The density of iron is $\pu{7.874 g/cm3}$ and its unit cell volume is $\pu{2.364E-23 cm3}$. These values indicate that there are: two Fe atoms per unit cell and the metal is body-centered cubic. I am not fully sure why this is correct or how you figure this out by using the data given. I am still very new to the unit cell concept in chem. Any explanation would be great, thanks! Also sorry I was not sure what to tag with this question.
$\begingroup$
$\endgroup$
4
-
4$\begingroup$ You're given the volume of the unit cell, and the density. So what is the mass of the unit cell? $\endgroup$– orthocresolCommented Nov 17, 2021 at 21:03
-
$\begingroup$ Work through the answers here, including the mentioned video. Though on first sight different, it contains the material you need. $\endgroup$– ButtonwoodCommented Nov 17, 2021 at 21:05
-
$\begingroup$ Atoms in the center of the cell count fully ... those on the side only half ... those in corners only a eighth, etc., if possible, have a look into the video which showcases the computation on Ne. $\endgroup$– ButtonwoodCommented Nov 17, 2021 at 21:38
-
4$\begingroup$ Try to invent more meaningful title // A good question title should be quite unique, resembling a magazine article headline, optionally formulated as a question. Avoid titles saying everything (like full formulation of the task) or nothing ( like "A question about chemistry") $\endgroup$– PoutnikCommented Nov 18, 2021 at 3:40
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
The density of iron is also the density of the unit cell : $\pu{7.874 g|cm3}$. Multiply this density by the volume of the unit cell. It gives you the mass of this unit cell : $\pu{7.874 g/cm3 · 2.314 · 10^{-23} cm3 = 1.861 · 10^{-22}} g$ . This is exactly twice the mass of an iron atom, as $1$ atom weighs : $\pu{\frac{56 ~g/mol}{6·10^{23} atom/mol}}$ = $\pu{9.33 · 10^{-23} g}$, So there are $2$ $\ce{Fe}$ atoms per unit cell.
This number $2$ can be justified by the structure of the body-centered cubic. This system is made of an infinite succession of identical cubes, piled along the three directions of space Ox, Oy and Oz. The unit cube has 8 vertexes or apexes, all occupied by the center of an iron atom. But around each of these iron atoms, there are 8 cubes starting from it : one above it at left-hand side and before, the second also above and at left-hand side, but behind ; the third is also above but at right-hand side and before ; the fourth is right-hand side, above and behind. The four remaining are obtained by replacing "above" by "under" in the preceding text.
So each vertex of the unit cell is occupied by one atom, but only $1/8$ of this atom is situated inside the first unit cell. In total, There are $8$ times $1/8$ atom in each corner of the cube. Well, in body-centered, there is still one iron atom in the very center of the unit cell. In total, each unit cell is occupied by $8/8 + 1$ iron atoms. So there is on average $2$ iron atoms per unit cell.
