# How to Calculate the Density of Aluminum in a Face-Centered Cubic Unit Cell

• Qube
In summary, the conversation discusses finding the density of aluminum in a face-centered cubic unit cell. The individual steps to solving the problem are outlined, including determining the mass of the atoms in a unit cell and calculating the volume of the unit cell. The correct formula for finding the density is also discussed, emphasizing the importance of using the basic definition of density as mass/volume.
Qube
Gold Member
(Skip to the bottom for my questions). I'm just sounding off my thoughts above.

## Homework Statement

Find the density of aluminum, which crystallizes in a face-centered cubic unit cell. The atomic radius is 143 pm.

## Homework Equations

edge length of FCC is (4/sqrt2)r.

## The Attempt at a Solution

Okay. I need to express density in grams per centimeter cubed so here goes.

1) There are 4 atoms per face-centered cubic unit cell. Therefore there are 4 aluminum atoms. What is the mass of these atoms?

The mass of one mole of Al is 26.98 grams. 26.98 grams divided by the number of atoms in a mole gives me the mass of each individual atom. 26.98 / (6.022 * 10^23) = 4.480 * 10^-23 grams per atom of Al.

Multiply by 4 and I have the mass of one unit cell of Al (1.79 * 10^-22 grams per unit cell).

2) I have mass. Now I need to find the volume.

10^12 pm = 1 m. Therefore 10^10 pm = 1 cm.

I can cube both sides and now I have 10^30 pm^3 = 1 cm^3. There we go. A measure of volume.

3) Volume of the unit cell.

Using the formula given above, the edge length of a unit cell is 404.46 pm, or 4.04 * 10^-8 cm.

Therefore the volume of a unit cell is simply the edge length (given above) cubed. Or 6.61 * 10^-23 cm^3.

4) We have grams and we have cm^3. We can divide them and get the (simplified) final result.

(1.79 * 10^-22 grams per unit cell) / (6.61 * 10^-23 cm^3) = 2.705 g/cm^3.

Questions:

1) I know my answer is correct. The second step was an intermediary step; I wasn't familiar with working with picometers. Is there, however, a simplified formula for the densities of the various unit cells?

2) How come dividing molar mass / moles = mass of one atom of the element?

g / moles = molar mass right?

(g/moles) / moles = g/moles^2 right?

Intuitively, however, dividing molar mass by moles seems to work. The mass of Avagadro's number of atoms divided by Avagadro's number is ... the mass of one atom.

Last edited:
Qube said:
Questions:

1) I know my answer is correct. The second step was an intermediary step; I wasn't familiar with working with picometers. Is there, however, a simplified formula for the densities of the various unit cells?

I suggest you not to look for it. The way you approached the problem is fine. It's always best to use the basic definition that density is mass/volume.

2) How come dividing molar mass / moles = mass of one atom of the element?

g / moles = molar mass right?

(g/moles) / moles = g/moles^2 right?

Intuitively, however, dividing molar mass by moles seems to work. The mass of Avagadro's number of atoms divided by Avagadro's number is ... the mass of one atom.

I don't get this. You agree that molar mass is mass of one mole? How many atoms are there in a mole? If there are N atoms, what is the mass of single atom?

1 person
Pranav-Arora said:
I suggest you not to look for it. The way you approached the problem is fine. It's always best to use the basic definition that density is mass/volume.

I don't get this. You agree that molar mass is mass of one mole? How many atoms are there in a mole? If there are N atoms, what is the mass of single atom?

Thanks! I think my problem was not realizing the correct units for Avogadro's number, which are atoms/mole.

I mistakenly thought the units for A's number was imply moles. If that were the case, then molar mass (units of which are grams/mole) divided by A's number would yield:

(g/mole) / mole = g/moles^2

But that's simply not the case.

The units of A's number are atoms / mole.

So grams / mole (molar mass) divided by atoms / mole is

grams / mole * (mole / atoms) = grams/atoms.

There we go :)!

## What is the definition of density of unit cells?

The density of unit cells refers to the mass per unit volume of a single unit cell in a crystal lattice. It is an important physical property that characterizes the compactness of a crystal structure.

## How is the density of unit cells calculated?

The density of unit cells can be calculated by dividing the mass of a single unit cell by its volume. The mass of a unit cell can be determined through experimental methods such as weighing and the volume can be calculated by measuring the dimensions of the unit cell.

## What factors affect the density of unit cells?

The density of unit cells can be affected by factors such as the type and mass of atoms or molecules present in the unit cell, the arrangement of these atoms or molecules, and the temperature and pressure conditions under which the crystal is formed.

## How does the density of unit cells relate to the overall density of a material?

The density of unit cells is an important factor in determining the overall density of a material. The overall density is a result of the combined density of individual unit cells and the number of unit cells present in a given volume.

## Why is the density of unit cells important in materials science?

The density of unit cells is an important physical property in materials science as it can provide valuable information about the structural and chemical properties of a material. It can also help in predicting the mechanical, thermal, and electrical properties of a material.

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