I have a piece of copper mesh from which I would like to remove all coatings/oils. The copper comes from a pot scrubber like this. In order to remove any residue, I held a standard bic lighter up to it. It gave off black smoke. I moved the lighter around the mesh, waiting for it to stop giving off any such smoke, but seemed to do so indefinitely. Even when I just held the lighter in one place, it would not stop giving off a steady stream of black smoke.

I then did the same thing with a torch lighter on a fresh piece of copper from the same source. This time it was hard to tell if there was any smoke at all-- and if there was, it stopped being produced very quickly.

With both the normal lighter and the torch lighter, the surface of the copper turned gray-black. I assume part of this is from the formation of copper oxides, but part of it may also come from the deposition of incompletely combusted carbon on the surface. In both cases I was able to restore the shiny surface by putting the mesh in $\ce{HCl}$-based toilet cleaner for a few minutes.

Why is the normal lighter producing black smoke where the torch lighter is not? What does this smoke consist of?


1 Answer 1


Common copper wire, especially used in electronic components like inductors and transformers, is usually covered with insulating varnish. This is the common source of black smoke, sometimes it may even burn on its own. I doubt, however, that kitchenware is coated with such varnish, there is no point in it.

There are two types of devices with flames: ones where the fuel is premixed with air, that usually produce blueish flame, and ones where fuel is not pre-mixed with air, but burns and mixes with it in same time, that may produce bright yellow flame.

For many organic compounds, especially ones that require large amount of oxygen per molecule to fully burn, in case the fuel is not premixed with enough air, the flame usually contains oxygen-deficient region where tiny particles of carbon and molecular hydrogen are formed. These carbon particles, since they are heated, give thermal light emission and are responsible for bright non-transparent orange-yellow organic flames. A typical example of such flame is a flame of a candle. It is a flame of heavy hydrocarbons or fatty acids (c20+), that require tens of molecules of oxygen per molecule of the fuel to fully burn.

In case gases from this yellow part of the flame are cooled without mixing with more air, this carbon particles may be obtained in form of deposits or black smoke. For this, a cold effective conductor of heat (i.e. metal, especially copper and silver) or a solid surface with high heat capacity introduced into yellow part of the flame may be used. In some cases, when too many of such particles are produced and the flame is not particularly hot, there is no need even to cool the flame. This is the usual case of hydrocarbons with big amount of carbon, like aromatics, and chlorinated hydrocarbons, like PVC (as long as they burn at all, which often is not the case even if there is no much chlorine in the compound).

Propane torches usually have a mean to regulate amount of air premixed with fuel. Torch lighters are small propane torches, usually without such means. When propane in such torch is premixed with excess of air, a very hot blueish short flame is formed. However, when air is not premixed with propane, the flame is organized in layers with more oxidized outer layers and more reduced inner layers. In inner layers, the dominating reactions are thermal decomposition

$\ce{C3H8 = 3C + 4H2}$

and incomplete combustion

$\ce{C3H8 + 2O2= 3C + 4H2O}$

(this is an oversimplification, as numerous other intermediates are formed, but that's advanced stuff)

In outer layers, propane combusts completely, just as if it was premixed with air

$\ce{C3H8 + 5O_2= 3CO2 + 4H2O}$

and carbon particles (and other reduced products) from inner layers also combust here.

Now, to get carbon particles from the flame in form of smoke the key is to cool them before they mix with more air. An ideal solution would be to suck them with tiny pipe and cool with water. However, it seems, that skein of copper mesh, at least until it is heated enough, works good for this purpose, both cooling the inner layers and hindering their mixing with air.

Since flames of fuels that are not pre-mixed with air have very efficient emitter of visible and and infrared light in it (many carbon particles), they tend to be cooler, especially the inner part. The flames with fuel pre-mixed with air (like in torch lighters with blueish flame like this), the flame is thin, quite short and much hotter.

Copper wire, when heated in air, also oxidizes. If done properly, it is a good way to produce pure black copper (II) oxide. However, it usually stay on the surface of the wire and indeed, can be washed with hydrochloric acid. If there are black particles that are not dissolved in hydrochloric acid, they are carbon particles.

  • $\begingroup$ Thanks for your answer. It has good information but my question is not really answered. Why do you think the normal lighter is producing the black smoke, whereas the torch lighter does not? Also, do you know what the chemistry of the varnish/smoke is? $\endgroup$ Sep 2, 2014 at 6:21
  • $\begingroup$ @SeanMackesey because in case of torch lighter with blue flame (is that the case?) like this southsmoke.com/pd-shark-series-torch-lighter.cfm the fuel is already pre-mixed with air, so no oxygen-deficient region arises and the flame is a good deal hotter. I doubt that in case of kitchenware any varnish was used, so the black smoke is likely to be plane soot (mostly carbon). Do I have to add it to the answer? $\endgroup$
    – permeakra
    Sep 2, 2014 at 9:32
  • $\begingroup$ OK, yes, please do add it to the answer-- could you also explain why the soot would make a stream of smoke? I understand the deposition of soot on the copper, but I don't understand why it would be put off as smoke. If it's mostly carbon, shouldn't it combust to make $\ce{CO_{2}}$ and $\ce{H_{2}O}$? $\endgroup$ Sep 2, 2014 at 15:43
  • $\begingroup$ @SeanMackesey I edited the answer. As for this question, see bold font. $\endgroup$
    – permeakra
    Sep 2, 2014 at 16:55

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