An element crystallizes in a body-centered cubic lattice. The edge of the unit cell is 3.5 Angstrom, and the density of the crystal is 7.39g/cm^3. Calculate the atomic weight of the element. A= ? g/mol
95.4g/mol
The bcc has two atoms/unit cell.
The edge is 3.5 x 10^-8 cm = a
Volume = a3 = 3.5 x 10^-8)^3 cc
mass = volume x density
mass = a3 x 7.49 g/cc = xx
2atoms/unit cell x atomic mass/6.022 x 10^23 = xx mass
Solve for atomic mass.
I made a typo. The density is 7.39, not 7.49 g/cc.
To calculate the atomic weight of the element, A, we need to use the formula:
A = (Z * M) / N_A
Where:
Z = number of atoms per unit cell
M = molar mass of the element
N_A = Avogadro's number (6.022 x 10^23 mol^-1)
First, let's find the number of atoms per unit cell (Z). For a body-centered cubic (BCC) lattice, there is one atom at each corner of the unit cell and an additional atom at the center of the cell.
The total number of atoms in a BCC unit cell (Z) is given by Z = 1 + 1/8 = 9/8.
Next, we need to find the molar mass of the element (M). To do this, we can use the density of the crystal and the edge length of the unit cell.
The volume of the unit cell (V) is given by V = a^3, where a is the edge length of the unit cell. In this case, a = 3.5 Angstrom = 3.5 x 10^-8 cm.
The mass of the unit cell (m) is given by m = V * density = a^3 * density.
Since the unit cell contains Z atoms, the mass of one atom (m_a) is given by m_a = m / Z.
Now, we can calculate the molar mass of the element (M) using the mass of one atom (m_a) and Avogadro's number (N_A):
M = m_a * N_A
Let's calculate the atomic weight (A) using the above formulas:
1. Calculate the volume of the unit cell (V)
V = (3.5 x 10^-8 cm)^3 = 4.91 x 10^-23 cm^3
2. Calculate the mass of the unit cell (m)
m = V * density = (4.91 x 10^-23 cm^3) * (7.39g/cm^3) = 3.62 x 10^-22 g
3. Calculate the mass of one atom (m_a)
m_a = m / Z = (3.62 x 10^-22 g) / (9/8) = 4.04 x 10^-22 g
4. Calculate the molar mass of the element (M)
M = m_a * N_A = (4.04 x 10^-22 g) * (6.022 x 10^23 mol^-1) = 243.68 g/mol
Therefore, the atomic weight of the element is approximately 243.68 g/mol.