Question

The coinage metals-copper, silver, and gold - crystallize in a cubic closest packed structure. Use the density of copper \(\left(8.95 \mathrm{~g} / \mathrm{cm}^{3}\right)\) and its molar mass \((63.55 \mathrm{~g} / \mathrm{mol})\) to calculate an approximate atomic radius for copper.

Short answer

Atomic radius for copper can be approximated using \( r = \frac{\sqrt{2}a}{4} \).

Step-by-step solution

Determine the number of atoms in a unit cell

A cubic closest packed (ccp) structure, also known as face-centered cubic (fcc), contains 4 atoms per unit cell.

Calculate the volume of the unit cell

Using the density formula, \ \[ \text{density} = \frac{\text{mass}}{\text{volume}} \], we rearrange to find the volume of the unit cell: \ \[ \text{volume} = \frac{\text{mass}}{\text{density}} \]. The mass of one unit cell can be found by \( 4 \times \text{molar mass of Cu} / \text{Avogadro's number} \).

Find the lattice parameter

The volume of the unit cell is equal to \( a^3 \) where \(a\) is the lattice parameter (edge length) of the fcc structure. Solve for \( a \) by taking the cube root of the unit cell volume.

Calculate the atomic radius

For a cubic closest packed structure, the face diagonal is \( 4r \) where \( r \) is the atomic radius. The face diagonal also equals \( \sqrt{2}a \) for a face-centered cubic structure. Set these two expressions equal to each other and solve for the atomic radius: \ \[ 4r = \sqrt{2}a \Rightarrow r = \frac{\sqrt{2}a}{4} \]

Key concepts

cubic closest packed structure

A cubic closest packed (ccp) structure, also known as a face-centered cubic (fcc) lattice, is one of the most efficient ways to pack equal spheres. This structure maximizes space utilization and minimizes gaps between atoms.
Each unit cell in this structure contains four atoms.
To visualize it, imagine a cube with one atom at each corner and one atom in the center of each face. This arrangement allows for atoms to be tightly packed together.
Another key feature of the ccp structure is its repeating layers. Atoms in one layer fit perfectly into the gaps of the layers above and below, creating a highly efficient packing method.
This concept is critical for understanding many properties of metals, such as their density and atomic radius.

density formula

Density is a fundamental property of materials defined by the formula:
\[ \text{density} = \frac{\text{mass}}{\text{volume}} \]
In problems involving atomic structures, we often need to rearrange this equation to find the volume:
\[ \text{volume} = \frac{\text{mass}}{\text{density}} \]
To calculate the volume of a unit cell, we must first determine the mass. For a ccp structure, the mass of one unit cell can be found by multiplying the number of atoms per unit cell (4 for fcc) by the molar mass of the element (e.g., Copper), and then dividing it by Avogadro's number.

• Molar mass of Cu (Copper): 63.55 g/mol
• Avogadro's number: 6.022 x 10^23 mol^-1

Thus, we get the mass of one unit cell:
\[ \text{mass of unit cell} = \frac{4 \times 63.55 \, \text{g/mol}}{6.022 \times 10^{23} \, \text{mol}^{-1}} \]
Next, using the density (8.95 g/cm^3 for Copper), we can find the volume. This volume corresponds to the cube of the unit cell's edge length (lattice parameter), denoted as \(a\).

face-centered cubic structure

The face-centered cubic (fcc) structure is synonymous with the cubic closest packed (ccp) arrangement. The distinguishing feature of this structure is the placement of atoms.
An fcc unit cell has atoms at the eight corners of the cube and one atom in the center of each of the six faces, making 4 atoms per unit cell in total.
One important geometric aspect of the fcc structure is its relationship with the atomic radius \(r\). The face diagonal of the cube is equal to 4 times the atomic radius since it spans across four atomic radii from one corner to the opposite corner of a face.
The face diagonal can also be expressed as the square root of 2 times the lattice parameter \(a\):
\[ \text{Face diagonal} = \sqrt{2}a = 4r \]
Reformatting this equation allows us to solve for the atomic radius:
\[ r = \frac{\sqrt{2}a}{4} \]
This formula is pivotal in calculating the atomic radius once the lattice parameter is known, which can be derived from the volume and properties of the unit cell.