Problem 90 - General Chemistry

Question

Chromium forms cubic crystals whose unit cell has an edge length of \(288.5 \mathrm{pm}\). The density of the metal is \(7.20 \mathrm{~g} / \mathrm{cm}^{3}\). Use these data and the atomic mass to calculate the number of atoms in a unit cell, assuming all atoms are at lattice points. What type of cubic lattice does chromium have?

Step-by-Step Solution

Answer

Chromium has 2 atoms per unit cell and a body-centered cubic lattice.

Step 1: Convert Edge Length to Centimeters

The edge length of the cubic unit cell is given in picometers (pm). To use it in density calculations, convert it to centimeters: \[ 1 \text{ pm} = 1 \times 10^{-12} \text{ cm} \]. Calculating the edge length in cm: \[ 288.5 \text{ pm} \times 1 \times 10^{-12} \text{ cm/pm} = 2.885 \times 10^{-8} \text{ cm} \].

Step 2: Calculate Unit Cell Volume

The unit cell is cubic, so its volume can be calculated by cubing the edge length: \[ V = (2.885 \times 10^{-8} \text{ cm})^3 = 2.40 \times 10^{-23} \text{ cm}^3 \].

Step 3: Determine Mass of Unit Cell Using Density

The mass of a unit cell can be found using the formula: \[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \]. Rearranging gives \[ \text{Mass} = \text{Density} \times \text{Volume} = 7.20 \text{ g/cm}^3 \times 2.40 \times 10^{-23} \text{ cm}^3 = 1.73 \times 10^{-22} \text{ g} \].

Step 4: Compute Number of Moles in Unit Cell

Using the atomic mass of chromium (approximately 51.996 g/mol), calculate the number of moles of chromium in the unit cell: \[ \text{Moles} = \frac{\text{Mass}}{\text{Atomic Mass}} = \frac{1.73 \times 10^{-22} \text{ g}}{51.996 \text{ g/mol}} = 3.33 \times 10^{-24} \text{ mol} \].

Step 5: Find Number of Atoms in Unit Cell

Using Avogadro's number \(6.022 \times 10^{23} \text{ atoms/mol} \), determine the number of atoms in the unit cell: \[ \text{Atoms} = 3.33 \times 10^{-24} \text{ mol} \times 6.022 \times 10^{23} \text{ atoms/mol} = 2 \text{ atoms} \].

Step 6: Identify the Type of Cubic Lattice

In cubic lattices, a body-centered cubic (BCC) structure has 2 atoms per unit cell. Therefore, chromium has a body-centered cubic lattice.

Key Concepts

Crystallography

Crystallography is the study of crystal structures and their properties. It explores how atoms are arranged in a solid and how these arrangements repeat to form a crystal lattice. This is fundamental to understanding the properties of materials in fields like chemistry, physics, and material science. Crystals are typically categorized by their lattice structures, which determine the shape, symmetry, and physical properties of the material. These structures fall into different categories known as crystal systems. The cubic system, which is relevant for chromium, includes three types: simple cubic, body-centered cubic (BCC), and face-centered cubic (FCC). In the example of chromium, we find that it forms a BCC structure. This means each cube in the lattice contains additional atoms in its body, with each cube corner contributing to multiple unit cells. Understanding crystallography is essential for predicting material behaviors, such as density and structural strength.

Unit Cell Volume

The unit cell volume is a critical concept in crystallography, capturing the amount of space a single unit of the crystal lattice occupies. This parameter is crucial for calculating other material properties, such as density and mass.For cubic lattices, the volume is found by cubing the edge length of the unit cell. In the case of chromium, where the edge length is originally given in picometers, it needs to be converted into centimeters for practical calculations. The formula used here is \[ V = (2.885 \times 10^{-8} \text{ cm})^3 = 2.40 \times 10^{-23} \text{ cm}^3 \] This step ensures a consistent unit application when plugged into subsequent density and mass equations.Calculating the unit cell volume aids in understanding the crystallographic layout of materials and prepares the data for determining the number of atoms, especially when the atomic size and spacing are at play.

Atoms in a Unit Cell

Determining the number of atoms in a unit cell involves several steps and is essential for characterizing the type of lattice structure. In a body-centered cubic (BCC) lattice, such as in chromium, each unit cell generally contains two complete atoms.To calculate this, one generally finds the mass of a unit cell using its known density and volume. Then, using the atomic mass, you can determine the number of moles in the cell and finally calculate the number of atoms using Avogadro's number:- Mass of cell: \( \text{Mass} = 7.20 \, \text{g/cm}^3 \times 2.40 \times 10^{-23} \, \text{cm}^3 = 1.73 \times 10^{-22} \, \text{g} \)- Moles of chromium in the cell: \( \text{Moles} = \frac{1.73 \times 10^{-22} \, \text{g}}{51.996 \, \text{g/mol}} = 3.33 \times 10^{-24} \, \text{mol} \)- Number of atoms: \( 3.33 \times 10^{-24} \, \text{mol} \times 6.022 \times 10^{23} \, \text{atoms/mol} = 2 \, \text{atoms} \)Each unit cell's arrangement reveals much about the material's macroscopic properties and the efficiency of atomic packing.

Density and Atomic Mass

Density and atomic mass are interrelated properties that help define a material's characteristics within its crystalline structure. The density gives an idea about how densely packed the atoms are within the crystal, informing various physical properties, including stability and strength.To find the mass of a unit cell, use the density formula rearranged to solve for mass: \[ \text{Mass} = \text{Density} \times \text{Volume} \]For chromium's unit cell, with a density of \(7.20 \, \text{g/cm}^3\) and a calculated volume of \(2.40 \times 10^{-23} \, \text{cm}^3\), the mass is \(1.73 \times 10^{-22} \, \text{g}\). This mass is then used alongside the atomic mass to determine how many atoms fit into the unit cell.The atomic mass, typically given in grams per mole for individual atoms, provides a bridge from mass to moles, completing the picture of atomic arrangement. This understanding allows precise calculations necessary for identifying the type of cubic lattice formed by a metal such as chromium. These measures are crucial for practical applications, offering insights into material composition and potential usage in industry.