In answering a question in an elementary chemistry text that asks for the density of gold ($\ce{Au}$) based on crystal dimensions and weight per atom, I found the approach I used was a little different from that of the book and wanted to see if this is reasonable.

Gold crystal cells are face-centered cubes of a given length edge and a given weight per atom. The book reasons that since each of the 8 corner atoms is shared by 8 cells and each of six face atoms is shared by 2 cells, each cell adds $8/8+6/2 = 4$ atoms to a crystal structure - which gives the correct density.

My approach was to count all the atoms in a cube of cells and disregard sharing, since as long as the count was accurate the average number of atoms per cell would be correct in the limit.

For example: in a cube of 8 cells, there are $3^3$ corner atoms and $3\cdot 2^2(2+1)$ face atoms. So the general formula is $(n+1)^3+3n^2(n+1)$ atoms in an $n \times n$ cube. But in $n^3$ cells the ratio $r$ of atoms per cell to cells is

$$r=\lim_{n\to\infty} \frac{(n+1)^3+3n^2(n+1)}{n^3}=4$$

So this also gives the correct density, and my question is whether there is anything wrong with this approach?

(In an earlier version of this question I failed to count all the face atoms; hopefully this is easier to follow.)

[Edited to correct formula.]


1 Answer 1


Yes of course, if you do the math right, that limes gives the exact same number.

Don't forget to look out for atoms that are completely within the unit cell.


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