Sciencemadness Discussion Board - Making copper nitrate Wikpedia misleading?

Making copper nitrate Wikpedia misleading?

CHRIS25 - 27-9-2012 at 09:49

I need to make some copper nitrate for photography use. Firstly I assume copper 2 nitrate to be the chemical and I want anhydrous. But my equation is: Cu + 2HnO3 = Cu(NO3)2. Thereby having the following stoichemetry: 63.5 + 63 = 187.5

However, I know that there will be nitric oxide gas and a little water from the nitric acid. But wikpedia says: use 4 times the amount of nitric acid in order to balance the equation stoichemetry speaking. But in practise is my equation more correct - practically speaking.

Also, despite all my notes and writings, I have forgotten how to convert the 63 g/mol into mLs nitric acid. 63g/mol is the 70% concentrated nitric acid, so when I need 2HnO3 i need 126 mLs of acid?

[Edited on 27-9-2012 by CHRIS25]

Vargouille - 27-9-2012 at 10:19

Cu + 2HNO3 -> Cu(NO3)2 (+ H2) doesn't work. Copper can't be oxidized by hydrogen ions, it's the acidified nitrate that does that.

Cu -> Cu⁺² + 2e^-
NO3^- + 2H⁺ + e^- -> NO2 + H2O
NO3^- + 4H⁺ + 3e^- -> NO + 2H2O

The concentration of the HNO3 determines which redox pathway takes precedence. Also, look up copper nitrate a little more for hydration states.

For exchanging the molar ratio, use densities. Not difficult.

kristofvagyok - 27-9-2012 at 10:48

Quote: Originally posted by CHRIS25

I need to make some copper nitrate for photography use. Firstly I assume copper 2 nitrate to be the chemical and I want anhydrous. But my equation is: Cu + 2HnO3 = Cu(NO3)2. Thereby having the following stoichemetry: 63.5 + 63 = 187.5

Basic chemistry vol 1: metals e.g. copper reacts with oxidizing acids e.g. HNO3 and the reaction will yield some NO2 and the it goes as the following: Cu + 4HNO3 ——> Cu(NO3)2 + 2NO2 + 2H2O

And this kind of reaction works with cc acids, so use high excess of HNO3 and distill the excess what didn't reacted.

P.S. nitric acid is HNO3 and not HnO3.

CHRIS25 - 27-9-2012 at 13:46

Vargouille: For exchanging the molar ratio, use densities. Not difficult......So 1.5g is the density. So if I understand you correctly (which I probably don't) then 4 x 63 = 252g. Then 252g / 1.5 = 168 mls of 70% nitric acid. Have I got this right? Just using what little logic I have since I am not a chemist at all. Just using chemistry because I like to, but it is a means to an end.

kristofvagyok: I do not have the means to distill anything, not even water. I have seen this in video, but it is way out of my reach. So I will have to allow the excess liquid to evaporate, use a home made dessicator bag to get the crystals. Can not think of any other way to do this. But excess acid has to be ruled out, hence the stoichemetry. That has to be precise then I minimize any excesses, well that is my theory. Has anyone made this before?

[Edited on 27-9-2012 by CHRIS25]

AJKOER - 27-9-2012 at 14:06

If you truly need anhydrous Cu(NO3)2, then per Wiki: "Anhydrous Cu(NO3)2 forms when copper metal is treated with N2O4:

Cu + 2 N2O4 → Cu(NO3)2 + 2 NO

Attempted dehydration of any of the hydrated copper(II) nitrates by heating instead affords the oxides, not Cu(NO3)2. At 80 °C, the hydrates convert to "basic copper nitrate" (Cu2(NO3)(OH)3), which converts to CuO at 180 °C.[2]"

Basically, you must burn thin Copper wire, or heat fine Cu powder, in an atmosphere of dry NO2 gas. When done, add some O2 to recapture some of the NO2 as:

2 NO + O2 --> 2 NO2

together with new Cu to form more Copper nitrate.

kristofvagyok - 27-9-2012 at 14:06

Quote: Originally posted by CHRIS25

Has anyone made this before?

Yes, I also made it with 92% HNO3, this was my first experiment with cc nitric acid

To minimalize the amount of acid needed then let it react for days and at the end heat it up until it stops fizzing. And use excess of copper, because to get rid of HNO3 is much worse then removing some copper.

Vargouille - 27-9-2012 at 15:07

Quote: Originally posted by CHRIS25

Vargouille: For exchanging the molar ratio, use densities. Not difficult......So 1.5g is the density. So if I understand you correctly (which I probably don't) then 4 x 63 = 252g. Then 252g / 1.5 = 168 mls of 70% nitric acid. Have I got this right? Just using what little logic I have since I am not a chemist at all. Just using chemistry because I like to, but it is a means to an end.

A few things, in no particular order. Density is not measured in grams, it uses a deriviative unit g/mL or g/cm^3. The density of 70% nitric is, straight from Wiki, 1.42 g/mL. The density of pure HNO3 is 1.51 g/mL. Assuming that most of the NOx formation is NO2, which it may or may not be, the stoichiometric amount of HNO3 to use is 177 mL. However, allowing an excess of Cu, with 168 mL, or even 150 mL, is not a bad idea.

CHRIS25 - 28-9-2012 at 03:29

AJKOER: yes thanks. I was aware that I could not heat it, but your example is perfect, unfortunately I do not have the means to carry that one out.
Kristof: you used 92%? I thought that 70% was the general accepted concentration in laboratories, 92% I read was a killer concentration to work with. Either way I have more work because I only have 30%.
Vargouille: Thanks, so I understood the maths correctly, just have to adjust for my 30% concentration. And add more copper than is actually required, does this then insure that ALL the acid has been reacted away so to speak so that there will only be excess water left in the solution? This might prove just as good if I could ensure that only water was left in the solution because for the photographic purposes I would have to dilute anyway and mix with another chemical that will also need dilution, so this extra water can be useful in the final calculations.
So at 30% I actually have 27.12 g/mol of nitric acid?

By the way guys, thankyou for your time in all this, as usual, you are real people behind a keyboard, not just a callsign on this forum.

So, to sum up. the final calculations based on 30% Nitric acid will be: 63g copper + 419ml of acid; this will yield (theory) 187g of Copper nitrate in solution, hopefully just water. If i have all this correctly understood, then I am on my way again with other calculations.

[Edited on 28-9-2012 by CHRIS25]

[Edited on 28-9-2012 by CHRIS25]

[Edited on 28-9-2012 by CHRIS25]

Hexavalent - 28-9-2012 at 08:03

Anhydrous copper nitrate is easier said than prepared;

http://www.sciencemadness.org/talk/viewthread.php?tid=7772

Vargouille - 28-9-2012 at 10:35

No, the molar mass of nitric acid is a constant value. It's the density that changes with concentration. Also, I must apologize, because I think that the way I had been going about it was incorrect. 1.42 g/mL is the weight of the solution, not the weight of the nitric acid. (1.42 g solution/mL)(70 g HNO3/ 100 g solution) gives the correct amount of nitric acid in an amount of 70% HNO3. The density of 30% HNO3 is found here. This site also has tables for many other chemicals, so it's quite useful. In any case, the amount of HNO3 to use is calculated as follows:

4 mol HNO3 x (63.01 g/mol) x (1 mL/ 1.1763 g sln.) X (100 g sln./ 30 g HNO3) = 714 mL 30% HNO3

Considering how much nitric is would take to react a mole of copper, perhaps a smaller batch size would be wise. A tenth of a mole of copper, perhaps. Same considerations as previously, use less than the stoichiometric amounts of HNO3.

[Edited on 28-9-2012 by Vargouille]

CHRIS25 - 28-9-2012 at 11:50

Vargouille thankyou for the table. Unfortunately you have lost me on that maths, I have literally no brain power left to enter into that sort of maths, it's beyond my abilities if you like. So please, if there is a simpler way, albeit less precise, then please show me. I am obviously way off in my own maths. Right now I feel I ought to just buy the stuff, but darn it, I have all the chemicals and actually the problem really arises from not being allowed to have concentrated nitric, otherwise I would not even be posting to the forum, it's that bloody 30% that causes all the problems. Just letting off steam. I don't understand my mistake. 70%/30% is 2.33. I needed 180ml of 70%. Therefore 180 x 2.33 = 419 ml of 30%. I really fail miserably to understand why this is wrong.

[Edited on 28-9-2012 by CHRIS25]

Vargouille - 28-9-2012 at 12:19

As I have said before, the maths that I used were incorrect. It would perhaps be simpler to use the molarity of a 30% solution of nitric acid, which is 5.60 M (I just calculated it from the percentage). M is the symbol for a "molar", which is calculated by dividing the moles of solute by the volume (in liters) of solution. You know that the stoichiometric amount of HNO3 that would (we assume) react completely with one mole of copper is 4 moles of HNO3. Now, you have the ratio of 5.60 moles per 1 liter of solution. This can form a fraction of 5.60 moles/L or approximately 0.1786 L/mole. By using dimensional analysis, you can apply mathematical functions to units. Multiplying two values with the same units gives a value of the unit squared, and dividing one of the values by the other gives an answer without units. If you have more questions, you can probably find a site that explains dimensional analysis better than I can, or if you U2U me, I can try to find out some way to explain it outside of the forum.

The reason why the 180 mL of 70% is incorrect is because the path I took only works if all of the 1.42 g/mL reacts. In fact, only the HNO3 reacts, which is 70% of the solution. Therefore, the density of HNO3 in a 70% solution is 70% of 1.42 g/mL.

CHRIS25 - 28-9-2012 at 12:36

Phew. Ok, I am too confused with all these terms and maths, and to be honest I will not be using this sort of thing much that would warrant me having to learn it. I am very very busy with many things that involve a lot of working out, so this maths is more than I can handle. Suffice it to say, i will just throw in some copper and nitric acid together at the best calculations that I can come up with. How did you work the mole thing out? 70% = 15.8m so 1% = 0.226m then 0.226 x 30% = 6.77m ????

[Edited on 28-9-2012 by CHRIS25]

[Edited on 28-9-2012 by CHRIS25]

Vargouille - 28-9-2012 at 13:03

The way you propose to solve for molarity of solutions doesn't quite work, because mixing 1 mL of water and 1 mL of HNO3 doesn't necessarily result in 2 mL of HNO3 solution. A more reliable way is by starting out knowing that the density of 30% HNO3 is 1.1763 grams of solution for every mL of solution. Since it's 30% HNO3 (by mass, as is most common), you know that for every 10 grams of solution, there are 3 grams of HNO3. Thus, it can also be said that there are 0.35289 g HNO3 in every mL of solution. It can also be said that in every liter of solution, there are 352.89 grams of HNO3. The molar mass of HNO3 is 63.01 g/mol. Dividing 352.89 g HNO3/ L solution by 63.01 g/mol gives 5.60 moles of HNO3 in every liter of solvent, thus 5.60 M.

CHRIS25 - 28-9-2012 at 14:11

This makes sense now. This I can follow. Thankyou. But I still haven't a clue how you arrived at 714ml in your previous post.
Here - 4 mol HNO3 x (63.01 g/mol) x (1 mL/ 1.1763 g sln.) X (100 g sln./ 30 g HNO3) = 714 mL 30% HNO3. There has to be, there is I am sure much more friendlier ways of doing this calculation. I am not an analyst or mathematician. I have seen this worked out in simpler maths, but I could not find where I had read it, hence my original post on this question. So you answered with this: It would perhaps be simpler to use the molarity of a 30% solution of nitric acid, which is 5.60 M ,so basically thanks for taking the time to help but we are going around now in circles because while I understand mols and such, I am at a loss as to how 5.6 M helps me to make the correct amount of nitric acid . I think its best to leave it, there are simpler ways of doing this, I do not need analytical maths, despite you taking the trouble to help. Which I appreciate, sometimes layman's explanations will suffice for everything that I need. And it does not need to be made so complex. I understand that you need to understand mols and grams and percentages and how solutions must be worked out stoichemetry wise, I have been able to grasp most of this, but a simple thing like how much 30% of a given solution do I need to have in order to equate to 'x' amount of 70% solution that is needed for a reaction to take place has to be child's play mathematically. Surely?

[Edited on 28-9-2012 by CHRIS25]

[Edited on 28-9-2012 by CHRIS25]

Vargouille - 28-9-2012 at 15:45

There are, of course, simpler ways to do the calculation I preformed from the density. That's why I calculated the molarity. In essence, it's the same calculations, just done in two parts, but splitting it in twain can reduce errors because the maths are just in smaller doses.

To answer how to use the value "5.6 M HNO3", remember what M stands for--moles per liter-- to writ, each liter of the solution has 5.6 moles of HNO3. Let's say that 4 moles of HNO3 react with one mole of copper. If you divide 4 moles of HNO3 by the molarity (the concentration in M), the answer is the number of liters of solution needed to "completely" react with the copper. Short version, for every mole of copper you want to react, use about 700 mL of the 30% HNO3.

In an ideal world, volumes would add perfectly (1 mL added to 1 mL results in 2 mL of solution), and the process of mathematically determining how much solution of 30% is equivalent to a given amount of solution of 70% would be trivial. Unfortunately, this isn't an ideal world, so that process isn't accurate. A similar process using molarities, however, is much more accurate, even if it isn't perfect. These kind of things are valuable to know, even if they aren't immediately in the scope.

CHRIS25 - 28-9-2012 at 16:03

Ok, I am following, it makes sense what you say, except (I'm infuriated by my own bloody ignorance), 4 moles divided by 5.63 ??? or of course 252 (4x63)/molarity, that is the concentration of the 30% so 252/5.63 = 44.7 ?? can you see why I am having problems cutting through to the light?

[Edited on 29-9-2012 by CHRIS25]

[Edited on 29-9-2012 by CHRIS25]

Vargouille - 28-9-2012 at 17:50

You're mixing your units. 252 is the number of grams of HNO3 in 4 moles, so you can't use the molarity on it directly. Molarity is used directly with volumes (of the solution) or moles (of the solute). If you do maths with dimensional analysis like this, you should always put your units.

CHRIS25 - 29-9-2012 at 01:15

You are not making sense. I understand moles and grams. But "If you divide 4 moles of HNO3 by the molarity (the concentration in M), "What on earth do you mean? divide 4 by what? The molarity is 5.63.

CHRIS25 - 29-9-2012 at 04:44

No I'm sorry, I think it has been made far too complicated by having to measure out to such micrscopic accuracy, there is no need to analyse to the 100th degree. It's simple: I need 177mls of 70% of HNO3. I have only 30%. Therefore 2.33 times this 177 is 412mls. It's really child's play, if I have a 50% solution of something and need to add 100% of 20mls. What do you do? You simply add twice the 50% which is 40mls of 50% to get equivalent 100% of 20mls, and remember the extra water content, it's so darn simple. I can understand your way of thinking for analytical chemistry and biochemical interactions, but I am making a toner for cyanotypes that I am doing. So accuracy is important to within 10% I suppose, subjectively speaking. Having said all this of course I will only be doing a small reaction, not involving litres of nitric acid.

[Edited on 29-9-2012 by CHRIS25]

Vargouille - 29-9-2012 at 05:25

No, that is incorrect. I've explained why. I've done a calculation, and the required amount of 70% HNO3 to "completely" react with a mole of copper is 254 mL. I've already calculated how much 30% solution it takes to react with a mole of copper (714 mL). If you multiply 254 mL by the ratio 70%/30% (equal to 2.33), you get 593 mL. An incorrect answer. I know how you got the answer, and it would be fine in an ideal system, but this is most certainly not ideal. Chemistry can be simple, but in practice, it's rarely that simple.

This is far from microscopic accuracy. It's not even analytical accuracy. Perhaps some applications will be fine with such a significant error as using 412 mL instead of the correct 714 mL, but I wouldn't risk it.

CHRIS25 - 29-9-2012 at 06:32

Hi, ok, earlier you gave this maths: 4 mol HNO3 x (63.01 g/mol) x (1 mL/ 1.1763 g sln.) X (100 g sln./ 30 g HNO3) = 714 mL 30% HNO3, when I asked you about this you switched to moles by saying: As I have said before, the maths that I used were incorrect. It would perhaps be simpler to use the molarity of a 30% solution of nitric acid, which is 5.60 M.

Now, both the first one and second maths threw me. Your answer here about using moles to calculate the amount of 30% I really do not understand your reasoning. So can we stick please to the first example where you have this: 4 mol HNO3 x (63.01 g/mol) x (1 mL/ 1.1763 g sln.) X (100 g sln./ 30 g HNO3) = 714 mL 30% HNO3. But just explain where you get the figures of 1.1763 from. And then why you use the 100g that is divided by 30g. (obviously I know what this has to do with) It's less about maths and so much more about understanding the WHY you use what figure. That's it. If I understand the reasoning behind this equation then I am on my way, I know it.

The 4 mole x the 63 is simple. It's the rest. If we stick with this then I should now be fine. And thankyou by the way, I am more annoyed at myself believe it or not because I simply have never studied maths or can even think in mathematically orientated brain patterns, that's a gift I think. So I am really battling way out of my comfortable zone - hope you patiently understand this.

Vargouille - 29-9-2012 at 07:24

1.1763 g/mL is the density of 30% HNO3 (at 25C), as from the site I linked previously. For analytical purposes, the ambient temperature would be used instead of the standard temperature of 25C. Unless ambient temperature is near 0C or near 50C, there isn't much deviance in the density. The significance of the difference between values of solution (sln) and values of HNO3 is that the solution as a whole doesn't react with copper, it is the HNO3 in the solution that does so. Because the specific solution is 30%, for every 100 grams of sln, there are 30 g of HNO3 (Hence the ratio 100 g sln/ 30 g HNO3). By using the two of these ratios, I can calculate the amount of HNO3 in a mL of the solution.

It sometimes helps to imagine that the values have a secondary unit defining what chemical or mixture it represents. Thus, 4 mol HNO3, 63.01 g HNO3/ mol HNO3, 1.1763 g sln/ mL sln, and etcetera.

CHRIS25 - 29-9-2012 at 07:47

Ok Finally, this now makes sense. Explains also why many of my previous reactions were off. nevermind. I have it. Though and no need to reply: It sometimes helps to imagine that the values have a secondary unit defining what chemical or mixture it represents. Thus, 4 mol HNO3, 63.01 g HNO3/ mol HNO3, 1.1763 g sln/ mL sln, and etcetera. Again too much info, I don't follow this at all. Actually I have a feeling that it is more the way that something is written rather than what you have written that I could not fathom. No need to reply though. That first calculation method that you gave, to which you have explained where the figures came from, if only that was said at the beginning I think you would have saved yourself some energy, however that is not to deny the fact that I do appreciate your effort and perseverence. thanking you. Hope you have a good weekend.

Just noticed something weird: you said:....I can calculate the amount of HNO3 in a mL of the solution. BUT, a very big BUT, you just calculated how many mLs 30% to use, how can this same equation equal the amount of HNO3 in 1 millilitre, or even 1 mole if you meant that, can you see how confusing you can be? It defies reasoning.

Just another thought. You said that 30g of HNO3 is in every 100g of 30% solution. This is obvious. But why then, if this be the case, are we still having to go way beyond the 2.3 x 30%? So in effect using TWICE the amount nitric acid content than what is asked for in the stoichemetry?

[Edited on 29-9-2012 by CHRIS25]

[Edited on 29-9-2012 by CHRIS25]

[Edited on 29-9-2012 by CHRIS25]

Vargouille - 29-9-2012 at 09:36

If you don't quite understand something, I'm going to at least attempt to explain it. There's no shame in the asking. The notation I use is a sort of mathematical short-hand. "4 mol HNO3" means "4 moles of HNO3". "63.01 g HNO3/ mol HNO3" is stating that there are 63.01 grams of HNO3 in every mole (of HNO3). This is to create a distinction between densities of solutions, such as the "1.1763 g sln/ mL sln" which states that there are 1.1763 grams of the solution in every mL (of the solution).

The same equation that finds the amount of HNO3 in a 30% solution is used to determine how much of that solution represents 4 moles of HNO3. Start with the density, 1.1763 g/mL. Note that both units refer to the solution. Multiplying by (30 g HNO3/ 100 g sln) results in a value with units of g HNO3/ mL sln. This value tells how many grams of HNO3 are in a mL of the solution. This value is used to find how many mL of the 30% solution to use to react with one mole of copper. Starting with one mole of copper, we assume that 4 moles of HNO3 will react completely with the 1 mole of copper. Next, to find how much mL of solution represents these four moles, we find the mass of 4 moles of HNO3, 252.04 g. We know how many grams of HNO3 are in a mL of the 30% solution from the previous equation from density, so you can plug that value into the equation, and find the volume of solution to use. This picture may clarify the process.

To answer your PPS, it's because 1 mL of 70% HNO3 doesn't have the same amount of HNO3 as 2.33 mL of 30% HNO3. There's a difference of about 0.17 g of HNO3, which doesn't sound like much at this scale, but at the scale of hundreds of milliliters of HNO3, it becomes a significant difference. If your process works, then 254 mL of 70% HNO3 should have the same amount of HNO3 as 593 mL of 30% HNO3 (593 is 254x2.33). 254 mL of 70% HNO3 is 252 g of HNO3, while 593 mL of 30% HNO3 is 209 g HNO3. So no, 714 mL of 30% HNO3 is not twice the stoichiometric amount of HNO3 required to react with one mole of copper. If anything, it's a little less than stoichiometric because of rounding.

CHRIS25 - 29-9-2012 at 10:32

254 mL of 70% HNO3 is 252 g of HNO3
593 mL of 30% HNO3 is 209 g HNO3.

How did you do that?

Also, since it has been wisely put forward that I should put in excess copper, then it seems to my logic that if I had 10ml of HNO3 and kept adding copper until there was no more reaction then I would have the same thing essentially, copper nitrate? And all this maths becomes null and void?

[Edited on 29-9-2012 by CHRIS25]

[Edited on 29-9-2012 by CHRIS25]

Vargouille - 29-9-2012 at 11:14

The density of a solution of 70% HNO3 solution is 1.42 g/mL at 25C. By multiplying it by the ratio (70 g HNO3/ 100 g sln), I received the answer of 0.994 g HNO3/ mL sln. Multiplying this by 254 mL gives 252 g HNO3. Previously, I calculated the amount of HNO3 in a mL of 30% solution (0.35289 g HNO3/ mL sln) and simply multiplied it by 593 mL.

You could do that, of course, but doing the maths gives you an idea of how much of the reagents you need, so you can have everything ready before the start of the reaction. One problem with your process is that once most of the HNO3 is depleted in the reactions, the final reactions are less visible. The solution, of course, is to simply leave it to react for a long time, but as from this video, the reaction can take an exceedingly long time to go to completion.

CHRIS25 - 29-9-2012 at 13:00

Hi thanks. Ok That is straightforward. By the way Im putting all this into a PDF. A lot of work here to note. So, I watched that video, two days? But that was 100ml of his unknown phosphoric and nitric mixture. Still I have time. Will perform the reaction and re-evaluate my other concoptions now that I have this new maths to be applying. But when you say excess copper needs to be added, what kind of percentage of the stoichemetry amount would you suggest?

Also, if you don't mind, I notice that Copper trihydrate is Cu(NO3)2.3H2O. Since my reaction theoretically ends with 2H2O, what will I actually be getting since there is no dihydrate that I have found. There is a monohydrate though, I suspect that since H2O can not exist, (its formula being 2H2O) then this is the monohydrate? Remember I am not a chemist?

[Edited on 29-9-2012 by CHRIS25]

Vargouille - 29-9-2012 at 14:44

I would suggest a 10% stoichiometric excess. It should be enough. As for the hydration states, I wouldn't presume that copper nitrate doesn't exist as a dihydrate, only that that hydration state hasn't been reported. However, the 2H2O formed from the reaction is unrelated to the hydration state of the crystallized copper nitrate, especially considering that the reaction takes place with water as a solvent. There are a few factors, including ambient temperature, and humidity, that impact the hydration state of crystallized salts. I can not honestly say that I am sure of the hydration state of the salt you will produce, because of those reasons, and due to the hygroscopy of copper nitrate, the tendency to absorb water from the atmosphere. My best guess is that it will be either in the trihydrate or hexahydrate form, both of which form blue rhombic crystals.

CHRIS25 - 29-9-2012 at 23:50

So the excess copper will, as was suggested earlier by someone, ensure that all of the Nitric acid part be used up. Leaving me with copper nitrate in water, though probably slightly acidic? Which I can then gently boil away and place the remaining wet solids into a dessicator bag. Is this a suitable method to proceed? I am not after the anhydrous, this seems impossible for someone like me and actually I don't need it since my end product will be re-dissolved again anyway. I have read that attempting to dehydrate any of the copper 2 nitrate hydration states produces CuO. But the action of boiling away until just the visible liquid is gone does not even come close to de-hydrating does it?

[Edited on 30-9-2012 by CHRIS25]

Vargouille - 30-9-2012 at 04:01

"Gently boiling" is still far above the decomposition temperature of copper nitrate, 80C. Applying heat above that temperature, even if the copper nitrate is in solution, will decompose it. Also, apparently, the hexahydrate decomposes at 26.4C, so it may be unlikely that you form that hydration state, depending on ambient temperature. I haven't yet found what the decomposition constitutes, mind you, but that's something that you can search up. In any case, the best solution is after filtering to remove the copper, to put it into a dessicator bag to dry, perhaps for weeks. If you keep the temperature low, around 50C, you should be able to remove a large amount of excess water, but I would prefer the "slow boat to China" route.

CHRIS25 - 30-9-2012 at 04:27

Right. I have been searching on google for the various decomposition rates at different temperatures: Frustrating when all you turn up is the obvious chit chat, and videos showing you what happens. Nothing around really that I can find listing various temperatures. Anyway this requires a sophisticated search for those tech documents - never know how to search for those, requires someone who knows exactly how to type in the correct words in the correct order - trouble with google is that unless you have one of those thick manuals that deliver you weeks of reading on how to target your query (phew - no time for that), people like me will always miss about 80% of what really is out there. I think I'll take a speed boat at first, gently heat keeping it below 40 degrees with my hot plate. I think that should do it, I saw how much heat it took to decompose that copper nitrate in a test tube, so 40 degrees or below should get rid of the water and keep the Cu(NHO3)2 in one piece. I can't wait weeks, I will probably put half away for weeks, and work on the other half. Many appreciations for your constant time on this, I have actually learned far more than just the copper nitrate question, which actually is what I want to do, I can never settle for how answers, I need to understand the answer otherwise the answer has no purpose for me and I can not use it.