Problem 47 - Chemistry and Chemical Reactivity

Question

Calcium fluoride is the well-known mineral fluorite. Each unit cell contains four \(\mathrm{Ca}^{2+}\) ions and eight \(\mathrm{F}^{-}\) ions. The \(\mathrm{F}^{-}\) ions fill all the tetrahedral holes in a face-centered cubic lattice of \(\mathrm{Ca}^{2+}\) ions. The edge of the CaF \(_{2}\) unit cell is \(5.46295 \times 10^{-8} \mathrm{cm}\) in length. The density of the solid is \(3.1805 \mathrm{g} / \mathrm{cm}^{3}\). Use this information to calculate Avogadro's number.

Step-by-Step Solution

Answer

Avogadro's number is approximately \(6.022 \times 10^{23}\) molecules/mol.

Step 1: Understanding the Unit Cell

The unit cell is face-centered cubic (fcc) with calcium ions at the lattice points. It contains 4 Ca\(^{2+}\) ions and 8 F\(^{-}\) ions. These contribute to the molar mass and calculate the volume.

Step 2: Calculate the Molar Mass of CaF\(_2\)

The molar mass of one mole of calcium (Ca) is approximately 40.08 g/mol, and that of fluorine (F) is about 19.00 g/mol. Thus, the molar mass of CaF\(_2\) is given by:\[ M = 40.08 + 2 \times 19.00 = 78.08 \, \text{g/mol} \]

Step 3: Calculate the Volume of the Unit Cell

Given that the edge length of the unit cell is \(5.46295 \times 10^{-8} \) cm, the volume \(V_c\) of the cubic unit cell is:\[ V_c = (5.46295 \times 10^{-8} \, \text{cm})^3 \approx 1.631 \times 10^{-22} \, \text{cm}^3 \]

Step 4: Calculate the Mass of the Unit Cell

The mass of one unit cell \( m \) can be obtained using the density \( \rho = 3.1805 \) g/cm\(^3\),\[ m = \rho \times V_c = 3.1805 \, \text{g/cm}^3 \times 1.631 \times 10^{-22} \, \text{cm}^3 \approx 5.187 \times 10^{-22} \, \text{g} \]

Step 5: Calculate the Number of Molecules per Unit Cell

Each CaF\(_2\) unit cell contains 4 formula units. Therefore, to find the number of molecules in one mole, we use Avogadro's number \(N_A\).

Step 6: Derive Avogadro's Number

The molar mass of CaF\(_2\) is 78.08 g/mol, and the mass of one unit cell is \( 5.187 \times 10^{-22} \, \text{g} \). Therefore:\[ N_A = \frac{78.08 \, \text{g/mol}}{5.187 \times 10^{-22} \, \text{g}} \times 4 \approx 6.022 \times 10^{23} \, \text{molecules/mol} \]

Step 7: Conclusion

The calculated Avogadro's number from the given data is approximately \(6.022 \times 10^{23}\) molecules/mol.

Key Concepts

Calcium Fluoride

Calcium fluoride, commonly known as fluorite, is a chemical compound consisting of calcium and fluorine. It is represented by the formula \( \text{CaF}_2 \) and appears as a naturally occurring mineral. In \( \text{CaF}_2 \), calcium exists as \( \text{Ca}^{2+} \) ions and fluorine as \( \text{F}^- \) ions. These ions are distributed within the crystal lattice to form a solid. The compound plays an essential role in various industries, including the production of hydrogen fluoride and as a flux in metallurgy.

• Calcium acts as the cation and forms the basis of the lattice structure.
• Fluorine ions fill the spaces in the lattice, contributing to its overall stability.

Face-Centered Cubic Lattice

In a face-centered cubic (fcc) lattice, the arrangement of atoms is such that each corner and the center of each face of the cube contains an atom. This structure is prevalent in many crystalline solids and provides a dense packing of atoms. In the case of calcium fluoride, \( \text{Ca}^{2+} \) ions form an fcc lattice.
This arrangement ensures maximum space utilization within the lattice

• The \( \text{F}^- \) ions occupy all the tetrahedral holes within this fcc structure.
• This configuration allows for four molecules of \( \text{CaF}_2 \) to exist per unit cell.

Unit Cell Calculation

To comprehend the volume and mass associated with a unit cell of calcium fluoride, we start by calculating its volume. The edge length of a unit cell is a specific measurement of the distance along one edge of the cube-shaped cell. For \( \text{CaF}_2 \), this edge length is \( 5.46295 \times 10^{-8} \text{ cm} \).
Using the formula for the volume of a cube, \( V = \text{edge}^3 \), we obtain:
\[ V_c = (5.46295 \times 10^{-8})^3 = 1.631 \times 10^{-22} \, \text{cm}^3 \]
This volume helps us further in determining the density and other physical properties of the mineral.

Molar Mass

The molar mass of a compound is crucial in converting between the number of moles and the amount of substance in grams. For calcium fluoride, the molar mass is derived from summing the atomic masses of calcium and fluorine. Since each formula unit of \( \text{CaF}_2 \) consists of one calcium atom and two fluorine atoms:

• Calcium (Ca): 40.08 g/mol
• Fluorine (F): 19.00 g/mol each

The molar mass is calculated as follows:

• \[ M = 40.08 + 2 \times 19.00 = 78.08 \, \text{g/mol} \]

This value is vital when dealing with calculations involving Avogadro's number.

Density Calculation

Density is a physical property that describes the mass per unit volume of a substance. For \( \text{CaF}_2 \), this is given as \( 3.1805 \, \text{g/cm}^3 \). By calculating the mass of a unit cell, we integrate the density into finding Avogadro's number. The mass of one unit cell can be determined by:
\[ m = \text{density} \times \text{volume} \] \[ m = 3.1805 \, \text{g/cm}^3 \times 1.631 \times 10^{-22} \, \text{cm}^3 \approx 5.187 \times 10^{-22} \, \text{g} \] Utilizing this mass, we relate it to the molar mass to derive Avogadro's number, emphasizing the linkage between microscopic properties of a unit cell and the macroscopic property of molar mass.