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I was given this problem recently to compute the number of ions in a unit cell of ferrous oxide $\ce{FeO}$. The data given was:

Side length $a = \pu{5 Å}$

Density $d = \pu{4 g/cc}$

Using the formula $\displaystyle d = \frac {ZM}{N_0 a^3}$, I obtained $Z = 4.18$. However, the answer had rounded $Z$ off to $4.$

Could a unit cell have a fractional number of atoms? Intuitively, it doesn't make sense for a cell to have $0.18$ of an atom. If, by experimental analysis, we obtain a fractional number of atoms in a unit cell, do we round them down as is done in this problem?

P.S. Another thought would be that such odd numbers are obtained due to stoichiometric defects.

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    $\begingroup$ 1. $Z \in \mathbb{N}$ (always, by the definition of the number of formula units); 2. Both $a$ and $d$ are given to the single significant figure. How did you manage to obtain the more precise answer with three significant figures using this data? $\endgroup$
    – andselisk
    Mar 1, 2020 at 7:49
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    $\begingroup$ @andselisk: Would you please elaborate your comment as an answer to the question? $\endgroup$ Mar 1, 2020 at 22:27

1 Answer 1

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Crystallographic background

A unit cell cannot contain parts of atoms from a formula unit, otherwise it would not be a geometric unit of repeatability. The simplest condition arising from translational symmetry is that the total number of atoms present in the unit cell must be either equal to or be a multiple of the number of atoms in the chemical formula, which is an equivalent to the integer number of formula units per unit cell: $Z \in \mathbb{N}.$

Note, however, that the asymmetric unit (crystallographically independent region) contains the $Z/n$th part of the atoms of the formula unit, where $n$ is the order of the crystallographic point group. Thus, the asymmetric region may contain an integer number of formula units, one formula unit or its fractional part, meaning that some atoms located on symmetry elements, are "divided" across several asymmetric regions.

Data precision

It appears you are using the correct formula for the crystallographic density $ρ$ (I strongly encourage not to use $d$ as a symbol for density as in crystallography $d$ is reserved for spacing between planes):

$$ρ = \frac{MZ}{N_\mathrm{A}V}$$

where $M$ is molar mass, $N_\mathrm{A}$ is Avogadro's number and $V$ is the unit cell volume. Although you can use the exact value for $N_\mathrm{A} = \pu{6.02214076E23 mol-1}$ and a precise tabulated value $M(\ce{FeO}) = \pu{71.844 g mol-1},$ note that you are given both $a$ and $d$ (experimental data) to the single significant figure, which prohibits your answer from being more precise than these parameters, and you must present a rounded number $Z = 4$ and not $Z = 4.18,$ which doesn't make either crystallographic or mathematical sense.

In real life situation there always will be a systematic difference between the measured and crystallographic densities due to the presence of defects, and the deviation of $Z$ from the integer value is often due to statistically filling of the proper point group by the atoms of a different kind.

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    $\begingroup$ The density presented by andselisk is the one based on the diffraction experiment and the subsequently built model (_exptl_crystal_density_diffrn in the .cif files you find in databases, iucr.org/__data/iucr/cifdic_html/1/cif_core.dic/…). You should at least check by an estimation, if this density is reasonable. The same files may include an entry about _exptl_crystal_density_meas, measured e.g., by floatation of your sample in liquids (iucr.org/__data/iucr/cifdic_html/1/cif_core.dic/…). $\endgroup$
    – Buttonwood
    Mar 2, 2020 at 6:29

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