Problem 111 - General Chemistry

Question

Iridium metal, Ir, crystallizes in a face-centered cubic (close-packed) structure. The edge length of the unit cell was found by \(x\) -ray diffraction to be \(383.9 \mathrm{pm} .\) The density of iridium is \(22.42 \mathrm{~g} / \mathrm{cm}^{3}\). Calculate the mass of an iridium atom. Use Avogadro's number to calculate the atomic mass of iridium.

Step-by-Step Solution

Answer

The mass of an iridium atom is approximately \(3.172 \times 10^{-22} \text{ g}\).

Step 1: Calculate Volume of the Unit Cell

To start, we need to calculate the volume of the unit cell. Since the unit cell has a face-centered cubic structure, the length of each edge is given as 383.9 pm. First, convert this length from **picometers** to **centimeters**: \[1 \, \text{pm} = 1 \times 10^{-12} \, \text{cm}\] Thus, the edge length is:\[383.9 \, \text{pm} = 383.9 \times 10^{-12} \, \text{cm}\]Then, calculate the volume using the formula for the volume of a cube:\[V_{\text{cell}} = a^3 = (383.9 \times 10^{-12} \, \text{cm})^3 = 5.6577 \times 10^{-23} \, \text{cm}^3\]

Step 2: Calculate Mass of the Unit Cell

To calculate the mass of the unit cell, use the density formula, which is Density = Mass/Volume. Rearrange to find mass:\[\text{Mass} = \text{Density} \times \text{Volume}\] Given the density of iridium: \[22.42 \, \text{g/cm}^3\] the mass of the unit cell is:\[ \text{Mass of cell} = 22.42 \, \text{g/cm}^3 \times 5.6577 \times 10^{-23} \, \text{cm}^3 = 1.2688 \times 10^{-21} \, \text{g} \]

Step 3: Determine Number of Atoms per Unit Cell

In a face-centered cubic (FCC) structure, there are 4 atoms per unit cell because there is 1 atom at each corner of the cube shared with 8 unit cells and 1 atom on each face shared with 2 unit cells. Therefore, the total number of atoms in an FCC unit cell is:\[\frac{8}{8} + \frac{6}{2} = 4\]

Step 4: Calculate Mass of a Single Iridium Atom

Using the mass of the unit cell and the number of atoms per unit cell, calculate the mass of a single iridium atom:\[ \text{Mass of one Ir atom} = \frac{\text{Mass of unit cell}}{\text{Number of atoms per unit cell}}\] \[= \frac{1.2688 \times 10^{-21} \, \text{g}}{4} = 3.172 \times 10^{-22} \, \text{g}\]

Step 5: Calculate Atomic Mass Using Avogadro's Number

To find the atomic mass, use Avogadro's number \(6.022 \times 10^{23}\text{ atoms/mol}\):Multiply the mass of one iridium atom by Avogadro’s number:\[\text{Atomic mass} = 3.172 \times 10^{-22} \text{ g/atom} \times 6.022 \times 10^{23} \text{ atoms/mol} = 190.2 \, \text{g/mol}\]

Key Concepts

Atomic Mass

Atomic mass is the mass of an atom, typically measured in atomic mass units (amu) or grams per mole (g/mol). It is a fundamental concept in chemistry that helps us understand the weight of an element at the atomic level. In a more practical sense, atomic mass allows us to compare the weight of one atom to another, providing insights into how different elements react and form compounds.
For instance, in the exercise provided, the atomic mass of iridium is calculated by first determining the mass of a single iridium atom in grams and then converting this to a molar mass using Avogadro's number. This calculation is crucial as it helps chemists know how much of an element they'll need for reactions or materials applications, especially when dealing with large scale productions.

Density Calculations

Density is defined as mass per unit volume and is commonly expressed in grams per cubic centimeter (g/cm³) for solids. It is a key property of materials, as it affects how substances mix, settle, or float relative to others. In the given exercise, it's crucial for understanding the compactness of iridium atoms in their face-centered cubic structure.
To find the mass of the unit cell in the exercise, the density formula is rearranged to solve for mass:

• Density = Mass / Volume
• Therefore, Mass = Density × Volume

By knowing the density and calculating the volume of the unit cell using its edge length, we can determine the mass of that cell, which subsequently allows us to calculate the mass of individual atoms within the cell. Density calculations like this are essential in materials science and various engineering disciplines, providing insights into material properties and behaviors.

X-ray Diffraction

X-ray diffraction (XRD) is a powerful technique used to determine the atomic and molecular structure of crystals. This method involves directing X-rays at a crystalline material and analyzing the diffraction pattern that results. Each pattern serves like a fingerprint, revealing the positions of the atoms within the crystal.
In the context of the exercise, X-ray diffraction was used to find the edge length of iridium's face-centered cubic unit cell, which is key for calculating the atomic mass and density. Understanding how crystals are structured helps scientists and engineers design better materials with specific properties or functionalities, making XRD an invaluable tool in fields like chemistry, physics, and materials science.

Avogadro's Number

Avogadro's number is a constant used to denote the number of atoms, molecules, or particles in one mole of a substance. The value is approximately 6.022 × 10^{23}. It provides a bridge between the macroscopic world that we can see and the microscopic world of atoms and molecules.
In this exercise, Avogadro's number is used to convert the mass of a single iridium atom to its molar mass. By multiplying the mass of one atom by Avogadro’s constant, we get the mass of one mole of iridium atoms, expressed in g/mol. This assists chemists in easily weighing out amounts of a chemical substance that can react with each other on a molecular scale, ensuring proper stoichiometric calculations and balances in reactions.