Lavoisier's principle of mass conservation states that the mass of a system must remain constant over time. But, as far as I know, there's no similar principle for density.

So, is it possible to develop an alloy as dense as gold but made of constituents less dense than gold?

  • $\begingroup$ For reference gold's density is $19.30 g/cm^3$ $\endgroup$ Commented Nov 13, 2015 at 15:56
  • $\begingroup$ Ok, also for reference, the tungsten density is 19.25 g/cm³ $\endgroup$
    – Mark Messa
    Commented Nov 13, 2015 at 15:58
  • $\begingroup$ Do you mean developing an alloy of gold and other materials that are less dense than gold? Or developing an alloy of any materials which results in a density equal to that of gold? $\endgroup$
    – Wasabi
    Commented Nov 13, 2015 at 16:33
  • 2
    $\begingroup$ Related (but not dupe): engineering.stackexchange.com/q/470/421 $\endgroup$ Commented Nov 13, 2015 at 17:07
  • $\begingroup$ @Wasabi The latter, an alloy of any material (less dense) which result in a density equal to that of gold. Probably it would be cheaper than gold, and difficult to detect that is not gold. $\endgroup$
    – Mark Messa
    Commented Nov 13, 2015 at 18:07

1 Answer 1


Unfortunately, your best bet is abusing the existence of isotopes of tungsten to produce "heavy" tungsten. A potential, but unexplored, alternative is to alloy tungsten with an interstitial atom other than carbon.

There is no way to make tungsten look or behave like gold beyond superficially, i.e. spraypainting it or electroplating it.

No other element or mixture of elements can make it to the density of gold without requiring denser constituents.


There are only a finite number of elements to play with, unfortunately, so options are very limited. I am going to assume when you say "less dense than gold" you mean without using any elements that have a density higher than gold at standard temperature and pressure, and in their as-typically-processed, pure states. To have an alloy with the same density of gold but be made of constituents that all have lower density than gold would require very special and unusual properties of those materials. Density of single crystals is controlled by four factors: crystal structure, atomic weight, interatomic spacing, and quantity of point defects.

Crystal Structure

Changing the crystal structure of an elemental metal can change its density. The change in density is usually very small, on the order of $1\%$, so we would need to start with either tungsten (W, $19.25\ \textrm{g/cc}$) or uranium (U, $19.05\ \textrm{g/cc}$). Other appropriate elements are simply too far below the density of gold.

Unfortunately tungsten occupies only the body-centered cubic (BCC) structure at all temperatures and pressures when solid (phase diagram). Uranium has an orthorhombic structure at room temperature, and changes first to tetragonal around $950\ \textrm{K}$, then to BCC at about $1050\ \textrm{K}$ (phase diagram).

It's worth noting that as the temperature increases, density decreases for two reasons: an increase in vacancy density, and an increase in interatomic spacing due to higher average atomic kinetic energy. It is unlikely uranium will have density $19.3\ \textrm{g/cc}$ at any temperature at atmospheric pressure (we'll get to elevated pressure momentarily). At any rate, you'd have a hard time fooling anyone with a radioactive slab of glowing hot metal. Or at the least you'd probably have folks in black suits knocking at your door.

Atomic Weight

We can change atomic weight by adding or removing neutrons from existing elements to create isotopes. How you would go about mass-producing specific isotopes is beyond the scope of this answer. The stable isotopes of Tungsten are listed here and have 182 to 186 nucleons, weighing between $181.95$ and $185.95\ \textrm{g/mol}$.

The atomic weight of Tungsten is $183.84\ \textrm{g/mol}$, and the value is an average of the isotopic content of typical samples of Tungsten containing the various stable isotopes. To increase the density of tungsten to that of gold, an increase of $0.25\%$ is required, so increasing the atomic weight to $184.3\ \textrm{g/mol}$ should be sufficient.

Uranium has no stable isotopes, but it does have two that last for billions of years, specifically 235 and 238. Unfortunately those are practically the only isotopes in existence, and virtually all of it is 238. The next isotope up, 239, only has a half life of 30 minutes, so by the time you made it it would already be decaying. Even still, that isn't enough of an increase in atomic weight to push it up to the density of gold.

It's also worth noting that, even if you went this route, neutron bombardment creates byproducts like hydrogen and helium which create density-reducing defects such as vacancies and voids, so you would have to reprocess your material after you make it to remove the defects. Alternately, you could process via atomic centrifuge set up to process tungsten, instead of uranium, and isolate the isotopes, then recombine them to the appropriate average atomic weight.

Point Defects

More vacancies would only reduce the density, which means we would need to start with something heavier than gold, so that won't work. More substitutional atoms would also generally require at least one denser component as density virtually always follows a linear rule of mixtures in alloys, so that won't work either. Because of the linear rule of mixtures issue, virtually all multi-metal alloys are out, and only pure elements have any real chance of approaching the density of gold without using elements that have higher density than gold.

The last point defect to consider is interstitial defects, where atoms small enough to fit between the metal atoms are used as alloying elements. Interstitials generally increase the density, as with carbon in steel. Unfortunately, tungsten does not dissolve carbon at standard temperature and pressure, as shown in the phase diagram, so that idea is out for carbon. I am unable to find a phase diagram for nitrogen, so it is unclear whether it would be possible to produce a tungsten-interstitial alloy. If it is up to the requisite $0.25\%$ atomic percent, then this method may also work.

Interatomic Spacing

Aside changing the magnitude of the electromagnetic force constant, the only way to change interatomic spacing is via the application of stress. In this case, we would want hydrostatic compressive stress. Again, we only need an increase of $0.25\%$ for tungsten, so a decrease in lattice constants of one-third that or $0.083\%$ would suffice. That is a strain of $0.00083$. The modulus of tungsten is $411\ \textrm{GPa}$, so the required hydrostatic compressive stress is $341\ \textrm{MPa}$, or a pressure of close to $50,000\ \textrm{psi}$. Fooling anyone by this method would require some interesting apparatus.

  • $\begingroup$ 1) first of all, excellent answer. Thks! $\endgroup$
    – Mark Messa
    Commented Nov 13, 2015 at 18:29
  • $\begingroup$ 2) Please, correct me if my conclusions are wrong. Interstitial atom increases the density, but not as much as you could take other element less dense than tungsten and make the correspondent alloy as dense as gold. $\endgroup$
    – Mark Messa
    Commented Nov 13, 2015 at 18:35
  • $\begingroup$ 3) It's not required that such alloy should look or behave like gold. If it could only pass the Archimedes density test, it would already cause a lot of trouble. $\endgroup$
    – Mark Messa
    Commented Nov 13, 2015 at 18:41
  • $\begingroup$ 2) Is only true if the desired alloy atom fits into interstitial sites in the crystal lattice. Larger atoms, as with most metals, occupy substitutional sites, meaning they replace one of the tungsten atoms. Assuming the lattice doesn't deform, replacing X density with <X density will reduce the overall density. So there is no way to increase it by this method. (The lattice does deform, but not nearly enough to overcome the difference in atomic weight) $\endgroup$ Commented Nov 13, 2015 at 19:04

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