Problem 95 - Chemistry: The Central Science

Question

Gold adopts a face-centered cubic structure with a unit cell edge of \(4.08 \AA\) (Figure 12.11). How many gold atoms are there in a sphere that is \(20 \mathrm{~nm}\) in diameter? Recall that the volume of a sphere is \(\frac{4}{3} \pi r^{3}\).

Step-by-Step Solution

Answer

First, we convert the diameter of the sphere (20 nm) to the radius and the unit cell edge length from Å to nm (4.08 Å = 0.408 nm). Next, we find the volume of the sphere and the volume of a unit cell in an FCC structure. Then, we calculate the number of unit cells in the sphere and the number of atoms per unit cell. In an FCC structure, there are four atoms per unit cell. Finally, to find the total number of gold atoms in the sphere, multiply the number of unit cells by the number of atoms per unit cell.

Step 1: Convert the diameter to radius and unit cell edge length from Å to nm

Given that the diameter of the sphere is 20 nm, we need to find the radius \(r\). The radius is half of the diameter, so: \[r = \frac{d}{2}\] We also need to convert the unit cell edge length from Å to nm, since 1 Å = 0.1 nm.

Step 2: Calculate the volume of the sphere

Now, we calculate the volume of the sphere using the formula: \[V_{sphere} = \frac{4}{3}\pi r^3\]

Step 3: Determine the volume of a unit cell

The volume of a unit cell in a face-centered cubic structure can be found using the formula: \[V_{unit~cell} = a^3\] Where "a" is the length of the unit cell edge.

Step 4: Calculate the number of unit cells in the sphere

Next, we find out how many unit cells are present in the sphere, by dividing the volume of the sphere by the volume of a unit cell: \[N_{unit~cells} = \frac{V_{sphere}}{V_{unit~cell}}\]

Step 5: Determine the number of atoms per unit cell in the FCC structure

In a face-centered cubic structure, there are four atoms per unit cell. This can be calculated as follows: - 1/8 of atoms for each of the 8 corner atoms: 8 * (1/8) = 1 atom - 1/2 of atoms for each of the 6 face atoms: 6 * (1/2) = 3 atoms Total number of atoms in an FCC unit cell = 1 + 3 = 4 atoms

Step 6: Calculate the total number of gold atoms in the sphere

Finally, we can calculate the total number of gold atoms in the sphere by multiplying the number of unit cells in the sphere by the number of atoms per unit cell: \[N_{gold~atoms} = N_{unit~cells} * 4\]

Key Concepts

Gold Atoms

Gold atoms are the basic building blocks of the precious metal known for its vibrant yellow color and excellent conductivity. These atoms exhibit unique properties that make gold one of the most sought-after materials for both decorative and industrial purposes. Gold has a face-centered cubic (FCC) atomic structure, which provides it with a high level of malleability and resistance to tarnishing.

In the context of this problem, knowing the number of gold atoms in a structure helps us determine the distribution of matter in the gold sphere. Calculating the total number of atoms within certain shapes, like a sphere, requires understanding the atomic arrangement and mass properties.

• The atomic mass of gold is approximately 197 grams per mole.
• Gold atoms pack in a crystalline structure that influences both the physical and chemical characteristics of gold.

A deeper understanding of atomic arrangements can lead to insights in fields such as nanotechnology and materials science.

Unit Cell

A unit cell is the smallest repeating structure that makes up the crystal lattice of a material. In the case of gold, this unit cell is cubic in shape, specifically a face-centered cubic structure. Understanding how these unit cells stack and interact is critical for understanding the macroscopic properties of the material.

The edge length of a unit cell is crucial for calculating its volume, allowing us to relate that to the overall structure. For gold, the edge length is given as 4.08 Ångströms, which is equivalent to 0.408 nm when converted. This conversion is essential for carrying out calculations in the same unit to ensure accuracy.

• Unit cells define the geometry and symmetry of the crystal lattice.
• Each unit cell contains several atoms, whose arrangement affects the material's properties.

Knowing the volume and number of unit cells helps in determining the density and atomic arrangements of metals like gold.

Volume of a Sphere

The volume of a sphere is a fundamental concept in geometry and is given by the formula: \[V_{sphere} = \frac{4}{3} \pi r^3\] where "\(r\)" is the radius of the sphere. The radius is half the diameter, so for a sphere with a 20 nm diameter, the radius would be 10 nm.

This calculation allows us to determine how much space a given volume occupies. It's especially useful when relating to atomic structures, as it helps us to understand spacing and density at an atomic level.

• The calculation of sphere volume forms the basis for many applications in physics, chemistry, and engineering.
• In this exercise, determining the volume allows us to find out how many unit cells fit inside the sphere.

Understanding the volume helps in visualizing how densely packed atoms are in a material, which is crucial for comprehending different properties.

Atomic Structure

Understanding atomic structure is key to grasping the properties of materials and how they interact. Atoms consist of a nucleus containing protons and neutrons, surrounded by electrons in orbitals. The way these atoms are arranged results in different materials having distinct characteristics.

In metals like gold, the atomic structure is organized in a repeating pattern forming a crystal lattice. This lattice, specifically face-centered cubic for gold, influences how atoms bond and how the material behaves under various conditions.

• Atomic arrangements dictate aspects such as conductivity, strength, and thermal properties.
• The FCC arrangement in gold leads to densely packed atoms, contributing to the metal's high density and malleability.

Understanding atomic structure is vital for numerous applications, ranging from materials development to electronic devices. This knowledge helps us innovate and improve many technologies that rely on specific material properties.