# Homework Help: How do I calculate density of calcium from atomic radius?

1. Sep 8, 2015

### Eclair_de_XII

1. The problem statement, all variables and given/known data
"Calcium crystallizes in a body-centered cubic structure. (a) How many Ca atoms are contained in each unit cell? (b) How many nearest neighbors does each Ca atom possess? (c) Estimate the length of the unit cell edge, a, from the atomic radius of calcium (1.97A). (d) Estimate the density of Ca metal."

2. Relevant equations
(a) two Ca atoms
(b) eight neighbors
Pythagorean theorem: a2 + b2 = c2
http://education.mrsec.wisc.edu/SlideShow/images/unit_cells/body_centered_cubic2.jpg
r = 1.97 A
DCa = 1.54 g/cm3

3. The attempt at a solution
Okay, I am sure I messed up at part (c), where I'm calculating the length of the unit cell's edges in relation to the atomic radius. The longest distance inside the cell from one end to another is 4r. I started out with the edge of the cube, which I presumed to be 2r + x (some unknown distance I needed to figure out). Then I would have to figure out the distance between two corners diagonally opposite each other on any given face of a cube, given by the square root of 2(2r +x)2. I would use the Pythagorean theorem to equate the squares of those two distances to 4r, which is the longest distance inside a cube. Then I would work backwards from there. This is how my distance calculations turned out, and I'm pretty sure I screwed something up.

(c)
(7.88)2 = (3.94 + x)2 + √2(3.94 + x)2)
62.0944 = 15.5236 + 7.88x + x2 + (3.94 +x)√(2)
x2 +7.88x - 46.5708 + 5.572 + x√(2) = 0
x2 + 9.3x - 41 = 0

x = (-9.3 ± √(86.49 + 164))/2
x = (-9.3 ± 15.827)/2
x = 3.264 (ignoring negative result)
d = 3.264 + 3.94 = 7.2 A

Next, I did the volume and mass calculations, the errors of which I am sure are the result of my distance calculations.

(d)

7.2 A = 7.2 ⋅ 10-8 cm
(7.2 ⋅ 10-8 cm)3 = 373.2 ⋅ 10-24 cm3

(40.078 g/mol)(1 mol/6.022 ⋅ 1023 atoms) = 6.66 ⋅ 10-23 g/atom
(2 atoms)(6.66 ⋅ 10-23 g/atom) = 13.32 ⋅ 10-23 g

D = 13.32 ⋅ 10-23 g/373.2 ⋅ 10-24 cm3
= 133.2 g/373.2 cm3
= 0.36 g/cm3

Now I know this isn't the correct answer. I checked, and the density of solid calcium is 1.54 g/cm3. Can someone tell me what I'm doing wrong with my calculations?

Last edited: Sep 8, 2015
2. Sep 8, 2015

### SteamKing

Staff Emeritus
How many calcium atoms in the body centered cubic structure?

3. Sep 8, 2015

### Eclair_de_XII

Two, right? That's the number I used to calculate the mass of the unit cell.

4. Sep 8, 2015

### SteamKing

Staff Emeritus
Two is right. I was thinking of something else...

5. Sep 8, 2015

### SteamKing

Staff Emeritus
Your determination of the edge length, d, of the crystal is suspect.

This image shows the geometry of the crystal:

Since you know r for calcium, do you think you can work out what d is?

6. Sep 9, 2015

### Eclair_de_XII

(4r)2 = d2 + d√(2)2
16r2 = d2 + 2d2
16r2 = d2(1+2)
16r2/3 = d2
√(d2) = √(16r2/3)
d = 4r/√(3)

I never thought of making d its own variable; I just thought to make d = 2r + x. But I think I saw something similar to the result I've obtained in my book. Now I'm wondering why my original result, with 2r + x, was wrong... Wait, turns out I actually wasn't; my way was just a longer way of executing the steps to achieve the desired volume. I saw that I made a mistake in my original determination of one of the sides of the cube. I actually forgot to square it. Anyways, orking back from my original problem...

(7.88)2 = (3.94 + x)2 + √2(3.94 + x)2)
or rather
(7.88)2 = (3.94 + x)2 + ((3.94 + x)√(2))2)

62.0944 = 15.5236 + 7.88x + x2 + 2(3.94 +x)2
x2 +7.88x - 46.5708 + 2(3.94 +x)2 = 0
3x2 + 23.64x - 15.5236 = 0
x2 + 7.88x - 5.174533 = 0

x = (-7.88 ± √(62.0944 + 20.6981))/2
x = (-7.88 ± 9.1)/2
x = 0.61 (ignoring negative result)

Adding this to the 2r occupying d...

d = 0.61 + 3.94 = 4.55 A

Which is not as far from the result of the orthodox method of obtaining the distance of the edge.

d = 4(1.97)/√(3) = 4.5495 A

Last edited: Sep 9, 2015
7. Sep 9, 2015

### Staff: Mentor

1.97Å is a circumference, not radius of the Ca atom.

8. Sep 9, 2015

### SteamKing

Staff Emeritus
A circumference is an unusual property to list:

https://en.wikipedia.org/wiki/Calcium

According to the article above, the atomic radius of the calcium atom is listed as 197 picometers, or 197×10-12 m = 1.97 Angstroms.

9. Sep 9, 2015

### SteamKing

Staff Emeritus
So what's the density now?

10. Sep 9, 2015

### Staff: Mentor

Oops, I misread the source, 0.99 is a radius for ionic calcium, hence the mistake.

11. Sep 9, 2015

### Eclair_de_XII

d = 4.55 A = 4.55 ⋅ 10-8 cm
d3 = (4.55 ⋅ 10-8 cm)3 = 94.2 ⋅ 10-24 cm3

(40.078 g/mol)(1 mol/6.022 ⋅ 1023 atoms) = 6.66 ⋅ 10-23 g/atom
(2 atoms)(6.66 ⋅ 10-23 g/atom) = 13.32 ⋅ 10-23 g

D = 13.32 ⋅ 10-23 g/94.2 ⋅ 10-24 cm3 = 133.2 g/94.2 cm3 = 1.41 g/cm3

Well, it's as close as it's going to get with these numbers...

12. Sep 9, 2015

### SteamKing

Staff Emeritus
Looks a lot closer to the handbook value.