It's a common problem that a mix which starts out well balanced will, after several hours, become Androstenone heavy after the other molecules have evaporated. Fixatives help, but are not perfect. Given issues like this, when looking at new pheromones it seems like it would be helpful to know whether the new molecule would fly away in a puff or hang around all night. I doubt that length of self effects are a reliable proxy, here.

There has to be some property in the scientific literature (or equation which we could feed several properties) which would lead us towards this information, but I could use some help finding it.

At first I thought that boiling point might be the key. But if that were true then androstenone (one of the slowest diffusing mones) should have a much higher boiling point than androstenol (one of the fastest diffusing mones.) Oddly, this doesn't seem to be supported by the hypothesized properties in the literature. Apparently actual experimental results aren't available...

Androstenone
371.6 °C at 760 mmHg
http://www.lookchem.com/Androst-16-en-3-one/

Androstenol
Boiling point 381°C at 760 mmHg
http://www.chemnet.com/dict/dict--12041-97-3--en.html

Would one of these other properties (or an equation connecting them) be more relevant? Would solubility suggest (vaguely) how likely a compound is to evaporate when mixed with another?

(In regards to enthalpy of vaporization, a "High enthalpy" means that vaporization is more energetically favorable, and thus more likely to happen.)

Some material properties of androstenol (predicted, not based on experimental data) as listed in chemSpider;

Polar Surface Area: 20.23 Å2
Enthalpy of Vaporization: 71.896 kJ/mol
Surface Tension: 41.1769981384277 dyne/cm
Boiling Point: 374.067 °C at 760 mmHg (predicted 354.92C based on episuite)
Vapour Pressure: 0 mmHg at 25°C
Density: 1.038 g/cm3
Molar Volume: 264.415 cm3
http://www.chemspider.com/Chemical-Structure.92136.html

Androstenone
Polar Surface Area: 17.07 Å2
Molar Volume: 263.926 cm3
Polarizability: 32.313 10-24cm3
Surface Tension: 36.8699989318848 dyne/cm
Density: 1.032 g/cm3
Flash Point: 157.654 °C
Enthalpy of Vaporization: 61.865 kJ/mol
Boiling Point: 371.61 °C at 760 mmHg
(Boiling Pt (deg C): 351.31 (Adapted Stein & Brown method)
Vapour Pressure: 0 mmHg at 25°C
http://www.chemspider.com/Chemical-Struc...54715.html

So I've looked up the enthalpies of vaporization to see how reliable they are for predicting evaporation rates of the other molecules that we're familiar with. These are theoretical, and apparently different models give different results. Androstadienone is also one of those more likely to hang around... low enthalpy of vaporization as expected. I don't know about Androsterone, but that's predicted to have a higher EoV than nols... Androstanone has an EoV equal to Androstenone. I've seen sites claim that Androstanone was more 30% volatile than Androstenone, though. I don't know how reliable those other sites were. And neno seems to top the charts with a whopping EoV of 80.885 kJ/mol

androstadienone
androsta-4,16-dien-3-one 63.97 kJ/mol

a Androsterone
5a-androstan-3a-ol-17-one 76.931 kJ/mol

Androstanone 61.511 kJ/mol
5a-androstan-3-one
http://www.chemspider.com/Chemical-Structure.92166.html

5a-pregnan-3a,21-diol-20-one 80.885 kJ/mol
http://www.chemspider.com/Chemical-Struc...2bc541ca2e

Edti:

Or maybe I'm completely off. Helium seems to have a very low EoV. But also a very low boiling point... and EoV is typically measured at the boiling point? *sigh* There has to be some info out there on this topic, though. Forgive me for posting while I educate myself instead of just laying everything out once I've gotten a complete grasp of the subject. I know there are some smart people on this board with some solid science backgrounds, and if any of you guys can straighten a wanderer's path, he'll be appreciative.

Quote:The enthalpy of vaporization, ( also known as the heat of vaporization or heat of evaporation, is the energy required to transform a given quantity of a substance from a liquid into a gas at a given pressure (often atmospheric pressure).

It is often measured at the normal boiling point of a substance; although tabulated values are usually corrected to 298 K, the correction is often smaller than the uncertainty in the measured value.


The enthalpy of vaporization can be viewed as the energy required to overcome the intermolecular interactions in the liquid (or solid, in the case of sublimation). Hence helium has a particularly low enthalpy of vaporization, 0.0845 kJ/mol, as the van der Waals forces between helium atoms are particularly weak

...Care must be taken, however, when using enthalpies of vaporization to measure the strength of intermolecular forces, as these forces may persist to an extent in the gas phase (as is the case with hydrogen fluoride), and so the calculated value of the bond strength will be too low. This is particularly true of metals, which often form covalently bonded molecules in the gas phase: in these cases, the enthalpy of atomization must be used to obtain a true value of the bond energy.

http://en.wikipedia.org/wiki/Enthalpy_of_vaporization

edit: p.s. If vapor pressure is hypothesized to be zero doesn't that mean zero hypothesized diffusion? So that has to be wrong...

Great researching, wiserd. That's something I've always wanted to know.

I think the best members to help us with this would be J. V. Khol and Chris@AD.

(06-18-2012 6:51 PM)Fly So Hi Wrote:  Great researching, wiserd. That's something I've always wanted to know.

I think the best members to help us with this would be J. V. Khol and Chris@AD.

Good call, FSH. I'll try JV Kohl. Chris has a lot on his plate right now. I'd actually written him an email on this topic shortly after this post, but don't really expect an answer and haven't gotten one. Not that I'm complaining.

If you crack this nut, I have been planning to create a computer simulation to show the diffusion rates for molecules in carrier/diluent/fixative - I pretty much slumped around the point you are at and haven't picked it back up, so I'm glad someone else is interested! I've been picturing the visualization for a while but haven't applied myself to gathering the research so maybe we can piggyback this!

(06-18-2012 8:29 PM)wiserd Wrote:  Good call, FSH. I'll try JV Kohl. Chris has a lot on his plate right now. I'd actually written him an email on this topic shortly after this post, but don't really expect an answer and haven't gotten one. Not that I'm complaining.

Try sending JV Khol a PM here notifying him of your e-mail. I know he's active on this forum and very willing to help.

Hey Ryan,

I see where you are going with this and I appreciate your enthusiasm and ambition towards the scientific quantification of mones...

Unfortunately however, while there is a way, there lacks the literature and physical experiments and essentially the data to quantify this for us.

In the meantime however there is a good way to guesstimate the volatility of a particular mone - I will get to this in a minute..

The vapor pressure cannot be used to determine this, since pheromones are solid substances at Standard Temperature and Pressure (STP), meaning room temperature given the atomspheric pressure at 1 bar. With a solid, there is a vapor pressure, but it is so minute and hard to measure that scientists essentially consider it zero. This is true since there is a liquid phase to these compounds over a wide temperature range - meaning the compounds do not sublime.

The enthalpy of evaporation described here describes enthalpy of evaporation as a function of energy in Joules Per Molar Mass. A mole is a good way of scientifically quantifying how many molecules are in a given space, and since the molar mass of androstenone is lower than that of androstenol, you get a lower enthalpy of vaporization which is lower...

What really needs to be done, is a test for vapor pressure (in millimeters of mercury (mmHg)) at the compound's Melting point, so that it will be liquid and therefore have a substantial vapor pressure that can be measured - then we can figure out the diffusion rate by the following equation ((Vapor pressure)/(melting point in C)) = ((vapor pressure)/(Temperature of the human body in C).. The equation won't work perfectly, as there is likely to be a steep drop in gradient once the point of solid is reached, however it will give us a good idea, approximately.

In essence, the way I factor for when considering the [effective] diffusion rate and dosing is the following:

1) - the melting point of the pheromone (the lower the melting point the more volatile the substance will be corresponding to intensity of effect for X hours)

2) - the relative potency of the compound (all pheromones have different potencies, which is why the androstenol / androstenone ratio in a formula is generally 5 : 1. Notwithstanding the volatility, the androstenone seems to have a much stronger effect on a per weight basis)

From my reading of the Erox literature, the effective dose of androstadienone was a few (less than 50) nanograms for an individual to feel the effect. So it can be safely assumed that about 1 / 1000th of what we apply is effectively diffused to other individuals.

Remember, the human nose can be extremely sensitive to chemical queues. Read about Geosmin Here, which the human nose can detect at parts per trillion!!


Cheers,

Chris @ Alpha

Hey Chris!

I was hoping you'd weigh in here. Thanks for taking the time. It's much appreciated.

I actually managed to find a site that did give vapor pressures for androstenol and androstenone.

(E-7 = 10 ^ -7, incidentally)



androstenol 2.27E-07mmHg at 25°C
http://www.chemnet.com/dict/dict--12041-97-3--en.html

http://www.chemnet.com/cas/supplier.cgi?...ubmit.y=14
1.02E-05mmHg at 25°C


So this is claiming that androstenone evaporates more rapidly than androstenol?

Quote:The enthalpy of evaporation described here describes enthalpy of evaporation as a function of energy in Joules Per Molar Mass. A mole is a good way of scientifically quantifying how many molecules are in a given space, and since the molar mass of androstenone is lower than that of androstenol, you get a lower enthalpy of vaporization which is lower...

Got it! Thanks. :-)

Quote:What really needs to be done, is a test for vapor pressure (in millimeters of mercury (mmHg)) at the compound's Melting point, so that it will be liquid and therefore have a substantial vapor pressure that can be measured

Hmm... Interesting. So is a liquid's vapor pressure more accurate because mones are used when dissolved? Or is melting the stuff just a matter of getting a bigger number?

Quote:1) - the melting point of the pheromone (the lower the melting point the more volatile the substance will be corresponding to intensity of effect for X hours)

Use melting point, not boiling point. Got it. ... not sure why that works but I can use that. Thanks.

Quote:2) - the relative potency of the compound (all pheromones have different potencies, which is why the androstenol / androstenone ratio in a formula is generally 5 : 1. Notwithstanding the volatility, the androstenone seems to have a much stronger effect on a per weight basis)

So if I'm reading this right.... We tend to focus on ratios of pheromones, but to some extent absolute quantities are relevant as well?

Some mones have sharper response thresholds than others. Like, androstenol might be noticed at 20 mcg, but not at 5mcg. While Androstenone
might be noticed at 2 mcg and also at .... say .2 nanograms. And this can give the illusion that one mone has vanished but another is still sticking around, because the relative doses of mones are not the only relevant factor? Or have I completely misrepresented what you've said?

Thanks again for the help!

think a little bit about polar molecules. or the polarity of groups in a molecule. and how they relate to the carrier and fixative in a spray.

Then I think you could base the diffusion rate on the rate the fixative evaporates and the affinity of the pheromones groups to the fixative.

The pheromones diffuse faster in ethanol, they diffuse slower in DPG.

I think the question is really, how well do the pheromones stick to the carrier/fixative.

It may turn out that TAA just sticks to ETOH like food stamps to a democrat.

(06-21-2012 7:23 PM)dexter Wrote:  think a little bit about polar molecules. or the polarity of groups in a molecule. and how they relate to the carrier and fixative in a spray.

Then I think you could base the diffusion rate on the rate the fixative evaporates and the affinity of the pheromones groups to the fixative.

The pheromones diffuse faster in ethanol, they diffuse slower in DPG.

I think the question is really, how well do the pheromones stick to the carrier/fixative.

It may turn out that TAA just sticks to ETOH like food stamps to a democrat.

I'm not sure I follow, Dexter. Smaller molecules diffuse faster than bigger molecules. So a molecule which is more attracted to the fixative will diffuse more slowly, correct?

Does this mean that we should compare the solubilty of a fixative in a polar solvent to the solubility of a mone in a polar solvent or the solubility of both in a non-polar solvent to get a sense of how well the two will stick together?

Should molecules which aren't highly soluble in a liquid also evaporate more rapidly from it, since they're less strongly attracted to it?