The question I have been given is:

Silver atoms in a metallic lattice only fill up $88\,\%$ of the space ($12\,\%$ is empty). The density of silver is $10.5\ \mathrm{g\cdot cm^{-3}}$. Assuming that silver atoms are hard spheres ($V=\tfrac43\cdot\pi\cdot r^3$, when $r$ is atomic radius), what is the radius of a silver atom? Give the answer in units of $10^{-12}$ meters.

The atomic mass of $\ce{Ag}$ is 107.8682.

My solution:

$$V=0.88\times V$$

$$V=\frac{0.88\times10.5\times6.022\times10^{23}}{107.8682}=5.158\times10^{22}\ \mathrm{cm^3}$$

$$V=\frac43\cdot\pi\cdot r^3 \Rightarrow r=\left(\frac34\cdot\frac V\pi\right)^{1/3}$$
Then I switched to the $10^{12}$ meters, the result was $4.953\times10^{17}$ and it is not correct. What am I doing wrong?

  • 2
    $\begingroup$ This is a homework question. We ‎have a policy which states that you should show your thoughts and/or efforts into solving the ‎problem. It'll make us certain that we aren't doing your homework for you. Otherwise, this ‎question may get closed.‎ $\endgroup$ – M.A.R. ಠ_ಠ Nov 20 '15 at 17:44
  • $\begingroup$ Lihi it's really hard to follow what you've written there. Please visit this page, this page and this ‎one on how to format your posts better and edit your question. I voted to reopen for now. $\endgroup$ – M.A.R. ಠ_ಠ Nov 21 '15 at 11:44
  • $\begingroup$ I've added the information about the atomic mass of $\ce{Ag}$ in an effort to clarify for you and others what information you'll need in order to do the problem. $\endgroup$ – Todd Minehardt Nov 21 '15 at 16:18
  • $\begingroup$ actually Ag crystallizes in FCC and the spheres fill up $$\dfrac{\pi}{3\sqrt{2}} \approx 0.74048$$ $\endgroup$ – MaxW Dec 27 '17 at 6:56

If you had included the units in your calculation, you would have noticed why your equation is not correct.

Molar mass $M$ is defined as $$M=\frac mn\tag1$$ where $m$ is mass and $n$ is amount of substance.
Since the Avogadro constant $N_\mathrm A$ is $$N_\mathrm A=\frac Nn\tag2$$ where $N$ is the number of particles, the mass $m$ of one atom $(N=1)$ is $$m=\frac M{N_\mathrm A}\tag3$$

Density $\rho$ is defined as $$\rho=\frac mV\tag4$$ where $V$ is volume.
Thus, the volume of a sample is $$V=\frac m\rho\tag5$$ Using Equation $\text{(3)}$, the volume $V$ can be calculated for a single atom: $$V=\frac M{N_\mathrm A\cdot\rho}\tag6$$

Assuming that a fraction of $88\,\%$ of the volume $V$ is filled with a hard sphere, the volume $V_\text{sphere}$ of the sphere is $$\begin{align} V_\text{sphere}&=0.88\times V\tag7\\[6pt] &=0.88\times\frac M{N_\mathrm A\cdot\rho}\tag8 \end{align}$$

Since the volume of a sphere is $$V_\text{sphere}=\frac43\pi r^3\tag9$$ where $r$ is the radius of the sphere, the radius $r$ is $$\begin{align} r&=\sqrt[3]{\frac{3V_\text{sphere}}{4\pi}}\tag{10}\\[6pt] &=\sqrt[3]{\frac{3\times0.88\times M}{4\pi \cdot N_\mathrm A\cdot\rho}}\tag{11}\\[6pt] &=\sqrt[3]{\frac{3\times0.88\times 107.86820\ \mathrm{g\ mol^{-1}}}{4\pi \times 6.022140857\times10^{23}\ \mathrm{mol^{-1}} \times 10.5\ \mathrm{g\ cm^{-3}}}}\\[6pt] &=1.53\times10^{-8}\ \mathrm{cm}\\[6pt] &=1.53\times10^{-10}\ \mathrm{m}\\[6pt] &=153\times10^{-12}\ \mathrm{m}\\ \end{align}$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.