So while I'm waiting for the emulsion to fully break on the salting of my last pull on this mescaline extraction I decided to play with the math
People often ask how many pulls will get all the mescaline. Well, like normal I titrated each pull on this run but this time I graphed the results and calculated the best-fit curve to extrapolate just where yield would get truly trivial.
Here you can see my yields, Y is grams of mescaline and X is the pull number.
Extending the X scale to the theoretical and we can see that pull 17 will just give you about a milligram of mescaline.
So the proper number of pulls: 17
I'm lazy so I stop when I only pull a fraction of one dose

My math teacher was right, math is worth knowing

Interesting. My measurings normalized for your values would be (3 pulls) :

0.95
0.32
0.157

I suppose the steeepness of the curve is given by NaCl I am using.

True. Lye concentration, salt concentration, viscosity, temperature, NPS used, and aqueous phase to organic phase ratio will all effect the slope of the curve, some times by quite a lot. The choices are often an issue of practicality.
For instance in that extraction I used the minimum effective dose of lye like always since lye is watched here, the aqueous phase was 5% salt because I often dont need more so I avoid waste, and my NPS [xylene] to aqueous phase ratio was 1:6 instead of the more customary 1:3 because I'll happily do an extra 2 pulls if it means I can fit more cactus in my 2 litre beaker. I've still got loads more to extract.
I think I'm going to throw a party when the chinese start mass producing 4 litre beakers

My god, lye watched chemical ... Always when it comes to having meth and entheogens in the same bag, I am sorry about our civilization!

Have you seen these? I love them!

At risk of myself sounding like a broken record, I'll reiterate here as elsewhere:
If you can get hold of or make calcium hydroxide, it's easy to make crude lye by mixing it with sodium carbonate solution. Calcium hydroxide can be made by roasting calcium carbonate rock using (best) an oxy-propane blowtorch.
Sodium carbonate should still be easy to get hold of - or has the world slid further into madness than previously imagined?

Thankfully in my locality there is still one brand of pure lye and I can get one container without much suspicion if I buy a variety of other general hardware supplies. I plan such shopping trips in advance so I can buy seasonal and repair project things. I think this is a precaution most of us should be taking, it just limits me to half a kilo of lye once or twice a year since there is only so much windshield washer fluid and wood sealant I need
I bought a box of sodium carbonate but it seems to be contaminated with some sort of surfactant, I have kilos and kilos of baking soda though
I've considered the calcium hydroxide/sodium carbonate route as a primary back-up plan, as stores that sell cement often have a crude lime in large bags, but I was afraid of the potential for any contaminant calcium or magnesium in the product to cause horrible emulsions.
In your experience, do these emulsions occur and if so are they remedied by potassium sodium tartrate?

I've not tried crude home-made lye with cactus yet - your concern about calcium is undoubtedly valid - but it certainly works with MHRB. Cactus experiments are in the pipeline, I'll keep the potassium sodium tartrate in mind.

For raw cactus STB, electrolysis would have to be the route for homemade NaOH if calcium contamination proves to be problematic.

Edit: Updating formulas from previous post earlier today which was not quite right.

Let's call Mt the total amount of mescaline in the water before the first pull. Each time a pull is done (under the same conditions), a fraction of this mescaline moves into the solvent. If we call that fraction p, then then during the first pull M1 = p*Mt will move into the solvent, and W1 = (p-1)*Mt will stay in the water. During the second pull, M2 = p*(p-1)*Mt will move into the solvent and W2 = (p-1)*(p-1)*Mt will stain in the water, and so on.

A practical way to estimate p is to divide the mescaline obtained from the first and second pulls since,

M1/M2 = 1/(1-p), or


And a practical way to estimate Mt is by remembering that M1 = p*Mt which can be rearranged to,

Mt = M1/(1-M2/M1), or


So in other words, by measuring what is obtained from each of the first two pulls, you can have an idea of the partition coefficient and the total amount of product in the water. But wait, there is more

After n pulls the amount of mescaline that moves into the solvent for that pull is:

Mn = p*Mt*(1-p)^(n-1)

And the amount remaining in the water after that pull is,

Wn = Mt*(1-p)^n

The total amount pulled after n pulls is simply the starting amount of mescaline minus what remains in the water,

Tn = Mt - Wn = Mt*[1-(1-p)^n]

The ratio of total mescaline pulled to starting mescaline after n pulls is,



And this last formula can be re-arranged to get the number of pulls for a certain ratio of total mescaline pulled,

n = log (1-Rn) / log (1-p), or


Let's go over an example that uses some of the formulas above. Suppose you are pulling DMT one day and measure what you get from the first two pulls. Let's say you get,

M1 = 350mg
M2 = 210mg

A lot of stuff can be estimated from these two numbers :

- Your pull coefficient p = 1- 210/350 ~ 0.40
- The total estimated DMT in the water Mt = 350^2/(350-210) = 875mg
- You can estimate that the water still has W2 = 875 - 350 - 210 = 315mg. (Note that alternatively you can use the general formula for Wn for the case n=2, W2 = 875*(1-.4)^2 = 315mg.)
- Lets say you want to get at least 90% of what is in the water, how many pulls do you need? That would simply be n = log (1-0.9) / log (210/350) ~ 4.5, which means go for 5 pulls (do 3 more). After the 5 pulls you can expect to have pulled R5 = 1-(0.6)^5 = 0.92, or 92% of the DMT that was originally in the water (which is above 90% as expected).

Now let's look at Elrik's fit for Mn,

Mn = p*Mt*(1-p)^(n-1) = 1.31*(0.67)^n

One can see from the formulas above that 1-p = 0.67, that is, p = 0.33
Also one can see from the rest of the formula that p*Mt/(1-p) = 1.31, since we now that p =0.33, that means that Mt = 1.31*0.67/0.33 = 2.66g.

Looks like Elrik did 5 pulls. What % of the starting mescaline has he pulled out?

R5 = 1-(1-p)^n = 1-(0.67)^5 = 0.86, or 86%.

Which means Elrik should have about 2.30g of mescaline total after the 5th pull and 360mg still in the water.

Loveall - I haven't read your post carefully, but from skimming through it, it seems to be based on the assumption that each pull extracts the same (i.e. constant across pulls) percentage of what's still in the soup. That also seems consistent with Elrik's empirical data.

But I wonder why it would be so? Is it determined by some sort of equilibrium between how much alkaloids stay in the aqueous layer and how much migrate into the solvent, which depends on the solubility constants?

My initial intuition (when I didn't know anything about chemistry) was that the solvent absorbs as much alks as it can hold.

Yes, this is called the partition coefficient, although saturation of the solvent with the alkaloids can also play a role (viz DMT/naphtha).

This is how I understand it intuitively: Stuff that is soluble in both water/solvent will disperse across the layers to maximize entropy. The end result is messy, with drug mixed in all over the place. Filling up the solvent as much as possible is not necessarily the state that maximizes entropy, kind of like when you connect two chamber and gasses get mixed/dispersed. One important example is the case where the entropy as a function of concentration is the same for both the water and solvent, then you would expect a 50/50 concentration distribution of the drug between the two layers (like when two identical gas chambers are connected).

downwardsfromzero is right, the drug will diffuse across the layers, but if the solvent saturates it will have to stop and wait for another opportunity to increase entropy. This special case is somewhat similar to your intuition, but it is happening because the solvent reaches saturation before entropy can be maximized. Under these conditions, the formulas are not valid but will kick in if more solvent is added or after doing few saturated pulls to move away from the saturated regime.

Finally, the old intuition that the drug will move into the solvent much as the solvent can hold runs into a symmetry problem. Imagine you do a pull and have your drug in the solvent. Then to that solvent, add fresh basic water. What do you expect to happen? A fraction of what is in the solvent will move into the water increasing the drug disorder and maximizing entropy. The alternative scenario where the solvent holds onto all the drug and the fresh water stays neatly free of drug would violate entropy increase, right?

Let's go through a specific example of this:

- You do a pull on water containing 875mg of DMT, and 350mg moves into the solvent, with 525mg staying in the water.
- Now, to test your intuition, you add fresh water without DMT (otherwise same volume, pH, salt, etc.) to the solvent you just pulled. What happens? Does the solvent keep hold of all the DMT it can absorve or does some DMT move into the water? It's the second option, otherwise it would be like connecting two gas chambers and the gas deciding to stay only in one chamber, not making sense from an entropy point of view. What would happen is that 210mg move into the fresh water, and 140mg stay in the solvent. This way the partition that maximizes the entropy is achieved again (since 350/525 = 140/210 = 2/3).

I hope this makes sense intuitively, sorry for the long post.

PS: In the technical literature the partition coefficient is defined at infinite dilution to avoid second order effects, which for our purposes we can ignore once we are not saturating the solvent. Also by knowing the functional form for the entropy of mixing as a function of drug concentration formulas predicting all this stuff can be developed. Take a look at this, they do relate the partition coefficient to the solubility in each layer - your current intuition was spot on .