Question

A cube of gold that is \(1.00 \mathrm{~cm}\) on a side has a mass of 19.3 g. A single gold atom has a mass of \(197.0 \mathrm{u} .(\mathbf{a})\) How many gold atoms are in the cube? (b) From the information given, estimate the diameter in \(\AA\) of a single gold atom. (c) What assumptions did you make in arriving at your answer for part (b)?

Short answer

Approximately 5.90 × 10^22 gold atoms are in the cube. The estimated diameter of a single gold atom is 2.88 Å, assuming spherical packing.

Step-by-step solution

Calculate Volume of the Gold Cube

The problem states the gold cube's side is 1.00 cm. Calculate its volume: \( V = \text{side}^3 = (1.00 \text{ cm})^3 = 1.00 \text{ cm}^3 \).

Find Number of Moles of Gold in the Cube

Given the mass of the gold cube is 19.3 g and the molar mass of gold is approximately 197 g/mol, calculate the number of moles: \[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{19.3 \text{ g}}{197 \text{ g/mol}} \approx 0.098 \text{ mol}. \]

Calculate Number of Gold Atoms in the Cube

Using Avogadro's number, \( 6.022 \times 10^{23} \) atoms/mol, calculate the number of atoms: \[ \text{Number of atoms} = 0.098 \text{ mol} \times 6.022 \times 10^{23} \text{ atoms/mol} \approx 5.90 \times 10^{22} \text{ atoms}. \]

Estimate the Volume of a Single Gold Atom

Assume each gold atom occupies an equal volume. Divide the cube's total volume by the number of atoms to find the volume per atom: \[ \text{Volume per atom} = \frac{1.00 \text{ cm}^3}{5.90 \times 10^{22}} \text{ atoms} \approx 1.69 \times 10^{-23} \text{ cm}^3. \]

Calculate Estimated Diameter of a Gold Atom

Assume the shape of each atom is spherical. Use the formula for the volume of a sphere, \( V = \frac{4}{3}\pi r^3 \), solve for the radius \( r \), and find the diameter: \[ r = \left( \frac{3V}{4\pi} \right)^{1/3} = \left( \frac{3 \times 1.69 \times 10^{-23}}{4\pi} \right)^{1/3} \approx 1.44 \times 10^{-8} \text{ cm}, \] \[ \text{Diameter} = 2r \approx 2.88 \times 10^{-8} \text{ cm} = 2.88 \text{ Å}. \]

State Assumptions

In estimating the diameter of a gold atom, we assumed that the atoms are perfectly spherical and packed densely in a simple cubic arrangement, where each atom occupies an equal volume within the cube.

Key concepts

Avogadro's Number

Avogadro's number, denoted as \(6.022 \times 10^{23}\), is a constant that represents the number of atoms, ions, or molecules contained in one mole of a substance. It is named after Amedeo Avogadro, who first hypothesized that equal volumes of gases, at the same temperature and pressure, contain an equal number of particles.
Understanding Avogadro's number is essential in converting moles to atoms. For example, if you have 0.098 moles of gold, you can find the number of gold atoms by multiplying the moles by Avogadro's number:

• \(0.098 \text{ mol} \times 6.022 \times 10^{23} \text{ atoms/mol} \approx 5.90 \times 10^{22} \text{ atoms}\).

This calculation tells you the total number of individual atoms present in the sample. It's an essential step in quantifying microscopic entities like atoms in macroscopic quantities of material.

Molar Mass

Molar mass is the mass of one mole of a given element or compound, expressed in grams per mole (g/mol). It provides a bridge between the macroscopic scale, where we weigh substances, and the atomic scale, where we count atoms or molecules.
For gold, the molar mass is approximately 197 g/mol. This value tells us that 197 grams of gold contain exactly one mole of gold atoms, or \(6.022 \times 10^{23}\) atoms.
In the exercise, you calculated the number of moles of gold from the given mass (19.3 g) using the formula:

• \( \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{19.3 \text{ g}}{197 \text{ g/mol}} \approx 0.098 \text{ mol} \).

By understanding molar mass, you can efficiently convert between mass and moles, providing a key to interpreting measurements and making further calculations on the number of atoms possible.

Volume Estimation

Volume estimation is crucial when determining the size or space that an object, such as an atom or molecule, occupies. In this exercise, you were asked to estimate the volume occupied by a single gold atom within a 1 cm³ gold cube.
To find this, you first calculate the total volume of the cube and then divide it by the number of atoms. With \(5.90 \times 10^{22}\) atoms within a 1.00 cm³ cube, the volume per atom is:

• \( \text{Volume per atom} = \frac{1.00 \text{ cm}^3}{5.90 \times 10^{22}} \approx 1.69 \times 10^{-23} \text{ cm}^3 \).

You can use this volume estimation as a basis to further deduce atomic dimensions, assuming spherical symmetry of the atoms, via the volume formula for a sphere: \( V = \frac{4}{3}\pi r^3 \). Solving for the radius and then the diameter: \( \text{Diameter} \approx 2.88 \times 10^{-8} \text{ cm} \), or in Angstroms, \( 2.88 \text{ Å} \).
This estimation provides insight into the spatial aspects of the atomic structure, connecting measurable quantities such as mass and volume to atomic properties.