Problem 60 - Chemistry: The Central Science in SI Units, Global Edition

Question

Silver chloride (AgCl) adopts the rock salt structure. The density of \(\mathrm{AgCl}\) at \(25^{\circ} \mathrm{C}\) is \(5.56 \mathrm{~g} / \mathrm{cm}^{3}\). Calculate the length of an edge of the AgCl unit cell.

Step-by-Step Solution

Answer

The edge length of the AgCl unit cell is approximately 5.63 Å, or 5.63 x 10⁻¹⁰ meters.

Step 1: Find the molecular weight of AgCl

To find the molecular weight of AgCl, we first need to know the atomic weights of silver (Ag) and chlorine (Cl). The atomic weight of Ag is 107.87 g/mol, and the atomic weight of Cl is 35.45 g/mol. Therefore, the molecular weight of AgCl is the sum of the two atomic weights: Molecular weight of AgCl = 107.87 g/mol (Ag) + 35.45 g/mol (Cl) = 143.32 g/mol

Step 2: Determine the number of AgCl units per unit cell

In the rock salt structure, each metal atom (Ag) is surrounded by six halide atoms (Cl), and each halide atom is surrounded by six metal atoms. This means there is one Ag atom and one Cl atom in each unit cell. Therefore, there are two AgCl units per unit cell.

Step 3: Use the formula for density to find the edge length

The formula for the density (ρ) of a unit cell is: ρ = (mass of one unit cell) / (volume of one unit cell) We can also write this formula in terms of the number of AgCl units, molecular weight, and edge length: ρ = (number of AgCl units × molecular weight of AgCl) / (volume of one unit cell) Since the volume of a cubic unit cell is equal to the cube of its edge length (a^3), we can write: ρ = (number of AgCl units × molecular weight of AgCl) / (edge length^3) Now we can plug in the given density (5.56 g/cm³), number of AgCl units (2 units), and molecular weight of AgCl (143.32 g/mol), and solve for the edge length (a): 5.56 g/cm³ = (2 × 143.32 g/mol) / (edge length^3) To find the edge length, first, divide both sides by the molecular weight of AgCl(286.64 g/mol): \( \frac{5.56 g/cm^3}{286.64 g/mol} = \frac{1}{a^3} \) Now, calculate the cube root of both sides to find the edge length: \( a = \sqrt[3]{\frac{286.64 g/mol}{5.56 g/cm^3}} \) Using a calculator, we obtain: a ≈ 5.63 Å

Step 4: Express the edge length in the appropriate units

Since we've found the edge length (a) in Ångström (Å), we can express it in the appropriate units. Recall that 1 Å = 10⁻¹⁰ meters. Edge length of the AgCl unit cell ≈ 5.63 Å ≈ 5.63 x 10⁻¹⁰ meters. Given that the edge length of the AgCl unit cell is approximately 5.63 Å, the response to the problem is complete.

Key Concepts

Molecular Weight

Understanding molecular weight is essential when dealing with compounds like silver chloride (AgCl). Molecular weight, often referred to as molar mass, represents the mass of a single molecule of a compound in terms of grams per mole (g/mol). To compute the molecular weight of AgCl, we need to add the atomic weights of its constituent elements: silver (Ag) and chlorine (Cl). The atomic weight of silver is 107.87 g/mol, while chlorine is 35.45 g/mol. Therefore, the molecular weight of AgCl is the sum of these two atomic weights:

• Molecular weight of AgCl = 107.87 g/mol (Ag) + 35.45 g/mol (Cl)
• = 143.32 g/mol

This calculation helps us understand how much one mole of AgCl weighs, which is a fundamental step for further calculations, especially in density estimation.

Density Calculation

Density is a critical parameter in understanding how tightly matter is packed within a certain volume. For crystalline structures like AgCl, density (\( \rho \)) is defined as mass per unit volume. For the task at hand, we consider the density of silver chloride as 5.56 g/cm³ at 25°C. The formula for density in the context of a crystal unit cell is \( \rho = \frac{\text{mass of one unit cell}}{\text{volume of one unit cell}} \).

• This can be reformulated using the number of AgCl formula units in the cell, the molecular weight, and the edge length: \[ \rho = \frac{(\text{number of AgCl units} \times \text{molecular weight of AgCl})}{(\text{edge length})^3} \]

In the case of AgCl, the cubic unit cell contains two formula units, and using the density and molecular weight, we can solve for the edge length as previously detailed. The density calculation provides insights into the structural arrangement and material properties at the molecular level.

Unit Cell

A unit cell is the simplest repeating unit in the crystal lattice of a compound, and it helps in defining the layout of atoms in a solid. For AgCl, its structure can be modeled as a "rock salt" type, which is critical to solving the exercise. In this arrangement, each atom is surrounded by six neighbor atoms, creating a characteristic cubic geometry.

• Understanding a unit cell involves recognizing that in AgCl's structure, there is one silver and one chlorine atom per unit cell, meaning there are effectively two AgCl units per cell.
• The volume of this cubic unit cell is calculated as \((\text{edge length})^3\), a vital part of establishing its dimensions.

Examining the unit cell contributes to comprehending how molecules are organized, which in turn affects the physical and chemical properties of the material, including its interaction with light, strength, and thermal conductivity.