SME
Mineralogy · Problem type
A mineral’s density from its unit cell
Use this for “Halite (58.44 g/mol) is cubic with a = 5.640 Å and 4 NaCl units per cell. Find its density.”
Diagram
Z = 4 formula units per cubic cell; the atoms inside are not drawn.
This is an example. Type your own values to start a new problem.
ρ (g/cm³)
=
Z
×
M (g/mol)
÷
(
N
A
×
(
a (Å)
)
3
)
V
Cell volume
calculated · ų
Formulas
V
=
a
3
179.4061
=
5.64
3
ρ
=
Z
×
M
6.022 ×
10
23
×
V
×
10
-24
2.1637
=
4
×
58.44
6.022 ×
10
23
× 179.4061 ×
10
-24
Assumptions
•
A cubic cell: its volume is V = a³; 1 ų is 10⁻²⁴ cm³.
•
The cell holds Z formula units, each of mass M ÷ 6.022 × 10²³ grams (Avogadro’s number).
Step-by-step
WE KNOW
Z = 4, M = 58.44 g/mol, a = 5.64 Å
FIND
cell volume (V), density (ρ)
Step 1 · Find cell volume (V)
Use V = a³
A
cube’s
volume
is
its
edge
cubed,
here
in
Å
3
.
V
=
5.64
3
V
=
179.4061
ų
Step 2 · Find density (ρ)
Use ρ = Z × M ÷ (6.022 × 10²³ × V × 10⁻²⁴)
The
cell’s
mass,
Z
units
of
M
6.022
×
10
23
grams,
over
its
volume
in
cm³.
ρ
=
4
×
58.44
6.022 ×
10
23
× 179.4061 ×
10
-24
ρ
=
233.76
108.04
ρ
=
2.1637
g/cm³
Check
Put the numbers back in. Both sides should match.
179.4061
=
5.64
3
✓
2.1637
=
4
×
58.44
6.022 ×
10
23
× 179.4061 ×
10
-24
✓
← Previous lesson
A cubic plane’s spacing and its peak
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A crystal system from its axes and angles
Topic check
See what you know: 5 questions on this topic
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Related lessons
Crystallography
Main lesson
A cubic plane’s spacing and its peak
Use this for “Halite is cubic with a = 5.640 Å. Find the spacing of its (200) planes and their 2θ peak with copper Kα.”
A crystal system from its axes and angles
Use this for “Zircon’s cell has a = b ≠ c with every angle 90°. Which crystal system is it?”