acid, which results in a viscous liquid. Depending on the method used in its preparation, it can contain a large amount of free SO₃. Actually, it is this free sulfur trioxide that really presents what makes oleum most interesting and useful in chemical synthesis.

JEE Main 2025: Chemistry Formula | Study Materials | High Scoring Topics | Preparation Guide

JEE Main 2025: Syllabus | Sample Papers | Mock Tests | PYQs | Study Plan 100 Days

NEET 2025: Syllabus | High Scoring Topics |  PYQs

This Story also Contains

1. Classifications and Types of Oleum
2. Oleum and its % labeling
3. Industrial Applications and Relevance in Real Life
4. Some Solved Examples
5. Summary

One of the most critical aspects used in labeling oleum is the percentage, related to the total mass of H₂SO₄ obtainable when a certain amount of oleum is diluted with water. For example, if an oleum sample was labeled as 109% H₂SO₄ then the addition of 9 grams of water to 100 grams of oleum is enough to result in 109 grams of pure H₂SO₄ This percentage labeling is very important to the chemist and industries that need specific concentrations due to various reactions and application purposes in such industries. The understanding of the implications of these percentages becomes quite important in ensuring safe handling of oleum and its effective use in industrial settings.

Classifications and Types of Oleum

Oleum can be classified based on the percentage label it carries that describes the amount of free SO₃ available in the solution. The percentage labeling is usually supra-100%, which describes the concentration of H₂SO₄ that a certain oleum will produce upon dilution. For instance, in 104%H₂SO₄ oleum, there will be 4 grams of free SO₃ in 100 grams of oleum while 118% oleum will contain 18 grams of free SO₃. The free SO₃ is important because it defines the reactivity and acidity of the oleum. The greater the percentage, the more water is required to convert free SO₃ into H₂SO₄ for industrial use.

Oleum and its % labeling

Oleum is a mixture of SO₃ dissolved in 100% H₂SO₄. The strength of the Oleum sample is expressed in terms of % labeling and it is defined as the grams of pure H₂SO₄ that can be obtained from 100 g of the Oleum sample upon dilution with water.

For example, if an Oleum sample is labeled as 109%, it means that upon the addition of 9g water to 100 g of Oleum sample, the amount of H₂SO₄ obtained is 109 g.

We can calculate the % of free SO₃ in the sample by simple stoichiometry as follows:

$\mathrm{SO}_3+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{H}_2 \mathrm{SO}_4$

Weight of H₂O added = 9 g

Moles of H₂O added = 0.5

Moles of SO₃ present = 0.5 (from reaction stoichiometry)

Weight of SO₃ in the 100g Oleum sample = 0.5 80 = 40 g

% of free SO₃ in Oleum = 40 %

Let us also discuss a general case

Let us suppose that the % labeling of the Oleum sample is y %. This means that (y-100) g of water is added to 100g Oleum sample

$\therefore$ moles of water $=\frac{\mathrm{y}-100}{18}=$ moles of $\mathrm{SO}_3$

$\therefore \%$ of free $\mathrm{SO}_3$ in Oleum $=\frac{80 \times(\mathrm{y}-100)}{18}$

In this manner, the % of free SO₃ in the Oleum sample can be calculated.

Industrial Applications and Relevance in Real Life

Oleum is in its greatest industrial use, especially in the manufacture of sulfuric acid, which represents one of the widest uses by any chemical agent throughout the world. It is used in the manufacture of fertilizers, explosives, and many chemical syntheses. Percentage labeling of oleum becomes very vital for the industry process, in which specific concentrations are called for. In producing detergents and dyes, specific amounts of oleum are required to get the kind of reaction desired.
Its properties make oleum find many applications in the teaching of some chemistry classes within the academic circle. These can be used to help students through acid-base reactions, stoichiometry, and some classes on industrial chemistry. Case studies of oleum could also be used to show the importance of precise measurements and concentration implications of chemicals for students.
Moreover, oleum cannot be dismissed when talking about environmental chemistry. It was implemented into those procedures that need tight control over the concentration of acids to neutralize the effect on the environment. Oleum is also a subject of constant research in terms of searching for methods of safer treatment and disposal since it is relevant both in industry and the environment.

Recommended topic video on(Oleum and its % labelling)

Some Solved Examples

Example 1

A sample of oleum is labeled as 104.5%. What is the percentage of free SO₃ in the sample?

1)20%

2)40%

3)60%

4)80%

Solution:

$\mathrm{SO}_3+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{H}_2 \mathrm{SO}_4$

Weight of H₂O added =4.5 mathrm{~g}

Moles of H₂O added = 0.25

therefore Moles of SO₃ present = 0.25

therefore Weight of SO₃ in the 100g Oleum sample = 0.25 times 80 = 20 g

therefore % of free SO₃ in Oleum = 20 %

Example 2

An Oleum sample contains 25% w/w of free SO₃. What is its % labelling?

1)105.625

2)107.25

3)114.525

4)122.5

Solution:
$\mathrm{SO}_3+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{H}_2 \mathrm{SO}_4$

% of free SO₃ in Oleum = 25 %

therefore Weight of SO₃ in the 100g Oleum sample = 25 g

therefore Moles of SO₃ present = 25/80 = 0.3125

therefore Moles of H₂O added = 0.3125 X 18

therefore Weight of H₂O added = 5.625 g

therefore % labelling of Oleum= (100+ 5.625) = 105.625 %

Example 3

The % of free SO3 in an Oleum sample is 20%. What is its labelling?

1) (correct)104.5

2)105.6

3)107.8

4)109.1

Solution: $\mathrm{SO}_3+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{H}_2 \mathrm{SO}_4$

% of tree SO₃ m Oleum =20 %

therefore Weight of SO₃ in the 100g Oleum sample = 20 g

therefore Moles of SO₃ present = 20/80 = 0.25

therefore Moles of H₂O added = 0.25 X 18

therefore Weight of H₂O added = 4.5 g

therefore % labelling of Oleum= (100+ 4.5) = 104.5 %

Example 4

What is the % of free SO₃ in oleum, that is labeled 118%?

1)20%

2)40%

3)60%

4) (correct)80%

Solution:
$\mathrm{SO}_3+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{H}_2 \mathrm{SO}_4$

Weight of H₂O added = 18 g

Moles of H₂O added = 1

therefore Moles of SO₃ present = 1

therefore Weight of SO₃in the 100g Oleum sample = 1 times 80 = 80 g

therefore % of free SO₃ in Oleum = 80 %

Summary

In the final sense, oleum is a solution of sulfur trioxide that is highly concentrated in sulfuric acid, characterized by its diverse unique properties and huge industrial applications. The percentage labeling of oleum is therefore important, as it represents the concentration of H₂SO₄ to be obtained after dilution; it is therefore very important for chemists and industries handling this substance. The percentage categorization allows for close monitoring of its reactivity, acidity, and precise conditions that would go into its proper working.