10
$\begingroup$

This thought came from researching a previous question. In the spirit of our get-rich-quick scheming, I looked up element densities and prices near gold.

Element Density ($g/cm^3$) Value ($\$/kg$)
Uranium $18.95$ $40$
Tungsten $19.25$ $24$
Gold $19.282$ $53 000$
Roentgenium $19.282$ No market
Plutonium $19.84$ $4 000 000$
Neptunium $20.25$ $660 000$
Rhenium $21.02$ $5 000$
Platinum $21.46$ $54 000$
Darmstadtium $21.46$ No market
Osmium $22.61$ $77 000$
Iridium $22.65$ $42 000$

It would be so easy if we could just substitute some of the gold for a cheaper metal, but it seems that gold is well-protected from this kind of forgery (which is perhaps part of the reason it's so special, to begin with). Almost all of the elements denser are vastly more expensive, or only producible in small quantities in a lab.

There is a very thin market for Rhenium, so we could buy some at a price lower than gold, but any more and we'll risk either bumping up the Remarket price or being foiled by the authorities. As thrifty criminals, we would like to go with Tungsten instead.

Can an allow increase the density of a metal? Qualitatively, I imagine that a lighter element could take up interstitial sites, and allow a higher density. But is there any known Tungsten compound that fits this, or any known method of alloying that would increase its density by that ever-so-slight amount?

$\endgroup$
7
  • $\begingroup$ There's a listed market value for plutonium??? $\endgroup$
    – user10851
    Jul 19, 2013 at 20:31
  • 1
    $\begingroup$ @ChrisWhite Yes! Of course there is. This is from the study of nuclear fuel cycle. It's extremely well established, because Plutonium can't be produced in its elemental form - only within fuel material. The Pu price is so much higher than U because of those multi-billion dollar reprocessing facilities. Advanced fuel cycles are often taken to be economic around $1,000/kg U price, but Pu burns maybe x50 times more thoroughly. That still doesn't get me to their price, so maybe it's the "lab" price. Of course virtually no one is allowed to buy it... $\endgroup$ Jul 19, 2013 at 20:44
  • $\begingroup$ Depleted uranium price must be low enough. Otherwise they would not be shooting it at people or their vehicles. The only problem is that it remains slightly radioactive, which would probably betray you. $\endgroup$
    – babou
    Jul 19, 2013 at 20:56
  • 4
    $\begingroup$ Interestingly, with its fission yielding $84\mathrm{TJ\,kg^{-1}}$, Pu's "market" price is way, way above any energy value, which would be of the order of tens of thousands of dollars. This number bespeaks a demand comprising almost wholly some other use (no prizes for guessing what). $\endgroup$ Jul 20, 2013 at 7:04
  • $\begingroup$ @WetSavannaAnimalakaRodVance Looks to me like your comment could be turned into an answer. $\endgroup$ Jul 26, 2013 at 21:20

4 Answers 4

7
$\begingroup$

I'm not sure whether this counts as an answer since it is just one more idea for a fraud, but your question is about the physics of alloying.

Actually there's no need to alloy to scam. You make up the filler mostly with tungsten, but add a little pellet of platinum. Neither of these materials will rouse the authorities' suspicion, since both have legitimate industrial uses in big quantities. At your quoted prices, the relative proportions of Pt and W needed to give gold density is 0,98552 : 0,0144796. This mixture, at your prices, would cost $806 a kilogram: far better than Rhenium. Of course an alloy could be more unseeable (to ultrasonic tomography and the like).

The method could be refined to a pseudo allow: you could distribute little pellets of Pt.

Alloying would likely still work, since there is quite a difference between the densities of W and Pt, but the proportions would likely change, for the very reasons you are asking about (physics of alloying). So you'd need to do some experimenting.

Even though Pt's price is quite volatile, the costing of this method would be very robust, since most of the cost of your filler is Pt itself and the filler's price per kg is much less than Au's (Pt would have to become outlandishly in demand and Au much, much cheaper than at present to upset the profit).

Lastly, once you have worked out the alloying, you would want to develop an acoustic matching multilayer to match the Au to the filler. You would need to find materials with acoustic refractive indices such that the layers would minimize ultrasound reflexions in the frequency bands most used by authorities. THis is the acoustic analogue of anti-reflexion coatings on optical instruments.

Since the matching layers also change the pverall density, you now have to change the proportions of Pt and W in your original mixture yet again to achieve the same overall density as Au. You now have some serious analysis and experimenting to do to find the right proportions of W and Pt. And that's before you work out a disposal scheme so that you won't get caught.

$\endgroup$
3
$\begingroup$

I noticed that no direct answers were provided to this question, so I decided to answer it from my own experience.

In general, alloy additives can shift the density of an alloy either up or down. I don't know prescisely what the mechanics of the density shift are, but high-speed steels are a good example because of their tungsten content. An extreme example of this example is AISI/SAE T1, which has much more tungsten than most steels that are still used today.

In relation to this question specifically, I simply searched for "rhenium" on MatWeb and browsed the single page of results. I was surprised to find that the increase in density that rhenium causes is so strong that the tungsten alloy with the lowest nominal rhenium content - tungsten with 3% rhenium - was significantly more dense than gold. Re-creating the density of gold could probably be done with a very small amount of rhenium.

For moral reasons I want to add that there are still many problems with this. Tungsten and rhenium are the two most difficult pure metals to melt, and they become even more difficult to melt when combined. A bar of tungsten-rhenium would be harder than many tempered steels and less conductive than gold, which will cause it to fail almost every other type of test, most of which are much more likely than a density test. And it wouldn't be yellow, obviously.

$\endgroup$
1
  • 2
    $\begingroup$ For moral reasons I am believing that the best way to commit this tungsten fraud is to bombard it with sub-atomic particles until it transmutes all the way up to gold... $\endgroup$
    – Michael
    Jul 23, 2015 at 19:17
3
$\begingroup$

Nobody would expect gold to be very pure so they would not attempt to distinguish it from tungsten by very close density. Pure tungsten would be obviously different from gold by colour, hardness, acid test etc., but tungsten embedded inside gold bars is harder to detect.

One way used to tell the difference relies on the slightly diamagnetic properties of gold compared to tungsten which is slightly ferromagnetic. You can tell the difference with a strong magnet and sensitive scales e.g. see this video.

Another method is ultrasound because tungsten is harder and so has a different speed of sound, see this video.

Here is a video to show that using tungsten to replace gold really happens.

Here is another news story on the subject.

$\endgroup$
2
  • $\begingroup$ Btw, I'm not sure if the first video would actually work with Tungsten, since he is using a paperclip and those are made out of steel. I have asked a related question: physics.stackexchange.com/questions/102463/… $\endgroup$
    – Roland
    Mar 7, 2014 at 17:15
  • $\begingroup$ The video "Here is a video to show that using tungsten to replace gold really happens." was blocked by "Seven.ONE on copyright grounds" according to youtube $\endgroup$ Aug 30, 2023 at 10:54
0
$\begingroup$

I think the answer is that pure tungsten is already denser than "gold", but let me tell you what I mean by that.

The london bullion market, LBMA, sets standards for "good delivery". When you sell gold bars "good delivery" is the set of rules for what you must deliver after you have agreed to sell some gold on the exchange. Good delivery only requires 99.5% pure gold under LBMA rules. So, there is 0.5 percent of wiggle room right there.

Next, the densities you quote for gold and tungsten are only 0.167 percent apart.

Lastly I will add that you need to be careful about densities because depending on how an element is crystallized its density as a solid may vary. There is a dispute over which is more dense, iridium or osmium, because of variations from sample to sample. Your table shows iridium being more dense by 0.177%. However if you use x-ray diffraction to calculate inter-atomic length and then apply that to theoretically ideal crystals osmium is denser. So, don't trust densities to these tiny degrees of accuracy.

So, I believe that even an alloy of pure typical tungsten would be denser than "gold" as defined within the limits of the LBMA standards for gold bars.

$\endgroup$
4
  • $\begingroup$ Um obviously a mix of tungsten and platinum... $\endgroup$ Jul 6, 2020 at 5:54
  • $\begingroup$ @AndreiPokrovsky - you probably intended this as a comment on the OP $\endgroup$
    – Paul Young
    Jul 6, 2020 at 16:15
  • $\begingroup$ @PaulYoung No I think he intended it as a comment to OP. Because I too think you, Paul, have confused tungsten and platinum: you said "I think the answer is that pure tungsten is already denser than 'gold'". But tungsten is less dense than gold, but platinum is more dense than gold. $\endgroup$
    – parsecer
    Jan 6, 2022 at 21:08
  • $\begingroup$ @parsecer - a big part of my post is that statements ranking tungsten against gold require more specific language. Because of the financial motive revealed in the body of the OP's question you will need to understand what is meant by "gold is 53000 USD/kg" and "tungsten is 24 USD/kg". At these types of commercial prices you will probably find "tungsten" is denser than "gold". You may find otherwise if you try to buy six 9's pure single crystal samples, but you won't be buying those at these types of prices. $\endgroup$
    – Paul Young
    Jan 8, 2022 at 20:19

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