Problem 89 - General Chemistry

Question

Gold has cubic crystals whose unit cell has an edge length of \(407.9 \mathrm{pm}\). The density of the metal is \(19.3 \mathrm{~g} / \mathrm{cm}^{3}\). From these data and the atomic mass, calculate the number of gold atoms in a unit cell, assuming all atoms are at lattice points. What type of cubic lattice does gold have?

Step-by-Step Solution

Answer

Gold has 4 atoms per unit cell and a face-centered cubic (FCC) lattice.

Step 1: Convert edge length from picometers to centimeters

The edge length of the cubic unit cell is given as 407.9 picometers. We need to convert this to centimeters because the density is given in g/cm³. Recall that 1 cm = 10¹² pm. Therefore, \[ 407.9 \text{ pm} = 407.9 \times 10^{-10} \text{ cm}. \]

Step 2: Calculate the volume of the unit cell

The volume of a cubic unit cell can be found using the formula \( V = a^3 \), where \( a \) is the edge length. Therefore, the volume of the unit cell is: \[ V = (407.9 \times 10^{-10} \text{ cm})^3. \] Evaluating this gives: \[ V \approx 6.78 \times 10^{-23} \text{ cm}^3. \]

Step 3: Calculate mass of a unit cell

The mass of a unit cell can be found using the density formula \( \text{Density} = \frac{\text{Mass}}{\text{Volume}} \). Rearranging, we get \( \text{Mass} = \text{Density} \times \text{Volume} \). The density is given as 19.3 g/cm³. Thus, the mass of the unit cell is: \[ \text{Mass} = 19.3 \text{ g/cm}^3 \times 6.78 \times 10^{-23} \text{ cm}^3 \approx 1.31 \times 10^{-21} \text{ g}. \]

Step 4: Calculate number of atoms in a unit cell using atomic mass

The atomic mass of gold (Au) is 197 g/mol. To find the number of atoms in a unit cell, use the formula: \( \text{Number of atoms} = \frac{\text{Mass of unit cell}}{\text{Atomic mass per atom}} \times \text{Avogadro's number} \), where Avogadro's number is \( 6.022 \times 10^{23} \text{ atoms/mol} \). Therefore: \[ \frac{1.31 \times 10^{-21} \text{ g}}{197 \text{ g/mol}} \times 6.022 \times 10^{23} \approx 4. \]

Step 5: Determine the cubic lattice type

Given that the number of gold atoms in a unit cell is 4, the cubic lattice type is face-centered cubic (FCC), since FCC unit cells contain 4 atoms per unit cell when all lattice points are considered.

Key Concepts

Unit Cell Volume

Understanding the unit cell volume is crucial when dealing with crystalline structures like gold. In a cubic lattice, every unit cell has a distinct volume, which can be calculated using its edge length. The formula to determine this volume is quite straightforward:

\[ V = a^3 \] where \( V \) represents the volume and \( a \) is the edge of the cube. When you have the edge length, as given in the exercise (407.9 pm), convert it to a standard unit like centimeters for consistency, since some other metrics like density might not use picometers. Use the conversion:

• 1 cm = \( 10^{12} \) pm

This would turn 407.9 picometers into centimeters, expressed as:

\[ 407.9 \times 10^{-10} \text{ cm} \] Calculating the cube of this edge length will give you the unit cell volume in cubic centimeters. This value is paramount for subsequent calculations like determining density and mass.

Density Calculation

The density of a material tells us how much mass is contained in a given volume, which is particularly useful in determining the mass of a unit cell. The basic formula for density is:

\[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \] Given by the problem, the density of gold is \( 19.3 \text{ g/cm}^3 \). To find the mass of the unit cell, we need to rearrange this formula to:

\[ \text{Mass} = \text{Density} \times \text{Volume} \]
With the volume calculated from the unit cell volume previously, simply insert it into this equation alongside the provided density. This multiplication gives us the mass of the cubic unit cell. Having this mass is essential for calculating the number of atoms in the unit cell since it connects atom-based mass to macro-scale density.

Atomic Mass

Atomic mass is a fundamental property in chemistry, providing us with the mass of one mole of a given atom. For gold, the atomic mass is specified as 197 g/mol. When dealing with individual atoms, it's crucial to use Avogadro's number, which helps to transition from mass-based measurements to an atomic scale. Avogadro's constant is:

• 6.022 \( \times 10^{23} \) atoms/mol

To find the number of atoms in a unit cell, use this relation:
\[ \text{Number of atoms} = \frac{\text{Mass of unit cell}}{\text{Atomic mass per atom}} \times \text{Avogadro's number} \] This allows us to calculate how many atoms, in reality, make up the mass in a single unit cell. By performing the calculation with the atomic mass and the previously determined mass of the unit cell, you determine how gold's atomic structure fits within the confines of each cubic arrangement.

Face-Centered Cubic

The face-centered cubic (FCC) is one of the most common lattice structures found in metals. In this structure, atoms are located at each of the cube's corners and the centers of each face but not along the edges. This particular arrangement results in a higher packing efficiency, meaning it occupies a larger fraction of space than some other lattice structures.

• FCC structures contain 4 atoms per unit cell.
• This stacking contributes to its distinctive geometric pattern.

Determining the type of cubic lattice involves understanding how atoms are positioned within the unit cell volume. In this exercise, once you calculated that there are 4 gold atoms per unit cell by using the known atomic mass and density, and confirmed the design suits the method of an FCC structure. The FCC attribute is significant for explaining the properties of metals related to their ductility and densification.