# Relationship between molar mass and density in solids

In ideal gases, it's pretty clear that 1 mole of gas occupies 22.4 L at STP. By knowing this, it's easy to calculate density given molar mass. However, this isn't true for solids. If I know molar mass of a solid, I cannot derive it's density.

I want to clarify whether my understanding of why we can't derive density in case of solids is correct. For solids, since each solid has its own lattice structure, knowing how the molecules are arranged is necessary, because if there is more space between molecules, it will have a lower density, which isn't really an issue in (ideal) gases where the arrangement of molecules/atoms is uniform.

• Density of a solid is given by d= Z.M/a^3.N_A where Z is the effective number of motif present in one unit cell, M is the mass of the motif, a= edge length of a cubic unit cell and N_A is the Avogadro number. Commented May 29, 2021 at 9:17
• 1 mol of an ideal gas does not occupy 22.4 l at STP. Apparently, you are using an old definition of STP that is obsolete since 1982. Commented May 29, 2021 at 14:53

If your sample is solid and crystalline, knowledge of the unit cell's dimension and symmetry (e.g., fcc) and density allows you to determine the molecular mass of your compound.

In fact, in crystallography, you may determine the macroscopic density of your sample (e.g., find a liquid which i) wets your crystals and ii) lets your crystals float) in first place. If you then know the dimensions of the unit cell by the diffraction experiment, knowledge of the chemical formula per unit cell may tell you the number of formula units per unit cell. See e.g, this demonstration about Ne.

The example above about Ne may be extended to crystals consisting of molecules for that you may assign atomic volumes to each element that add up to a molecular volume. An early approximation for organic compounds[1] was to assign $$\pu{18 Å^3}$$ for any atom of $$(\ce{C, N, O})$$ and $$\pu{2 Å^3}$$ for any atom of $$\ce{H}$$. Decades later,[2] backed by much more experimental data, it was possible to refine these assignments, e.g., to average volumes of $$\pu{13.87 Å^3}$$ ($$\ce{C}$$), $$\pu{11.8 Å^3}$$ ($$\ce{N}$$), $$\pu{11.39 Å^3}$$ ($$\ce{O}$$), and $$\pu{5.08 Å^3}$$ ($$\ce{H}$$), respectively. Large deviations from these guesses from the then experimentally determined molecular volume may be caused by solvent-accessible voids in the crystal structure.[3]

The following quotes the averaged atomic volumes (at $$T = \pu{298 K}$$) about typical elements of organic molecules (table 2 in Hofmann's publication, a selection of the published data):

|---------+--------|
| element | volume |
|         | [A**3] |
|---------+--------|
| H       |   5.08 |
| C       |  13.98 |
| N       |   11.8 |
| O       |  11.39 |
| F       |  11.17 |
| P       |   29.5 |
| S       |   25.2 |
| Cl      |   25.8 |
| Br      |   32.7 |
| I       |   46.2 |
|---------+--------|


References:

1 Kempster, C. J. E.; Lipson, H., A rapid method for assessing the number of molecules in the unit cell of an organic crystal in Acta. Cryst. B28, 1972, 3674; doi: 10.1107/S056774087200857X.

2 Hofmann, D. W. M., Fast estimation of crystal densities in Acta. Cryst. B58, 2002, 489-493; doi: 10.1107/S0108768101021814.

3 Florence, A. J., The use of supplementary information to solve crystal structures from powder diffraction in International Tables for Crystallography (2019). Vol. H. ch. 4.4, pp. 433-441; doi: 10.1107/97809553602060000959.

Buttonwood has given an excellent answer with a nice demonstrational video. However, there is one point I'd like to emphasize. In ideal gases, the molar mass is directly proportional to the density:

$$PV = nRT \ \Rightarrow \ \frac{n}{V} = \frac{P}{RT} \tag1$$

If the molar mass of a gas is $$M$$ and its mass is $$m$$, $$n = \frac{m}{M}$$ and $$\frac{n}{V} = \frac{m}{MV}=\frac{m}{V} \cdot \frac{1}{M} = \frac{\rho}{M}$$ where $$\rho$$ is the density of the gas. Thus, applying this in the equation $$(1)$$ you get:

$$\frac{n}{V} = \frac{\rho}{M} = \frac{P}{RT} \tag2$$

Thus, under known temperature and pressure, $$\frac{\rho}{M} = \text{constant}$$ for given ideal gas.

However, for solids (and liquids), this is not true. To calculate density for a given solid, you need more than its molar mass as Buttonwood correctly pointed out (e.g., use Archimedes' principle). You need at least information about crystal packing of the solid (if it is crystalline). My point is some solids pack more tightly compared to others even if they have the same crystal packing structure (e.g., face-centered cubic or fcc). The following example shows how to calculate the density if you know the unit cell dimensions and molar mass, or vise versa.

The molar masses of gold $$(\ce{Au})$$ and platinum $$(\ce{Pt})$$ are $$\pu{196.967 g mol-1}$$ and $$\pu{195.084 g mol-1}$$, respectively. In solid state, both $$\ce{Au}$$ and $$\ce{Pt}$$ consist of the same crystal packing structure called face-centered cubic (fcc). Crystal studies have revealed that the lengths of cubic unit cells ($$a$$) of $$\ce{Au}$$ and $$\ce{Pt}$$ are $$\pu{407.82 pm}$$ and $$\pu{392.31 pm}$$, respectively. Prove that $$\ce{Pt}$$ is denser than $$\ce{Au}$$.

If you inspect the fcc unit cell closely, you realize that each of the eight corner atoms is shared with eight unit cells while each of the six face-centered atoms are shared with only two unit cells. Thus:

$$\text{Total atoms per unit cell} = 8 \times \frac{1}{8} + 6 \times \frac{1}{2} = 4$$

Since the unit cell is cubic, its volume is $$a^3$$. Based on the value of $$a$$ and number of atoms per unit cell for each metal, you can calculate the molar volume of each metal ($$V_\ce{Au}$$ or $$V_\ce{Pt}$$). Since $$a_\ce{Au} = \pu{407.82 pm}$$:

$$V_\ce{Au} = a_\ce{Au}^3 \times \frac{1}{\pu{4 atoms}} \times N_A \\ = (\pu{407.82 pm})^3 \times \left(\frac{\pu{1 cm}}{\pu{10^10 pm}}\right)^3 \times \frac{1}{\pu{4 atoms}} \times \pu{6.022 \times 10^{23} atoms\:mol-1} = \pu{10.21 cm3 mol-1}$$

and similarly, since $$a_\ce{Pt} = \pu{392.31 pm}$$:

$$V_\ce{Pt} = a_\ce{Pt}^3 \times \frac{1}{\pu{4 atoms}} \times N_A \\ = (\pu{392.31 pm})^3 \times \left(\frac{\pu{1 cm}}{\pu{10^10 pm}}\right)^3 \times \frac{1}{\pu{4 atoms}} \times \pu{6.022 \times 10^{23} atoms\:mol-1} = \pu{9.09 cm3 mol-1}$$

Now, since we know the molar volume of each metal, we can calculate the density ($$\rho$$) by using $$\rho = \frac{M_\ce{M}}{V_\ce{M}}$$:

$$\rho_\ce{Au} = \frac{\pu{196.97 g mol-1}}{\pu{10.21 cm3 mol-1}} = \pu{19.29 g cm-3}$$

$$\rho_\ce{Pt} = \frac{\pu{195.08 g mol-1}}{\pu{9.09 cm3 mol-1}} = \pu{21.46 g cm-3}$$

Therefore, platinum is denser than gold $$(\rho_\ce{Pt} \gt \rho_\ce{Au})$$ even though their molar masses are reversed $$(M_\ce{Pt} \lt M_\ce{Au})$$.

Note: If you know the densities without knowing molar masses, you can calculate molar masses using the same strategy (only one unknown).