I am trying to calculate the theoretical density for lithium oxide using the following formula $$\rho= \frac{n^{'}\sum M_c+\sum M_A}{V_c N_{AV}}$$ where

$n^{'}$ is the number of formula units in unit cell

$M_c$ is the sum of atomic weights of all cations in unit cell.

$M_A$ is the sum of atomic weights of all anions within unit cell.

$V_c$ is the unit cell volume

$N_{AV}$ Avogadro's number.

Here is what I have so far: The structure for lithium oxide is similar to that of FCC so the number of oxygem atoms in this unit cell is $8/8+6/2=4$.

If the unit cell is to have an electrically neutral charge, it must have a total of $8$ lithium anions. The radius of a oxygen ion with a coordination number of 8 is about $128$pm and for lithium with a coordination number of 4 is about $73$ pm

Since the crystal structure is cubic, the volume is $a^3$ but we need to write $a$ in terms of the radius of the lithium and oxygen ions.

From the structure we can deduce that $a=2\sqrt{2}(r+R)$ (since lithium and oxygen ions touch along the diagonal of a square).

The atomic weight for oxygen is $16$g/mol and the atomic mass for lithium is $6.491$g/mol

$$\rho = \frac{4\times 16+8\times 6.491}{16\sqrt{2}(73\times10^{-10}+128\times10^{-10})^3\ (6.022 \times 10^{23})}=1.05 \text{ g/cm}^3$$

The theoretical density however is closer to $2$ am I missing something?

  • 2
    $\begingroup$ I'm not too sure about your relation $a = 2\sqrt{2}(r+R)$. If you check the structure of the antifluorite unit cell, I suspect it should be $r+R = \sqrt{3}a/4$. That would lead to a correction factor of $[2\sqrt{2}/(4/\sqrt{3})]^3$, and a final density of $\pu{1.929 g cm-3}$. I'll be happy to post this as an answer a bit later. $\endgroup$ Commented Sep 17, 2017 at 17:05
  • 1
    $\begingroup$ Interesting how did you come up with that relation? I would like to see the drawing if you can. Thanks! $\endgroup$
    – adam
    Commented Sep 17, 2017 at 17:18
  • $\begingroup$ Welcome to ChemistrySE! Feel free to take a tour of this site. $\endgroup$ Commented Sep 17, 2017 at 17:32
  • $\begingroup$ By the way, if you would like to get your accounts merged, please use this link and select "I need to merge user profiles" from the dropdown list. @adam $\endgroup$ Commented Sep 17, 2017 at 18:46

1 Answer 1


Here's the unit cell of $\ce{Li2O}$, which adopts an antifluorite structure. The image is adapted from this webpage; technically it shows the fluorite structure but it doesn't matter.

diagram of fluorite/antifluorite unit cell

The cation-anion contacts occur along the diagonal of the cube (black dotted line). By Pythagoras' theorem, the length of the black dotted line is $a\sqrt{3}/2$, as I indicated on the diagram. The sum of the ionic radii is half the length of this line, so

$$r + R = \frac{a\sqrt{3}}{4}$$

Using this I obtained a value of $\pu{1.929 g cm-3}$ for the density. By way of comparison, Wikipedia cites a density of $\pu{2.013 g cm-3}$. By the way, regarding your final equation, it's a good idea to leave the units in. At first I was fairly confused why you were using $10^{-10}$ instead of $10^{-12}$, until I realised it was meant to be in centimetres.

  • $\begingroup$ Very good representation of the unit cell geometry. It's hard to see these relationships $\endgroup$
    – adam
    Commented Sep 17, 2017 at 18:51
  • $\begingroup$ It has been too long since I studied things like this, but that structure could be reduced to a diamond-like tetragonal cell with a cation in the middle, no? Is that the standard way of visualizing this structure? $\endgroup$
    – Stian
    Commented Sep 18, 2017 at 8:06
  • $\begingroup$ @StianYttervik I think, for the purposes of calculating density, that would be ok and it would certainly be easier to visualise. But that wouldn't be a unit cell, as it cannot generate the lattice by appropriate translations. The above is the full unit cell for the structure (and how it's nearly always portrayed, in my own limited experience) $\endgroup$ Commented Sep 18, 2017 at 9:01

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