Find the percent of void in the crystal structure of $\ce{NaCl}$.

I am trying to calculate the percent void of the packing structure of $\ce{NaCl}$. Let's denote radius of anion as $r$, radius of cation as $R$. What I know

$$\text{side length} = 2R + 2r$$

$$R + r = \pu{2.50 Å}$$

The ratio of cation radius to the anion radius is $0.600$. The equation to find void space:

$$100 \times \frac{\text{total volume} - \text{volume of spheres}}{\text{total volume}}$$

Yes, this is a homework question, but I'm stuck and I just need some help to get started.


closed as off-topic by airhuff, MaxW, Mithoron, Jon Custer, A.K. Dec 16 '17 at 4:14

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    $\begingroup$ Homework is fine, you just have to explain what you have tried and what specifically is giving you trouble. For example, you need to find the void space. It looks like you have the formula for it, so are you having trouble obtaining the total volume, volume of spheres, both? $\endgroup$ – Tyberius Dec 15 '17 at 21:54
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    $\begingroup$ For the assignment I can't use decimals. I have not figured out how to set up the equation without solving for both of the variables. Would I just call one R and the other 0.6R? $\endgroup$ – Elijah Dec 15 '17 at 23:52
  • $\begingroup$ Yes of course! The absolute value of R+r is irrelevant, the percentage of voids is the same in any arbitrarily scaled model or in actual NaCl. $\endgroup$ – Karl Dec 16 '17 at 8:05
  • $\begingroup$ Ok, I tried that I ended up with 18%. Is that right? $\endgroup$ – Elijah Dec 16 '17 at 21:08