7
$\begingroup$

How do you determine the chemical formula of a compound given its crystal structure?

$\endgroup$
  • $\begingroup$ Presumably we have some other information as well? I mean, if I gave you a sample and told you it's a CsCl crystal structure, you can't really say much, since a good number of compounds assume the CsCl structure. $\endgroup$ – chipbuster Jan 9 '14 at 4:58
  • 2
    $\begingroup$ Presumably, if you know the crystal structure, you know the unit cell? Count all the atoms in the unit cell taking into account fractional atoms (that is account for shared atoms) and adjust so the ratios are integers and you have the answer. $\endgroup$ – matt_black Jan 9 '14 at 23:23
  • $\begingroup$ If you mean something more like "given its X-Ray diffraction pattern", you might want to edit the question to reflect that. $\endgroup$ – Aesin Jan 10 '14 at 1:17
10
$\begingroup$

Knowledge of the crystal structure implies that you are also able to describe the unit cell of the compound. To determine its formula, count all sorts of atoms inside the unit cell, also considering fractional atoms which belong to more than one unit cell. For example, an atom at the corner of the unit cell counts as $\frac{1}{8}$ atom because it is shared between 8 adjacent unit cells. Likewise, an atom at the edge of the unit cell counts as $\frac{1}{4}$ as it belongs to 4 unit cells etc. When you obtain fractional values for the atom numbers, multiply the values of all atoms with a certain number so that all values are integers.

Let us take sodium chloride ($\ce{NaCl}$) as a simple example. Its unit cell, which is shown below, contains $\ce{Na+}$ (violet) and $\ce{Cl-}$ ions (green).

enter image description here

Counting the number of chloride ions in the unit cell gives

$$12 \cdot \frac{1}{4}+1=4 \ \ce{Cl-}$$

because there are 12 $\ce{Cl-}$ on the edges and 1 $\ce{Cl-}$ in the center of the unit cell. For the sodium ions, which are situated at the corners and on the faces (count as $\frac{1}{2}$ because they are shared between 2 cells) of the unit cell, we obtain

$$8 \cdot \frac{1}{8}+6 \cdot \frac{1}{2}=4 \ \ce{Na+}$$

The composition of the unit cell is therefore $\ce{Na4Cl4}$. Since the number of formula units per unit cell is 4, the formula of the compound is $\ce{NaCl}$.

For compounds in which the building blocks of the crystal structure are molecules, the procedure is essentially the same.

$\endgroup$

protected by orthocresol Jul 30 '17 at 16:32

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.