Pure gold adopts a face-centered cubic crystal structure. It also has a density of 19.30g/mL. What is the edge length of gold's unit cell? What is the atomic radius of gold?
I understand how to do this problem the other way: when you have the unit cell and are looking for the mass, but I'm not sure where to start here.
To find the edge length of gold's unit cell and the atomic radius, we can use the concepts of crystal structure and density.
Step 1: Determine the number of atoms per unit cell
In a face-centered cubic (FCC) crystal structure, there are 4 atoms per unit cell. Each corner atom contributes 1/8th of its volume to the unit cell, and each face-centered atom contributes 1/2 of its volume.
Step 2: Calculate the volume of the unit cell
Since the unit cell of an FCC crystal structure consists of cube-shaped cells, the volume can be calculated using the formula:
Volume of unit cell = edge length^3
Step 3: Determine the mass of the unit cell
Using the given density of gold (19.30 g/mL), we can calculate the mass of the unit cell by multiplying the volume of the unit cell by the density:
Mass of unit cell = Volume of unit cell x Density
Step 4: Calculate the mass of one gold atom
The mass of one gold atom can be determined by dividing the mass of the unit cell by the number of atoms per unit cell (4 in this case).
Step 5: Calculate the atomic radius
The atomic radius can be found by using the formula:
Atomic radius = (0.5 x edge length) / √2
Let's apply these steps to solve the problem:
Given: Density of gold = 19.30 g/mL
We need to find: Edge length of the unit cell and atomic radius of gold.
Step 1: Number of atoms per unit cell = 4
Step 2: Calculate the volume of the unit cell:
Volume of unit cell = edge length^3
Step 3: Calculate the mass of the unit cell:
Mass of unit cell = Volume of unit cell x Density
Step 4: Calculate the mass of one gold atom:
Mass of one gold atom = Mass of unit cell / Number of atoms per unit cell
Step 5: Calculate the atomic radius:
Atomic radius = (0.5 x edge length) / √2
Let's proceed with the calculations.
To find the edge length of gold's unit cell and the atomic radius of gold, we can use the information that gold adopts a face-centered cubic (FCC) crystal structure and has a density of 19.30 g/mL.
1. Find the number of gold atoms per unit cell in the FCC structure:
In FCC structure, each corner of the unit cell contains 1/8th of an atom, and each face contains 1/2 of an atom. So, the total number of gold atoms per unit cell is:
Number of atoms = (8 corners * 1/8) + (6 faces * 1/2) = 4 atoms.
2. Calculate the volume of the unit cell:
The unit cell of a FCC structure can be thought of as a cube with atoms positioned at each corner and face center. The volume (V) of such a cube is given by:
V = a^3, where "a" is the edge length of the cube.
3. Calculate the mass of the unit cell:
The mass of the unit cell (M) can be calculated by multiplying the density (ρ) by the volume (V):
M = ρ * V.
4. Calculate the atomic mass of gold:
The atomic mass of gold (Au) is 197 g/mol.
5. Calculate the number of moles of gold atoms in the unit cell:
The number of moles (n) can be calculated by dividing the mass of the unit cell by the atomic mass of gold:
n = M / atomic mass.
6. Calculate the edge length of the unit cell:
To find the edge length (a) of the unit cell, we can rearrange the formula for volume (V = a^3) and solve for "a":
a = V^(1/3).
7. Calculate the atomic radius of gold:
To find the atomic radius (r) of gold, we can use the formula:
r = a / 2.
Now, let's calculate the edge length of the unit cell and the atomic radius of gold using these steps.