Oleum and its % labelling

Classifications and Types of Oleum

Oleum can be classified based on the percentage label it carries that describes the amount of free $\mathrm{SO}_3$ 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 $\mathrm{SO}_3$ 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 $\mathrm{SO}_3$ into H₂SO₄ for industrial use.

Oleum and its % Labeling

Oleum is a mixture of $\mathrm{SO}_3$ dissolved in $100 \% \mathrm{H}_2 \mathrm{SO}_4$. 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 $\mathrm{SO}_3$ present = 0.5 (from reaction stoichiometry)

Weight of $\mathrm{SO}_3$ in the 100g Oleum sample = $0.5 \times 80$= 40 g

% of free $\mathrm{SO}_3$ 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 $(\mathrm{y}-100) \mathrm{g}$ of water is added to 100 g Oleum sample

$
\begin{gathered}
\text { moles of water }=\frac{y-100}{18}=\text { moles of } \mathrm{SO}_3 \\
\text { of free } \mathrm{SO}_3 \text { in Oleum }=\frac{80 \times(y-100)}{18}
\end{gathered}
$

In this manner, the % of free $\mathrm{SO}_3$ in the Oleum sample can be calculated.

Industrial Applications and Relevance in Real Life

Oleum is crucial in modern industry, especially in the production of sulfuric acid—one of the most widely used chemicals globally. As a concentrated solution of sulfur trioxide in sulfuric acid, oleum enables precise control over acid strength by allowing titration of free $\mathrm{SO}_3$ content, or effectively supplying “over‑100%” sulfuric acid.

In chemical education, oleum offers precious classroom applications. Its role in acid–base titrations, stoichiometric calculations, and demonstrations of industrial chemistry principles provides a absorbing case study on the importance of precise measurement, concentration control, and understanding exothermic reactions in practical laboratory or industrial scenarios.

Also Read:

• Some Basic Concept In Chemistry Formula
• Laws Of Chemical Combination For Elements And Compounds

Some Solved Examples

Example 1

A sample of oleum is labeled as 104.5%. What is the percentage of free $\mathrm{SO}_3$ 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 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 $\mathrm{SO}_3$. 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) 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 free SO₃ in 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) 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 %

Practice More Questions From the Link Given Below:

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. Close control over acidity and stoichiometry ensures proper reaction conditions, safe handling, and consistency in industrial and laboratory processes.