metallic iron crystallizes in a cubic lattice. The unit cell edge length is 287 pm. The density of iron is 7.87 g/cm^3. Calculate the mass (in grams) of one unit cell

Fe is a bcc.

mass unit cell = 2*atomic mass Fe/6.02E23 = ?

To calculate the mass of one unit cell of metallic iron, we need to determine the volume of the unit cell and then use the density to calculate the mass.

First, we need to determine the volume of the unit cell. In a cubic lattice, the unit cell is in the shape of a cube, and the edge length is given as 287 pm (picometers).

To convert picometers (pm) to centimeters (cm), we can use the conversion factor 1 pm = 1 x 10^(-12) cm.

Therefore, the edge length in centimeters is:
287 pm x (1 x 10^(-12) cm/pm) = 2.87 x 10^(-8) cm.

Since the unit cell is cubic, the volume can be obtained by raising the edge length to the power of three:

Volume of the unit cell = (2.87 x 10^(-8) cm)^3 = 2.87 x 10^(-8) cm^3.

Now, we can use the density of iron, which is given as 7.87 g/cm^3, to calculate the mass. The density is defined as mass per unit volume.

Therefore, mass = density x volume:

mass = 7.87 g/cm^3 x 2.87 x 10^(-8) cm^3.

Calculating this gives:

mass = 2.26169 x 10^(-7) g.

Thus, the mass of one unit cell of metallic iron is approximately 2.26169 x 10^(-7) grams.