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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?

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

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