Problem 66 - Chemistry The Molecular Science

Question

Tungsten has a body-centered cubic unit cell and an atomic radius of \(141 \mathrm{pm} .\) Calculate the density of solid tungsten.

Step-by-Step Solution

Answer

The density of solid tungsten is approximately 17.68 g/cm³.

Step 1: Understand the Body-Centered Cubic Structure

First, recall the characteristics of a body-centered cubic (BCC) unit cell. It has one atom at each corner of the cube and one atom in the center. Each corner atom is shared by eight unit cells. Thus, a single BCC unit cell contains 2 atoms: 1 atom fully in the center and 8 corners, each contributing 1/8 of an atom (8 corners * 1/8 = 1 atom).

Step 2: Determine the Relationship Between Atomic Radius and Unit Cell Edge Length

For a BCC structure, the body diagonal of the cube is equal to four times the atomic radius. The equation for the body diagonal in terms of the edge length (\(a\)) is \(\sqrt{3}a = 4r\). Using this formula, solve for \(a\): \[ a = \frac{4r}{\sqrt{3}} \] Given that the atomic radius \(r\) is 141 pm, substitute this value to find \(a\).

Step 3: Calculate the Edge Length

Substitute the given atomic radius into the formula:\[ a = \frac{4 \times 141 \text{ pm}}{\sqrt{3}} \approx \frac{564}{1.732} \approx 325.29 \text{ pm} \] So, the edge length of the unit cell is approximately 325.29 pm.

Step 4: Calculate the Volume of the Unit Cell

Convert the edge length from pm to cm for density calculations (1 pm = \(1 \times 10^{-10}\) cm):\[ a = 325.29 \text{ pm} = 325.29 \times 10^{-10} \text{ cm} \] Calculate the volume of the cubic unit cell using: \[ V = a^3 = (325.29 \times 10^{-10})^3 \text{ cm}^3 = 3.45 \times 10^{-23} \text{ cm}^3 \]

Step 5: Find the Mass of Atoms in the Unit Cell

Tungsten's atomic mass is approximately 183.84 g/mol. Since there are 2 atoms per unit cell, calculate the mass of atoms in a unit cell:1 mole of tungsten atoms corresponds to Avogadro's number, \(6.022 \times 10^{23}\) atoms.The mass of 2 atoms is calculated as follows:\[ \text{Mass} = 2 \times \frac{183.84 \text{ g/mol}}{6.022 \times 10^{23} \text{ atoms/mol}} \approx 6.10 \times 10^{-22} \text{ g} \]

Step 6: Calculate the Density of Tungsten

Density is mass per unit volume. Use the mass of the unit cell and the volume calculated:\[ \text{Density} = \frac{\text{Mass of atoms in unit cell}}{\text{Volume of unit cell}} = \frac{6.10 \times 10^{-22} \text{ g}}{3.45 \times 10^{-23} \text{ cm}^3} \approx 17.68 \text{ g/cm}^3 \]

Step 7: Conclusion

The calculated density of tungsten is approximately 17.68 g/cm³. This is a check with known values that verify our process and assumptions.

Key Concepts

Body-Centered Cubic Structure

The Body-Centered Cubic (BCC) structure is one of the common types of crystal lattice arrangements that you can find in metallic solids. Understanding this structure is crucial when calculating the density of a material. Unlike the simple cubic structure that only has one atom per unit cell, the BCC arrangement adds an additional atom at the center of the cube, giving a total of two atoms per unit cell.

• Each corner atom in the BCC is shared with other adjacent unit cells. Specifically, each corner atom is shared by eight neighboring cells.
• The atom at the center of the cube is not shared with any other cells, making it unique to its particular unit cell.
• This unique configuration means that the BCC effectively consists of 2 atoms per unit cell: 1 from the corners and 1 standing alone in the middle.

This understanding helps in calculating the number of atoms per unit cell, which is essential for determining the mass of the unit cell and thus the density of the material.

Atomic Radius

The atomic radius is a fundamental concept in chemistry that refers to the average distance from the nucleus of an atom to the boundary where the electrons spend most of their time. In a BCC structure, the atomic radius directly influences the computation of the unit cell's dimensions.

• In a BCC, the relationship between the atomic radius and the cell dimension is determined by the diagonal of the cube. The diagonal is calculated using the formula: \( \sqrt{3}a = 4r \), where \(a\) represents the edge length of the cube and \(r\), the atomic radius.
• Given the atomic radius of tungsten as 141 pm, this formula allows us to calculate the cube’s edge length precisely. This is crucial for understanding how tightly packed the atoms are and their spatial arrangement, which influences the density.
• Thus, by knowing the atomic radius, we can not only compute the edge length but also delve deeper into how the atoms within the material are arranged spatially, providing insights into the density and other physical properties.

Unit Cell Volume

The unit cell volume is a key aspect in arithmetic related to the density of substances in solid-state chemistry. It refers to the total space occupied by atoms within a single unit of a crystal lattice.

• In the context of the BCC structure, once you know the edge length \(a\), the volume \(V\) can be readily calculated since the unit cell is cubic.
• The formula to determine the volume of the unit cell is \( V = a^3 \), providing a cube with each side equal to the calculated length \(a\).
• This calculation is fundamental because the unit cell volume, when used in conjunction with the mass of atoms per cell, allows us to compute the density. The density \(\rho\) is ultimately calculated with \( \rho = \frac{\text{mass of atoms in unit cell}}{\text{volume of unit cell}} \)
• Accurate measurement of this volume ensures reliable calculation of the density. As seen in tungsten, precise unit cell dimensions lead to a clear understanding of its density and the efficient packing of atoms in its BCC structure.