I think water has the atomic weight of 22, and that this density is defined by it being made up by two halves with the atomic weight of 11. The formula for water, where C=3, is: 2(H2C3)
I think that the common air not only contains water vapour, I think that it IS water vapour. Under the phlogiston theory, nitrogen was considered to be air saturated by phlogiston. I suspect phlogiston is another name for carbon.
Common air, being water vapour, will also have the same atomic weight as water: 22, and therefore, they also share the same formula: 2(H2C3). In an attempt to derive atomic structure from the formula, we might say that the hydrogen ions present in H2C3, actually represent the cyclones that make up the dipolar vortices which are a part of atomic structure. Based on this idea, H2C3 shall therefore represent 2 donutoms. This being the case, then the value for C needs to change in order for us to dismantle the formula. I think this entire process would be more effective if we just include the ACTUAL value for C in the formula, for example:
half vol water = H2C9 = 2(H C4.5) = 11
Whereas a full volume of water, atomic weight 22, shall be:
4(H C4.5) = 22
The atomic weight of nitrogen is 14. We've seen that the weight ratio of hydrogen to carbon in nitrogen is 1:6. The formula for nitrogen might look something like:
H2C12 = 2(H C6) = 14
This formula for nitrogen is conspicious because it looks like we've pumped up a water-donutom with more carbon to give us our nitrogen-donutom.
Which brings me, somewhat abruptly, to carbon monoxide. The atomic weight of carbon monoxide is 28. Classically, carbon monoxide is made up by one volume of carbon (12), and one volume of oxygen (16). In previous posts we have come to understand oxygen a little differently. Oxygen is not an element at all, but a compound of hydrogen and carbon. Oxygen now has the formula H4C12. The overall formula for carbon monoxide will be:
carbon + oxygen = C12 + H4C12 = H4C24 = 28
In the past, we've also played around with the idea that carbon monoxide is made up by water vapour and carbon. Happily, this formula concurs nicely with the above formula:
water vapour + carbon = H4C18 + C6 = H4C24 = 28
H4C24 might also be represented as: 4(H C6) = 28
Remember, that the formula for nitrogen was found to be: 2(H C6)
The formulas for carbon monoxide and nitrogen share a very obvious similarity - the weight ratio of hydrogen to carbon (1:6) appears to be the same in both substances. It remains to be seen though that carbon monoxide is twice as dense as nitrogen, and perhaps the formula should do more to reflect that:
carbon monoxide = H4C24 = 4(H C6) = 2(2H C12) = 28
The value for carbon here is 12, the largest value that we have seen so far, and which might explain why carbon in the form of soot is more conspicious in carbon monoxide. The mass of the hydrogen ion (protium) has effectively doubled, and so the next appropriate question that I am now going to ask is - is that actually deuterium hanging out in there?
I think that deuterium (2H) is a cyclone just like protium, but that it has twice the mass. As far as I'm aware, deuterium does not make up a donutom - it is protium which makes up donutoms. From what little I have managed to learn about deuterium thus far, I think it generally hangs-out on its lonesome. At this stage mind, it does appear that deuterium and C12 have evolved somekind of partnership. Are we looking at a fat donutom?
Carbon dioxide is made up by carbon monoxide and oxygen, hence the formula CO2, giving it the atomic weight of 44. We know oxygen, atomic weight 16, has the formula H4C12, which might also be written as 4(H C3). The overall formula for carbon dioxide is therefore:
carbon monoxide + oxygen = 4(H C6) + 4(H C3) = 28 + 16 = 44
According to Priestley, half the weight of carbon dioxide was made up by water. The formula for water can be written as H4C18, or as 4(H + C4.5) . In order for water to make up half the weight of carbon dioxide, then it is necessary for the value of carbon to be spread out over the formula, so that the new value for carbon is 4.5, something like:
CO2 = 4(H C6) + 4(H C3) = 8(H C4.5) = 44
The value for carbon in carbon dioxide is therefore the same as it is in water. I am now weighing-up the idea that carbon dioxide could be somekind of water vapour which manages to retain twice the density of ordinary water. Just as we did previously with carbon monoxide and nitrogen, let's try and reflect the difference in density between carbon dioxide and water in formula:
CO2 = 8(H C4.5) = 4(2H C9) = 44
There are a few things to be gathered from this formula then. From what little experience I have of nuclear fusion, there does not appear to be a higher value for a proton than deuterium, so I'm pretty sure that the value of 9 for carbon in carbon dioxide is the right one.
I wonder if we can now apply all that we have learned into something which is intriguing me. If I run my hand through the yellow part of a flame (not that we do that much round here, we're not in a biker gang or anything) it will certainly get hot, but it will also get black from where it picks up soot. The soot is evidence of carbon monoxide.
If I now run my hand over above the flame, the presence of soot is far less obvious because it has effectively been vapourized. If you imagine that we have a flame burning in pure oxygen, we have carbon monoxide reacting with the oxygen to produce carbon dioxide:
carbon monoxide + oxygen = carbon dioxide
2(2H C12) + 4(H C3) = 4(2H C9)
2H + 2H + C24 + H + H + H + H + C12
= 2H + 2H + C24 + 2H + 2H + C12
= 4(2H) + C24 + C12
= 4(2H) + C36
= 4(2H C9)
The thing which strikes me the most about the reaction is how the hydrogen ions are converted into deuterium. I think we've seen this happen someplace else - nuclear fusion! The very first step in the Proton-Proton process is where two protons form to make one deuterium atom, with the release of a positron and neutrino. Can something as exotic as a positron and neutrino be found in a simple flame?
Obviously, a flame emits light and heat. These wavelengths of EMR are not as active as those gamma rays released under nuclear fusion, but they are present nonetheless. I think these formulas pave the way for a greater understanding of how exactly ALL the energy is released from a flame.