Question

Network Covalent Atomic Solids The density of an unknown metal is \(12.3 \mathrm{g} / \mathrm{cm}^{3},\) and its atomic radius is 0.134 \(\mathrm{nm} .\) It has a face-centered cubic lattice. Find the atomic mass of this metal.

Short answer

The atomic mass of the metal is obtained by dividing the mass of a unit cell by the number of atoms within it.

Step-by-step solution

Calculate the Volume of the Unit Cell

For a face-centered cubic (fcc) lattice, the edge length 'a' of the unit cell can be found by multiplying the atomic radius by 4 and then dividing by the square root of 2, since the diagonal of the face of the cube which passes through the center of four atoms will be equal to 4 times the atomic radius. The formula for the length of the edge of the cube (a) is: \( a = \frac{4 \times r}{\sqrt{2}} \). After finding 'a', calculate the volume of the unit cell using the volume formula for a cube: \( V = a^3 \).

Determine the Number of Atoms per Unit Cell

In a face-centered cubic lattice, there are 4 atoms per unit cell. This is because each corner atom is shared with 8 adjacent cells, and there are 8 corners, each contributing 1/8 of an atom, and there are 6 face atoms, each of which is shared with an adjacent cell, each contributing 1/2 of an atom. Therefore, the total number of atoms per unit cell is \( 8 \times \frac{1}{8} + 6 \times \frac{1}{2} = 4 \).

Calculate the Mass of the Unit Cell

Using the given density (D) and the volume of the unit cell (V) found in Step 1, calculate the mass (m) of the unit cell using the density formula: \( m = D \times V \).

Find the Atomic Mass

Since there are 4 atoms per unit cell and you have the mass of the unit cell from Step 3, divide the mass of the unit cell by the number of atoms to find the atomic mass (M) of the metal. The formula is: \( M = \frac{m}{\text{number of atoms}} \).

Key concepts

Atomic Mass Calculation

The quest to find the atomic mass of an element is fundamental in understanding its properties, and it becomes intriguing when dealing with a face-centered cubic (fcc) lattice structure. Atomic mass calculation involves determining the mass of the individual atoms that compose a substance. In the context of an fcc lattice, the calculation is slightly nuanced due to the arrangement of atoms within the unit cell.

In the given exercise, the challenge presented is to deduce the metal's atomic mass from its density and the geometric configuration of its lattice. In an fcc lattice, each unit cell's unique arrangement contributes precisely four atoms to the structure's overall composition. The clever trick of this step-by-step solution lies in connecting the lattice's macroscopic density with the microscopic mass of the individual atoms.

• Understanding the coordination of atoms in an fcc lattice is key to this calculation.
• The accurate determination of the unit cell's mass from the overall density is a critical stage.
• With the knowledge of how many atoms are present in each unit cell, a simple division yields the atomic mass.

By following these logical leaps, students develop a deeper comprehension of the way atoms constitute matter and the interplay between microscale atomic masses and macroscale physical properties like density.

Unit Cell Volume

Calculating the volume of a unit cell serves as the bridge between the size of an atom and the larger crystalline structure it forms. In the context of a face-centered cubic lattice, the unit cell is a cube defined by its edge length 'a', which can be determined using the metal's atomic radius. The relationship between the atomic radius and the edge length is rooted in the geometry of the cube, where the face diagonal is equivalent to four atomic radii.

The formula to calculate the edge length from the atomic radius is pivotal and can be mathematically expressed as

Edge Length (a) = \( \frac{4 \times r}{\sqrt{2}} \)

Once the edge length is found, the volume

Volume (V) = \( a^3 \)

connects the micro (atomic radius) to the macro (unit cell volume) scale.

• Remembering that the volume of a unit cell is simply cube geometry can ease students’ understanding.
• Visualization tools or models can greatly aid in grasping how atomic radii correlate to the cube’s dimensions.
• Unit cell volume is a cornerstone in calculating other physical properties and transcends its basic geometry to provide insight into the structure of the crystal lattice.

By simplifying the concept and focusing on the relationships between dimensions, educators can ensure students internalize the geometric principles governing a crystal's architecture.

Density Formula Application

Density, a key physical property, relates an object's mass to its volume. The application of the density formula is particularly enlightening in the realm of crystal lattices, where it unlocks the mystery of atomic mass. For a face-centered cubic lattice, understanding how to apply the density formula to a unit cell's volume and to the mass of the metal contained therein enables the calculation of the elusive atomic mass.

In the step-by-step exercise, it’s shown how to calculate the mass of the unit cell using its volume and the given density by the formula

Mass (m) = Density (D) × Volume (V)

From this mass and the knowledge that there are four atoms per unit cell, the atomic mass is uncovered through a straightforward division.

• It’s crucial to comprehend how unit cell mass differs from atomic mass.
• Density is not just a concept, but a tool that bridges the microscopic and macroscopic worlds.

This application sheds light not just on a single property, but on the inherent beauty of the material world, the consistency of physical laws, and the interconnectedness of the various realms of physics and chemistry.