Calculating Density Bravais Lattices

[PsychonautQQ]

Homework Statement

Calculate the density of Iron with the following information.
side length = 2.86 angstroms
Atomic weight = 55.85
Body centered cubic, so 2 atoms per unit cell.

Homework Equations

Density = ((# of atoms)(Atomic Weight)) / ((volume of cell)/Avogadro's Number))

The Attempt at a Solution

So this should be really easy, but I can't get the correct number with this formula, what am I doing wrong?

Atomic Weight of 55.85 means a mass of 9.274117017e-26 kg.
Side Length of 2.87 angstroms mean that the side length is 2.87E-10 Meters.
Cube this side length to get the total volume.

so using the formula...
(2)(9.274117E-26) / (2.87E-10)^3*(6.022^23) = 1.302915E-20 kg/m^3.. which is obviously very incorrect... Help meh?

 

[Dick]

Homework Statement

Calculate the density of Iron with the following information.
side length = 2.86 angstroms
Atomic weight = 55.85
Body centered cubic, so 2 atoms per unit cell.

Homework Equations

Density = ((# of atoms)(Atomic Weight)) / ((volume of cell)/Avogadro's Number))

The Attempt at a Solution

So this should be really easy, but I can't get the correct number with this formula, what am I doing wrong?

Atomic Weight of 55.85 means a mass of 9.274117017e-26 kg.
Side Length of 2.87 angstroms mean that the side length is 2.87E-10 Meters.
Cube this side length to get the total volume.

so using the formula...
(2)(9.274117E-26) / (2.87E-10)^3*(6.022^23) = 1.302915E-20 kg/m^3.. which is obviously very incorrect... Help meh?

You already converted the atomic weight to mass per atom. I don't see why you are dividing by Avogadro's number again. Density is just mass/volume.