# How do I calculate the density of calcium from atomic radius?

• Eclair_de_XII
In summary, the student attempted to solve a problem with volume and mass calculations for calcium, but was unsuccessful.
Eclair_de_XII

## Homework Statement

"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."

## Homework 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

## 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:
How many calcium atoms in the body centered cubic structure?

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

Eclair_de_XII said:
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

Eclair_de_XII said:
Two, right? That's the number I used to calculate the mass of the unit cell.
Two is right. I was thinking of something else...

Eclair_de_XII said:

## Homework Statement

"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."

## Homework 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

## 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?
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?

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

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

Eclair_de_XII said:
(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
So what's the density now?

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

SteamKing said:
So what's the density now?

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...

Eclair_de_XII said:
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...
Looks a lot closer to the handbook value.

## 1. How do I calculate the density of calcium from atomic radius?

To calculate the density of calcium from its atomic radius, you will need to use the formula: density = mass/volume. First, find the mass of one calcium atom using its atomic mass from the periodic table. Then, calculate the volume of one calcium atom by using the formula for the volume of a sphere (4/3 x π x radius^3). Finally, divide the mass by the volume to get the density of calcium.

## 2. What is the atomic radius of calcium?

The atomic radius of calcium is 0.197 nanometers (nm).

## 3. Does the density of calcium change with temperature?

Yes, the density of calcium can change with temperature. As temperature increases, the atoms in calcium will vibrate faster, causing the atoms to take up more space and decreasing the density. However, this change is usually small and can be considered negligible for most calculations.

## 4. How does the density of calcium compare to other elements?

The density of calcium is relatively low compared to other elements. It has a density of 1.55 grams per cubic centimeter (g/cm^3), which is lower than elements such as iron (7.87 g/cm^3) and lead (11.34 g/cm^3).

## 5. Can the density of calcium be measured experimentally?

Yes, the density of calcium can be measured experimentally by using various techniques such as Archimedes' principle or pycnometry. These methods involve measuring the mass and volume of a sample of calcium and using the formula density = mass/volume to calculate the density.

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