CaO has a face-centered cubic unit cell in which the O2- anions occupy corners and face centers, while the cations fit into the hole between adjacent anions. What is its density if the ionic radii of Ca2+, O2- ions are 57.63 pm and 184.0 pm, respectively?
To calculate the density of CaO, we need to determine the volume of the unit cell and the total mass.
The face-centered cubic (FCC) unit cell consists of 4 atoms: 1 calcium ion (Ca2+) and 3 oxygen ions (O2-). The calcium ion occupies the center of the unit cell and the oxygen ions occupy the corners and face centers.
Let's calculate the volume of the unit cell:
Since the O2- ions are on the corners, there are 8 corners per unit cell. Each corner is shared by 8 adjacent unit cells, so each corner contributes 1/8 of an O2- ion to the unit cell.
Since there are 4 Ca2+ ions in each unit cell, the formula unit is Ca4O4.
The contribution of the cations (Ca2+) can be calculated by the number of cations per unit cell multiplied by their volume contribution. Since Ca2+ is at the center of the unit cell, there is only one Ca2+ per unit cell, and its volume contribution is the volume of a whole ion.
The contribution of the anions (O2-) can be calculated by the number of anions per unit cell multiplied by their volume contribution. Since O2- is on the corners, there are 8 corners per unit cell, and each corner contributes 1/8 of an O2- ion. Additionally, there are 6 face-centered anions per unit cell, and each face-centered anion contributes 1/2 of an O2- ion. So, we have a total of (8/8) + (6/2) = 8 O2- ions per unit cell, and each O2- ion has the same volume contribution.
Now let's calculate the volume of the unit cell:
1. Volume of the calcium ion (Ca2+):
V_Ca2+ = (4/3) * π * (r_Ca2+)³
= (4/3) * π * (57.63 pm)³
2. Volume of the oxygen ion (O2-):
V_O2- = (4/3) * π * (r_O2-)³
= (4/3) * π * (184.0 pm)³
3. Total volume of the unit cell:
V_unit cell = (1 * V_Ca2+) + (8 * V_O2-)
Now let's calculate the total mass of the unit cell:
The total mass of the unit cell will be the sum of the masses of all the atoms in the unit cell.
M_Ca2+ = molar mass of Ca2+
M_O2- = molar mass of O2-
Now we can calculate:
M_unit cell = (M_Ca2+ * 1) + (M_O2- * 8)
Finally, we can calculate the density of the unit cell:
Density = M_unit cell / V_unit cell
To determine the density of CaO, we need to know the formula unit, the volume of the unit cell, and the total mass of the unit cell.
In a face-centered cubic (FCC) unit cell, there are 4 atoms or ions in total. The formula unit of CaO consists of one Ca2+ ion and one O2- ion, so we have 2 ions in the unit cell.
First, let's calculate the volume of the unit cell:
The length of the unit cell edge (a) can be determined using the ionic radii of Ca2+ and O2- ions. In an FCC arrangement, the length of the unit cell edge is related to the ionic radii as:
a = 4 * r_cation
In this case, the cation is Ca2+, so we have:
a = 4 * (57.63 pm) = 230.52 pm = 230.52 x 10^-10 m (converting to meters)
The volume of an FCC unit cell can be calculated as:
volume = a^3
Therefore, the volume of the unit cell is:
volume = (230.52 x 10^-10 m)^3 = 12.702 x 10^-27 m^3
Now, we need to determine the mass of the unit cell:
The mass of the unit cell is the sum of the masses of the cation and anion present in the formula unit.
The molar mass of CaO can be obtained from the periodic table. The molar mass of Ca is approximately 40 g/mol, and the molar mass of O is approximately 16 g/mol.
So, the mass of the unit cell is:
mass = (molar mass of Ca) + (molar mass of O)
= 40 g/mol + 16 g/mol
= 56 g/mol
Now, let's convert the mass to kg:
mass = 56 g/mol = 56 x 10^-3 kg/mol
Finally, we can calculate the density using the formula:
density = mass / volume
density = (56 x 10^-3 kg/mol) / (12.702 x 10^-27 m^3) = 4.402 x 10^3 kg/m^3
Therefore, the density of CaO is approximately 4.402 x 10^3 kg/m^3.