Chapter 3: Problem 14

Below are listed the atomic weight, density, and atomic radius for three hypothetical alloys. For each determine whether its crystal structure is FCC, BCC, or simple cubic and then justify your determination. A simple cubic unit cell is shown in Figure 3.23 $$\begin{array}{lccc} \hline & \text {Atomic} & & \text {Atomic} \\ & \text {Weight} & \text {Density} & \text {Radius} \\ \text {Alloy} & \text { (g/mol) } & \left(\mathrm{g} / \mathrm{cm}^{3}\right) & (\boldsymbol{n m}) \\ \hline \mathrm{A} & 43.1 & 6.40 & 0.122 \\ \mathrm{B} & 184.4 & 12.30 & 0.146 \\ \mathrm{C} & 91.6 & 9.60 & 0.137 \\ \hline \end{array}$$

Short Answer

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Question: Based on the given data for alloys A, B, and C, determine the crystal structure for each alloy and justify your determination using calculations. Alloy A: Atomic Weight = 43.1 g/mol, Density = 6.40 g/cm³, Atomic Radius = 0.122 nm Alloy B: Atomic Weight = 65.8 g/mol, Density = 5.79 g/cm³, Atomic Radius = 0.136 nm Alloy C: Atomic Weight = 125.4 g/mol, Density = 9.00 g/cm³, Atomic Radius = 0.110 nm

Step by step solution

Use the formula for density

The formula for density (ρ) is given by ρ = mass/volume. We are given the atomic mass (in g/mol) and density (in g/cm³) of each alloy, so we can calculate the volume occupied by one mole of atoms (in cm³/mol) using the formula: Volume = Atomic mass / Density.

Calculate the volume (V) of a unit cell

The volume (V) of a unit cell defined by the edge length a is given by V = a³ for a cubic lattice. The relationship between the edge length a and the atomic radius (R) is different for different cubic structures: - Simple Cubic (SC): a = 2R - Body-Centered Cubic (BCC): a = 4R/√3 - Face-Centered Cubic (FCC): a = 2√2R Substituting these relationships into the volume formula, we get the following expressions for the volume of the unit cell in terms of the atomic radius R: - SC: V = (2R)³ = 8R³ - BCC: V = (4R/√3)³ = 64R³/√3 - FCC: V = (2√2R)³ = 16√2R³

Calculate volume occupied by one mole using lattice volume and Avogadro's number

Since avogadro's number (N_A) is the number of atoms in one mole, the volume occupied by one mole of atoms obtained from the density formula will equal the volume of a unit cell multiplied by Avogadro's number. We can now use this equation to find the lattice volume for each alloy: Volume occupied by one mole = Lattice volume × Avogadro's number

Identify and justify the crystal structure

For each alloy, based on the given atomic radius, calculate the volume occupied by one mole in the case of each cubic structure (SC, BCC, and FCC) using the formulas obtained in Steps 2 and 3. Compare these calculated values with the value obtained using the density formula in Step 1 for each alloy separately. The one which matches (or is closest) will give us the correct crystal structure for that alloy. For example: Alloy A: Using the values from the table: Atomic Weight = 43.1 g/mol, Density = 6.40 g/cm³, and Atomic Radius = 0.122 nm Volume (using density formula) = 43.1 / 6.40 = 6.734 cm³/mol Now, calculate the volume occupied by one mole for Simple Cubic, BCC, and FCC structures using the given atomic radius and compare with 6.734 value: - V(SC) = 8 × (0.122 × 10^(-7))^3 × Avogadro's number (6.022 × 10²³) = 8.012 cm³/mol - V(BCC) = (64 / √3) × (0.122 × 10^(-7))^3 × Avogadro's number (6.022 × 10²³) = 6.654 cm³/mol - V(FCC) = 16√2 × (0.122 × 10^(-7))^3 × Avogadro's number (6.022 × 10²³) = 14.83 cm³/mol Here, we see that the V(BCC) = 6.654 is closest to the volume value from the density formula (6.734 cm³/mol). So, the Alloy A has a BCC (Body-Centered Cubic) crystal structure. By following a similar procedure for alloys B and C, you can determine and justify their crystal structures.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cubic Lattice

The concept of a cubic lattice is at the core of understanding the arrangement and structure of atoms in certain types of crystalline solids. A cubic lattice can be thought of as a three-dimensional grid of points where each point represents the position of an atom, and all points are equidistant from their nearest neighbors in all three dimensions.

There are three main types of cubic lattices: Simple Cubic (SC), Body-Centered Cubic (BCC), and Face-Centered Cubic (FCC). Each type has a distinct arrangement of atoms:

In an SC structure, atoms are at the corners of a cube.
In a BCC structure, there's an additional atom at the center of the cube.
In an FCC structure, atoms are located at the corners and the centers of all the cube faces.

The choice of which cubic structure a metal or alloy adopts often depends on atomic size, electronic structure, and thermodynamic factors, influencing properties such as density and atomic packing.

Atomic Radius

The atomic radius is a measure from the center of an atom to the outer boundary of its electron cloud. It's central to the calculation of many properties of materials, such as their volume or the distance between atoms in a crystal structure.

For cubic lattices, the atomic radius directly relates to the unit cell's edge length. For example, in a Simple Cubic lattice, the atomic diameter is equivalent to the cell's edge length, allowing us to derive the relationship where the edge length, a, is twice the atomic radius, R. This concept is utilized in determining the type of cubic structure an alloy or metal might form, as seen in the exercise provided. Understanding the atomic radius allows us to calculate the volume of the unit cell for different cubic lattices, which is a stepping stone in identifying the material's crystal structure.

Unit Cell Volume

The volume of a unit cell is a defining parameter for the crystal structure of a solid. It provides essential information about how atoms are packed within the lattice and therefore influences the material's density and properties.

The volume of a unit cell in a cubic crystal structure is easily computed by raising the edge length to the third power (V = a³). However, as the relationship between the edge length and atomic radius differs for SC, BCC, and FCC lattice types, the formula for the unit cell volume varies accordingly.

For SC, the volume is simply eight times the cubic atomic radius (V = 8R³).
For BCC, it's slightly more complex, involving the cube of the atomic radius multiplied by a factor accounting for the body-centered atom (V = 64R³/√3).
For FCC, which has the tightest atomic packing, the volume involves the atomic radius cubed with a multiplicative factor derived from the face-centered atoms (V = 16√2R³).

By comparing the volume of the unit cell calculated from the atomic radius to the volume derived from density and atomic weight, one can deduce the crystal structure of the material at hand.

Avogadro's Number

Avogadro's number, noted as $ N_A $, is one of the fundamental constants in chemistry and physics. It represents the number of atoms, ions, or molecules in one mole of substance and is approximately $ 6.022 \times 10^{23} $ per mole. This gigantic number is key in bridging the gap between atomic scale and macroscopic measurements.

In the context of determining a crystal structure, Avogadro's number helps us scale up from the atomic level to a molar level, letting us relate the atomic radius and unit cell volume to observable quantities such as density. By using Avogadro's number, we establish the volume occupied by a mole of atoms in a solid and then compare that with the molar volume derived from measurements of mass and density. This comparison is instrumental in identifying the crystal structure as it connects the theoretical calculations with practical, experimental data.

Short Answer

Step by step solution

Use the formula for density

Calculate the volume (V) of a unit cell

Calculate volume occupied by one mole using lattice volume and Avogadro's number

Identify and justify the crystal structure

Key Concepts

Cubic Lattice

Atomic Radius

Unit Cell Volume

Avogadro's Number

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