The combustion of methanol tends to generate flames that are less visible than the flames generated by the combustion of other substances?

My guess is that methanol burns quickly as it has a relatively simple structure that provides some of the oxygen needed for combustion. I am assuming that the flame comes from a mixture of fuel and air that is undergoing reaction. If the reaction is fast enough, then there will be less flame. However, methane and hydrogen both have visible flames, much more visible than methanol, leading me to wonder how the one oxygen in methanol makes so much of a difference.


3 Answers 3


Here's a link to a YouTube video showing a dish of methanol burning next to a dish of ethanol. The methanol flame is more difficult to see. Methanol was also widely used as a fuel in the auto racing circuit, until fires that resulted from crashes or fuel filling accidents resulted in "invisible fires" that were difficult to see during daylight, resulting in severe burns to crews and drivers ("you can't fight the fire, if you can't see it", this is one reason why it was replaced by ethanol as a fuel for Indy racing).

Flame color (visibility) is dependent upon the elemental composition of the material being heated and the temperature to which it is heated. This is because heat will excite electrons in the element(s) to an excited state; when these electrons return to the ground state visible light (flame test) will often be emitted

If we look at the carbon content of methanol and ethanol we find the following weight percents:

methanol - 38% carbon by weight

ethanol - 52% carbon by weight

If we compare the heats of combustion for methanol and ethanol we find:

methanol $\ce{~~}$ -726 kJ/mol (exothermic)

ethanol $\ce{~~}$ -1368 kJ/mol (much more exothermic)

There's just not much carbon there to burn in methanol! Consequently burning methanol doesn't get that hot and its flame will not be as colored as materials that burn more exothermically and result in higher temperatures.

  • $\begingroup$ Am I correct in thinking that carbon burns hotter because of the higher heat of formation of CO2 vs H2O and the higher heat capacity of H2O vs CO2? $\endgroup$ Aug 23, 2014 at 18:49
  • $\begingroup$ "Am I correct in thinking that carbon burns hotter because of the higher heat of formation of CO2 vs. H2O", Yes -394 vs. -242 kJ/mol for CO2 vs. H2O. I hadn't though about the heat capacities (0.844 vs. 1.93 for CO2 vs. H2O), but that makes sense too. $\endgroup$
    – ron
    Aug 23, 2014 at 19:04
  • $\begingroup$ >There's just not much carbon there to burn in methanol ! | while it is true that it is the reason in this case, the way it happens is completely different from what you described. $\endgroup$
    – permeakra
    Aug 23, 2014 at 19:11
  • $\begingroup$ Perhaps I should elaborate. It is well known and easily provable at home that pale blue flame of propane hand torch is much hotter, than bright yellow flame of same torch. $\endgroup$
    – permeakra
    Aug 23, 2014 at 19:13
  • $\begingroup$ @permeakra My explanation was based on 1) lower carbon content leading to 2) lower heat output leading to 3) fewer excited states being produced leading to 4) less coloration of the flame. I don't understand your comment, specifically which step in my explanation do you disagree with? Soot is not an issue when adequate oxygen is present. $\endgroup$
    – ron
    Aug 23, 2014 at 19:21

The yellow 'flame' is actually an aerosol of black particles heated to temperatures around 1000-1500 Celsium. Depending on temperature, they can emit deep red, orange, yellow and almost white light. So, for yellow flame to occur, black particles must be present. In case of carbon containing fuels the particles are usually carbon particles (soot). Soot is a product of incomplete combustion of carbon containing molecules. Now, let's see the difference between ethanol and methanol, specifically, at amount of oxygen required for combustion of same volume of vapors. Equations of burning:

$\ce{2CH3OH + 3O2 = 2CO2 + 4H2O}$ $\ce{C2H5OH + 3O2 = 2CO2 + 3H2O}$

As it can be seen, same volume of vapors in case of ethanol requires twice as much of air to fully burn, so in case of ethanol incomplete combustion is more common. It is even more common in case of larger molecules, say, higher hydrocarbons, like solid paraffines used in some candles. In case the flame of organic substance has right amount of oxygen (incoming fuel-air mixture contains proper amount of air), it burns with pale flame. It is easily observable in case of hand propane torches, where user regulates amount of air in incoming mixture, depending on amount of air allowed, it is possible to have bright yellow or pale blue flame and everything in between.

The pale blue emission has completely different nature, it is atomic emission of excited radicals and molecules. For example, it is known that $CH$ particle emits a blue light.

Yellow flame may also happen if another source of particles is available and the temperature is right. For example, $\ce{CuO + Al}$ mixture burns explosively, producing bright yellow flash. However, if temperature is too high, bright white flame should be observed. Indeed, magnesium burns with blinding white flame, especially if it is premixed with solid oxidizer (potassium nitrate).

Now, atomic/molecular emission is usually low in common flames. However, even small amounts of sodium in flame makes it bright yellow even if it normally almost colorless. Some other elements may color flame into different colors, ranging from pale violet to deep red. This, however, is a different situation from what you considered.

  • $\begingroup$ Under what cases would incomplete combustion of methanol yield soot? I would expect the combustion to comprise three reactions. First H3COH->CO+2H2, and then 2CO+O2->2CO2 and 2H2+O2->2H2O. Given that coal gas is formed by dissociating water to produce hydrogen and carbon monoxide, I would see no reason to expect any elemental carbon to be produced as a temporary decomposition byproduct. $\endgroup$
    – supercat
    Aug 24, 2014 at 0:05
  • $\begingroup$ @supercat I never cared to experiment, but guess if a drastically low amount of oxygen is present, it is possible. It is usually easier to achieve with larger flames. The reaction of combustion of ethanol is definitely radical in nature, with many branches. It is likely include such particles as $\ce{.OH , CH2O, .CH2OH, .HCO, CO, HO2, .O.}$ etc., and I'm not ready to discuss it. However, I must note that at low-medium temperatures $\ce{CO}$ is prone to disproportionation into carbon dioxide and elementary carbon. $\endgroup$
    – permeakra
    Aug 24, 2014 at 0:47

The spectra of the yellow flame of burning alcohols is continuous. It is not made of lines. This means that it is produced by a black body, which is a hot solid substance. It is not produced by a gas. As a consequence, in order for such a flame to be seen, it has to produce from time to time some relatively big particules of solid carbon, containing a huge amount of $\ce{C}$ atoms. Let's suppose that the solid particules are made of $100$ Carbon atoms. To get to such a particule, it is necessary to join the $\ce{C}$ atoms from $100$ methanol molecules. It is not easy. It is much easier to use ethanol to do the same job, as only $50$ molecules are required to "collaborate" to this result. That is why there are more carbon particules in the flame of ethanol. And of course the flame of a higher alcohol, like butanol, will still be more visible, as only 25 molecules has to interact to produce carbon particules. For higher alcohols, the flame produces so much carbon particules that they cannot be burned in the upper part of the flame, and it yields soot.


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