Sciencemadness Discussion Board

Ammonia Flower Chemistry

Ioxoi - 17-11-2008 at 16:56

I made a toy I call an "ammonia flower." I have no idea of someone else has already done this and studied the chemistry and I simply don't know about it, but I'll ask here.

It came as an accident from the prepublication article on the Copper Sulfate synth from ammonia, (NH4)2SO4 and copper scrap bits. I have lots of copper scrap and wanted to make salts from it, and found that simple submersion in dilute H2SO4 and bubbling did not seem to dissolve it very well. (Is there an electrochemical reason oxidation and dissolution of copper in alkali happens more readily than that of oxidation and dissolution in acid medium?)

I took a 1oz glass bottle, added household ammonia halfway, scrap copper wire, and a pea-size piece of ammonium sulfate, and sealed it off and shook. A strong lavender color (tetraammine copper) appeared as expected, and became strongest after about 30sec shaking. I continued shaking the jar and did not notice that after ~20min it had become almost clear again, and soon became completely clear.

Opening the bottle, blowing in fresh air, capping and shaking made the blue reappear in seconds. After another 30min however, it was gone. It bloomed and faded like a flower, so I call it an ammonia flower.

I believe, but would like to verify, that the mechanism is first the fast formation of [Cu(NH3)4]+2, and then slow reaction with copper metal to form [Cu(NH3)2]+ ion, which is colorless. (Ie Cu+2 + Cu --> 2Cu+). Air oxidizes the cuprous complex to the cupric one, evidenced by suction in the bottle. Is this more or less a correct mechanism?

I also have another question. After about 10-15 bloom-fades, the blue color persisted and would not go away. Adding more air made it darker and darker, and no amount of time makes the color lighten. I have two possible reasons for this:
1. All available ammonia is complexed. However, this might not explain why the blue color did not fade.
2. Maximum solution concentration of [Cu(NH3)2]+ has been reached. Thus any cupric complexes could not convert to cuprous ones because this would perturb equilibrium the wrong way.

Any ideas on this?

I will next week see if the ammonium sulfate is really necessary, and try it with plain ammonia and copper wire.

Granted, it's not the highest tech of chem, but still a beautiful toy and demonstrator of the equilibrium of three different oxidation states of copper, and cheap to maintain (I can get 0.5 gallon of household ammonia for a dollar, which is good enough for, oh, 10,000 bloom-fades).

If you have a pretty glass bottle, it does make a neat conversation starter.

I also am wondering if the solubilities of cuprous diamine hydroxide (formed when only ammonia is used) and cuprous diamine sulfate (formed with ammonia and ammonium sulfate) are different. If the second explanation for why the flower stops working is right, whichever is more soluble is the way to make the flower last the longest time before refills. Has anyone ever crystallized cuprous diamine salts before? I don't have access to vacuum drying equipment; wonder what shape the crystals are?

bfesser - 17-11-2008 at 17:39

Please note that clear and colorless have two distinct and non-synonymous meanings in the chemistry lexicon.

<strong>Clear - not cloudy, transparent
Colorless - lacking color</strong>

Also, be sure to familiarize yourself with the definitions of <a href="" target="_blank"><em>transparent</em></a> and <a href=""><em>translucent</em></a>, and <a href=""><em>opaque</em></a>.

While titrating HCl with NaOH in water with phenolphthalein as the indicator, the solution being titrated is clear and colorless until the endpoint, after which it becomes pink, but remains clear (although most would just write that it is pink, and omit <em>clear</em>;).

(Interesting note about this post... I took out my Thesaurus to look up the antonyms of 'synonymous'--it doesn't even list synonymous. And I think this is the only time I've ever actually used the book. Useless. So non-synonymous it shall be.)

chemrox - 17-11-2008 at 19:56

The best I can do is a WAG. With extra OH- in solution the Cu++ is stabilized .. maybe tetrahedral as in 3d84s1 so that the tetrahedron has two unfilled orbitals at opposite ends of the tetrahedron and these are stabilized by association with hydroxyl ions.

12AX7 - 18-11-2008 at 02:14

Cu(2+) is almost always square planar. Hence (NH3)4, (H2O)4, Cl4, etc.

I agree, the color change is due to oxidation (or reduction).

Copper is indifferent to OH, although Cu(OH)2 will dissolve slightly in strong NaOH (it's amphoteric).

Copper doesn't much dissolve in H2SO4 or NaOH because neither complexes it very strongly. NH3 and Cl are excellent ligands, and it helps that the Cu(I) intermediate is stable and soluble. (It dissolves in HNO3, but of course just about everything is simply by brute force...)


blogfast25 - 18-11-2008 at 09:33

I saw such a bottle at one of those 'chemical magic' shows I took my 12 year old daughter to once (organised by the Mad Chemist that features on Brainiac). On shaking, the blue appears, on standing it disappears to clear and colourless. I don't recall the guy opening the bottle though, only shaking it up. I don't recall copper wire in there either but that could easily have been concealed (I wasn't close to the experiment). I've never figured out how the trick was done, now I have a possible explanation.

If the amount of initial Cu(NH<sub>3</sub>;)<sub>4</sub><sup>2+</sup> and the amount of oxygen in the system were carefully chosen, there should be enough oxygen for a few 'bloom-fade' cycles (w/o adding fresh oxygen). The amount of copper (surface) would determine how fast the solution fades. The Cu(NH<sub>3</sub>;)<sub>4</sub><sup>2+</sup> blue is very intense, so even dilute solutions would work well...

12AX7 - 18-11-2008 at 10:43

You can do something like that with methylene blue indicating Na2CO3 titration. The solution changes color, then changes back as H2CO3 equilibriates with CO2(aq) and CO2(g). You can appear to reach the endpoint numerous times and then not, until it finally stops changing.


not_important - 18-11-2008 at 17:02

Originally posted by blogfast25
I saw such a bottle at one of those 'chemical magic' shows I took my 12 year old daughter to once (organised by the Mad Chemist that features on Brainiac). On shaking, the blue appears, on standing it disappears to clear and colourless....

Might be "The Blue Bottle Reaction", usually done with the bottle 1/2 to 2/3 full:


How to Do the Blue Bottle Chemistry Demonstration -
Introduction and Materials
In this chemistry demonstration, a blue solution gradually becomes clear. When the flask of liquid is swirled around, the solution becomes blue again. Instructions are given for performing the reaction, the chemistry is explained, and options for making red -> clear -> red and green -> red/yellow -> green color change reactions are explained. The blue bottle reaction is easy to perform and uses readily-available materials.


* tap water
* two 1-liter Erlenmeyer flasks, with stoppers
* 7.5 g glucose (2.5 g for one flask; 5 g for the other flask)
* 7.5 g sodium hydroxide NaOH (2.5 g for one flask; 5 g for the other flask)
* 0.1% solution of methylene blue (1 ml for each flask)


1. Half-fill two one-liter Erlenmeyer flasks with tap water.
2. Dissolve 2.5 g of glucose in one of the flask (flask A) and 5 g of glucose in the other flask (flask B).
3. Dissolve 2.5 g of sodium hydroxide (NaOH) in flask A and 5 g of NaOH in flask B.
4. Add ~1 ml of 0.1% methylene blue to each flask.
5. Stopper the flasks and shake them to dissolve the dye. The resulting solution will be blue.
6. Set the flasks aside (this is a good time to explain the chemistry of the demonstration). The liquid will gradually become colorless as glucose is oxidized by the dissolved dioxygen. The effect of concentration on reaction rate should be obvious. The flask with twice the concentration uses the dissolved oxygen in about half the time as the other solution. A thin blue boundary can be expected to remain at the solution-air interface, since oxygen remains available via diffusion.
7. The blue color of the solutions can be restored by swirling or shaking the contents of the flask.
8. The reaction can be repeated several times.

Safety & Clean-Up

Avoid skin contact with the solutions, which contain caustic chemicals. The reaction neutralizes the solution, which can be disposed of by pouring it down the drain.

How the Blue Bottle Reaction Works

In this reaction, glucose (an aldehyde) in an alkaline solution is slowly oxidized by dioxygen to form gluconic acid:


Gluconic acid is converted to sodium gluconate in the presence of sodium hydroxide. Methylene blue speeds up this reaction by acting as an oxygen transfer agent. By oxidizing glucose, methylene blue is itself reduced (forming leucomethylene blue), and becomes colorless.

If there is a sufficient available oxygen (from air), leucomethylene blue is re-oxidized and the blue color of solution can be restored. Upon standing, glucose reduces the methylene blue dye and the color of the solution disappears. In dilute solutions the reaction takes place at 40-60°C, or at room temperature (described here) for more concentrated solutions.

How to Do the Blue Bottle Chemistry Demonstration - Other Colors
In addition to the blue -> clear -> blue of the methylene blue reaction, other indicators may be used for different color-change reactions. For example, resazurin (7-hydroxy-3H-phenoxazin-3-one-10-oxide, sodium salt) produces a red -> clear -> red reaction when substituted for methylene blue in the demonstration. The indigo carmine reaction is even more eye-catching, with its green -> red/yellow -> green color change.

How to Perform the Indigo Carmine Color Change Reaction

1. Prepare a 750 ml aqueous solution with 15 g glucose (solution A) and a 250 ml aqueous solution with 7.5 g sodium hydroxide (solution B).
2. Warm solution A to body temperature (~98-100°F). Warming the solution is important.
3. Add a 'pinch' of indigo carmine, the disodium salt of indigo-5,5’-disulphonic acid, to solution A. You want a quantity sufficient to make solution A visibly blue.
4. Pour solution B into solution A. This will change the color from blue -> green. Over time, this color will change from green -> red/golden yellow.
5. Pour this solution into an empty beaker, from a height of ~60 cm. Vigorous pouring from a height is essential in order to dissolve dioxygen from the air into the solution. This should return the color to green.
6. Once again, the color will return to red/golden yellow. The demonstration may be repeated several times.


blogfast25 - 19-11-2008 at 09:07

Tim and not_important:

Interesting, something to consider for X-mas, I feel...