Question

An unknown metal is found to have a density of \(7.8748 \mathrm{~g} / \mathrm{cm}^{3}\) and to crystallize in a body-centered cubic lattice. The edge of the unit cell is \(0.28664 \mathrm{nm}\). Calculate the atomic mass of the metal.

Short answer

The atomic mass of the metal is calculated using the formula \( M = \frac{\text{density} \times a^3}{\text{number of atoms per unit cell}} \) where \( \text{density} \) is 7.8748 g/cm^3, \( a \) is 0.28664 nm (converted to cm), and there are 2 atoms per unit cell in a BCC lattice.

Step-by-step solution

Calculate the volume of the unit cell

The volume of the unit cell can be calculated using the formula for the volume of a cube, which is the edge length cubed. Here, we need to convert the edge length from nanometers to centimeters before cubing it: \( V = a^3 \), where \( V \) is the volume and \( a \) is the edge length. Since \( a = 0.28664 \times 10^{-7} \text{cm} \), the volume \( V \) is \( (0.28664 \times 10^{-7} \text{cm})^3 \).

Determine the number of atoms per unit cell

In a body-centered cubic (BCC) lattice, there is 1 atom at each corner of the cube and 1 atom in the center. The corner atoms are shared among 8 adjacent unit cells, so each corner contributes \( 1/8 \) of an atom to the unit cell. With 8 corners, this equates to 1 atom from the corners. The center atom belongs entirely to the unit cell. Thus, there are a total of \( 1 + 1 = 2 \) atoms per unit cell.

Calculate the mass of the unit cell

The mass of the unit cell can be determined by multiplying the density of the metal by the volume of the unit cell. The mass of the unit cell \( m_{\text{cell}} \) is given by \( m_{\text{cell}} = \text{density} \times V \), where \( \text{density} = 7.8748 \text{g/cm}^3 \) and \( V \) is the volume from Step 1.

Calculate the atomic mass of the metal

The atomic mass of the metal can be calculated by dividing the mass of the unit cell by the number of atoms per unit cell. This is given by the formula \( M = \frac{m_{\text{cell}}}{\text{number of atoms per unit cell}} \) where \( m_{\text{cell}} \) is the mass of the unit cell from Step 3 and the number of atoms per unit cell is 2 from Step 2.

Key concepts

Body-Centered Cubic Lattice

The body-centered cubic (BCC) lattice is one of the ways that atoms can be arranged in a solid. In this arrangement, atoms are located at each corner of a cube and one additional atom is located at the very center of the cube, creating a distinct pattern that repeats throughout the material. This structure is notably efficient in terms of space utilization within the lattice.

A critical aspect of the BCC lattice is how atoms are shared between adjacent unit cells. Each corner atom is shared by eight different unit cells, so only one eighth of each corner atom actually belongs to a single unit cell. When you add up all the contributions from the corner atoms and the single atom in the center, a BCC unit cell contains a total of 2 atoms. This is vital for calculating properties such as density and atomic mass, as seen in the given exercise.

Unit Cell Volume

Understanding unit cell volume is essential when dealing with crystalline structures. The unit cell is the smallest repeating unit that displays the full symmetry of a crystal structure, and its volume is simply the space enclosed by a unit cell.

To calculate the unit cell's volume, we typically use the formula for the volume of a cube (\(V = a^3\)), where 'a' represents the edge length of the unit cell. Since crystals are often measured at the atomic or nanoscale levels, edge lengths are usually given in nanometers or Angstroms, and converting them to centimeters is essential before performing volume calculations for purposes such as determining density, which is based on the unit of grams per cubic centimeter (\text{g/cm}^3).

Density Properties of Metals

Density is a fundamental property that relates the mass of a substance to its volume. For metals, which can form crystalline structures, such as BCC, density provides insights into their atomic mass and the compactness of their atomic arrangement. By knowing the density, one can derive important information, including the atomic mass, if the unit cell dimensions are known.

Calculating the density of a metal crystallizing in a specific lattice structure involves determining the mass of a unit cell and dividing it by the volume of the cell. The mass of the cell is influenced by the number of atoms in the unit cell and their individual atomic masses. In the proposed exercise, the number of atoms per unit cell and the volume of the unit cell, combined with the given density, are used to calculate the atomic mass of the metal. This calculation is fundamental in materials science and chemistry, providing an understanding of the metal's atomic structure and its implications on physical properties.