Problem 7 - A Complete Resource Book in Chemistry for JEE Main

Question

The number of oxygen atoms in \(4.4 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) is (a) \(6 \times 10^{23}\) (b) \(11 \times 10^{23}\) (c) \(12 \times 10^{23}\) (d) \(1.2 \times 10^{23}\)

Step-by-Step Solution

Answer

(d) \(1.2 \times 10^{23}\)

Step 1: Calculate Molar Mass of \(\mathrm{CO}_2\)

The molar mass of \(\mathrm{CO}_2\) can be calculated by adding the atomic masses of one carbon atom and two oxygen atoms. The atomic mass of carbon (C) is approximately \(12 \text{ g/mol}\) and that of oxygen (O) is \(16 \text{ g/mol}\). Thus, the molar mass of \(\mathrm{CO}_2\) is \(12 + 2(16) = 44 \text{ g/mol}\).

Step 2: Calculate Moles of \(\mathrm{CO}_2\) in \(4.4 \text{ g}\)

The number of moles of a substance is calculated by dividing its mass by its molar mass. For \(\mathrm{CO}_2\), we have \(\frac{4.4 \text{ g}}{44 \text{ g/mol}} = 0.1 \text{ moles}\).

Step 3: Determine Number of \(\mathrm{CO}_2\) Molecules

The number of molecules in a mole is given by Avogadro's number, which is \(6.022 \times 10^{23}\). Thus, \(0.1\) moles of \(\mathrm{CO}_2\) contains \(0.1 \times 6.022 \times 10^{23} = 6.022 \times 10^{22}\) molecules.

Step 4: Calculate Number of Oxygen Atoms

Each \(\mathrm{CO}_2\) molecule contains 2 oxygen atoms. Therefore, the total number of oxygen atoms in \(6.022 \times 10^{22}\) molecules of \(\mathrm{CO}_2\) is \(2 \times 6.022 \times 10^{22} = 1.2044 \times 10^{23}\) atoms.

Key Concepts

Avogadro's Number

Avogadro's Number is a fundamental concept in chemistry that helps us bridge the microscopic world of atoms and molecules with the macroscopic world we interact with. It is simply the number of entities, usually atoms or molecules, that are contained in one mole of any substance. Avogadro's Number is represented as \(6.022 \times 10^{23}\).

This large number helps chemists work with manageable quantities of substances in the laboratory by relating the mass of a sample to the number of particles it contains. When we say we have one mole of a substance, it means we've gathered enough material to count \(6.022 \times 10^{23}\) of its molecules. Understanding Avogadro's Number allows us to determine how many moles are present in a given mass and to perform conversions between mass, moles, and molecules.

• Connects macroscopic and microscopic calculations
• Essential in solving stoichiometric problems
• Facilitates conversions in chemical reactions

Molar Mass Calculation

Calculating the molar mass of a compound is a crucial step in solving many chemistry problems. The molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol).

To calculate it, one must add up the atomic masses of all the elements in a compound. For instance, in carbon dioxide \(\mathrm{CO}_2\), the molar mass is determined by adding the atomic masses of one carbon atom (\(12 \text{ g/mol}\)) and two oxygen atoms (each \(16 \text{ g/mol}\)), resulting in \(44 \text{ g/mol}\).

Knowing the molar mass is essential when converting the mass of a sample to moles, which is often needed to understand how chemicals will react and in what proportions.

• Key to determining the amount of a substance
• Involves summing atomic masses
• Converts mass to moles for reactions

Stoichiometry

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. This concept allows us to predict how much of a reagent we need or how much product we can obtain from a given amount of reactant.

It relies on balancing chemical equations and using the coefficients to understand the proportions of substances involved. By knowing the mole amounts, achieved through the use of Avogadro's Number and Molar Mass Calculations, we can accurately calculate how substances will combine chemically.

• Enables calculation of reactant and product amounts
• Requires balancing equations for precise proportions
• Uses moles to predict reaction outcomes

Stoichiometry ensures that resources are used efficiently, and it plays a vital role in industries and research, allowing chemists to scale up reactions from the lab bench to large-scale production.