Problem 58 - Chemistry An Atoms First Approach

Question

Barium has a body-centered cubic structure. If the atomic radius of barium is \(222 \mathrm{pm},\) calculate the density of solid barium.

Step-by-Step Solution

Answer

The density of solid Barium with a body-centered cubic structure and an atomic radius of 222 pm is approximately 3.62 g/cm³.

Step 1: Determine the unit cell edge length for a body-centered cubic structure

In a body-centered cubic (BCC) structure, the relationship between the atomic radius (r) and the unit cell edge length (a) is given by the formula: \(a = 4r / \sqrt{3}\) We are given that r (the atomic radius of Barium) is 222 pm.

Step 2: Calculate the unit cell edge length

Plug in the value of r into the formula to find the unit cell edge length: \(a = \dfrac{4(222 \,\text{pm})}{\sqrt{3}} \approx 512.66 \,\text{pm}\)

Step 3: Find the number of atoms in a body-centered cubic unit cell

In a BCC structure, there is one atom at each corner of the cube and one atom at the center of the cube: - 8 corner atoms × (1/8) contribution per atom = 1 atom - 1 center atom × 1 contribution per atom = 1 atom Total atoms in the unit cell: 1 (corner atoms) + 1 (center atom) = 2 atoms

Step 4: Find the volume of the unit cell

Now that we have the unit cell edge length, we can find the volume (V) of the unit cell by using the formula for the volume of a cube: \(V = a^3 ≈ (512.66 \,\text{pm})^3 ≈ 1.35 × 10^8 \,\text{pm}^3\) Note that we'll need to convert the volume to \(cm^3\) in the next step, so let's do that now: \(V ≈ 1.35 × 10^8 \,\text{pm}^3 \cdot \left(\dfrac{1 \,\text{cm}}{10^7 \,\text{pm}}\right)^3 ≈ 1.35 × 10^{-23} \,\text{cm}^3\)

Step 5: Calculate the density

To find the density (ρ), we need the atomic weight of Barium and Avogadro's number. The atomic weight of Barium is 137.33 g/mol. Avogadro's number (N_A) is \(6.022 \times 10^{23} \,atoms/mol\). We'll use the following formula: \(ρ = \dfrac{2\,atoms \cdot 137.33 \,\text{g/mol}}{6.022 \times 10^{23} \,atoms/mol \cdot 1.35 \times 10^{-23} \,\text{cm}^3}\)

Step 6: Calculate the density of solid Barium

Plug the values into the formula and calculate the density: \(ρ ≈ \dfrac{2 \cdot 137.33 \,\text{g/mol}}{6.022 \times 10^{23}\,\text{atoms/mol} \cdot 1.35 \times 10^{-23} \,\text{cm}^3} ≈ 3.62 \,\text{g/cm}^3\) Therefore, the density of solid Barium is approximately 3.62 g/cm³.

Key Concepts

Body-Centered Cubic Structure

The body-centered cubic (BCC) structure is a common crystalline structure found in metals like barium. In this arrangement, atoms are positioned at the eight corners of a cube and a single atom is located at the center. This setup results in each atom at the corners sharing its position with eight other neighboring unit cells, contributing a fraction of itself to each.

• The central atom is unique to that unit cell.
• Total number of atoms contained in a BCC unit cell is two: one from the corners and one in the center.

This orientation maximizes the packing efficiency and stability of the structure, though not as tightly packed as some other structures like face-centered cubic. Understanding this layout is crucial for calculations involving density, as it directly affects the formula used for assessing the volume and mass of material in the unit cell.

Unit Cell Edge Length

In BCC structures, the cell's edge length, denoted by 'a', is not straightforwardly the same as the atomic diameter. Instead, it involves a geometric relationship with the atomic radius 'r'.
In a BCC unit cell, the atoms along the cube diagonal touch each other in sequence, and the edge length is tied directly to this diagonal arrangement:

• Formula: \[a = \frac{4r}{\sqrt{3}}\]
• This formula arises because the diagonal distance across the cube links four radii of the central and corner atoms.

By inserting the known atomic radius into this relation, you can determine 'a', which is crucial for further calculations, such as unit cell volume and indirectly, the material's density.

Atomic Weight

Atomic weight, sometimes referred to as atomic mass, is fundamental in various chemical and physical calculations. For barium, it is around 137.33 g/mol. This value tells us the average mass of an atom of barium compared to 1/12th of a carbon-12 atom.

• It is essential when calculating the mass of atoms within a unit cell.
• The atomic weight is used with Avogadro's number to convert the mass of single atoms to kilograms or grams.

Given the average atomic weight, in density-related calculations, the use of atomic weight helps bridge the collection of atoms in a macroscopic sample, enabling the calculation of tangible quantities like density.

Avogadro's Number

Avogadro's number, denoted as \(N_A\), is a fundamental constant in chemistry, representing the number of constituent particles (usually atoms or molecules) per mole of substance. It has a value of approximately \(6.022 \times 10^{23}\) atoms/mol.

• This large number is used to relate the macroscopic quantities like grams to the microscopic scale of atomic measurements.
• In density calculations, Avogadro's number allows us to convert the mass of a mole of atoms to the mass of individual atoms in a unit cell.

By facilitating the conversion from atoms per mole to mass, Avogadro's number is essential in anchoring atomic scale measurements to practical laboratory scales, playing a crucial role in calculating material densities and understanding the composition of materials on an atomic level.