Problem 96 - Chemistry: An Atoms-Focused Approach

Question

Crystalline potassium bromide (KBr) has a rock salt structure and a density of \(2.75 \mathrm{g} / \mathrm{cm}^{3} .\) Calculate the edge length of its unit cell.

Step-by-Step Solution

Answer

Answer: The edge length of the unit cell of crystalline potassium bromide is approximately \(3.050 \times 10^{-8} \mathrm{cm}\).

Step 1: Calculate the mass of one KBr molecule

To begin, we need to find the mass of one molecule of KBr. The molar mass of potassium (K) is \(39.0983 \mathrm{g/mol}\) and the molar mass of bromine (Br) is \(79.904 \mathrm{g/mol}\). Therefore, the molar mass of KBr is \(M = 39.0983 + 79.904 = 119.0023 \mathrm{g/mol}\). To find the mass of one molecule, we'll use the molecular weight formula: $$ m = \frac{M}{N_A} = \frac{119.0023 \mathrm{g/mol}}{6.022 \times 10^{23} \mathrm{mol}^{-1}} = 1.976 \times 10^{-22} \mathrm{g} $$

Step 2: Calculate the volume of one KBr molecule

Next, we need to find the volume occupied by one KBr molecule. We can use the density formula to do this: $$ V = \frac{m}{\rho} = \frac{1.976 \times 10^{-22} \mathrm{g}}{2.75 \mathrm{g/cm^3}} = 7.185 \times 10^{-23} \mathrm{cm^3} $$

Step 3: Calculate the edge length of the unit cell

Finally, we can use the volume of one KBr molecule to find the edge length of the unit cell. Since KBr has a face-centered cubic structure, there are four KBr molecules in one unit cell. Therefore, the volume of one unit cell is four times the volume of one KBr molecule: $$ V_{unit\ cell} = 4 \cdot V = 4 \cdot 7.185 \times 10^{-23} \mathrm{cm^3} = 2.874 \times 10^{-22} \mathrm{cm^3} $$ Now we can calculate the edge length of the unit cell using the cube root of the unit cell volume: $$ a = \sqrt[3]{V_{unit\ cell}} = \sqrt[3]{2.874 \times 10^{-22} \mathrm{cm^3}} = 3.050 \times 10^{-8} \mathrm{cm} $$ Thus, the edge length of the unit cell of crystalline potassium bromide is approximately \(3.050 \times 10^{-8} \mathrm{cm}\).

Key Concepts

Rock Salt Structure

The rock salt structure is a specific arrangement seen in ionic compounds. It gets its name from the mineral rock salt, also known as halite, which has the same crystalline arrangement. This structure is a type of face-centered cubic (FCC) lattice.
In the rock salt structure, each cation (positively charged ion) is surrounded by six anions (negatively charged ions), and vice versa. This arrangement results in a highly stable structure due to maximized electrostatic attraction between the ions. For example, in potassium bromide (KBr), each K+ ion is surrounded by six Br- ions, forming an octahedral geometry. This alternating pattern of ions creates the three-dimensional lattice that is characteristic of the rock salt structure.

Face-Centered Cubic Lattice

The face-centered cubic (FCC) lattice is a common type of crystalline structure. In an FCC lattice, atoms or ions are placed at each corner of a cube and in the center of each face, but not inside the cube itself. This arrangement allows for a highly efficient packing of particles.
In the face-centered cubic structure, each unit cell contains a total of four atoms or ions, contributing to its density and stability. This is because each corner atom is shared among eight adjacent unit cells, and each face atom is shared with two. In the context of the rock salt structure, the face-centered cubic lattice is modified by the presence of two different types of ions, such as K+ and Br- in KBr, which alternate positions in the lattice.

Unit Cell Calculation

Unit cell calculation is crucial in determining the dimensions and properties of a crystalline material. The unit cell is the smallest repeating structure in a crystal lattice that, when repeated, builds up the entire crystal. Calculating the unit cell involves understanding the mass, volume, and density of the material.
For solids like potassium bromide (KBr), knowing the density and the number of formula units per unit cell allows us to find the edge length of the unit cell. The unit cell's edge length can be calculated by determining the volume of one unit cell and then taking the cube root of that volume, as was done in the provided solution. This process helps us understand the size and structure of the crystals on an atomic level.