Isotopic mass to density

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?

• 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. – Mithoron Nov 25 '17 at 21:04
• 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? – Bishop Nov 25 '17 at 21:30
• Number of particles/cm³ can be derived from particle mass and material density, it can't be measured directly. – Mithoron Nov 25 '17 at 21:46
• I see. Well thats no good for me I need a exact solution. Will keep on looking, thanks anyway. – Bishop Nov 25 '17 at 22:00
• @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. – andselisk Nov 26 '17 at 0:29

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 ($a^3; a = \pu{0.409nm}$).