Aqua Regia process with nitric acid

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Scavenger,

Inquarting, or inquartation , involves alloying the gold with a base metal, preferably silver, to 25% gold content. So if the undissolved nugget weighs 5 grams simply melt it together with 15 grams of silver (stir with a carbon rod when molten), slowly pour the molten alloy into a deep stainless steel dish of cold water, and dissolve the resulting 'cornflakes' in a diluted solution of 1 part 70% nitric to 1 part distilled water.

My video on separating silver from gold shows the dissolving in detail. I have since learned from Harold and GSP that it's better to drop the silver from the Silver Nitrate solution using metallic copper once the gold powder is separated out. I have a new video demonstrating this 'cementing' process. I'll post it to my site soon. It is in the editing stage now.

Steve
 
I have a jar of what I assume is dropped silver. It is a grey powder and it was filtered from acid peroxide after stripping foils and cpus. Do I use the same technique to melt into a button as in the video?
 
Scavenger,

I assume you are using a torch.

You won't need borax. Just perform the melt in a shallow metling dish and be sure everything is good and molten. Be careful of any smoke that this may produce as silver gases are very dangerous. Silver will volatize very easily so use the torch on the lowest possible setting to get the metals melted together. The standard circular motion of the torch will work well for this. I like to preheat the metals to redness with a MAPP gas bottle torch to conserve my OXY/ACT rig gas.

I am not sure that the gray powder you mentioned from the AP process is a silver compound. I know what you are talking about, but I'm not convinced it's silver at all. If you don't have any known silver or sterling silver on hand many other base metals will work. Choose one that has a low melting point and dissolves easily in nitric acid. Try to avoid anything with tin or lead in it as these can interfere with the inquarting process. Copper works but has a higher melting point and will require more heat as well as consuming more nitric acid than silver.

Harold and GSP should have some good tips here as well.

Steve
 
While doing some research on the reaction of copper and nitric acid, I was searching for the concentration of nitric acid that the dilute nitric was used to react with copper in this popular formula:

3Cu + 8HNO3 --> Cu(NO3)2 + 2NO(g) + 4H2O

I found the concentration in a chemistry book to be 20% HNO3:
http://books.google.com/books?id=svv_AgAAQBAJ&pg=PA163&lpg=PA163&dq=3Cu+%2B+8HNO3,+nitric+concentration,+amount+of+nitric+needed&source=bl&ots=cLt8CiRXV6&sig=U7iFf1kAeT2y_Ir_J8vNDsTux6g&hl=en&sa=X&ei=noRYU5K0HoOayASflIGYBA&ved=0CFIQ6AEwBA#v=onepage&q=3Cu%20%2B%208HNO3%2C%20nitric%20concentration%2C%20amount%20of%20nitric%20needed&f=false

With this in mind and the fact the density of 20% HNO3 is about 1.11g/ml

If my math is correct this is what I get:

3Cu (190g of copper for 3 moles) + 8HNO3 (504g of HNO3 for 8moles{20%HNO3 sp.gr 1.11g/ml}) --> yields
Cu(NO3)2 (one mole of copper nitrate 187.5g/mol) +2NO(g) (30 grams or 2moles of nitrogen monoxide gas) + 4H2O (72g of water or 4moles)

If my math is correct we need 559 ml of 20% HNO3 to dissolve 190grams of copper metal or about 0.41 pound of copper in this reaction with dilute HNO3.
 
butcher said:
While doing some research on the reaction of copper and nitric acid, I was searching for the concentration of nitric acid that the dilute nitric was used to react with copper in this popular formula:

3Cu + 8HNO3 --> 3Cu(NO3)2 + 2NO(g) + 4H2O

I found the concentration in a chemistry book to be 20% HNO3:
http://books.google.com/books?id=svv_AgAAQBAJ&pg=PA163&lpg=PA163&dq=3Cu+%2B+8HNO3,+nitric+concentration,+amount+of+nitric+needed&source=bl&ots=cLt8CiRXV6&sig=U7iFf1kAeT2y_Ir_J8vNDsTux6g&hl=en&sa=X&ei=noRYU5K0HoOayASflIGYBA&ved=0CFIQ6AEwBA#v=onepage&q=3Cu%20%2B%208HNO3%2C%20nitric%20concentration%2C%20amount%20of%20nitric%20needed&f=false

With this in mind and the fact the density of 20% HNO3 is about 1.11g/ml

If my math is correct this is what I get:

3Cu (190g of copper for 3 moles) + 8HNO3 (504g of HNO3 for 8moles{20%HNO3 sp.gr 1.11g/ml}) --> yields
Cu(NO3)2 (one mole of copper nitrate 187.5g/mol x 3= 562,5g ) +2NO(g) (30 grams or 2moles of nitrogen monoxide gas) + 4H2O (72g of water or 4moles)

If my math is correct we need 559 ml of 20% HNO3 to dissolve 190grams of copper metal or about 0.41 pound of copper in this reaction with dilute HNO3.
Click to expand...
 
Thanks I have been tired, and missed that 3 in front of the copper salt in my writing of the original formula, which I used for calculation, I am still unsure if I done the math right.
thanks for catching the mistake.

Them doggone moles, just like in my garden always giving me troubles.
 
60g NO and 2520,4g 20wt% HNO3
the other values are correct

*I love it, when I can give some help back to my mentors*
 
not to forget:

2NO+O2-->2NO2
and
4NO2+O2+2H2O-->4HNO3

...so under ideal conditions you gain 126,02g HNO3 or you need less HNO3 depending on the setup
 
but if I remember right, this one is better (please double check):

Cu+H2O2-->CuO+H2O
CuO+2HNO3-->Cu(NO3)2+H2O

63gCu+34gH2O2 (102g at 33%, 283g at 12%, 1133g at 3%) + 127g HNO3 --> 187g Cu(NO3)2 + 36g H2O (theoretically!!)

so instead of using max. 2520g 20wt% HNO3 you would use at least 1905g 20wt% HNO3 and 306g 33wt% H2O2 to dissolve 189g Cu, don't know,if this is cheaper for you, at least less fumes to handle

by using HNO3 AND H2O2 in small increments to dissolve the copper
 
Although others may disagree, I think this is the most straight-forward way of looking at this problem.

In the link that butcher gave, there is a typo on the left side of the equation for strong (68%) nitric. Instead of Cu + HNO3, it should read Cu + 4HNO3. Also, for very weak (1%) nitric, it is assumed that the left side would be Cu + 2HNO3, although that equation isn't given in the link. For 20%, it is assumed to be, as stated, 3Cu + 8HNO3.

Notes:
(1) For this particular problem, the left sides of the equations, assuming they are correct, is all we need worry about. The left side shows what goes into the reaction and the right side shows the end products. In this case, we are only interested in what goes into the reaction and not what comes out.
(2) The left side of the equation for 68% nitric is Cu + 4HNO3. The number before these 2 items indicates the Mole relationship of each that are required for this particular reaction. Therefore, 1 Mole of Cu reacts with 4 Moles of nitric acid. Therefore, the Mole ratio of Cu/4HNO3 is 1:4 or 1/4.
(3) Very simply put, one Mole of a substance is the addition of all the atomic weights of the elements in the substance, expressed in grams. These atomic weights are found in most any chemistry text or handbook. If you have a copy of CRC or Lange's, you can simply look up the total atomic weight of most compounds without having to add up all the individual atomic weights of the elements involved.
(4) The molecular atomic weight of HNO3 is 63.02. Therefore, one mole of HNO3 weighs 63.02g. One Mole of copper is 63.54g. Therefore, when using 68% and the left side of the equation is Cu + 4HNO3, it takes 4 X 63.02 = 252.08g of HNO3 to dissolve 63.54g of copper.
(5) For those interested, here's the chemistry definition of a Mole: "a chemical mass unit, defined to be 6.022 x 1023 (called Avogadro's number) of molecules, atoms, or some other unit. The mass of a mole is the gram formula mass of a substance." In other words, 63.54g of Cu contains 6.022 X 1023 atoms and 63.02g of HNO3 contains 6.022 X 1023 molecules. That's 602,200,000,000,000,000,000,000 or, .602 billion billion.

According to the left sides of these equations, the Mole ratios of Cu/HNO3:
@1%, it is 1/2
@20%, it is 3/8
@68%, it is 1/4

According to the nitric chart on handymath.com
http://www.handymath.com/cgi-bin/nitrictble2.cgi?submit=Entry
The specific gravities (g/cc or ml), at 20C, of these nitric solutions:
@1%, it is 1.00364
@20%, it is 1.115
@68%, it is 1.4048

Therefore, the weight of HNO3 and the number of Moles of nitric (1 Mole of HNO3 weighs 63.02g) contained in one liter of these solutions:
@1%, it is 1.00364 X 1000 X .01 = 10.0g. This is 10.0/63.02 = 0.158 Moles/liter
@20%, it is 1.115 X 1000 X .20 = 223g. This is 223/63.02 = 3.54 Moles/liter
@68%, it is 1.4048 X 1000 X .68 = 955g. This is 955/63.02 = 15.15 Moles/liter

One Mole of copper weighs 63.54g. Therefore, based on the Mole ratios given above, 1 liter of these 3 solutions would dissolve these weights of copper:
@1% = 0.158 x 1/2 X 63.54 = 5.02g
@20% = 3.54 X 3/8 X 63.54 = 84.3g
@68% = 15.15 X 1/4 X 63.54 = 241g

The 241g/l for 68% figures out almost exactly to 2 pounds of copper/gallon [(241 X 3.785)/454], the figure I have always promoted. Easy to memorize. For silver, it comes out a little under 7 pounds/gallon (actually 6.8#/gal) or 100 tr.oz./gallon (actually, 99 TO/gal), again easy to memorize.

In my experience, when using hot 68% nitric, cut 50/50 with water, in an open top container, the ratio of the copper dissolved to the amount of 68% nitric used conforms very closely to the 68% numbers above.

Besides using weaker nitric, there are several things that will swing the ratio closer to the other strengths: The use of hydrogen peroxide, ethylene glycol, etc., in the solution; Keeping the solution cool; Covering the vessel; Refluxing the off gases; etc.

Edited and added to several times.
 
solar_plasma said:
@GSP

do those 20% mean vol% ? I have never seen vol% used other than for drinking alcohol :shock:
Click to expand...

Surely not. They are all w/w. In any scientific literature I have seen, it is assumed to be by weight unless declared otherwise.

I might mention that, at 20C, 68% nitric, by weight, cut 50%, by volume, with water, ends up as about 38.5%, by weight.
 
I first thought I did make a mistake in my calculation, but my calculation was based on dissolving 3 mole Cu.


I might mention that, at 20C, 68% nitric, by weight, cut 50%, by volume, with water, ends up as about 38.5%, by weight.
Click to expand...

welll, yesss....this can easily get lost by mistake. That's why I never think in volume but in weight, if possible.
 
Thanks goldsilverpro, and solar_plasma, I have always had problems with math, and numbers, and it looks like the post above is no exception to that. I really messed that calculation up, not taking time to check my work.

I was up about 20 hours yesterday working on that problem, after spending most of the day and about half the night, to get the wrong answer, I think I will just enjoy my coffee and study gold silver pro's solution to this problem.

I was going to try to redo that calculation, but I had better not post more mistakes, at this time.

GSP I really envy your skills with those numbers, and really appreciate you doing the math with explanations above, hopefully with reading and studying it, maybe some of that skill with the numbers will rub off on me, although for now it looks like I have a lot more studying to do.

Thanks again
 

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