# Question #dcf44

##### 1 Answer

#### Answer:

Here's what I got.

#### Explanation:

Your strategy here will be to determine the *volume* of the unit cell by using the given edge length, use gold's density to determine the *mass* of the unit cell, then use the mass of a *single* gold atom to determine how many atoms you get *per unit cell*.

So, the first important thing to notice here is that the density of gold is given in *grams per cubic centimeters*, but that the length of the unit cell is given in *picometers*.

This means that you're going to have to convert the length of the cell to match that used for density. Use the fact that

#"1 pm" = 10^(-12)"m" = 10^(-10)"cm"#

to get

#407.9color(red)(cancel(color(black)("pm"))) * (10^(-10)"cm")/(1color(red)(cancel(color(black)("pm")))) = 4.079 * 10^(-8)"cm"#

As you know, the *volume* of a cube is given by the formula

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

Plug in the value you have for the length of the cubic unit cell to get

#V = (4.079 * 10^(-8))^3 "cm"^3 = 6.787 * 10^(-23)"cm"^3#

Use the density of gold to determine the *mass* of a cubic unit cell

#6.787 * 10^(-23)color(red)(cancel(color(black)("cm"^3))) * "19.3 g"/(1color(red)(cancel(color(black)("cm"^3)))) = 1.31 * 10^(-21)"g"#

Now all you have to do is determine *how many atoms* of gold you could fit in the unit cell. To do that, use gold's **molar mass**, which tells you what the mass of **one mole** of gold atoms is.

Since you're interested in finding the mass of a *single* gold atom, divide the molar mass by **Avogadro's number**, which tells you the number of atoms you get per mole of an element.

#196.97"g"/color(red)(cancel(color(black)("mole of Au"))) * overbrace((1color(red)(cancel(color(black)("mole Au"))))/(6.022 * 10^(23)"atoms"))^(color(green)("Avogadro's number")) = 3.271 * 10^(-22)"g/atom"#

Now, if your unit cell has a mass of

#1.31 * 10^(-21)color(red)(cancel(color(black)("g"))) * "1 atom of Au"/(3.271 * 10^(-22)color(red)(cancel(color(black)("g")))) = 4.0049 ~~ color(green)("4 atoms of Au")#

Now, when it comes to **cubic lattice structures**, you can only deal with three possible sub-types

asimple cubic structureabody-centeredcubic structureaface-centeredcubic structure

Now, take a look at those three possible unit cells. Notice that they all have **lattice points** in the corners of the cube.

Each of these lattice points contains **faces** of the cube.

For a face-centered cubic cell, each face contains **total number of atoms** that can fit into a face-centered cubic cell will be

#"no. of atoms" = overbrace(1/8 xx 8)^(color(red)("8 corners")) + overbrace(1/2 xx 6)^(color(green)("6 faces")) = "4 atoms"#

And there you have it, gold has a **face-centered cubic** lattice structure.