You are given a small bar of an unknown metal, X. You find the density of the metal to be 11.7 g/cm3. An X-ray diffraction experiment measures the edge of the unit cell as 5.12 angstroms. Assuming the metal crystallizes in a face-centered cubic lattice, what is the atomic weight of X in g/mole? (1 angstrom = 1 x 10^-8 cm)

Question

You are given a small bar of an unknown metal, X. You find the density of the metal to be 11.7 g/cm3. An X-ray diffraction experiment measures the edge of the unit cell as 5.12 angstroms. Assuming the metal crystallizes in a face-centered cubic lattice, what is the atomic weight of X in g/mole? (1 angstrom = 1 x 10^-8 cm)

Answer 1

edge unit cell = 5.12E-8 cm

volume unit cell = edge^3
mass unit cell = volume x density = ?
Substitute into
? mass = 4atoms/unit cell x atomic mass/6.022E23
Solve for atomic mass.

Answer 2

To determine the atomic weight of element X, we need to use the formula:

Atomic weight = (Molar mass / Avogadro's number)

To calculate the molar mass, we need to find the number of atoms in a unit cell. For a face-centered cubic (FCC) lattice, there are four atoms in each unit cell.

Next, we need to calculate the volume of the unit cell. Since the edge of the unit cell is given as 5.12 angstroms, we can find the volume using the formula:

Volume = (Edge length)^3

Convert the edge length from angstroms to centimeters:
Edge length = 5.12 angstroms * (1 x 10^-8 cm/1 angstrom) = 5.12 x 10^-8 cm

Calculate the volume:
Volume = (5.12 x 10^-8 cm)^3 = 1.32768 x 10^-23 cm^3

Now, divide the molar mass by Avogadro's number to get the atomic weight:

Atomic weight = Molar mass / Avogadro's number

Since we are given the density of the metal, we can use the equation:

Density = (Molar mass / Volume) x (Avogadro's number / Atomic volume)

However, we know that the atomic volume for FCC lattice is given by:

Atomic volume = (Edge length)^3 / 4

Substituting the values into the equation:

Density = (Molar mass / (1.32768 x 10^-23 cm^3)) x ((6.022 x 10^23) / (5.12 x 10^-8 cm))^3 / 4

To solve for the atomic weight, rearrange the equation:

Molar mass = (Density x Atomic volume) / (Avogadro's number/4)

Substitute the given values:

Molar mass = (11.7 g/cm^3 x (5.12 x 10^-8 cm)^3 / 4) / (6.022 x 10^23/4)

Simplifying the equation:

Molar mass = 11.7 g/cm^3 x (5.12 x 10^-8 cm)^3 / 6.022 x 10^23

Now, plug in the values and calculate the molar mass:

Molar mass = 11.7 g/cm^3 x (5.12 x 10^-8 cm)^3 / 6.022 x 10^23
= 48.81 / 6.022 x 10^23
≈ 8.1137 x 10^-23 g/mol

Therefore, the atomic weight of element X is approximately 8.1137 x 10^-23 g/mol.