I made this precious-metal version of the truncated icosahedron using *Stella 4d*, a program which is available here: http://www.software3d.com/Stella.php.

# Tag Archives: gold

# One Dozen Precious Metal Cubes: A Problem Involving Geometry, Chemistry, and Finance (Solution Provided, with Pictures)

The troy ounce is a unit of mass, not weight, and is used exclusively for four precious metals. At this time, the prices per troy ounce, according to this source for current precious metal prices, for these four elements, are:

- Gold, $1,094
- Palladium, $600
- Platinum, $965
- Silver, $14.82

(As a side note, it is rare for platinum to have a lower price per troy ounce than gold, as is now the case. I would explain the reasons this is happening, except for one problem: I don’t understand the reasons, myself, well enough to do so. Yet.)

A troy ounce equals 31.1034768 grams, but, for most purposes, 31.103 g, or even 31.1 g, works just fine.

Also, as you can see here, these “troy elements” are all in one part of the periodic table. This is related to the numerous similarities in these elements’ physical and chemical properties, which is itself related, of course, to the suitability of these four elements for such things as jewelry, coinage, and bullion.

To determine the volume of a given mass of one of these metals, it is also necessary to know their densities, so I looked them up, using Google (they are not listed on the periodic table above):

- Gold, 19.3 g/cm³
- Palladium, 11.9 g/cm³
- Platinum, 21.46 g/cm³
- Silver, 10.49 g/cm³

In chemistry, of course, one must often deal with elements (as well as other chemicals) in terms of the numbers of units (such as atoms or molecules), except for one problem: this is absurdly impractical, due to the outrageously small size of atoms. Despite this, though, it *is* necessary to count such things as atoms in order to do much chemistry at all, so chemists have devised a “workaround” for this problem: when counting units of pure chemicals, they don’t count such things as atoms or molecules directly, but count them a mole at a time. A mole is defined as a number of things equal to the number of atoms in exactly 12 grams of pure carbon-12. To three significant figures, this number is 6.02 x 10²³. To deal with moles, since atoms have differing masses, we need to know the *molar mass* (mass of one mole) of whatever we are dealing with to convert, both directions, between moles and grams. Here are the molar masses of the four troy-measured elements, as seen on the periodic table above, below each element’s symbol.

- Gold, 196.97 g
- Palladium, 106.42 g
- Platinum, 195.08 g
- Silver, 107.87 g

I’ve given these numbers as the information needed to solve the following problem: rank one dozen precious metal cubes (descriptions follow) by ascending order of volume. There are three cubes each of gold, palladium, platinum, and silver. Four of the twelve (one of each element) have a mass of one troy ounce each. Another four each have a value, at the time of this writing, of $1,000. The last set of four each contain one mole of the element which composes the cube, and, again, there is one of each of these same four elements in the set.

If you would like to do this problem for yourself, the time to stop reading is now. Otherwise (or to check your answers against mine), just scroll down.

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In the solutions which follow, a rearrangement of the formula for density (d=m/v) is used; solved for v, this equation becomes v = m/d. In order, then, by both volume and edge length, from smallest to largest, here are the twelve cubes:

**Smallest cube: one troy ounce of platinum**

One tr oz, or 31.103 g, of platinum would have a volume of v = m/d = 31.103 g / (21.46 g/cm³) = 1.449 cm³. A cube with this volume would have an edge length equal to the its volume’s cube root, or 1.132 cm. (This explanation for the calculation of the edge length, given the cube’s volume, is omitted in the items below, since the mathematical procedure is the same each time.)

**Second-smallest cube: $1000 worth of gold**

Gold worth $1000, at the time of this posting, would have a troy mass, and then a mass in grams, of $1000.00/($1,094.00/tr oz) = (0.914077 tr oz)(31.103 g/tr oz) = 28.431 g. This mass of gold would have a volume of v = m/d = 28.431 g / (19.3 g/cm³) = 1.47 cm³. A cube with this volume would have an edge length of 1.14 cm.

**Third-smallest cube: $1000 worth of platinum**

Platinum worth $1000, at the time of this posting, would have a troy mass, and then a mass in grams, of $1000.00/($965.00/tr oz) = (1.0363 tr oz)(31.103 g/tr oz) = 32.231 g. This mass of platinum would have a volume of v = m/d = 32.231 g / (21.46 g/cm³) = 1.502 cm³. A cube with this volume would have an edge length of 1.145 cm.

**Fourth-smallest cube: one troy ounce of gold**

One tr oz, or 31.1 g, of gold would have a volume of v = m/d = 31.1 g / (19.3 g/cm³) = 1.61 cm³. A cube with this volume would have an edge length of 1.17 cm.

**Fifth-smallest cube: one troy ounce of palladium**

One tr oz, or 31.1 g, of palladium would have a volume of v = m/d = 31.1 g / (11.9 g/cm³) = 2.61 cm³. A cube with this volume would have an edge length of 1.38 cm.

**Sixth-smallest cube: one troy ounce of silver **

One tr oz, or 31.103 g, of silver would have a volume of v = m/d = 31.103 g / (10.49 g/cm³) = 2.965 cm³. A cube with this volume would have an edge length of 1.437 cm.

**Sixth-largest cube: $1000 worth of palladium**

Palladium worth $1000, at the time of this posting, would have a troy mass, and then a mass in grams, of $1000.00/($600.00/tr oz) = (1.6667 tr oz)(31.103 g/tr oz) = 51.838 g. This mass of palladium would have a volume of v = m/d = 51.838 g / (11.9 g/cm³) = 4.36 cm³. A cube with this volume would have an edge length of 1.63 cm.

**Fifth-largest cube: one mole of palladium**

A mole of palladium, or 106.42 g of it, would have a volume of v = m/d = 106.42 g / (11.9 g/cm³) = 8.94 cm³. A cube with this volume would have an edge length of 2.07 cm.

**Fourth-largest cube: one mole of platinum**

A mole of platinum, or 195.08 g of it, would have a volume of v = m/d = 195.08 g / (21.46 g/cm³) = 9.090 cm³. A cube with this volume would have an edge length of 2.087 cm.

**Third-largest cube: one mole of gold**

A mole of gold, or 196.97 g of it, would have a volume of v = m/d = 196.97 g / (19.3 g/cm³) = 10.2 cm³. A cube with this volume would have an edge length of 2.17 cm.

**Second-largest cube: one mole of silver**

A mole of silver, or 107.87 g of it, would have a volume of v = m/d = 107.87 g / (10.49 g/cm³) = 10.28 cm³. A cube with this volume would have an edge length of 2.175 cm.

**Largest cube: $1000 worth of silver**

Silver worth $1000, at the time of this posting, would have a troy mass, and then a mass in grams, of $1000.00/($14.82/tr oz) = (67.48 tr oz)(31.103 g/tr oz) = 2099 g. This mass of gold would have a volume of v = m/d = 2099 g / (10.49 g/cm³) = 200.1 cm³. A cube with this volume would have an edge length of 5.849 cm.

Finally, here are pictures of all 12 cubes, with 1 cm³ reference cubes for comparison, all shown to scale, relative to one another.

A third of these cubes change size from day-to-day, and sometimes even moment-to-moment during the trading day, if their value is held constant at $1000 — which reveals, of course, which four cubes they are. The other eight cubes, by contrast, do not change size — no precious metal prices were used in the calculation of those cubes’ volumes and edge lengths, precisely because the size of those cubes is independent of such prices, due to the way those cubes were defined in the wording of the original problem.

# My Aqua Regia Story

This is my twentieth year teaching, but only the first year when I have not taught at least one class in chemistry, and I miss it. One of my fondest memories of chemistry lab involves the one time I experimented with aqua regia — a mixture of acids which, unlike any single acid, can dissolve both gold and platinum, the “noble metals.” I had read a story of a scientist’s gold Nobel Prize being protected from the Nazis by dissolving it in aqua regia, and then recovering the gold from solution after World War II had ended. Having read about this, I wanted to try it myself, and also thought it would make an excellent lab for classroom use — if I could figure out how to recover the gold, and also learn what precautions would be needed to allow high school students to perform this experiment safely. For sensible and obvious reasons, I conducted a “trial run” without students present, but with another chemistry teacher nearby, since aqua regia, and the gases it produces when dissolving gold, are quite dangerous. Someone else has put a video on YouTube, showing aqua regia dissolving gold, so you can see something much like what I saw, simply by watching this video.

First, I obtained one-tenth of a troy ounce of gold, which cost about $80 at the time. I had read about the extreme malleability of gold, one of the softest metals, and wanted to see evidence for it for myself — so, before I prepared the the aqua regia, I used a hammer to try flattening the gold sample into a thin sheet. That didn’t work, but it didn’t take long for me to figure out why — I had accidentally bought gold coin-alloy, which is 10% copper, not pure gold. Since this alloy is far less malleable than pure gold, my attempt to flatten it had failed, but I also knew this would not pose a problem for my primary experiment — the one involving aqua regia. Also, I didn’t have another spare $80 handy, to purchase another 1/10 troy ounce of pure gold, so I proceeded to make, for the first time in my life, a small amount of aqua regia — Latin for “royal water.”

Unlike what is shown in the video above, I prepared the acid-mixture first, before adding the gold, using a slightly-different recipe: the traditional 1:4 ratio, by volume, of concentrated nitric acid to concentrated hydrochloric acid. Both these acids look (superficially) like water, but the mixture instantly turned yellow, and started fuming, even before anything was added to it. Wearing full protective gear, I watched it for a few minutes — and then, using tongs held by gloved hands, lowered my hammer-bashed sample of gold into the fuming, yellow mixture of concentrated acids.

It worked. It was a fascinating reaction, and a lot of fun to watch. At approximately the same time that the last of my gold sample dissolved, something occurred to me: I had failed to research how to recover the dissolved gold from the resulting solution! No problem, I thought — I can figure this out. (I am seldom accused of lacking self-confidence, even when I’m wrong.)

My first idea was to use a single-replacement reaction. Many times, I have had students extract pure silver from a solution of silver nitrate by adding a more-active metal, such as copper. The copper dissolves, replacing the silver in the silver nitrate solution, and silver powder forms, as a precipitate, on the surface of the copper. Thinking that a similar process could be used to precipitate out the gold from my gold / aqua regia mixure, I simply added come copper to the reaction beaker. The corrosive properties of my aqua regia sample had not yet been exhausted, though, and so the remaining aqua regia simply “ate” the copper. The result was a mess — I had only succeeded in turning an already-complicated problem into an even-more-complicated problem, by adding more chemicals to the mixture. More attempts to turn the gold ions back into solid gold dust, using other chemicals, followed, but all of them failed. Finally, I used a strong base, sodium hydroxide, to neutralize the still-acidic mixture, and then, disgusted by my failure to recover the gold, found a way to safely dispose of the mixture, and did so.

In retrospect, I think I know where I messed up — I should have neutralized the remaining acids in the mixture with sodium hydroxide *first*, before adding copper to cause the gold to precipitate out, in a no-longer-acidic solution of ions with *much* less hydronium present. That, I think, will work, and I do intend to try it sometime — after doing more research first, to increase my level of certainty, and also after waiting for the current price of gold to drop to less-expensive levels. Right now, after all, a tenth of a troy ounce of gold costs roughly $120, not a mere $80.

As for the lost $80, I’m not upset about that anymore. I definitely learned things while doing this, and now view the $80 spent as simply the cost of tuition for an educational experience.