An element crystallizes in a body-centered cubic lattice. The edge of the unit cell is 3.13Å , and the density of the crystal is 7.78g/cm3. Calculate the atomic weight of the element. Please help, thanks
3.13 A = 3.13E-8 cm
volume = (3.13E-8)^3 = ?
m = volume x density = ?
bcc is 2 atoms/unit cell.
2 atoms/unit cell*atomic mass/6.02E23 = mass from above.
Solve for atomic mass.
To calculate the atomic weight of an element, we need to know the mass of one atom and the number of atoms in a unit cell.
In a body-centered cubic (BCC) lattice, each corner atom contributes 1/8 to the unit cell, while the atom in the center contributes the full atom to the unit cell. So, there are 8 corner atoms and 1 center atom in a BCC unit cell.
The edge length of the unit cell (a) is given as 3.13 Å.
To find the mass of one atom, we can follow these steps:
Step 1: Convert the edge length from Ångstroms to centimeters.
1 Ångstrom (Å) = 1 x 10^(-8) cm
3.13 Å = 3.13 x 10^(-8) cm
Step 2: Calculate the volume of the unit cell.
The volume of a cube is given by the formula V = a^3, where a is the edge length.
V = (3.13 x 10^(-8) cm)^3 = 3.13^3 x (10^(-8))^3 = 3.13^3 x 10^(-24) cm^3
Step 3: Calculate the mass of the unit cell using the density information.
Density (ρ) = mass (m) / volume (V)
Rearranging the equation to solve for mass: mass (m) = density (ρ) x volume (V).
Given density (ρ) = 7.78 g/cm^3 and volume (V) = 3.13^3 x 10^(-24) cm^3:
mass (m) = 7.78 g/cm^3 x 3.13^3 x 10^(-24) cm^3
Step 4: Calculate the mass of one atom.
Since there are 8 corner atoms and 1 center atom in a BCC unit cell, the mass of one atom is:
Mass of one atom = mass (m) / (8 + 1) = mass (m) / 9
Finally, you can calculate the atomic weight by multiplying the mass of one atom by the number of atoms in 1 mole.
Atomic Weight = Mass of one atom x Avogadro's number
Avogadro's number, N_A = 6.022 x 10^23 atoms/mole
Please plug in the values and compute the final result.