Question

Argon has a cubic closest packed structure as a solid. Assuming that argon has a radius of 190. pm, calculate the density of solid argon

Short answer

The density of solid argon is approximately 0.1361 g/cm³.

Step-by-step solution

Identifying the cubic closest packed structure

A cubic closest packed (ccp) structure is also called a face-centered cubic (fcc) structure. In an fcc structure, there are 8 corner atoms and 6 face atoms totally contributing to just one unit cell.

Calculate the total number of atoms per unit cell

For the fcc structure, we have: - 8 corner atoms with 1/8 of each inside the unit cell: \(8\cdot \frac{1}{8} = 1\) - 6 face atoms with 1/2 of each inside the unit cell: \(6\cdot \frac{1}{2} = 3\) Therefore, the total number of atoms per unit cell is: \(1+3 = 4\)

Calculate the lattice constant

In an fcc structure, the lattice constant 'a' can be found by relating it to the atomic radius 'r'. The relationship is given by the equation: \(a = 4r / \sqrt{2}\) We are given the radius of argon, r = 190 pm. Using this, we can calculate the lattice constant (a): \(a = \frac{4 \cdot 190\, \text{pm}}{\sqrt{2}} \approx 269.53\, \text{pm}\)

Calculate the volume of the unit cell

The volume of a cubic unit cell can be calculated using the formula: \(V = a^3\) Now, we will calculate the volume using the lattice constant (a) obtained in step 3: \(V = (269.53\, \text{pm})^3 \approx 1.955\times 10^7\, \text{pm}^3\)

Calculate the molar mass of argon

To proceed to the calculation of density, we need the molar mass of argon which is already known. The molar mass of argon (Ar) = 39.948 g/mol

Calculate the density of solid argon

To calculate the density, we can use the formula: \(\rho = \frac{\text{mass}}{\text{volume}}\) The mass can be calculated using: Mass = (number of atoms per unit cell) x (molar mass of argon) / Avogadro's number The volume will be the volume of the unit cell calculated in step 4. First we find the mass for one unit cell, by dividing the molar mass by Avogadro's number and then multiplying by the number of argon atoms in one unit cell: Mass per unit cell = \(\frac{4\times 39.948\,\text{g/mol}}{6.022\times10^{23}\, \text{ atoms/mol}} \approx 2.660\times10^{-22}\,\text{g}\) Now, we calculate the density by dividing the mass per unit cell by the volume of the unit cell: \(\rho = \frac{2.660\times10^{-22}\,\text{g}}{1.955\times 10^7\, \text{pm}^3} = 1.361\times10^{-29}\, \text{g/pm}^3\) Finally, converting the density from g/pm³ to g/cm³ (1 pm = 1 x 10⁻¹² cm), we get: \(\rho \approx 1.361\times10^{-1}\, \text{g/cm}^3\) Thus, the density of solid argon is approximately 0.1361 g/cm³.

Key concepts

Cubic Closest Packed Structure

The cubic closest packed (ccp) structure is a highly efficient way of arranging atoms in a crystal. It's synonymous with the face-centered cubic (fcc) structure, where atoms are densely packed to occupy the least space while maintaining the highest coordination. This structure is known for its optimal usage of space, closely stacking layers of atoms. In a ccp arrangement, each atom is equally surrounded by 12 neighboring atoms, delivering a coordination number of 12. This configuration is prevalent in metals and some elements like argon, particularly in their solid state. Understanding the ccp structure is pivotal for calculating and predicting the properties of crystals, especially when determining densities.

Face-Centered Cubic

The face-centered cubic (fcc) structure is a variation of the cubic lattice system. It's characterized by atoms positioned at each cube's face center and corners. In an fcc crystal, you'll find 8 atoms at the corners shared among 8 unit cells, and 6 atoms at the faces. Each face atom is shared between two unit cells, equating to a total of 4 full atoms per unit cell. This efficient packing results in a filling factor of 74%, making it one of the most tightly packed structures. Understanding fcc helps in comprehending the structural formation of various elements, which is essential when performing density calculations for these elements.

Atomic Radius

The atomic radius is a key component when evaluating the density of elements like solid argon, which assumes a cubic closest packed structure. Defined as the average distance from the nucleus of the atom to its outermost electron shell, the atomic radius provides insight into the size of an individual atom. In solid argon, this is particularly simple to measure given its inert, singular element nature. For this calculation, the given atomic radius is 190 pm (picometers), which is crucial for determining the lattice constant – a step integral to uncovering the unit cell's characteristics and ultimately calculating the element's density.

Lattice Constant

To understand a crystalline structure, you need to calculate the lattice constant, an essential parameter indicating the edge length of the cube. In the case of an fcc configuration, the lattice constant 'a' is related to the atomic radius 'r' through the expression: \[a = \frac{4r}{\sqrt{2}}\]This equation derives from the geometric placement of atoms within the cube. Knowing the atomic radius of argon is 190 pm, the lattice constant can be calculated to be approximately 269.53 pm. This measurement is pivotal in determining the actual volume of the unit cell, as it's the length of one side of the cube.

Molar Mass

The molar mass is a fundamental property for density calculations in chemistry, as it represents the mass of one mole of a substance. For argon, the molar mass is 39.948 grams per mole, which is a constant recognized in scientific tables. This value is essential for converting from atomic mass units to a more tangible scale, facilitating the calculation of the element's density. By knowing the molar mass, you can determine the total mass of the atoms within the unit cell, which is subsequently used in the density formula: \[\rho = \frac{\text{mass}}{\text{volume}}\]This relationship allows scientists and students alike to find the precise density of the solid state of argon.

Unit Cell Volume

The volume of the unit cell, a cube in the face-centered cubic system, is vital for determining densities. To find this, you use the formula for the volume of a cube,\(V = a^3\), where 'a' is the lattice constant defined in the earlier steps. Using our lattice constant for argon at 269.53 pm, the volume of the unit cell can be calculated as approximately \(1.955 \times 10^7 \, \text{pm}^3\). The accurate volume of the unit cell allows you to understand how much space the argon occupies, serving as the foundation for further calculations in determining solid argon's comprehensive properties.

Avogadro's Number

Avogadro's number is an essential constant in chemistry, defined as \(6.022 \times 10^{23}\) atoms or molecules per mole. This value links the atomic scale to the macroscopic scale, providing a bridge from atoms to grams. When calculating density, Avogadro's number allows you to determine the mass of a single atom in the unit cell and relate it to its macroscopic properties. By dividing the molar mass by Avogadro's number and multiplying by the number of atoms per unit cell, you obtain the mass of the unit cell. This step is key in reaching an accurate calculation of an element's density, such as solid argon.