Crystal Packing: How Full Is a Metal?

Choose a metal, or type a structure, a lattice parameter and a molar mass. The page finds where the atoms touch, takes the radius from that, and works out how much of the crystal the atoms fill, how much room each one owns, and the density that follows. The atoms are treated as hard spheres, which is the assumption two of the three insights below take apart.

Worked example — Crystal Packing: How Full Is a Metal?

Atoms in one cell	4
Nearest neighbours	12
Nearest-neighbour distance	2.5562 Å
Radius of the touching spheres	1.2781 Å
Volume of the cell	47.240 Å³
Fraction of space filled	74.05%
Volume each atom owns	11.810 Å³
Density from the lattice	8.935 g/cm³

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — Crystal Packing: How Full Is a Metal?

A crystal repeats one unit cell. Counting its atoms takes sharing into account: a corner atom belongs to eight cells, a face atom to two. The packing fraction is the share of the cell inside the atoms, with each atom a sphere just touching its nearest neighbours: φ = n · (4/3)πr³ / V, where the radius r is half the nearest-neighbour distance. The same cell gives the density, ρ = nM / (N_AV), since it holds n atoms of known mass.

What each symbol means

a: the lattice parameter: the edge of the cubic cell, or the edge of the hexagonal base, measured by X-ray or neutron diffraction
n: atoms per cell, counting a corner atom as ⅛ and a face atom as ½
r: the radius of the touching spheres, half the shortest distance between two atom centres
φ: the packing fraction, the part of the cell's volume inside the spheres

Assumes: Atoms are hard spheres of one radius, touching along the shortest distance in the lattice. Lattice parameters are room-temperature values from Wyckoff's 1963 compilation, except iron at 912 °C and polonium at 283 K. The crystal is perfect: no missing atoms, no impurities, one isotope mixture, the standard atomic weight. Every density the page prints is the X-ray density of such a crystal; pores and vacancies in a real sample can only lower it.
Breaks when: Wherever the atom is not a hard sphere, which the page itself shows twice: iron's radius changes by 2.52% with its neighbours, and zinc's c/a says its bonds between layers differ from those within a layer. For compounds, whose two kinds of ion have two radii and pack by a different rule. And at the level of parts per billion, where vacancies, impurities and the isotope mix all have to be measured, as the Avogadro project did.

Face-centred cubic and hexagonal packing fill the same 74.05%, and no packing of equal spheres fills more 🖖

Lay spheres in a flat layer and each touches six. The next layer sits in the hollows, and the third has a choice: directly over the first, ABAB, which is hexagonal close packing, or over the hollows the second left open, ABCABC, which is face-centred cubic. Either way every sphere touches twelve, and the spheres fill π/(3√2) = 74.05% of space. Press Copper, then switch the structure to hexagonal, which the page starts at the ideal c/a: the same figure. Kepler guessed in 1611 that nothing does better. Thomas Hales's proof appeared in 2005 and ran partly on a computer, and a formal version, every step checked by machine, followed in 2017. Body-centred cubic looks nearly as tight, with eight neighbours and six more only 15% farther off, and fills 68.02%.

Iron gets denser when heated through 912 °C, by 1%, where hard spheres say 8% 🖖

Press Iron below 912 °C, then Iron above 912 °C. Body-centred iron has a = 2.9058 Å and gives each atom 12.268 Å³; face-centred iron at the same temperature has a = 3.6487 Å and gives each atom 12.143. That is 1.02% less room, so heating iron through the change makes it denser, and a dilatometer sees the bar shorten while the furnace is still getting hotter. Spheres of one fixed size would have shrunk by 8.14%, the gap between 68.02% and 74.05%. They do not, because the atom is not one fixed size: taken from each lattice, iron's radius is 1.2583 Å in the body-centred form and 1.2900 Å in the face-centred one, 2.52% larger with twelve neighbours than with eight. X-ray measurements from 1955 give the same step, 1.04%.

Zinc is called hexagonal close-packed, and packs less than iron 🖖

Hexagonal layers are close-packed only when the cell's height is √(8/3) = 1.6330 times its width, so that an atom's six neighbours in the next layer are as near as its six in its own. Magnesium comes close, at 1.6235, and fills 73.62%. Press Zinc: its c/a is 1.8563. The layers stand apart, the atoms in the next one sit 2.9128 Å away against 2.6648 Å within the layer, 9.31% farther, and as spheres touching in the layer zinc fills 65.14%. That is less than body-centred iron's 68.02%, a structure nobody calls close-packed. Cadmium is the same, 64.12%. The name records the stacking order, ABAB. Whether the stack is close is set by c/a, and zinc's says its bonds between layers are not the same as its bonds within one, which a sphere has no way to express.

Problem solved in full

Counting Avogadro's number with a silicon sphere 4 steps
In 2017 the International Avogadro Coordination reported a 1 kg sphere of silicon enriched to 99.9985% silicon-28. Its molar mass was 27.976942674 g/mol, its lattice parameter 543.0996218 pm and its density 2320.068748 kg/m³. Silicon has the diamond structure. How many atoms are there in a mole?

Open this problem in the tool
1. The diamond cell holds eight atoms: eight corners at ⅛, six faces at ½, and four inside.
  
  $n = 8 \times \frac{1}{8} + 6 \times \frac{1}{2} + 4 = 8$
2. Its volume is a³, with a = 5.430996218×10⁻⁸ cm: 1.6019114×10⁻²² cm³.
  
  $a^{3} = (5.430996218 \times 1 0^{- 8} cm)^{3} = 1.6019114 \times 1 0^{- 22} c m^{3}$
3. Its mass is density times volume, 2.320068748 g/cm³ × 1.6019114×10⁻²² cm³ = 3.7165447×10⁻²² g, so one atom weighs an eighth of that, 4.6456808×10⁻²³ g.
  
  $m_{cell} = ρ a^{3} = 2.320068748 \times 1.6019114 \times 1 0^{- 22} = 3.7165447 \times 1 0^{- 22} g m_{atom} = \frac{m _{cell}}{8} = 4.6456808 \times 1 0^{- 23} g$
4. A mole of these atoms weighs 27.976942674 g. Dividing by the mass of one atom counts them.
  
  $N_{A} = \frac{M}{m _{atom}} = \frac{8 M}{ρ a ^{3}} = \frac{27 . 976942674}{4 . 6456808 \times 1 0 ^{- 23}}$
Answer

N_A = 6.022140516×10²³ per mole, the paper's own figure to all ten digits it gives for this sphere. The page runs the same arithmetic forwards: press Silicon and natural silicon's 5.4307 Å, eight atoms and 28.085 g/mol give 2.329 g/cm³. Since 2019 Avogadro's constant is fixed by definition at 6.02214076×10²³, so the measurement now runs the other way, from counted atoms to a mass, and it is one of the two most accurate methods by which the kilogram is realised.

$N_{A} = 6.022140516 \times 1 0^{23} mo l^{- 1}$

References (3)

Every room-temperature lattice parameter on the page, entry by entry, from Wyckoff's compilation as the Crystallography Open Database transcribes it: R. W. G. Wyckoff, Crystal Structures, vol. 1, 2nd edition. Interscience, New York, 1963; entries 9008460-9008565 of the Crystallography Open Database (crystallography.net). Polonium: W. H. Beamer and C. R. Maxwell, J. Chem. Phys. 17, 1293 (1949), COD 1509138.
Iron's two lattice parameters at 912 °C, from neutron diffraction in the furnace: M. Onink, C. M. Brakman, F. D. Tichelaar, E. J. Mittemeijer, S. van der Zwaag, J. H. Root and N. B. Konyer, "The lattice parameters of austenite and ferrite in Fe-C alloys as functions of carbon concentration and temperature." Scripta Metallurgica et Materialia 29(8), 1993, pp. 1011-1016. DOI 10.1016/0956-716X(93)90169-S
The silicon-28 sphere of the worked problem, and the Avogadro constant it gave: G. Bartl, P. Becker, B. Beckhoff et al. (33 authors), "A new ²⁸Si single crystal: counting the atoms for the new kilogram definition." Metrologia 54(5), 2017, pp. 693-715. DOI 10.1088/1681-7575/aa7820

By the 42Math team · Reviewed against published sources. Corrections welcome.