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.
