Problem 92 - General Chemistry

Question

Lead has a face-centered cubic lattice with all atoms at lattice points and a unit-cell edge length of \(495.0 \mathrm{pm}\). Its atomic mass is \(207.2\) amu. What is the density of lead?

Step-by-Step Solution

Answer

The density of lead is approximately \(11.34 \mathrm{g/cm^3}\).

Step 1: Understand the Structure

In a face-centered cubic (FCC) lattice, atoms are located at each corner and the centers of all the cube faces. There are eight corner atoms and six face atoms in total, but due to sharing, each unit cell of an FCC lattice effectively contains four atoms.

Step 2: Calculate the Volume of the Unit Cell

The unit-cell edge length of lead is given as \(495.0 \mathrm{pm}\). To find the volume, convert this length into centimeters: \(495.0 \mathrm{pm} = 495.0 \times 10^{-10} \mathrm{cm}\). The volume \(V\) of the cube is \((495.0 \times 10^{-10} \mathrm{cm})^3\).

Step 3: Calculate the Mass of the Atoms in the Unit Cell

The atomic mass of lead is given as \(207.2\) amu (where 1 amu = \(1.66053906660 \times 10^{-24} \mathrm{g}\)). The mass of four lead atoms in grams is \(4 \times 207.2 \times 1.66053906660 \times 10^{-24} \mathrm{g}\).

Step 4: Calculate the Density of Lead

Density \(\rho\) is calculated using the formula \(\rho = \frac{\text{mass of unit cell}}{\text{volume of unit cell}}\). Substitute the values of mass from Step 3 and the volume from Step 2 to find the density of lead.

Key Concepts

Face-Centered Cubic Lattice

In crystallography, the organization of atoms within a crystal structure is crucial. A face-centered cubic lattice is one of the most common arrangements. Each unit cell of this type consists of atoms positioned at each of the eight corners of a cube as well as in the centers of all six face sides. However, because the corner and face atoms are shared with adjacent cells, each unit cell effectively contains four atoms.

• Each corner atom is shared by eight adjacent cells, so it contributes only one-eighth of itself to the unit cell.
• Each face atom is shared by two adjacent cells, contributing half of itself to each unit cell.

This arrangement, making it a tightly packed lattice, allows for maximum space efficiency and contributes to the density of materials like metals.

Atomic Mass

Atomic mass is a fundamental property that reflects the mass of a single atom of a chemical element. It's usually expressed in atomic mass units (amu). For lead, this is given as 207.2 amu.To put this in perspective, 1 amu is defined as one twelfth of the mass of a carbon-12 atom and is numerically equivalent to approximately \(1.66053906660 \times 10^{-24}\) grams.

Knowing the atomic mass helps in calculating the mass of a set of atoms or a specific crystal structure. When dealing with unit cells in a lattice, the atomic mass allows us to ascertain how much the group of atoms in each unit cell weighs, crucially affecting calculations like density.

Unit-Cell Edge Length

The unit-cell edge length is a measurement of the dimensions of a single unit cell, which forms the basic repeating structure of a crystal lattice. For lead's face-centered cubic lattice, the unit-cell edge length is provided as 495.0 pm (picometers).To use this measurement in calculations, converting it to a more conventional unit of measure such as centimeters is necessary: \[495.0\ \text{pm} = 495.0 \times 10^{-10}\ \text{cm} \]The converted unit-cell edge length can then be used to determine the volume of the unit cell. Volume is calculated as the cube of the edge length, \(a^3\), highlighting the importance of this dimension in density calculations.

Mass and Volume Relationship

Understanding the mass and volume relationship is central to density calculations. Density (\(\rho\)) is defined as mass per unit volume and serves as an indicator of how much matter is packed within a given space.To calculate the density of a substance:

• First, ascertain the total mass of the atoms in the unit cell. This requires knowing the number of atoms per unit cell and their atomic mass.
• Next, calculate the volume of the unit cell using the unit-cell edge length.
• Once both mass and volume are determined, use the formula: \[\rho = \frac{\text{mass of unit cell}}{\text{volume of unit cell}}\] to find the density.

By combining these measurements, one can determine how densely packed the atoms are within the crystal structure, giving insights into the material's physical properties.