I have a small (but growing) collection of chemistry books from the 1800s and early 1900s. If you’re a student of chemistry, it’s nice to have an idea of the science’s roots. These old books help to show how the science has matured over the years.
J.L Comstock’s Elements of Chemistry textbook (1851 edition, pages 388-389) provides us with a method for analyzing the composition of native gold – that is, gold as you might find it in the ground. If you’re very, very lucky. (I’d wager that finding native gold was a lot easier in 1851 than in 2006!)
ANALYSIS OF NATIVE GOLD
Native gold, or gold as it occurs in its natural state, is usually alloyed with various proportions of metallic silver and copper. The proportions of each are found by the following method:
Process 1 – Digest a given quantity of the metal, say 100 grains, with so much nitro-muriatic acid as to dissolve the whole. During the process, a white flocculent precipitate will fall to the bottom of the vessel, which is the silver in the form of a chloride of that metal. The clear liquid must be decanted, leaving this to be collected, washed, and dried on a filter, and then weighed. The proportion of pure silver may be estimated at three quarters the weight of the chloride.
100 grains is about six and a half grams. That’s about $120 worth, at the time I typed this. (That is why I’m not going to be able to do this particular bit of chemistry in our lab. 🙂 )
Comstock refers to nitro-muriatic acid, which is a mixture of nitric and hydrochloric (muriatic) acids. This mixture is also referred to as aqua regia, and is one of the few solvent mixtures that will dissolve metallic gold.
The chloride of silver produced is commonly known as silver chloride (AgCl), which is indeed three quarters silver by mass. My intro and freshman chemistry students should be able to verify that percentage with no difficulty.
(107.87 g Ag) / (143.12 g AgCl) * 100% = 75.35% Ag
Back to Comstock …
Process 2 – The remaining solution to which the washings of the precipitated silver was added, contains the solutions of gold and copper. On adding a solution of the proto-sulphate of iron, the gold will be precipitated, when the clear liquid must be decanted, and the precipitate washed and dried, and afterwards reduced to the metallic state, by fusion with poitash and borax, as above directed.
We certainly don’t want to throw away the gold, so Comstock advises us to precipitate the gold out of solution using “the proto-sulphate of iron”, known today as iron(II) sulfate (FeSO4). I suspect that the iron(II) reduces the gold ions to metallic gold, forming a precipitate of gold and leaving iron(III) behind in solution.
3Fe2+(aq) + Au3+(aq) --> 3Fe3+(aq) + Au(s)
Comstock then has us mix the gold with potash (potassium hydroxide) and borax (sodium borate), then heat in a silver crucible. I’m not entirely certain how this one works, having never had the opportunity to try it.
Process 3 – The liquor now remaining contains the copper and the little iron which was added for the separation of the gold. Of the iron, no account is to be taken, but the copper is to be precipitated by inserting in the liquor clean plates of iron, and heating the solution, when the plates will be covered with metallic copper, the weight of which may be ascertained by first weighing the plates, and then finding out how much they have gained. If any of the copper falls to the bottom of the vessel, this must, after washing and drying, be added to that on the plates.
The weight of each metal thus obtained, will, of course, show the proportions in the mass.
To get the copper out of solution, Constock relies on good old iron. Iron, being a more active metal than copper, will replace copper in solution, causing the copper metal to fall out of solution and coat the iron plates. Freshman chemistry and high school students will be familiar with this kind of reaction too – a single replacement reaction, which might go something like this.
Fe(s) + Cu2+(aq) --> Fe2+(aq) + Cu(s)
Here’s what that might look like, using an iron nail in a solution containing copper ion.
Before heating
![[Iron nail in copper solution, before heating]](http://rickandpatty.com/blogfiles/ironcopper1.jpg)
Copper(II) ion in solution gives it the blue color you see now. The reaction is slow. You can see a little copper metal coating the nail, but there is still a lot of copper ion left in the solution. This was taken about 15-20 minutes after the nail was dropped into the tube.
There seems to be a problem here, though. Since some of the iron from the plates replaces copper in the solution, the plates contain less iron after the reation than they did before. Merely subtracting the final and initial weight of the iron plates neglects the loss of iron to the solution. Since copper, atom for atom, weighs more than iron does, the plates will always gain mass when put into the solution, but the overall amount of copper reported will always be too low, unless I’m missing something.
Still, a rather interesting (if expensive) analysis, using some fairly simple and clever chemistry for separating the components of native gold.