# Packing density question (HCP/CCP/BC)

1. Sep 25, 2006

### berry2

Hi! I'm doing a Grade 12 Chem course by correspondance, so I don't have a teacher to answer my questions about the module. I had to do an experiment about hexagonal closest packing, cubic closest packing and body-centred space lattice. Here is one of the questions: "If a box fit tightly around the three layers, what would its volume be? Assuming a mass of 1 unit per sphere, calculate the density in units/cm^3."

This question has already been addressed here: https://www.physicsforums.com/archive/index.php/t-126524.html, but after reading through it I still did not understand where everything came from. The textbook that came with the course is "Foundations of Chemistry" by Harcourt and I cannot find any information on calculating the density anywhere in the book.

I am absolutely confused.

Here is part of question that was addressed in the link above:

"You have an FCC lattice. Look at the unit cell - it has 8 corner atoms and 6 face centers {I'm looking at some pictures of the FCC lattice and I cannot figure out how it has 8 cornor atoms or where 6 face centres came from} . Each corner atom is shared by 8 neighboring unit cells and each face center atom is shared by two. So the total number of atoms per unit cell in the FCC lattice = (8*1/8) + (6*1/2) = 1+3 = 4

Nearest neighbor atoms ("touching" each other) may be found along a face diagonal. If the side of the unit cell is 'a', the length of the face diagonal is a*sqrt(2) = about 1.4a. Going from one corner to another, along the face diagonal, you encounter the radius of the first corner atom, then the diameter of the face center atom, and finally the radius of the second corner atom. In terms of the atomic radius, the length of the face diagonal is then r + 2r + r. So, we have 1.4a = 4r or roughly a = 2.83r

That establishes a relationship between the sphere radius and the unit cell size. Next you say that the unit cell volume is a^3, and the volume occupied by spheres is 4*(4/3)*pi*r^3. The 4 comes from the number of atoms per unit cell, calculated abouve. The packing density is then the ratio of the second volume to the first.

PD = [(16/3)*pi*r^3]/[a^3] = 16.75*r^3/a^3 = [16.75]/[2.83^3] = 16.75/22.62 = 0.74"

I have never been very good with chemistry, and this has completely stumped me. If someone wouldn't mind just explaining how to do this question, and why certain numbers are in certain places, I would be really grateful. Also, what is the equation that is supposed to be used for this question (I can't find it anywhere in the material provided)?

Thanks so much!

2. Sep 25, 2006

### big man

Well you just need some good pictures to demonstrate this.

Here are some notes I found a while back from a uni near me on the basics. They're pretty decent. This is just a start and I'm sure other people will chime in with a lot of useful information.

http://www.physics.curtin.edu/teaching/units/Materials_Science_Units/Materials_Science_201/

Lectures 1 and 2 cover what you want to know I think.

EDIT

Lecture 2 from the above link gives you the formula for atomic packing factor (or packing density)

$$APF = \frac {n V_s} {V_c_e_l_l}$$

Where Vs is the volume of a sphere and n is the number of atoms per unit cell for a particular structure. The volume of a cubic unit cell (Vcell) (simple-cubic, body-centred or Face-centred) is just $$a^3$$ (a is the lattice parameter). To solve for the APF you need to express a in terms of the atomic radius. Lecture 2 from my link I think clearly demonstrates how you do this.

The volume of an HCP unit cell (Vcell) is obviously slightly different. The base of an HCP structure is a hexagon. The hexagon can be divided into 6 equal triangles. So you find the area of one of those triangles (using the fact that you know the lattice parameter) and multiply by 6 to get the area of the base of the HCP structure. Then multiply this area by the height (c, where c=1.633a) to obtain the volume of an HCP structure. The notes don't explain how to calculate the volume of the HCP structure, but if you use this explanation and look at the diagram in the notes maybe you can make some sense out of it???

The notes that I linked to (lecture 2) also give the formula for caculating the density of an element given the information of the crystal structure and the atomic mass.

I'm probably not the best person to explain this though, because I'm not too good at clearly making a point, so if you don't understand what I've tried to say then just say so so that the others will help you. If you don't then they might think that your problem is sorted out.

Last edited: Sep 26, 2006