Question

Argon crystallizes in the face-centered cubic arrangement at \(40 \mathrm{~K}\). Given that the atomic radius of argon is \(191 \mathrm{pm},\) calculate the density of solid argon.

Short answer

The density of solid argon is \(1.69 \text{ g/cm}^3\).

Step-by-step solution

Understand the Structure

Argon crystallizes in a face-centered cubic (FCC) lattice, meaning there are atoms at each corner and one atom at the center of each face of the cube.

Calculate the Edge Length of the Unit Cell

In an FCC lattice, the edge length \(a\) is related to the atomic radius \(r\) by the formula \(a = 2\sqrt{2}r\). Substituting \(r = 191 \text{ pm}\), we calculate \(a = 2\sqrt{2} \times 191 \text{ pm} = 539.37 \text{ pm}\).

Find the Volume of the Unit Cell

Convert the edge length from picometers to centimeters: \(a = 539.37 \text{ pm} = 539.37 \times 10^{-10} \text{ cm}\). The volume \(V\) of the unit cell is then \(a^3 = (539.37 \times 10^{-10})^3 \text{ cm}^3 = 1.568 \times 10^{-22} \text{ cm}^3\).

Determine the Number of Atoms per Unit Cell

In an FCC lattice, there are 4 atoms per unit cell (1/8 of each corner atom shared by 8 cells plus 1/2 of each face atom shared by 2 cells).

Calculate the Mass of Atoms in a Unit Cell

The molar mass of argon is approximately 39.95 g/mol. Therefore, the mass of one atom of argon is \(\frac{39.95}{6.022 \times 10^{23}} \text{ g} = 6.63 \times 10^{-23} \text{ g/atom}\). For 4 atoms per unit cell, the total mass is \(4 \times 6.63 \times 10^{-23} \text{ g} = 2.65 \times 10^{-22} \text{ g}\).

Calculate the Density

The density \(\rho\) is given by the mass divided by the volume of the unit cell: \(\rho = \frac{2.65 \times 10^{-22} \text{ g}}{1.568 \times 10^{-22} \text{ cm}^3} = 1.69 \text{ g/cm}^3\).

Key concepts

Face-Centered Cubic Lattice

In crystallography, a face-centered cubic (FCC) lattice is one of the most common types of crystal structures. An FCC lattice is characterized by its unique arrangement of atoms. Here, atoms are positioned at every corner of the cube as well as at the center of each face. This arrangement ensures that each atom is surrounded in the most efficient way possible by neighboring atoms, maximizing the structure's density and stability.
In solid argon, this kind of lattice arrangement means that for each cubic unit cell, there are a total of four atoms. To understand this, consider that each of the eight corner atoms is shared among eight unit cells, and each of the six face-centered atoms is shared between two unit cells.

• Corner atoms contribute 1/8th of each atom per unit cell.
• Face-centered atoms contribute 1/2 of each atom per unit cell.

This arrangement is particularly important for understanding how atoms pack together and how we calculate related properties such as density.

Atomic Radius

The atomic radius is a measure of the size of an atom, crucial in determining how atoms pack in a crystal structure. For argon, the atomic radius is given as 191 picometers (pm). This radius represents the distance from the center of an atom's nucleus to the outer boundary of its electron cloud.
Knowing the atomic radius is key in calculating dimensions and properties of the unit cell in crystal structures like the FCC. For instance, the length of the edges of the unit cell (denoted as 'a') is directly related to the atomic radius in an FCC lattice. The formula used is: \[ a = 2\sqrt{2}r \]where \(r\) is the atomic radius. This relationship allows us to calculate the edge length when we know the atomic radius, facilitating further calculations for volume and density.

Unit Cell Volume

The unit cell is the smallest repeating unit in a crystal lattice, defining its structure. For an FCC lattice like that of solid argon, understanding the unit cell's geometry allows us to calculate its volume. Knowing the edge length 'a', which we calculate using the atomic radius, we find the volume of the unit cell with the equation: \[ V = a^3 \]To convert picometers (pm) to centimeters (cm), we use the equivalence of \(1\text{ pm} = 10^{-10}\text{ cm}\). Thus, the unit cell edge length in centimeters is \(539.37\times 10^{-10}\text{ cm}\).
Consequently, the resulting volume is: \[ V = (539.37 \times 10^{-10})^3\text{ cm}^3 = 1.568 \times 10^{-22}\text{ cm}^3 \]By calculating this volume, we gain insight into how much space the atoms in the lattice effectively occupy, which is crucial for determining other properties like density.

Molar Mass of Argon

The molar mass is the mass of one mole of a substance, a key concept in chemistry. For argon, this value is approximately 39.95 grams per mole. Understanding molar mass allows us to relate macro-scale measurements of substances to the atomic scale. In solid argon, we can use the molar mass to determine the mass of individual atoms.
To find the mass of a single argon atom, we divide the molar mass by Avogadro's number, \(6.022 \times 10^{23} \text{ atoms/mol}\). This gives:\[ \text{Mass of one argon atom} = \frac{39.95}{6.022 \times 10^{23}} \text{ g} = 6.63 \times 10^{-23} \text{ g/atom} \]This calculation is vital for predicting how much space a group of such atoms will occupy and hence determining the density of the crystal. With four atoms in an FCC unit cell, totaling this mass gives the total mass per unit cell, allowing us to calculate the overall density of the solid.