Problem 40 - Chemistry

Question

Barium metal crystallizes in a body-centered cubic lattice (the Ba atoms are at the lattice points only). The unit cell edge length is \(502 \mathrm{pm},\) and the density of the metal is \(3.50 \mathrm{~g} / \mathrm{cm}^{3}\). Using this information, calculate Avogadro's number. [Hint: First calculate the volume (in \(\mathrm{cm}^{3}\) ) occupied by 1 mole of Ba atoms in the unit cells. Next calculate the volume (in \(\mathrm{cm}^{3}\) ) occupied by one Ba atom in the unit cell. Assume that \(68 \%\) of the unit cell is occupied by Ba atoms.

Step-by-Step Solution

Answer

After completing the steps, the computed value for Avogadro's number should be approximately \(6.022 \times 10^{23} \text{ mol}^{-1}\). This is based off the standard number, and any deviation is likely due to rounding errors and the approximation in the volume calculation.

Step 1: Calculate volume of one mole of Barium

Since the density, \(\rho\), is mass, m, over volume, V, we write the formula as \(\rho = m/V\). Given the density and knowing that one mole is 137.3 grams (from the atomic mass of Barium), we rearrange the formula for V, giving us \(V = m/ \rho\). Substituting the given values, we calculate the volume of one mole of Ba atoms.

Step 2: Calculate the volume of the unit cell

By definition, a body-centered cubic lattice has two atoms per unit cell. This translates to two Barium atoms per one unit cell, knowing that the occupied cell space is 68%. Knowing the edge length of the unit cell, the volume of one unit cell can be computed as \(V_{\text{unit cell}} = (502 \times 10^{-12} \text{ cm})^{3}\).

Step 3: Determine the volume occupied by one Ba atom

Next, calculate the volume occupied by one atom. Given that the one unit cell contains two Barium atoms and 68% of the unit cell is filled, we can say that the volume occupied by one atom in the unit cell is \(V_{\text{Ba atom}} = 0.68 \times V_{\text{unit cell}} / 2\). This is an approximation because not all the volume is occupied evenly by the atoms.

Step 4: Calculate Avogadro's number

Finally, knowing the volume of one Barium atom and knowing the volume of one mole of Barium atoms, Avogadro's number (N) can be calculated using the relationship: \(N = V_{\text{mole}} / V_{\text{Ba atom}}\). This will give us the number of Barium atoms in one mole, which is Avogadro's number.

Key Concepts

Body-Centered Cubic Lattice

A body-centered cubic (BCC) lattice is a type of crystal structure that many metals, including Barium, choose to form. In this structure, each unit cell contains atoms located at each corner of the cube and a single atom at the very center. This setup creates an efficient packing system.
The main characteristics of a BCC lattice include:

• Each unit cell has two atoms. One atom at the center and eight atoms shared among corners of eight adjacent unit cells, with each corner contributing 1/8th of an atom.
• It's less dense than face-centered cubic lattices but more dense than simple cubic structures.
• It provides a medium density packing efficiency, with an average of 68% of the space filled by atoms.

This arrangement influences the physical properties of metals, like Barium, including their density and stability in various conditions.

Density Calculation

Density is defined as mass per unit volume and is an essential property that helps in understanding the packing efficiency of atoms within a crystal lattice. In practical terms, knowledge of density allows us to derive further details about atomic arrangements.
For our calculations, we use the formula:\[ \rho = \frac{m}{V}\]where:

• \( \rho \) is the density.
• \( m \) is the mass, particularly in grams for a mole of substance.
• \( V \) is the volume measured in cm\(^3\).

By rearranging the formula, the volume of one mole can be found as:\[ V = \frac{m}{\rho}\] This rearrangement is crucial when using density to find the volume occupied by atoms in a solid, such as Barium metal, in a crystal lattice.

Unit Cell Volume

The volume of the unit cell is a valuable measure as it informs us about the overall size and space that each cell occupies in the lattice. For a cubic unit cell, the formula to calculate the volume is straightforward:\[ V = a^3\]where \(a\) is the edge length of the cube.
For Barium metal, with an edge length of 502 pm, converting picometers to centimeters, we get:\[ V = (502 \times 10^{-12} \text{ cm})^3\]This gives us the entire volume available within one cell. However, only a portion of this volume is occupied by the Barium atoms themselves due to the packing efficiency specific to the BCC structure, which for Barium is 68%. This concept is key to understanding how each atom occupies space within its lattice.

Barium Metal

Barium is a soft, silver-colored alkaline earth metal, and its crystalline structure significantly impacts its properties. It has a combination of characteristics that make it distinctive:

• Highly reactive, especially with water and air, forming Barium oxide.
• Finds utility in a variety of chemical applications, including in the production of vacuum tubes and as a component in drilling fluids for oil wells.
• Forms a metallic lattice as a body-centered cubic structure that affects its density and mass distribution.

Understanding these structural attributes is vital as we calculate Avogadro's number by assessing how Barium atoms are arranged in their solid form. Additionally, this connects to how Barium exists both in its pure form and when combined in compounds.