If 80 G Of $SO_3$ Is Added To Water To Form Oleum, And The Ratio Of Oxygen Atoms In $SO_3$ To The Number Of Oxygen Atoms In $H_2SO_4$ Is 1/4, What Is The Mass Of The Acid ($H_2SO_4$) In The Oleum?

Jun 21, 2025 by ADMIN 197 views

79.(2018) Determining the Mass of Acid in Oleum Formed by Adding SO3 to Water

This article delves into the intricate chemical problem presented in question 79 from the 2018 chemistry Olympiad, focusing on the reaction between sulfur trioxide ( $SO_3$ ) and water ( $H_2O$ ) to form oleum, a solution of sulfur trioxide in sulfuric acid ( $H_2SO_4$ ). This exploration will not only elucidate the step-by-step solution to the problem but also enhance the understanding of the underlying chemical principles, including stoichiometry, mole ratios, and the composition of oleum. We aim to provide a comprehensive guide that aids students and chemistry enthusiasts in grasping the concepts involved and applying them effectively in problem-solving scenarios.

Reworded Problem Statement

If the ratio of the number of oxygen atoms in $SO_3$ in oleum, obtained by adding 80 g of $SO_3$ to water, is 1/4 to the number of oxygen atoms in $H_2SO_4$ , determine the mass of the acid in the oleum.

A) 24.5 B) 29.4 C) 73.5 D) 49

Initial Analysis

The core of this problem lies in understanding the reaction between $SO_3$ and water to form sulfuric acid and subsequently, oleum. Oleum is essentially a solution of $SO_3$ in $H_2SO_4$ , and its composition is crucial for solving the problem. The given ratio of oxygen atoms in $SO_3$ to that in $H_2SO_4$ provides a critical link to determine the amounts of each compound present in the oleum. To solve this problem effectively, we will first calculate the moles of $SO_3$ added and then use the oxygen atom ratio to find the moles of $H_2SO_4$ formed. From there, calculating the mass of the acid in the oleum becomes straightforward.

Step-by-Step Solution

Calculate the moles of $SO_3$ added:

The molar mass of $SO_3$ is (32.07 + 3 * 16.00) = 80.07 g/mol. Given that 80 g of $SO_3$ is added, the number of moles of $SO_3$ is approximately:
$n(SO_3) = \frac{80 \text{ g}}{80.07 \text{ g/mol}} ≈ 1 \text{ mol}$
Understand the formation of Oleum:

When $SO_3$ is added to water, it reacts to form sulfuric acid:
$SO_3 + H_2O → H_2SO_4$
If excess $SO_3$ is present, it dissolves in the sulfuric acid to form oleum, represented as $H_2SO_4 \cdot xSO_3$ . In this case, we can consider the oleum as a mixture of $H_2SO_4$ and unreacted $SO_3$ .
Use the oxygen atom ratio:

The ratio of oxygen atoms in $SO_3$ to $H_2SO_4$ is given as 1/4. Let's denote the moles of $SO_3$ in the oleum as $n(SO_3)$ and the moles of $H_2SO_4$ as $n(H_2SO_4)$ . Each molecule of $SO_3$ contains 3 oxygen atoms, and each molecule of $H_2SO_4$ contains 4 oxygen atoms. Thus, the ratio of oxygen atoms can be expressed as:
$\frac{3 \cdot n(SO_3)}{4 \cdot n(H_2SO_4)} = \frac{1}{4}$
Simplifying this gives:
$12 \cdot n(SO_3) = 4 \cdot n(H_2SO_4)$ $n(H_2SO_4) = 3 \cdot n(SO_3)$
Determine the moles of $H_2SO_4$ formed:

Let's assume that 'x' moles of $SO_3$ reacted with water to form $H_2SO_4$ . So, x moles of $H_2SO_4$ are formed, and (1 - x) moles of $SO_3$ remain unreacted in the oleum. Thus:
$n(SO_3) = 1 - x$ $n(H_2SO_4) = x$
Using the relationship $n(H_2SO_4) = 3 \cdot n(SO_3)$ , we get:
$x = 3 \cdot (1 - x)$ $x = 3 - 3x$ $4x = 3$ $x = \frac{3}{4} = 0.75 \text{ mol}$
Therefore, 0.75 moles of $H_2SO_4$ are present in the oleum.
Calculate the mass of $H_2SO_4$ in the oleum:

The molar mass of $H_2SO_4$ is (2 * 1.01 + 32.07 + 4 * 16.00) = 98.09 g/mol. The mass of $H_2SO_4$ in the oleum is:
$m(H_2SO_4) = n(H_2SO_4) \cdot M(H_2SO_4) = 0.75 \text{ mol} \cdot 98.09 \text{ g/mol} ≈ 73.57 \text{ g}$
This is approximately 73.5 g.

Conclusion

By meticulously following the steps of stoichiometric calculations and applying the given oxygen atom ratio, we have successfully determined that the mass of sulfuric acid in the oleum is approximately 73.5 g. Therefore, the correct answer is:

C) 73.5

Key Takeaways

Stoichiometry is crucial: Understanding mole ratios and stoichiometric relationships is fundamental for solving chemical problems involving reactions.
Oleum Composition: Recognizing oleum as a mixture of $H_2SO_4$ and $SO_3$ is key to setting up the problem correctly.
Ratio Application: Utilizing given ratios effectively helps in linking the amounts of different substances involved in the reaction.

Additional Practice

To further solidify your understanding, consider solving similar problems involving oleum formation and composition. Experiment with different masses of $SO_3$ added to water and varying oxygen atom ratios. This will help you become more proficient in tackling complex chemistry problems.

Understanding chemical reactions and stoichiometry is not just about solving problems; it's about grasping the fundamental principles that govern the behavior of matter. By mastering these concepts, you'll be well-equipped to excel in chemistry and related fields.

Delving into the complexities of chemical reactions requires a strong foundation in stoichiometry and a keen understanding of the substances involved. This article will serve as a comprehensive guide to mastering oleum chemistry, with a focus on solving intricate problems related to its formation and composition. We will dissect the key concepts, provide step-by-step solutions, and offer valuable insights to enhance your problem-solving skills in chemistry. Our goal is to empower students and chemistry enthusiasts with the knowledge and confidence to tackle even the most challenging stoichiometry problems related to oleum.

Understanding Oleum and Its Formation

At the heart of oleum chemistry lies the understanding of what oleum is and how it is formed. Oleum, also known as fuming sulfuric acid, is a solution of sulfur trioxide ( $SO_3$ ) in sulfuric acid ( $H_2SO_4$ ). Its formation is a critical process in the industrial production of sulfuric acid. The reaction involves the absorption of $SO_3$ into $H_2SO_4$ , which is an exothermic process. The resulting solution can contain varying concentrations of $SO_3$ , making the composition of oleum a key factor in its chemical behavior.

The formation of oleum begins with the reaction of sulfur trioxide ( $SO_3$ ) with water ( $H_2O$ ) to produce sulfuric acid ( $H_2SO_4$ ). This is a highly exothermic reaction, releasing a significant amount of heat. The chemical equation for this reaction is:

SO_3(g) + H_2O(l) → H_2SO_4(l) \quad ΔH < 0

In industrial processes, $SO_3$ is often absorbed into concentrated sulfuric acid rather than directly reacting with water. This is because the direct reaction with water can be too vigorous and difficult to control. When $SO_3$ is absorbed into sulfuric acid, it forms oleum. Oleum is essentially a mixture of sulfuric acid and dissolved sulfur trioxide. The general formula for oleum can be represented as $H_2SO_4 \cdot xSO_3$ , where 'x' indicates the number of moles of $SO_3$ dissolved in one mole of $H_2SO_4$ .

The formation of oleum can be represented by the following equation:

H_2SO_4(l) + xSO_3(g) → H_2SO_4 \cdot xSO_3(l)

The value of 'x' determines the concentration of $SO_3$ in the oleum. For example, if x = 1, the oleum is $H_2SO_4 \cdot SO_3$ , which can also be written as $H_2S_2O_7$ , known as pyrosulfuric acid or disulfuric acid. Understanding the value of 'x' is crucial for solving problems related to oleum composition and reactions. The ability to calculate the value of 'x' based on given data, such as the mass of $SO_3$ added or the concentration of oxygen atoms, is a key skill in oleum chemistry.

Key Concepts in Oleum Chemistry

Stoichiometry: The quantitative relationship between reactants and products in a chemical reaction.
Mole Ratios: The ratio of the number of moles of different substances in a chemical reaction.
Molar Mass: The mass of one mole of a substance.
Composition of Oleum: Understanding oleum as a mixture of $H_2SO_4$ and $SO_3$ , and how the ratio of these components affects its properties.

Practical Applications of Oleum

Oleum is a crucial industrial chemical with a wide range of applications. Its primary use is in the production of sulfuric acid, which is one of the most widely used chemicals in the world. Sulfuric acid is used in the manufacture of fertilizers, detergents, synthetic resins, and many other chemical products. Oleum is also used in various industrial processes, including:

Sulfonation: A chemical process where a sulfonic acid group ( $−SO_3H$ ) is introduced into a molecule. Sulfonation is used in the production of detergents, dyes, and pharmaceuticals.
Dehydration Reactions: Oleum's strong affinity for water makes it an effective dehydrating agent in various chemical reactions.
Petroleum Refining: Oleum is used as a catalyst and a reagent in certain petroleum refining processes.

Example Problem and Solution

To illustrate the application of these concepts, let's consider a problem similar to the one presented in the introduction.

Problem: If 100 g of $SO_3$ is added to water, and the resulting solution is oleum, with the ratio of oxygen atoms in $SO_3$ to $H_2SO_4$ being 1/5, determine the mass of sulfuric acid in the oleum.

Solution:

Calculate moles of $SO_3$ added:
- Molar mass of $SO_3$ = 80.07 g/mol
- Moles of $SO_3$ = 100 g / 80.07 g/mol ≈ 1.25 mol
Set up the reaction:
- $SO_3 + H_2O → H_2SO_4$
- Assume 'x' moles of $SO_3$ react with water to form $H_2SO_4$ .
- Moles of $H_2SO_4$ formed = x
- Moles of unreacted $SO_3$ = 1.25 - x
Apply the oxygen atom ratio:
- Ratio of oxygen atoms in $SO_3$ to $H_2SO_4$ = 1/5
- (3 * (1.25 - x)) / (4 * x) = 1/5
- 15 * (1.25 - x) = 4x
- 18.75 - 15x = 4x
- 19x = 18.75
- x ≈ 0.99 mol
Calculate the mass of $H_2SO_4$ :
- Molar mass of $H_2SO_4$ = 98.09 g/mol
- Mass of $H_2SO_4$ = 0.99 mol * 98.09 g/mol ≈ 97.11 g

Therefore, the mass of sulfuric acid in the oleum is approximately 97.11 g.

Tips for Solving Oleum Chemistry Problems

Understand the Reaction: Clearly understand the reaction between $SO_3$ and water to form $H_2SO_4$ and the subsequent formation of oleum.
Stoichiometric Calculations: Master the art of stoichiometric calculations, including mole ratios, molar masses, and balancing chemical equations.
Apply Given Ratios: Utilize given ratios effectively to establish relationships between the amounts of different substances involved.
Break Down Complex Problems: Divide complex problems into smaller, manageable steps.
Practice Regularly: Consistent practice is key to mastering any subject, including oleum chemistry.

Advanced Concepts in Oleum Chemistry

As you delve deeper into oleum chemistry, you's encounter more advanced concepts that build upon the fundamentals. These concepts include:

Equilibrium Reactions: Understanding the equilibrium between $SO_3$ and $H_2SO_4$ in oleum solutions.
Thermodynamics of Oleum Formation: Analyzing the enthalpy, entropy, and Gibbs free energy changes associated with oleum formation.
Reaction Kinetics: Studying the rates of reactions involving oleum and the factors that influence them.
Industrial Processes: Exploring the industrial methods for producing oleum and sulfuric acid.

Mastering these advanced concepts will provide you with a comprehensive understanding of oleum chemistry and its applications in various fields.

Resources for Further Learning

To enhance your knowledge of oleum chemistry, consider exploring the following resources:

Textbooks: Consult chemistry textbooks that cover topics such as stoichiometry, chemical reactions, and industrial chemistry.
Online Courses: Enroll in online courses that provide in-depth coverage of oleum chemistry and related concepts.
Scientific Articles: Read research articles and publications that discuss the latest advancements in oleum chemistry.
Practice Problems: Solve a variety of practice problems to reinforce your understanding and problem-solving skills.

By leveraging these resources, you can expand your knowledge and expertise in oleum chemistry.

To truly master oleum chemistry, it is essential to move beyond rote memorization and develop a deep understanding of the underlying chemical principles. This involves not only solving numerical problems but also grasping the conceptual framework that governs the behavior of oleum. This article aims to provide a comprehensive guide to mastering oleum chemistry by focusing on both problem-solving techniques and conceptual clarity. We will explore the key concepts in detail, provide step-by-step solutions to various types of problems, and offer valuable insights to help you develop a strong foundation in this fascinating area of chemistry.

Conceptual Understanding of Oleum Chemistry

Before diving into problem-solving, it is crucial to have a solid conceptual understanding of oleum chemistry. This involves grasping the nature of oleum, its formation, and its chemical properties. Oleum as we've established, is a solution of sulfur trioxide ( $SO_3$ ) in sulfuric acid ( $H_2SO_4$ ). The concentration of $SO_3$ in oleum can vary, leading to different grades of oleum with varying properties. Understanding this variability is key to solving problems involving oleum composition and reactions.

Key Chemical Properties

Strong Acid: Oleum is a strong acid due to the presence of sulfuric acid and the acidic nature of dissolved $SO_3$ .
Dehydrating Agent: Oleum is a powerful dehydrating agent, capable of removing water from various substances.
Sulfonating Agent: Oleum is used as a sulfonating agent, introducing sulfonic acid groups ( $−SO_3H$ ) into organic molecules.
Exothermic Reactions: The reactions involving oleum, especially its formation and reactions with water, are highly exothermic.

Stoichiometry and Mole Ratios

The foundation of solving any quantitative problem in chemistry, including oleum chemistry, is stoichiometry. Stoichiometry deals with the quantitative relationships between reactants and products in chemical reactions. Mole ratios are derived from balanced chemical equations and provide the key to converting between the amounts of different substances involved in a reaction. To master oleum chemistry, it is essential to be proficient in stoichiometric calculations and the use of mole ratios. For example, in the reaction between $SO_3$ and water to form $H_2SO_4$ , the mole ratio between $SO_3$ and $H_2SO_4$ is 1:1, meaning that one mole of $SO_3$ reacts with water to produce one mole of $H_2SO_4$ .

Applying the Concepts to Problem-Solving

To effectively apply the concepts of oleum chemistry to problem-solving, it is essential to follow a systematic approach. This involves:

Understanding the Problem: Read the problem carefully and identify the given information and what needs to be determined.
Balancing Chemical Equations: Write and balance the relevant chemical equations to establish the mole ratios between the substances involved.
Converting to Moles: Convert the given masses or volumes to moles using the molar masses or molar volumes of the substances.
Applying Mole Ratios: Use the mole ratios from the balanced equations to determine the moles of the desired substances.
Converting Back to Desired Units: Convert the moles back to the desired units, such as mass or volume.
Checking the Answer: Verify the answer for reasonableness and units.

Common Problem Types in Oleum Chemistry

Composition Problems: Determining the composition of oleum, such as the mass percentage of $SO_3$ and $H_2SO_4$ .
Reaction Stoichiometry Problems: Calculating the amounts of reactants and products involved in reactions with oleum.
Titration Problems: Determining the concentration of oleum solutions using titration methods.
Acid-Base Reactions: Understanding the acid-base properties of oleum and its reactions with bases.

Strategies for Success in Oleum Chemistry

Mastering oleum chemistry requires a combination of conceptual understanding, problem-solving skills, and consistent practice. Here are some strategies to help you succeed:

Build a Strong Foundation: Ensure that you have a solid understanding of basic chemistry concepts, such as stoichiometry, chemical reactions, and acid-base chemistry.
Practice Regularly: Solve a variety of problems to reinforce your understanding and develop your problem-solving skills.
Seek Clarification: Don't hesitate to ask questions and seek clarification from teachers, professors, or online resources if you encounter difficulties.
Use Visual Aids: Use diagrams, flowcharts, and other visual aids to help you visualize the concepts and processes involved in oleum chemistry.
Work in Groups: Collaborate with classmates or study groups to discuss concepts and solve problems together.

Resources for Further Learning

To enhance your knowledge of oleum chemistry, consider exploring the following resources:

Chemistry Textbooks: Consult chemistry textbooks that cover topics such as stoichiometry, chemical reactions, and industrial chemistry.
Online Chemistry Courses: Explore online courses that provide in-depth coverage of oleum chemistry and related concepts.
Chemistry Websites: Utilize chemistry websites and online resources that offer tutorials, practice problems, and interactive simulations.
Scientific Journals: Read research articles and publications that discuss the latest advancements in oleum chemistry and its applications.

Mastering oleum chemistry is a rewarding endeavor that requires a combination of conceptual understanding and problem-solving skills. By understanding the key concepts, mastering stoichiometric calculations, and practicing regularly, you can develop a strong foundation in this fascinating area of chemistry. Remember, the key to success is not just memorizing formulas and equations but truly understanding the underlying principles and how they apply to real-world problems.