Question

Which of the following is the odd one with regard to mass? (a) \(1 \mathrm{~g}\) atom of sulphur (b) \(0.5\) moles of \(\mathrm{CO}_{2}\) (c) 1 mole of \(\mathrm{O}_{2}\) (d) \(3 \times 10^{23}\) molecules of \(\mathrm{SO}_{2}\)

Short answer

The odd one out is (b): 0.5 moles of CO2.

Step-by-step solution

Calculate the mass of each option

First, let's calculate the mass of each of the given options:- **Option (a):** The mass of 1 g atom of sulfur. Sulfur has an atomic mass of 32 grams/mole, so 1 g atom of sulfur is 32 g.- **Option (b):** The mass of 0.5 moles of \( \text{CO}_2 \). The molar mass of \( \text{CO}_2 \) is \( 44 \text{ grams/mole} \). Therefore, 0.5 moles is \( 0.5 \times 44 = 22 \text{ grams} \).- **Option (c):** The mass of 1 mole of \( \text{O}_2 \). The molar mass of \( \text{O}_2 \) is \( 2 \times 16 = 32 \text{ grams/mole} \).- **Option (d):** The number of molecules in 1 mole is approximately \( 6.022 \times 10^{23} \). So, \( 3 \times 10^{23} \) molecules represents \( 0.5 \) moles. The molar mass of \( \text{SO}_2 \) is \( 32 + 2 \times 16 = 64 \text{ grams/mole} \). Hence, \( 0.5 \times 64 = 32 \text{ grams} \).

Identify the odd one out

Now we need to identify the option with a different mass. - Option (a) has a mass of 32 g. - Option (b) has a mass of 22 g. - Option (c) has a mass of 32 g. - Option (d) also results in 32 g. Clearly, option (b), with 22 g, differs from the others which all have a mass of 32 g.

Key concepts

Moles and Mole Concept

Understanding the mole and mole concept is key in chemistry. A mole is a fundamental unit in chemistry that represents a specific quantity of substance. A mole of any substance contains Avogadro's number of atoms, molecules, or ions, which is approximately \(6.022 \times 10^{23}\). This incredibly large number allows chemists to count and group small particles like atoms and molecules in measurable amounts.

For example, when we say 1 mole of sulfur, it means there are \(6.022 \times 10^{23}\) sulfur atoms. Similarly, 1 mole of \(\text{CO}_2\) means there are \(6.022 \times 10^{23}\) carbon dioxide molecules. This helps convert between the number of particles and the amount of substance in grams, making it easier to work with in laboratory settings.

In the given exercise, understanding how many moles or atoms correspond to each option helps in determining their mass by linking the microscopic world of molecules to a tangible scale.

Molar Mass Calculation

Molar mass is a crucial concept when dealing with chemical calculations and equations. It refers to the mass of one mole of a substance, allowing you to easily convert between grams and moles.

To calculate the molar mass, you sum up the atomic masses of all elements in a compound. For example:

• For oxygen \(\text{O}_2\), each oxygen atom has an atomic mass of 16. Therefore, \(\text{O}_2\) would have a molar mass of \(2 \times 16 = 32\) grams/mole.
• In \(\text{CO}_2\), carbon's atomic mass is 12, and oxygen is 16. Hence, \(\text{CO}_2\) is calculated as \(12 + 2 \times 16 = 44\) grams/mole.

In practice, mastering molar mass calculations can help determine how much of a substance will react or be produced in a chemical reaction. In the original exercise, the molar masses are used to find the mass of each option, enabling you to compare them directly.

Chemical Equations

Chemical equations are vital for describing chemical reactions. They show the reactants and products of a reaction using their chemical formulas and provide a visual representation of how substances interact.

A balanced chemical equation has equal numbers of atoms for each element on both the reactant and product sides. This is crucial due to the law of conservation of mass which states that matter cannot be created or destroyed in a chemical reaction.

When solving the exercise, the proficiency in reading chemical symbols and understanding the basics of chemical equations allow us to determine the molar masses of compounds like \(\text{CO}_2\), \(\text{O}_2\), and \(\text{SO}_2\). It emphasizes the importance of proper mole and mass calculations to match the stoichiometry of the reaction, ensuring that all products and reactants are accounted for in terms of their mass and mole ratio.