The density of an unknown metal is 1.55 and its atomic radius is 0.197 . It has a face-centered cubic lattice. What is the atomic weight of this metal?

Question

The density of an unknown metal is 1.55 and its atomic radius is 0.197 . It has a face-centered cubic lattice. What is the atomic weight of this metal?

Answer 1

I could work this problem if it had units. No units on density, none on radius (that could be pm, nm, what?).

Answer 2

Sorry about that - the density is 1.55 g/cm3 and the radius is 0.197 nm.

Answer 3

There are 4 atoms to the unit cell in a fcc.

(4*atomic mass /6.02 x 10^23) = mass of unit cell. You need the mass.

4*radius atom = a*2^1/2
Solve the last equation for a

volume of unit cell = a³

Then plug into mass = volume x density to calculate mass of unit cell. Plug mass into the top equation to calculate atomic mass.

Answer 4

By the way, I would not use the radius in pm. I would change that to cm first thing out of the cage; otherwise you must convert volume in pm³ to cc and that's a little more difficult and one more step to make an error. Same thing for your next post.

Answer 5

To calculate the atomic weight of the metal, we first need to determine its molar volume, which is the volume occupied by one mole of atoms.

In a face-centered cubic (FCC) lattice, each lattice point is fully occupied by an atom, with additional atoms at the centers of each face. This means that there are eight atoms contained within the unit cell.

The volume of the unit cell in a face-centered cubic lattice can be calculated using the formula:
V = [(4/3)πr^3] * (8/8)

Given the atomic radius (r) of 0.197, we plug it into the formula:
V = [(4/3)π(0.197)^3] * (8/8)

V ≈ 0.004879

Next, we calculate the molar volume (Vm) by dividing the volume of the unit cell by the Avogadro's number (6.022 x 10^23):
Vm = V / (6.022 x 10^23)

Vm ≈ 8.109 x 10^-27 m^3

The density of the metal (ρ) is given as 1.55 g/cm^3. To convert this to kg/m^3, we need to multiply by 1000 (since there are 1000 grams in a kilogram):
ρ = 1.55 * 1000 kg/m^3

ρ = 1550 kg/m^3

Now, we can calculate the atomic weight (M) using the equation:
M = (ρ * Vm) / Na

Substituting the known values:
M = (1550 kg/m^3 * 8.109 x 10^-27 m^3) / 6.022 x 10^23

M ≈ 2.09 x 10^-23 kg

Therefore, the atomic weight of the unknown metal is approximately 2.09 x 10^-23 kg.

The density of an unknown metal is 1.55 and its atomic radius is 0.197 . It has a face-centered cubic lattice. What is the atomic weight of this metal?

I could work this problem if it had units. No units on density, none on radius (that could be pm, nm, what?).

Sorry about that - the density is 1.55 g/cm3 and the radius is 0.197 nm.

There are 4 atoms to the unit cell in a fcc.

By the way, I would not use the radius in pm. I would change that to cm first thing out of the cage; otherwise you must convert volume in pm3 to cc and that's a little more difficult and one more step to make an error. Same thing for your next post.

To calculate the atomic weight of the metal, we first need to determine its molar volume, which is the volume occupied by one mole of atoms.

By the way, I would not use the radius in pm. I would change that to cm first thing out of the cage; otherwise you must convert volume in pm³ to cc and that's a little more difficult and one more step to make an error. Same thing for your next post.