Info:
The atoms of crystalline solid pack together into a three-dimensional array of many small repeating units called unit cells. The simplest of the unit cells are those that have cubic symmetry, with atoms positioned at the corners of a cube. Atoms can also be found in the sides (faces) of the cube, or centered within the body of the cube. It is important to realize that a unit cell is surrounded by other unit cells in every direction. Therefore, face and corner atoms are shared with neighboring unit cells. The fraction varies with the type of atom as shown in the following table.
Type of atom / Fraction in unit cell
corner 1/8
face 1/2
body 1
The size of a unit cell in any given solid can be calculated by using its density. This and the reverse calculation are common test questions in general chemistry courses.
**I am having problems with the following 2 questions:
1. Gallium crystallizes in a primitive cubic unit cell. The length of an edge of this cube is 362pm. What is the radius of a gallium atom?
Express your answer numerically in picometers.
"radius = _____pm"
&
2. The face-centered gold crystal has an edge length of 407pm. Based on the unit cell, calculate the density of gold.
Express your answer numerically in grams per cubic centimeter.
density = ______g/cm^3
For a primitive unit cell, a = 2r where a is the length of the edge of the cell and r is the radius.
For the fcc gold atom, density = mass/volume.
volume calculation:
Volume = (edge)3. Don't forget to change the pm to cm if you want density in g/cc.
mass calculation:
There are 4 atoms per unit cell along the face of one cube.
mass = (4atoms)(atomic mass Au)/6.02 x 10^23 atoms mol-1
Then density = mass/volume
1. To find the radius of a gallium atom, we need to first determine the length of the diagonal of the unit cell.
In a primitive cubic unit cell, the length of the edge is equal to twice the radius of the atom.
Given that the edge length of the cube is 362 pm, we can calculate the radius as follows:
Radius = Edge Length / 2 = 362 pm / 2 = 181 pm
Therefore, the radius of a gallium atom is 181 pm.
2. To calculate the density of gold, we need to know the mass of the unit cell and its volume.
The face-centered gold crystal has an edge length of 407 pm. Since it is face-centered, there are atoms at the corners and in the center of each face, but none in the body.
First, let's calculate the volume of the unit cell:
Volume = Edge Length^3 = (407 pm)^3 = 67,974,791 pm^3
Next, we need to determine the mass of the unit cell. This can be done using the information provided in the question.
Gold has a density of 19.32 g/cm^3. Given that the unit cell consists of both corner and face atoms, we need to consider the fraction of atoms in the unit cell.
Density = (Mass of Unit Cell) / (Volume of Unit Cell)
To calculate the mass of the unit cell, we can use the given fraction of atoms in the unit cell.
Using the table provided in the information:
Mass of Unit Cell = (Number of Corner Atoms x Atomic Mass) + (Number of Face Atoms x Atomic Mass)
= (1/8 x Atomic Mass) + (1/2 x Atomic Mass)
= 5/8 x Atomic Mass
Now we can calculate the density:
Density = (5/8 x Atomic Mass) / (Volume of Unit Cell)
= (5/8 x Atomic Mass) / (67,974,791 pm^3)
Since we want the density in grams per cubic centimeter, we need to convert the volume to cm^3 and the atomic mass to grams.
1 pm = 1E-10 cm (conversion factor)
Atomic Mass of Gold = 196.97 g/mol
Volume of Unit Cell (cm^3) = Volume of Unit Cell (pm^3) x (1E-10 cm/pm)^3
= 67,974,791 pm^3 x (1E-10 cm/pm)^3
Now we can substitute the values and calculate the density:
Density = (5/8 x 196.97 g/mol) / (67,974,791 pm^3 x (1E-10 cm/pm)^3)
Simplifying the equation will give the density in g/cm^3.
Once the calculation is complete, the answer will be expressed as the density in grams per cubic centimeter (g/cm^3).