Calculating the theoretical Density of a Lithium oxide

Question

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?

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.

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.