Percentage Labeling of Oleum — Formula & Solved Practice Problems
Oleum, commonly known as fuming sulfuric acid, is a mixture of pure sulfuric acid (H2SO4) and free sulfur trioxide (SO3). Its chemical formula is written as H2SO4 · x SO3 or H2S2O7.
In Physical Chemistry, particularly for exams like JEE Main, JEE Advanced, and NEET, the percentage labeling of oleum is a high-yield stoichiometric concept. Understanding how to calculate the strength of oleum and convert it to its equivalent H2SO4 percentage is essential for solving numericals.
What is >100% Labeling of Oleum?
Unlike standard solutions, oleum is labeled with percentages greater than 100% (e.g., 104.5%, 109%, 118%). But how can a percentage exceed 100?
This labeling represents the total mass of pure H2SO4 obtained when 100 grams of the oleum sample is diluted with sufficient water. The added water reacts with the free SO3 to form extra H2SO4.
SO3 + H2O → H2SO4
(80g SO3 reacts with 18g H2O to produce 98g H2SO4)
The Master Formula for Oleum Labeling
If an oleum sample is labeled as (100 + y)%, it means that y grams of H2O are required to combine with the free SO3 present in 100g of the oleum.
To find the mass percentage of free SO3 in the oleum, use this direct shortcut formula:
% of free SO3 = (80 / 18) × y
Where 'y' is the value above 100 in the labeling (e.g., for 109% oleum, y = 9).
Solved Example 1: Decoding 109% Oleum
Question: An oleum sample is labeled as 109%. Calculate the percentage of free SO3 in the sample.
Solution:
- Label = 109%, which means 100 + 9. So, y = 9g.
- This means 9g of H2O is required to react with all free SO3 in 100g of oleum.
- Using the stoichiometry (18g H2O reacts with 80g SO3):
- Mass of SO3 = (80 / 18) × 9 = 40 grams.
- Since this is out of 100g of oleum, the sample contains 40% free SO3 (and 60% H2SO4).
Practice Problems for JEE & NEET
Problem 1: Find the maximum possible percentage labeling of an oleum sample.
View Solution
The maximum labeling occurs when the sample is 100% pure SO3 (0% initial H2SO4).
For 100g of pure SO3, the mass of water required is:
(18 / 80) × 100 = 22.5 grams of H2O.
Therefore, Maximum Label = 100 + 22.5 = 122.5%.
Problem 2: What is the percentage of free SO3 in a 104.5% labeled oleum sample?
View Solution
Label = 104.5%. Thus, y = 4.5g (mass of water needed).
Mass of free SO3 = (80 / 18) × y
Mass of free SO3 = (80 / 18) × 4.5 = 20 grams.
Answer: The sample contains 20% free SO3.
Problem 3: 100g of 113.5% oleum is mixed with 100g of water. What will be the mass of pure H2SO4 formed, and will any water be left over?
View Solution
Label = 113.5%, meaning 100g of this oleum requires exactly 13.5g of water to convert all SO3 into H2SO4, giving a total of 113.5g of pure H2SO4.
We are adding 100g of water. The SO3 will consume 13.5g of it.
Mass of pure H2SO4 formed = 113.5g.
Water left over = 100g - 13.5g = 86.5g. (The final solution is aqueous sulfuric acid).
Multiple Choice Questions (MCQs)
Q1. What is the percentage of free SO3 in an oleum sample labeled as 118%?
- A) 18%
- B) 80%
- C) 20%
- D) 98%
View Answer & Solution
Correct Answer: B) 80%
Label = 118%, so y = 18g (mass of water required).
Using the formula: % free SO3 = (80 / 18) × y
% free SO3 = (80 / 18) × 18 = 80%.
Q2. Which of the following represents the theoretical maximum possible labeling of an oleum sample?
- A) 100%
- B) 109%
- C) 118%
- D) 122.5%
View Answer & Solution
Correct Answer: D) 122.5%
Maximum labeling occurs when the sample is 100% pure SO3. 100g of SO3 requires 22.5g of water to completely convert to H2SO4, making the label 100 + 22.5 = 122.5%.
Q3. If an oleum sample contains 20% free SO3 by mass, what is its percentage labeling?
- A) 104.5%
- B) 109%
- C) 120%
- D) 111.1%
View Answer & Solution
Correct Answer: A) 104.5%
Mass of free SO3 in 100g = 20g.
Water required (y) = (18 / 80) × 20 = 4.5g.
Labeling = 100 + y = 100 + 4.5 = 104.5%.
Q4. 100g of 109% oleum is diluted with water. What is the exact amount of water required to convert all the free SO3 into H2SO4?
- A) 9g
- B) 18g
- C) 80g
- D) 109g
View Answer & Solution
Correct Answer: A) 9g
By definition, a labeling of (100 + y)% means y grams of water are needed for 100g of the sample. Here, y = 9g.
Q5. An oleum sample is labeled as 113.5%. What is the mass of pure H2SO4 present initially in 100g of this sample (before any water is added)?
- A) 60g
- B) 13.5g
- C) 40g
- D) 86.5g
View Answer & Solution
Correct Answer: C) 40g
Label = 113.5%, so y = 13.5g of water needed.
Mass of free SO3 = (80 / 18) × 13.5 = 60g.
Since the total mass is 100g, initial H2SO4 = 100g - 60g = 40g.
Frequently Asked Questions (FAQs)
What does 109% oleum mean?
A labeling of 109% means that 100g of the oleum sample requires 9g of water to completely convert all the free SO3 into pure H2SO4, resulting in exactly 109g of pure sulfuric acid.
What is the maximum percentage labeling of oleum?
The maximum possible labeling of oleum is 122.5%. This theoretical state occurs when the sample is 100% pure SO3, requiring 22.5g of water per 100g of sample.
Why is oleum called fuming sulfuric acid?
It is called fuming sulfuric acid because the free, unreacted sulfur trioxide (SO3) escapes as a gas and reacts with moisture in the air to produce thick, dense white fumes of sulfuric acid.
Related Study Resources
- Inorganic Chemistry – Notes & PYQs
- Chapter-wise Practice Tests – JEE/NEET Chemistry
- Short Revision Notes – Class 11 & 12
Conclusion: Understanding the percentage labeling of oleum simplifies complex stoichiometric calculations. Remember the master formula: Label = 100 + (18/80) × %SO3, and practice analyzing the mass basis to avoid mistakes in exams.
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