Class: General Chemistry II
Unit: Crystal Structures
Introduction
In today’s post we will learn how to perform calculations to solve for properties of a unit cell, the smallest repeating unit of a crystal structure. As an example, we will use the following problem:
Problem Statement
Vanadium crystallizes in a body-centered cubic lattice (bcc), and the length of the edge of a unit cell is 305 pm. What is the density of V (in g/cm^3)? (1 pm = 1×10-12 m; Atomic Mass of V = 54.94 g/mol).
Background
Definitions
Unit Cell – Smallest repeating structure of a crystalline solid.
Lattice Point – Each atom or ion within the unit cell.
Common Unit Cells (1)
Cubic Cell Type | Atoms per Unit Cell |
Simple cubic (sc) | 1 |
Body-centered cubic (bcc) | 2 |
Face-centered cubic (fcc) | 4 |
Solving the Problem
The first thing that we always want to do when solving a complex problem is determine what information is provided to us and what information the problem is asking us to solve.
1. We will start by highlighting useful information
Original: Vanadium crystallizes in a body-centered cubic lattice (bcc), and the length of the
edge of a unit cell is 305 pm. What is the density of V (in g/cm^3)? (1 pm = 1×10-12 m; Atomic Mass of V = 54.94 g/mol).
Orange – Information Provided
Green – Information Requested
Highlighted: Vanadium crystallizes in a body-centered cubic lattice (bcc), and the length of the edge of a unit cell is 305 pm. What is the density of V? (1 pm = 1×10-12 m; Atomic Mass of V = 54.94 g/mol).
2. After highlighting useful information, we want to create an index of the information and describe how that information can be used.
Type of Information | Information Provided (Extracted directly or indirectly from the problem) | Interpretation (thought process) | Extracted Information (What we will use to solve the problem) |
Unit Cell Type | body-centered cubic lattice (bcc) | bcc contains 2 atoms per unit cell. Typically, this is something one memorizes. This can also be calculated as the volume of the central atom + 8 (1/8th corner atoms around the central body atom) | 2 atoms |
Unit Cell Edge Length | length of the edge of a unit cell is 305 pm | In a cubic unit cell all edge lengths are the same, therefore the volume is a cube and can be represented as L(length)^3 = V (volume) or L^2 = A (area) | Unit Cell Length = 305 pm |
Length Conversion | 1 pm = 1×10-12 m | Common conversion, provided to ensure students unfamiliar with the conversion can solve the problem without external resources. | 1 pm = 1×10-12 m |
Atomic Mass of Vanadium (V) | Atomic Mass of V = 54.94 g/mol | The atomic mass or molar mass of a single Vanadium atom is 54.94 g/mol or 9.123E-23 g/atom | V = 50.94 g/mol V = 8.459E-23 g/atom |
Density of V? | ? |
| ? g/cm^3 |
3. After documenting all provided information, we will build a problem-solving strategy using the information provided.
The problem is asking us to solve for the density of Vanadium.
Density (Greek rho, ρ) is by definition a unit cells mass divided by the same unit cells volume:
Density (D) = Mass (m)/Volume (V)
Density(ρ) = Mass Unit Cell (m)/Volume Unit Cell (V)
Neither mass nor volume have directly been provided to us, however we can use information provided to solve for both of those values. Let’s start with mass.
4. Solving for Mass of Unit cell
The mass of a unit cell is equal to the mass of the atoms contained within the unit cell. We know that for a body-centered cubic (bcc) we have 2 atoms per unit cell. Therefore, with the knowledge of the number of atoms per unit cell and the mass of each atom we can calculate the mass of a unit cell.
5. Solving for Volume of Unit cell
The volume of a unit cell is equal to the product of the three dimensions (x, y and z) composing the unit cell. In a cubic cell these three dimensions are equivalent. Therefore, a single side length (S) can be used to calculate Area (A) = S^2 or Volume (V) = S^3.
We will call the edge length a = 305 pm
The first thing we want to do is convert this value to the units requested in the problem.
The problem tells us that 1 picometer (pm) is equal to 1x10-12 m.
We also know that 1 m = 100 cm.
Therefore,
a = 305x10-10 cm
V = a^3
Volume = (305x10-8 cm)^3 = 2.838x10-23 cm3
6. Density Calculation
Using the mass and volume calculated we can now plug into our density equation and solve for our answer.
ρ = 5.96 g/cm^3
Conclusion
By following these steps, you can calculate the density of any crystalline material given its unit cell dimensions and atomic mass. For vanadium, with a BCC lattice and an edge length of 305 pm, we found the density to be . This method is applicable to a wide range of materials and crystallographic structures. We can also use this method to back solve for any of these inputs given the appropriate starting information.
References and Attributions:
Acknowledgements
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