Diamond is the hardest substance on earth which can be attributed to its tightly-bonded crystal lattice. Hence, assuming there were a substance more dense than diamond (which there are), mustn't it be harder than Diamond?

  • 2
    $\begingroup$ The tightly-packed crystal structure of diamond is only one of the factors involved in determining its hardness, but I do recall reading that diamond is the material with the highest atomic density (atoms per cubic centimeter) in ambient conditions, though graphene probably beats it when it comes to area density (atoms per square centimeter). $\endgroup$ Aug 16, 2022 at 14:22
  • 7
    $\begingroup$ Mercury is about 4 times denser than diamond and it it a liquid. Graphite is also made of carbon, but is relatively soft. Gold is nearly six times denser but is also so soft when pure you can mark it with your fingernail. Perhaps hardness is more complex than you think. $\endgroup$
    – matt_black
    Aug 16, 2022 at 15:18

1 Answer 1


There is no general relation between hardness and density. One thing is the packing and bonding character ( what matters mostly for hardness), other is the idealized atomic ball radius and the atomic mass, what matters mostly for density.

Osmium has the highest density among chemical elements, being more than 6 times denser than diamonds with $\pu{22.6 g cm-3}$ versus $\pu{3.52 g cm-3}$. Gold with $\pu{19.5 g cm-3}$ should be one of hardest materials, what is obviously not true as it is very soft.

  • $\begingroup$ I have asked in physics.SE: Is there any empirical relation between hardness and density? (cc @CannedOrgi) $\endgroup$ Aug 21, 2022 at 6:28
  • $\begingroup$ @NilayGhosh There will be close relation hardness vs density within a very narrow group of similar materials, in special cases. Like binary alloys as extreme example. But not generally. $\endgroup$
    – Poutnik
    Aug 21, 2022 at 6:35
  • 1
    $\begingroup$ @NilayGhosh There is obviously many soft and hard materials across the density scale. Perhaps more interesting would be comparing hardness and molar amount density, with the latter addressing density of atomic packing. $\endgroup$
    – Poutnik
    Aug 21, 2022 at 6:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.