I Converting Unit Cell Volumes of Minerals to Density

[oshah]

TL;DR Summary

To implement various EoS for planetary modelling in an Astrophysical context I need to know how unit cell volumes of minerals are converted into density.

I'm modelling the interiors of core-dominated (exo)planets. The EoS I use in my calculations are mostly either a Birch-Murnaghan formulation or a Mie-Grüneisen-Debye formulation. In either case, the ambient density ρ₀ at ambient pressure and temperature are required for the implementation. However, in most of the literature the ambient volume V₀ is given instead. E.g. Sun 2018 (https://www.sciencedirect.com/science/article/pii/S0012821X18301195) give the unit cell volume for Perovskite and post-Perovskite as 162.37 A³ and 164.22 A³ respectively (see their Table 1). As I have a purely Astrophysical background I do not know how to convert the unit cell volume into density as the conversion would of course depend on the lattice structure of the material which itself depends on the mineral phase at hand. To avoid loosing too much time diving into the world of solid state physics, I would appretiate it if, using the example from Sun 2018, someone more familier with the field could provide me with the proceedure to convert V₀ into ρ₀.

 

[Baluncore]

Welcome to PF.
Identify the unit cell and it's geometry.
Evaluate the volume of the cell using the identified geometry.
Find the molecular weight of one unit cell.
Divide weight by the Avagadro constant = 6.02214076×10²³
You now have mass and volume, the ratio is density.

 

[crashcat]

I can't see the paper right now and it looks like your crystal structure is complicated, but here's a simple example to illustrate: the density of alpha iron (BCC structure).

(image from wikipedia)

It's a cube with lattice constant 2.867 Angstrom, or volume 2.358E-23 cc. This cube contains two Fe atoms (one whole atom in the center, and eight 1/8th atoms from each corner). The mass of two Fe atoms is $2\times 55.845 amu \times 1.673\text{x}10^{-24} \text{g/amu} = 1.868\text{x}10^{-22} \text{g}$. Then density would be $1.868\text{x}10^{-22} \text{g} / 2.358\text{x}10^{-23} \text{cc} = 7.92 \text{g/cc}$. This compares ok with the true value of 7.87 g/cc.

 

[oshah]

Thank you very much for your quick answers. I have indeed come up with a very similar value for the pPv density in the mean time. I stumbled upon materialsproject.org where using the structural information for pPv (MgSiO3) from Wikipedia (orthorombic, space group Cmcm) I got a density of 3.96 gcc for a volume of 84.198 A³ (I don't know what that value is supposed to represent). Scaling this to 164.22 A³ from Sun 2018 would yield a value of 7.72 gcc, in good agreement. However, the density contrast of Pv and pPv would then be unrealistically large. For instance, Dorfmann et al. 2014 (https://www.researchgate.net/publication/263030768_Effect_of_Fe-enrichment_on_seismic_properties_of_perovskite_and_post-perovskite_in_the_deep_lower_mantle) point to a density contrast between Fe-free Pv and pPv at lower mantle conditions (125 GPa) of roughly 2%. Using the EoS parameters from table 1 in Sun 2018 (see original question) and the corresponding values for the ambient densities of 4.107 gcc and 7.72 gcc for Pv and pPv respectively yields a density contrast at 125 GPa of ~50%. So there is clearly an inconsistency somewhere. The alleged contrast of ~2% would be roughly reached for a density of pPv of ~4.14 gcc keeping the other EoS parameters as they are. As a brute force approach I could of course just adjust the value of ρ₀ for pPv to match the density contrast at 125 GPa from Dorfmann et al. 2014. This is, however, not very clean and I would prefer to actually calculate a consistent value from the measured unit cell volume. Any suggestions on where the problem could be?