1. The element gold, Au, has a face-centered cubic structure. (Density - 19.3 g/cm^3)
(a) What are the # of gold atoms in 1 unit cell?
(b) What are the # of unit cells in 1 mol?
(c) What is the volume of 1 mol gold unit cells?
(d) What is the volume of 1 unit cell?
(e) what is the length of 1 side of a unit cell?
(f) What is the metallic diameter of 1 atom?
You don't look for answers? I responded to your first post below. If you have questions about it follow up with them but just posting again is no help to either of us. Show your work if you have questions.
To answer these questions, we need to know some details about the face-centered cubic (FCC) structure of gold.
In a face-centered cubic lattice, each atom is present at the center of each face of a cube and at the corners of the cube. Each edge of the cube has one atom, and each corner atom is shared by eight unit cells.
Now, let's go through each question step by step:
(a) Number of gold atoms in 1 unit cell:
In an FCC structure, there are 4 atoms per unit cell. Each corner atom contributes 1/8th of itself to the unit cell, and each face-centered atom contributes 1/2. So, the total contribution of atoms from corners is 1 * (1/8) = 1/8, and the total contribution of atoms from face centers is 6 * (1/2) = 3. Adding them together, we get 1/8 + 3 = 4 atoms per unit cell.
(b) Number of unit cells in 1 mol:
Avogadro's number (6.022 x 10^23) represents the number of atoms or molecules in one mole of a substance. Since there are 4 atoms per unit cell, the number of unit cells in 1 mole is given by (6.022 x 10^23)/(4).
(c) Volume of 1 mol gold unit cells:
To calculate this, we need the density of gold. The density is given as 19.3 g/cm^3. From this, we can infer that 19.3 g of gold occupies 1 cm^3 of space. Since the molar mass of gold is approximately 197 g/mol, we divide the molar mass by the density to find the volume occupied by 1 mole of gold atoms.
(d) Volume of 1 unit cell:
The volume of 1 unit cell can be calculated by taking the volume of the cube. The length of one side of the cube is required to find the volume.
(e) Length of 1 side of a unit cell:
To find the length of one side of the unit cell, we can use the formula for the FCC structure, which relates the radius of the atoms to the side length of the unit cell. It can be expressed as follows:
Side length (a) = 4 * radius (r) * sqrt(2)
So, by rearranging the formula, we can find the length of one side of the unit cell.
(f) Metallic diameter of 1 atom:
The metallic diameter of an atom can be calculated using the side length of the unit cell. Since each atom is present at the center of a face of the cube, the distance from one face center to the opposite face center is equal to twice the metallic diameter.
By calculating using the length of one side of the unit cell, we can find the metallic diameter of one atom.
Keep in mind that you'll need to know the atomic radius of gold and use appropriate units in your calculations (e.g., cm for lengths and cm^3 for volumes) to get the correct results.