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I want to figure out/calculate the density ($\pu{g/cm3}$) of the various isotopes, here $\ce{^109Ag}$. I know it's isotopic mass is $\approx109$ and calculated the number of particles/gram. But I couldn't find out how to get the density.

From what I could find out over some mole maps I need to multiply the molar mass with a factor to get to the density but is that even possible or am I wrong?

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    $\begingroup$ o.O It seems you don't know what you are talking about. Density is found experimentally, is variable and does not depend on other physical parameters in any simple way; it's other parameters that depend on it. $\endgroup$ – Mithoron Nov 25 '17 at 21:04
  • $\begingroup$ Uhm, yes. Thats the reason I'm asking it here. My thoughts were that one could calculate it by: number of particles/cm³ multiplied by the atomic mass. So, the density is like a approximation through testing then? $\endgroup$ – Bishop Nov 25 '17 at 21:30
  • $\begingroup$ Number of particles/cm³ can be derived from particle mass and material density, it can't be measured directly. $\endgroup$ – Mithoron Nov 25 '17 at 21:46
  • $\begingroup$ I see. Well thats no good for me I need a exact solution. Will keep on looking, thanks anyway. $\endgroup$ – Bishop Nov 25 '17 at 22:00
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    $\begingroup$ @Mithoron There is unit cell and crystalline solid, "crystalline cell" doesn't make much sense. I don't understand how the crystallographic density determination is "dumb". It's no dumber as any other method, and typically provides quite good results. $\endgroup$ – andselisk Nov 26 '17 at 0:29
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This is possible, just use the density equation for a crystal lattice $$ d = \frac{z\times M}{V_c \times N_\mathrm A}$$ Where $d$ is density; $z$ is the number of atoms per unit cell; $M$ ist he molar mass of the material (isotope in this case); $V_c$ is the volume of a unit cell; and $N_\mathrm A$f is avogadro's number. Since silver is FCC, $z= 4$ ($\frac{1}{8}\times 8 + \frac{1}{2}\times 6 = 4$) and $V_c$ is equal to the lattice parameter squared ($a^3; a = \pu{0.409nm}$).

This method will give you a very close approximate value but is limited in accuracy to the experimentally observed density as it does not account for free space in the crystal due to vacancies in the lattice or voids at grain boundaries. For example, if you calculated the densities of iridium and osmium you find that iridium should be denser, but experimentally you will not observe this result because iridium has a sufficiently high concentration of vacancies that osmium is actually denser.

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  • $\begingroup$ Thanks for your answer. I have tested that equation and came the expected result. $\endgroup$ – Bishop Nov 26 '17 at 20:32

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