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

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  • $\begingroup$ 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
    Commented Jan 9, 2014 at 4:58
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    $\begingroup$ 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
    Commented Jan 9, 2014 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
    Commented Jan 10, 2014 at 1:17

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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.

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