Question

An element crystallizes in a face-centered cubic lattice, and it has a density of \(1.45 \mathrm{~g} / \mathrm{cm}^{3}\). The edge of its unit cell is \(4.52 \times 10^{-8} \mathrm{~cm}\) (a) How many atoms are in each unit cell? (b) What is the volume of a unit cell? (c) What is the mass of a unit cell? (d) Calculate an approximate atomic mass for the element.

Short answer

There are 4 atoms per unit cell. Volume: \(9.22 \times 10^{-23} \text{ cm}^3\). Mass: \(1.34 \times 10^{-22} \text{ g}\). Approximate atomic mass: 20.18 u.

Step-by-step solution

Determine the number of atoms per unit cell

In a face-centered cubic (FCC) lattice, there are 4 atoms per unit cell. Use this knowledge to understand the number of atoms in each unit cell is 4.

Calculate the volume of the unit cell

The volume of a unit cell in a cube is calculated by cubing the edge length. \[ V_{\text{cell}} = a^3 \] where \( a = 4.52 \times 10^{-8} \text{ cm} \).Substitute the edge length into the formula: \[ V_{\text{cell}} = (4.52 \times 10^{-8} \ \text{cm})^3 \] Calculate the volume: \[ V_{\text{cell}} = 9.22 \times 10^{-23} \ \text{cm}^3 \]

Determine the mass of a unit cell

Density (\( \rho \)) of the element is given as 1.45 g/cm³. Use the formula for density: \[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \] Rearrange the formula to solve for mass (m): \[ m = \rho \times V \] Substitute the given values: \[ m = 1.45 \text{ g/cm}^3 \times 9.22 \times 10^{-23} \text{ cm}^3 \] Calculate the mass: \[ m = 1.34 \times 10^{-22} \ \text{g} \]

Calculate the atomic mass of the element

Use the fact that there are 4 atoms per unit cell and divide the mass of the unit cell by the number of atoms to get the atomic mass:\[ \text{Atomic Mass} = \frac{\text{Mass of unit cell}}{\text{Number of atoms per unit cell}} \] Substitute the values: \[ \text{Atomic Mass} = \frac{1.34 \times 10^{-22} \ \text{g}}{4} \] Calculate the atomic mass: \[ \text{Atomic Mass} = 3.35 \times 10^{-23} \ \text{g/atom} \]

Convert atomic mass to more familiar form (if needed)

1 Dalton (Da) is equivalent to approximately \(1.66 \times 10^{-24} \ \text{g} \). So the approximate atomic mass in atomic mass units (u) is obtained by dividing: \[ \text{Atomic Mass (in u)} = \frac{3.35 \times 10^{-23} \ \text{g}}{1.66 \times 10^{-24} \ \text{g/u}} \] Calculate: \[ \text{Atomic Mass (in u)} = 20.18 \ \text{u} \]

Key concepts

Unit Cell Calculations

A unit cell is the smallest repeating structure in a crystal lattice. In a face-centered cubic (FCC) lattice, each unit cell contains 4 atoms. Here's how you can understand and calculate various properties of a unit cell:
First, to find how many atoms are in each unit cell, remember that in FCC structures, there are atoms at the corners and centers of each face. Each corner atom is shared among 8 unit cells and each face-centered atom is shared between 2 cells. Using this, the equation would be:

• Corners: 8 atoms x (1/8) = 1 atom
• Faces: 6 atoms x (1/2) = 3 atoms.

Adding these together, we get 4 atoms per unit cell in an FCC lattice.
This basic understanding helps in dealing with subsequent calculations such as determining the volume, mass, and atomic mass of the unit cell.

Density in Crystallography

Crystallography involves understanding and measuring the density of these repetitive units. In this context, density (\( \rho \)) is calculated as mass per unit volume. The formula is:
\( \text{Density} = \frac{\text{Mass}}{\text{Volume}} \)
For a unit cell in a FCC lattice, determining the volume is straightforward. If the edge length (a) is given, the volume (\( V_{\text{cell}} \)) can be found by:
\( V_{\text{cell}} = a^3 \)
Let’s say the edge length a = 4.52 x 10^-8 cm. Cubing this value yields the unit cell volume:
\( V_{\text{cell}} = (4.52 \times 10^{-8} \, \text{cm})^3 = 9.22 \times 10^{-23} \, \text{cm}^3 \)
Then, the mass (\( m \)) of the unit cell can be determined using the given density.
Rearranging the density formula to solve for mass gives us:
\( m = \rho \times V \)
Plug in the known values:
\( m = 1.45 \, \text{g/cm}^3 \times 9.22 \times 10^{-23} \, \text{cm}^3 \)
This simplifies to:
\( m = 1.34 \times 10^{-22} \, \text{g} \)
By using these calculations, one can determine the density and, subsequently, the mass of intricate crystallographic structures.

Atomic Mass Determination

Determining the atomic mass of an element within a crystal structure involves calculating the mass of the unit cell and dividing it by the number of atoms in the cell.
From the previous steps, we have the mass of one unit cell, 1.34 x 10^-22 g. Since the FCC unit cell contains 4 atoms, the atomic mass (M) can be found using:
\( M = \frac{\text{Mass of unit cell}}{\text{Number of atoms per unit cell}} \)
Plugging in the numbers:
\( M = \frac{1.34 \times 10^{-22} \, \text{g}}{4} = 3.35 \times 10^{-23} \, \text{g/atom} \)
To convert this mass into more familiar atomic mass units (u), use the relation between grams and Daltons: 1 Dalton (Da) ≈ 1.66 x 10^-24 g. The atomic mass in atomic mass units (u) is:
\( M_{\text{u}} = \frac{3.35 \times 10^{-23} \, \text{g}}{1.66 \times 10^{-24} \, \text{g/u}} \)
Which simplifies to:
\( M_{\text{u}} = 20.18 \, \text{u} \)
This shows how the atomic mass of an element in a face-centered cubic lattice can be systematically found by using the mass of the unit cell and the number of atoms within it.