Question

Withoutdoing any detailed calculations (but using a periodic table to give atomic weights), rank the following samples in order of increasing number of atoms: \(3.0 \times 10^{23}\) molecules of \(\mathrm{H}_{2} \mathrm{O}_{2}, 2.0 \mathrm{~mol} \mathrm{CH}_{4}, 32 \mathrm{~g} \mathrm{O}_{2}\)

Short answer

The ranking in order of increasing number of atoms for the given samples is: \(1.2 \times 10^{24}\) atoms (\(\mathrm{H}_{2} \mathrm{O}_{2}\)), \(1.2 \times 10^{24}\) atoms (\(\mathrm{O}_{2}\)), and \(7.22 \times 10^{24}\) atoms (\(\mathrm{CH}_{4}\)).

Step-by-step solution

Find Atomic Weights

Using the periodic table, the atomic weights for each element involved in the samples are: - Hydrogen: H ≈ 1 g/mol - Carbon: C ≈ 12 g/mol - Oxygen: O ≈ 16 g/mol

Calculate the Number of Atoms in Each Sample

: Sample 1: \(3.0 \times 10^{23}\) molecules of \(\mathrm{H}_{2} \mathrm{O}_{2}\) Each molecule of \(\mathrm{H}_{2} \mathrm{O}_{2}\) has 2 atoms of hydrogen and 2 atoms of oxygen, meaning 4 atoms per molecule. The total number of atoms is: Atoms ≈ \(3.0 \times 10^{23}\) molecules × 4 atoms/molecule = \(1.2 \times 10^{24}\) atoms Sample 2: 2.0 mol of \(\mathrm{CH}_{4}\) Each molecule of \(\mathrm{CH}_{4}\) has 1 atom of carbon and 4 atoms of hydrogen, meaning 5 atoms per molecule. Since 1 mole = \(6.022 \times 10^{23}\) molecules: Atoms ≈ 2.0 mol × \(6.022 \times 10^{23}\) molecules/mol × 5 atoms/molecule ≈ \(12.0 \times 6.022 \times 10^{23}\) atoms = \(7.22 \times 10^{24}\) atoms Sample 3: 32 g of \(\mathrm{O}_{2}\) Each molecule of \(\mathrm{O}_{2}\) has 2 atoms of oxygen, and the molar mass of \(\mathrm{O}_{2}\) is 32 g/mol. Therefore: Moles of O₂ = \(\frac{32 \: g}{32 \: g/mol} \approx 1 \: mol\) Atoms ≈ 1 mol × \(6.022 \times 10^{23}\) molecules/mol × 2 atoms/molecule = \(1.2 \times 10^{24}\) atoms

Rank the Samples

Now that we have the approximate number of atoms in each sample, we can rank them in order of increasing number of atoms: 1. Sample 1: \(1.2 \times 10^{24}\) atoms (\(\mathrm{H}_{2} \mathrm{O}_{2}\)) 2. Sample 3: \(1.2 \times 10^{24}\) atoms (\(\mathrm{O}_{2}\)) 3. Sample 2: \(7.22 \times 10^{24}\) atoms (\(\mathrm{CH}_{4}\))

Key concepts

Avogadro's Number

Understanding Avogadro's Number is key to grasping many concepts in chemistry, particularly when dealing with atoms and molecules at the microscopic level. Avogadro's Number, denoted as \(6.022 \times 10^{23}\), is the number of atoms, ions, or molecules in one mole of substance. It's named after Amedeo Avogadro, who first proposed that the volume of a gas (at a given pressure and temperature) is proportional to the number of atoms or molecules, regardless of the gas's nature.

One practical application of Avogadro's Number is converting between the number of moles and the number of particles. This conversion is especially useful when ranking substances in order of increasing number of atoms. For example, in the given exercise, Sample 2 involves \(2.0\) moles of \(\mathrm{CH}_{4}\), which translates to \(2.0 \times 6.022 \times 10^{23}\) molecules. Knowing the number of molecules in one mole allows you to then calculate the total number of atoms in the sample, taking into account the molecular composition of methane.

Molar Mass

Significance of Molar Mass in Calculations

Molar mass provides a bridge between the atomic scale and the macroscopic scale. It's defined as the mass of one mole of a given substance and is measured in grams per mole (g/mol). The molar mass is numerically equal to the atomic (or molecular) weight of a substance but takes on different units for practical measurement purposes.

In the exercise, the molar mass of oxygen (\(\mathrm{O}_{2}\)) is critical for converting grams to moles. With a molar mass of 32 g/mol, we determine that 32 grams of \(\mathrm{O}_{2}\) are equivalent to one mole. Knowing that one mole contains Avogadro's number of molecules allows us to calculate that 32 grams of \(\mathrm{O}_{2}\) contains approximately \(1.2 \times 10^{24}\) atoms. It's important to remember that molar mass is specific to each element or compound, thus having a periodic table at hand is essential for such calculations.

Stoichiometry

Applying Stoichiometry to Determine Composition

The concept of stoichiometry is central in chemistry for understanding the quantitative relationships between the amounts of reactants and products in a chemical reaction. Stoichiometry is based on the conservation of mass and the law of definite proportions, meaning the ratio of elements in a chemical compound is constant.

In the exercise, stoichiometry is applied not to a chemical reaction but to determine the number of atoms present in different molecules. For example, by knowing the stoichiometry of \(\mathrm{H}_{2} \mathrm{O}_{2}\), with two atoms of hydrogen and two of oxygen per molecule, you can determine the total number of atoms from the given number of molecules. Similarly, understanding that \(\mathrm{CH}_{4}\) has a fixed ratio of one carbon atom to four hydrogen atoms per molecule is crucial to calculating the total count of atoms within the 2.0 moles provided. This stoichiometric perspective is essential for solving problems in both theoretical and applied chemistry.