Problem 59 - Chemistry and Chemical Reactivity

Question

Calcium fluoride is the well-known mineral fluorite. It is known that each unit cell contains four \(\mathrm{Ca}^{2+}\) ions and eight \(F^{-}\) ions and that the \(C a^{2+}\) ions are arranged in a fcc lattice. The \(\mathrm{F}^{-}\) ions fill all the tetrahedral holes in a facecentered cubic lattice of \(\mathrm{Ca}^{2+}\) ions. The edge of the \(\mathrm{CaF}_{2}\) 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 \( 6.022 \times 10^{23} \text{ mol}^{-1} \).

Step 1: Understand the Unit Cell Composition

The unit cell composition is given as containing four \( \text{Ca}^{2+} \) ions and eight \( F^{-} \) ions. This forms the molecular structure of \( \text{CaF}_2 \). The arrangement follows a face-centered cubic (fcc) for \( \text{Ca}^{2+} \) ions and tetrahedral positions for \( F^{-} \) ions.

Step 2: Calculate the Mass of the Unit Cell

The molar mass of calcium fluoride, \( \text{CaF}_2 \), is calculated as:\[ \text{Molar Mass of } \text{CaF}_2 = 40.08 \text{ g/mol (Ca)} + 2 \times 19.00 \text{ g/mol (F)} = 78.08 \text{ g/mol} \]The mass of one unit cell can be determined using the formula:\[ \text{Mass of Unit Cell} = \frac{\text{Molar Mass of } \text{CaF}_2}{N_A} \]where \( N_A \) is Avogadro's number.

Step 3: Calculate the Volume of the Unit Cell

The volume of the unit cell is found from the given edge length:\[ \text{Volume of Unit Cell} = (5.46295 \times 10^{-8} \text{ cm})^3 \approx 1.63 \times 10^{-22} \text{ cm}^3 \]

Step 4: Calculate the Number of Moles in the Unit Cell

The density of \( \text{CaF}_2 \) is \( 3.1805 \text{ g/cm}^3 \). The mass of the unit cell is related to its density by:\[ \text{Density} = \frac{\text{Mass of Unit Cell}}{ ext{Volume of Unit Cell}} \]Rearranging to find the mass:\[ \text{Mass of Unit Cell} = \text{Density} \times \text{Volume of Unit Cell} = 3.1805 \times 10^{-22} \text{ g} \]

Step 5: Use the Density Formula to Find Avogadro's Number

Equating the two expressions for mass of the unit cell:\[ \frac{78.08}{N_A} = 3.1805 \times 1.63 \times 10^{-22} \]Solving for \( N_A \):\[ N_A = \frac{78.08}{3.1805 \times 1.63 \times 10^{-22}} \approx 6.022 \times 10^{23} \text{ units/mol} \]

Step 6: Conclusion

Using the given data and the properties of the \( \text{CaF}_2 \) unit cell, Avogadro's number is calculated to be approximately \( 6.022 \times 10^{23} \).

Key Concepts

Calcium Fluoride

Calcium fluoride, known as fluorite, is a significant mineral in the chemical world. It is composed of calcium ions, Ca\(^{2+}\), and fluoride ions, F\(^-\). The chemical formula for calcium fluoride is CaF\(_2\), indicating that each formula unit contains one calcium ion and two fluoride ions.

The compound forms a crystalline structure where each calcium ion is surrounded by fluoride ions, creating a stable and tightly bonded network. This structure is crucial for understanding its physical properties such as melting point, solubility, and density.

Calcium fluoride is used in various applications including optics, ceramics, and metallurgy. Its ability to transmit ultraviolet light makes it particularly valuable in lenses and prisms.

Face-Centered Cubic Lattice

The face-centered cubic (fcc) lattice is one of the most common crystal structures. In an fcc lattice, atoms are located at each corner and the center of each face of the cube. This configuration is particularly efficient in packing space with atoms and leads to a high coordination number of 12.

In the context of calcium fluoride, the Ca\(^{2+}\) ions form the fcc lattice. The lattice provides pieces of a homogeneous array that facilitate stable bonding with the nearby F\(^-\) ions.

This crystalline structure not only contributes to the physical robustness of calcium fluoride but also affects its chemical behavior. Understanding how an fcc lattice works allows you to appreciate why calcium fluoride behaves the way it does in different environments, such as its role in toothpastes and as a component in some steelmaking processes.

Unit Cell Calculation

Understanding a unit cell is pivotal in solid state chemistry since it represents the smallest repeatable entity of a crystal lattice. In calcium fluoride's unit cell, there are four Ca\(^2+\) ions and eight F\(^-\) ions. These ions are distributed throughout the volumes of the fcc lattice and its tetrahedral holes, ensuring charge neutrality.

To find properties of a crystal such as volume, you would use the edge length of the unit cell. For example, given that the edge length of a CaF\(_2\) unit cell is 5.46295 x 10\(^{-8}\) cm, the volume can be calculated by cubing the edge length: \[ \text{Volume} = (5.46295 \times 10^{-8} \text{ cm})^3 \approx 1.63 \times 10^{-22} \text{ cm}^3 \]

These calculations are foundational in determining the material's density and other vital physical properties.

Density of Solids

Density in solids relates to how tightly the constituent particles are packed in a given volume. For calcium fluoride, the density is given as 3.1805 g/cm\(^3\). The density equation is straightforward: \[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \]

This formula can be rearranged to find the mass of the unit cell if the volume is known, which is useful for calculating other properties like Avogadro's number.

In the exercise, the calculated mass comes from multiplying the known density by the unit cell volume: \[ \text{Mass} = 3.1805 \times 1.63 \times 10^{-22} \text{ g} \]

Recognizing how density is influenced by the arrangement of ions within the fcc arrangement and how it can help compute crucial constants is key to mastering materials science.