Question

Copper crystallizes in a face-centered cubic unit cell with an edge length of \(362 \mathrm{pm}\). What is the radius of a copper atom in picometers? What is the density of copper in \(\mathrm{g} / \mathrm{cm}^{3}\) ?

Short answer

The radius of a copper atom is approximately 128 pm and its density is 8.91 g/cm³.

Step-by-step solution

Understanding the FCC structure

In a face-centered cubic (FCC) unit cell, each corner atom is shared among eight unit cells and each face-centered atom is shared between two unit cells. There are 4 atoms per unit cell in an FCC structure.

Calculate the atomic radius

In an FCC structure, the diagonal across the face of the cube equals four times the atomic radius, \(4r\). The face diagonal for a cube of edge length \(a\) is \(\sqrt{2}a\). So, \(4r = \sqrt{2}a\). To find \(r\), rearrange this to get \(r = \frac{\sqrt{2}a}{4}\). Substitute \(a = 362 \ \mathrm{pm}\) to find \(r = \frac{\sqrt{2} \times 362}{4} \approx 128 \ \mathrm{pm}\).

Convert the edge length to cm

First, we need to convert the edge length from picometers to centimeters to calculate density. Since \(1 \ \mathrm{pm} = 10^{-12} \ \mathrm{m}\) and \(1 \ \mathrm{m} = 100 \ \mathrm{cm}\), \(362 \ \mathrm{pm} = 362 \times 10^{-10} \ \mathrm{cm}\).

Calculate the volume of the unit cell

The volume of the cubic unit cell \(V\) is given by \(V = a^3\). So, \(V = (362 \times 10^{-10})^3 \ \mathrm{cm}^3\). Evaluate \(V = 4.74 \times 10^{-23} \ \mathrm{cm}^3\).

Determine the molar mass of copper

The atomic mass of copper is approximately \(63.55 \ \mathrm{g/mol}\).

Calculate the density of copper

Density \(\rho = \frac{\text{mass of atoms in the unit cell}}{\text{volume of the unit cell}}\). The mass of 4 copper atoms is computed by the ratio \( \frac{4 \ \mathrm{atoms} \times 63.55 \ \mathrm{g/mol}}{6.022 \times 10^{23} \ \mathrm{atoms/mol}} = 4.22 \times 10^{-22} \ \mathrm{g} \). Finally, \(\rho = \frac{4.22 \times 10^{-22} \ \mathrm{g}}{4.74 \times 10^{-23} \ \mathrm{cm}^3} = 8.91 \ \mathrm{g/cm}^3\).

Key concepts

Face-Centered Cubic

In the fascinating world of crystal structures, one popular arrangement is the face-centered cubic (FCC) structure. This structure stands out because it is both simple and efficient.
But what does it really mean? Imagine a cube. In a traditional cubic structure, atoms are placed at each of the eight corners of the cube. Now, in a face-centered cubic structure, aside from these corner atoms, there is also an atom located at the center of each of the six faces of the cube.
It's important to note how atoms are shared in this structure:

• Corner atoms are shared by eight different cubes.
• Face-centered atoms are shared between two cubes.

Due to this arrangement, each FCC unit cell contains a total of 4 whole atoms. This high level of packing efficiency is why many metals, like copper, adopt the FCC structure.

Atomic Radius

The atomic radius in a crystal structure is an important measure, representing the distance from the center of the atom to its outermost shell of electrons.
In an FCC structure, the atomic radius can be determined using the cube's face diagonal. In mathematical terms, the face diagonal of a cube is equivalent to four times the atomic radius of the atom.
Using the formula, for a cube with an edge length of 'a', the relationship is:

• Face diagonal = \( \sqrt{2} \times a \)
• Total face diagonal = \( 4r \)

Thus, we can derive the radius \( r \) using the formula: \[ r = \frac{\sqrt{2} \times a}{4} \]For copper, with an edge length of 362 pm, the radius is calculated to be approximately 128 pm. This simple rearrangement of the formula allows us to understand the spatial layout of atoms within the cubic cell.

Density Calculation

Calculating the density of a crystal structure like copper involves understanding both the mass of the atoms in a unit cell and the volume of that cell. Density, in its most basic form, is mass divided by volume.
For an FCC structure:

• The mass involves understanding how many atoms are in our unit cell and their contribution to the total mass. For copper, there are 4 atoms per unit cell.
• The mass of the atoms can be obtained from their molar mass and Avogadro's number. This is done by using the formula: \[ \text{Mass} = \frac{\text{Number of atoms} \times \text{Molar mass}}{\text{Avogadro's number}} \]
• For the mass of 4 copper atoms, this equals approximately \( 4.22 \times 10^{-22} \ \mathrm{g} \).
• The unit cell volume is found by cubing the edge length (converted to centimeters): \( (362 \times 10^{-10})^3 \ \mathrm{cm}^3 \approx 4.74 \times 10^{-23} \ \mathrm{cm}^3 \).

Finally, the density is determined by dividing the total mass by the unit cell volume, leading to a density of \( 8.91 \ \mathrm{g/cm}^3 \). This method allows for accurately understanding how densely packed the atoms are within the structure.

Molar Mass of Copper

The molar mass of copper is a crucial value in chemistry, symbolizing the mass of one mole of copper atoms. In simpler terms, it's a measure that relates atomic mass to macroscopic quantities we can measure physically.
Copper has an atomic mass of approximately 63.55 grams per mole. This means that one mole of copper atoms, which is \( 6.022 \times 10^{23} \) atoms (Avogadro's number), will weigh 63.55 grams.

• This molar mass allows scientists to link molecular levels to physical amounts.
• It plays a significant role in various calculations, such as converting between atoms and grams or in density calculations for structures like FCC.

It's essential for understanding how much copper we are dealing with at the atomic and bulk levels, making it a fundamental concept in both chemistry and material science.