# Why is platinum denser than gold?

The atomic masses of gold and platinum are 196.96657 u and 195.084 u respectively, meaning that (on average) an individual gold atom is heavier than an individual platinum atom.

At the same time, the van der Waals radius of gold is 166 pm, whereas that of platinum is 175 pm. So an individual gold atom is smaller than an individual platinum atom.

Given these data, how is it possible that platinum metal can actually be denser than gold (21.5 g/cm3 vs 19.3 g/cm3)?

• At the atomic scale, "size" is a concept that requires careful consideration of definitions and context. – Nicolau Saker Neto Apr 11 '20 at 12:58
• They crystallize in the same face-centered cubic packing, with a unit cell length of 407 pm for gold and 392 pm for platinum. This length goes into the density calculation as cubed, so gold uses 12% more space per atom, but is only 1% heavier (196.97 g/mol vs 195.08 g/mol). – Karsten Theis Apr 11 '20 at 13:33
• What is important is not the atomic size, but the geometry of the metallic lattice. – Poutnik Apr 11 '20 at 13:42
• To add just a bit of additional information, the tabulated atomic radii is an "average" size. So the radii of Na in each of the salts NaF, NaCl, NaBr, and NaI is not exactly the same. – MaxW Apr 11 '20 at 18:07
• Shubhang Walavalkar - I don't disagree, but the gold and platinum atoms in the metal aren't free atoms either, even though the formal charge on them would be 0. – MaxW Apr 12 '20 at 7:09

Since OP is still in the high school, I'll try to explain it simply as possible using mathematical manipulation (hoping OP is more familiar with mathematics than chemistry).

Both gold and platinum consist of same crystal packing called body-centered cubic, which is illustrated in following diagram:

Crystal studies of both gold and platinum has revealed that the lengths of their cubic unit cells ($$a$$) are $$\pu{407.82 pm}$$ and $$\pu{392.31 pm}$$, respectively. If you inspect closely, you would realize that each of eight corner atoms shares with eight unit cells while each of six face-centered atoms shares with only two unit cells. Thus, total atoms per unit cell $$= 8 \times \frac{1}{8} + 6 \times \frac{1}{2} = 4$$. Keep in mind that $$a$$ is depend on the atomic radius of each metal and how closely they packed against each other against some concerning forces (e.g., van der Waal's). For example, the experimental finding has been reported that the closest $$\ce{Au-Au}$$ separation is $$\pu{288.4 pm}$$ while that of $$\ce{Pt-Pt}$$ is $$\pu{277.5 pm}$$. These values imply that the metallic radius of gold is $$\pu{144.2 pm}$$ and that of platinum is $$\pu{139 pm}$$. This would cover the quote "the radii of the atoms and the bond lengths within the crystal lattice."

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 ($$V_\mathrm{Metal}$$):

$$V_\ce{Au} = a_\ce{Au}^3 \times \frac{1}{\pu{4 atoms}} \times N_A \\ = (\pu{407.82 pm})^3 \times \left(\frac{\pu{1 cm}}{\pu{10^10 pm}}\right)^3 \times \frac{1}{\pu{4 atoms}} \times \pu{6.022 \times 10^{23} atoms\:mol-1} = \pu{10.21 cm3mol-1}$$

and similarly,

$$V_\ce{Pt} = a_\ce{Pt}^3 \times \frac{1}{\pu{4 atoms}} \times N_A \\ = (\pu{392.31 pm})^3 \times \left(\frac{\pu{1 cm}}{\pu{10^10 pm}}\right)^3 \times \frac{1}{\pu{4 atoms}} \times \pu{6.022 \times 10^{23} atoms\:mol-1} = \pu{9.09 cm3mol-1}$$

Once we know the molar volume of each metal, we can calculate the density ($$d$$) by using $$d=\frac{\text {molar mass}}{\text {molar volume}}$$:

$$d_\ce{Au} = \frac{\pu{196.97 gmol-1}}{\pu{10.21 cm3mol-1}} = \pu{19.29 gcm-3}$$

$$d_\ce{Pt} = \frac{\pu{195.08 gmol-1}}{\pu{9.09 cm3mol-1}} = \pu{21.46 gcm-3}$$

Therefore, it is true that $$d_\ce{Pt} \gt d_\ce{Au}$$.

Relevant resources:

1. John W. Arblaster, In Selected Values of the Crystallographic Properties of Elements; ASM International: Material Park, OH, 2018 (ISBN-13: 978-1627081542).
2. Arnulf Maeland, Ted B. Flanagan, “Lattice Spacings of Gold–Palladium Alloys,” Canadian Journal of Physics 1964, 42(11), 2364-2366 (https://doi.org/10.1139/p64-213).
3. https://www.webelements.com/gold/crystal_structure.html
4. https://www.webelements.com/platinum/crystal_structure.html
• Thank you Sir. I understood everything. But one thing I couldn't grasp is on what basis does the length of cubic unit cell depends upon? – Shubhang Walavalkar Apr 12 '20 at 6:20
• I can't give a direct answer to that, but one of our crystallography persons would do. I'm pure organic chemist. :-) – Mathew Mahindaratne Apr 12 '20 at 8:04
• @ShubhangWalavalkar In a nutshell, unless you are looking at crystalline noble gases the atoms will be in some way bonded to each other. How exactly these bonds are formed in different elements is a rather complex topic but as a rule of thumb if all other things (such as number of shells) are equal a shorter bond length means stronger or more bonds. Comparing gold and platinum, the shorter $\ce{Pt-Pt}$ distance can be rationalised as a stronger or more effective bond – in an extremely simplified manner. Once you know the bond length, the length of the unit cell is simple trigonometry. – Jan Jul 3 '20 at 6:47
• The stronger bonding in metallic platinum is confirmed by its higher melting point (1768 C for Pt, 1064 C for Au). – James Gaidis Jul 3 '20 at 13:26