Iron crystallizes in a face-centered cubic system. If the radius of an iron atom is 1.26 A (angstroms), what is the edgelength of the unit cell? What is the density of iron if its atomic weight is 55.847 g/mole?

1 Answer
Dec 19, 2015

Answer:

Here's what I got.

Explanation:

In order to be able to calculate the edge length of the unit cell, you need to start from the characteristics of a face-centered cubic system.

As you know, a face-centered cubic system is characterized by a unit cell that has a total of #14# lattice points

  • one lattice point for every one of the eight corners of the unit cell
  • one lattice point for every one of the six faces of the unit cell

Now, these lattice points will contain

  • #1/8"th"# of an atom in every corner lattice point
  • #1/2# of an atom in every face lattice point

https://www.studyblue.com/notes/note/n/ceramics/deck/5172784

To calculate the edge length of the unit cell, you need to focus on one side of the cube. This face will look like this

http://www.chemteam.info/Liquids&Solids/WS-fcc-AP.html

Let's say that #r# represents the radius of an iron atom and #x# represents the edge length of the unit cell. Notice that you can use Pythagoras' Theorem to express #x# in terms of the diagonal of the cell, #d#

#d^2 = x^2 + x^2#

#d^2 = 2x^2#

The diagonal of the cell will be equal to one radius from the top-corner atom, two radii from the face atom, and one radius from the lower-corner atom

#d = r + 2r + r = 4r#

This means that you have

#(4r)^2 = 2x^2#

#16 * r^2 = 2x^2 implies x = sqrt(8r^2)#

This is equivalent to

#x = 2sqrt(2) * r = 2sqrt(2) * "1.26 A" = color(green)("3.56 A")#

So, the edge length of the unit cell is equal to #"3.56 A"#.

As you know, the density of a substance is defined as mass per unit of volume. This means that your goal now will be to find

  • the mass of a unit cell
  • the volume of a unit cell

Start by using Avogadro's number to determine the mass of a single atom of iron

#55.847"g"/color(red)(cancel(color(black)("mol"))) * (1color(red)(cancel(color(black)("mol"))))/(6.022 * 10^(23)"atoms") = 9.274 * 10^(-23)"g/atom"#

Next, use the aforementioned lattice points to determine how many atoms of iron you get per unit cell

#"no. of atoms" = overbrace(1/8 xx 8)^(color(blue)("corner lattices")) + overbrace(1/2 xx 6)^(color(red)("face lattices")) = "4 atoms"#

This means that the mass of a unit cell will be

#9.274 * 10^(-23)"g"/color(red)(cancel(color(black)("atom"))) * (4color(red)(cancel(color(black)("atoms"))))/"unit cell" = 3.710 * 10^(-22)"g/unit cell"#

Next, focus on finding the volume of a unit cell. As you know, the volume of a cube is given by the formula

#color(blue)(V = l xx l xx l = l^3)" "#, where

#l# - the edge length of the cube

Now, convert the edge length of the unit cell from Angstroms to meters first, then from meters to centimeters by using the conversion factor

#"1 A" = 10^(-1)"m"#

#3.56 color(red)(cancel(color(black)("A"))) * (10^(-10)color(red)(cancel(color(black)("m"))))/(1color(red)(cancel(color(black)("A")))) * "1 cm"/(10^(-2)color(red)(cancel(color(black)("m")))) = 3.56 * 10^(-8)"cm"#

The volume of the unit cell will thus be

#V = (3.56 * 10^(-8))^3"cm"^3#

#V = 4.512 * 10^(-23)"cm"^3#

Finally, the density of iron will be equal to

#color(blue)(rho = m/V)#

#rho = (3.710 * 10^(-22)"g")/(4.512 * 10^(-23)"cm"^3) = color(green)("8.22 g/cm"^3)#

The answers are rounded to three sig figs, the number of sig figs you have for the radius of an iron atom.