Hello!

A general question about (crude) extracts/tinctures that may sound stupid: How does filtering inbetween pulls affect final yield, given the same total amount of time and solvent used?
So let's say I boil my plant material with 500ml ethanol for an hour, filter and then boil it again with 500ml of fresh solvent for an hour, filter and combine. Shouldn't the total amount of alkaloids pulled be approx. the same as when simply doing one pull with 1l for two hours?

No, there are principles in chemistry that dictate how likely a chemical is to move in to a solvent, the more free space there is (new water) the more room there is for the DMT to migrate. Same principle with solvent pulls hence hwy the FAQ says multiple pulls are better, same with multiple cooks/tincture sextracts.

Ok, makes sense. Thanks for replying!

Edit 4/22: Updating notation to better match with this post.


This is a good question.

If you can measure what you get after the first two identical pulls you can roughly predict how many pulls are needed to get to 90% pulled. Here is a plot intended to help with that.

As for your question of 2x500 or 1x1000 pull, here is what would happen.

What is constant is the partition, that is the drug concentration in the pull divided by the drug concentration in what we are pulling from. Let's call this important ratio k. This is our constant from which everything is derived. Why is this constant? That's because it maximizes the disorder in the system. Formally, this is defined at low concentrations where saturation does not occur. In practice if the solvent is saturated then the formulas don't kick in until we are not in the saturation regime.

Think of disorder as ink in water. Assume this ink can dissolve in both water and oil. If you add some oil to the water, some of the ink will move to the oil, right? However it would be weird and too orderly if all the ink moved/stayed to one side. If the ink was more soluble in oil than in water you would expect the oil to have a higher ink concentration (and vice versa), but once equilibrium is reached, the ratio of these concentrations is always the same. This is true for all pulls.

OK, so now let's get to your question.

Let's assume the source has a total amount of drug Mt, and an effective volume Vi (small when pulling from solid plant material and larger when pulling from a liquid solution). We will assume that Vi does not change to first order as we pull. Using a solvent with a partition p, Let's compare 1x1000 to 2x500 pulls.

1x1000 case, let's call M1000 the amount of product that moves from the source to the pull. simply write down p in this situation as the drug moves from one place to the other by an amount M1000,

k = [concentration in pull] / [concentration in source] = [M1000/1000] / [(Mt-M1000)/Vi]

Bingo, now we have M1000 as a function of p, Mt, and Vi. All we need to do is rearrange terms,

k * [(Mt-M1000)/Vi] = M1000/1000,
k * Mt / Vi = M1000 *(1/1000+k/Vi),
M1000 = Mt*k/Vi/[1/1000+k/Vi], finally giving,

M1000 = Mt / [ 1 + Vi/1000 * 1/k]

Seems legit. As k grows by a lot, 1/k goes to zero, making M1000 go to Mt. Also, as the pull volume grows, Vi/1000 would go to zero, also making the pulled mass go to Mt.

Now let's do the same for 2x500. Simply have to use the above equation twice, but this time use 500ml as the volume.

The first time is simply,

M500 = Mt / [ 1 + Vi/500 * 1/k]

Where we have used the previous result and changed the pull volume. For the second 500ml pull, Mt has now become Mt - M500. Let's call the extracted material in the second pull M500', then:

M500' = [Mt - M500] / [ 1 + Vi/500 * 1/k] = [Mt - {Mt / [ 1 + Vi/500 * 1/k]} ] / [ 1 + Vi/500 * 1/k]

So, the total of the two 500 ml pulls is (let's call this M2x500),

M2X500 = M500 + M500'

We can see it is going to be different than the M1000 result.

We can't compare numerically at this time since we need to know a little more. Let's assume a liquid-liquid extraction with p = 0.75 and Vi = 1500ml (or if you prefer to think in terms of a solid-liquid extraction assume Vi = 100ml and p = 0.05, result will be the same since Vi/p is the same). What would be the ratio of M2x500 and M1000? I'll leave this as an open question, see if anyone answers. Otherwise I'll give what I think is the answer next week.

Edit: Bonus question. Is there a Vi/k value where the ratio of M2x500/M1000 is maximized? If yes, what is it?

Lololol. I think Loveall wins the prize!!

Short answer: depends on surface area exposure, contact time, volume, solubility of the target molecule, temperature, and concentration.

Long answer: experiment some and decide for your self. If yah figure out something that works, see what different conditions cause varying results.

Keep in mind, everyone has something that is important to them when they are extracting. Some are time, efficiency, resources (solvent, lye, etc), qualitative expectations, or quantitative expectations.

For me - it really depends on the day. Some days one thing is important; other days other things seem important. Eventually, I found a compromise that allows me to get a nice balance of all of the above.

So, yeah. Experiment some.

Take Care!
ACY

Substituting the k/Vi = 0.75 gives a ratio of M2x500/M1000 = 1.08. That is, 8% more is extracted by doing two pulls with 500ml of solvent instead of one with 1000ml in this particular case.

There is a more "practical" way to do this. There is a link between k and p. Comparing the equation for R after one pull

R = 1 - (1-p)^1 = p from this post

R = 1/[1 + Vi/Vp*1/k], from post #3 above where we have called the volume of solvent simply Vp.

Combining these we trivially get,

p = 1/[1 + Vi/Vp*1/k]

Now, we can get p for each situation strategy. If using 1000ml pulls, then,

p1000 = 1/[1 + 1500/1000*1/0.75] = 0.33

And if doing 500ml pulls,

p500 = 1/[1 + 1500/500*1/0.75] = 0.20

And the ratio is simply,

M2X500/M1000 = [1-(1-.25)^2]/.33 = 1.08

That is, we get 8% more by doing two pulls.

Finally, there is an even faster way to estimate this, starting with p500 = 0.2 and only using the tables from this post.

From the first table we see that for p= 0.2 and n=2, R =0.36.

Since we double the volume, alpha = 2. We can't get the exact value for the new p, but we can see from table 2 that it is between 0.30 and 0.35 (and closer to the higher value), therefore we can reasonably select p'=0.33. Table 1 shows that for this n=1, R' = p' = 0.33 is expected.

The answer to the question is simply M2X500/M1000 =. 36/.33 = 1.08, or 8% moar. Again, this is for our particular case, under different extraction efficiency conditions there are different answers.

I'll leave the bonus question unanswered to see if anyone bites on that one. Notice that it is asking, what is the maximum boost to total yield theoretically possible if doing two pulls instead of one pull (same total amount of solvent)?

Well, so much for a "possibly stupid question" LOL.

There is now a general look at this here. Disclaimer is that formulas have not been cross-checked and could be wrong, but at least an attempt was made to answer.

In practice doing two instead of one pulls improves the yield by roughly 8% depending on the situation. Theoretical max is 12.5%. This is for pull conditions where only the solvent volume and number of pulls changes from 2 pulls at half solvent volume to one pull at full solvent volume (all other steps the same).

I dived into the math as well but hit a bit of a brick wall trying to generalize it.
I might give it another try, though. My goal was to have a formula that tells you the optimal number of pulls for a given amount of total solvent available. There has to be some sweet spot.

I think I know the answer. It is not practical, but it may help understand the situation.

Each solvent volume now depends on the pull number. Let's call it Vsn (Vs1, Vs2,...). Also the extraction power is changing with beach pull if we allow the volumes to change, so let's introduce the notation, pn.

First, if possible, do a saturated pull. That is, add the max amount of solvent that will stay saturated. Let's call the solvent saturated concentration c.

In the saturated condition,

c = M1/Vs1

Which can be written as, Vs1 = M1/c

Also, we know from this post that,

M1 = pn * M0 = 1 / [1+(Va/Vs1)*(1/k)] * M0

Substituting into this last equation Vs1 = M1/c and rearranging,

M1 = M0 - Va*c/k

If M0 - Va*c/k > 0 then we can do this saturated pull using a volume,

Vs1 = M1/c = M0/c - Va/k

Now the drug that remains in the solvent after this saturated pull is,

N1 = M0-M1 = M0 - (M0 - Va*c/k) = Va*c/k

And we are no longer able to do saturated pulls (note that if we try to use the formula for a saturated pull for this value of mass in the solvent, we get a 0 volume).

If the total amount of starting solvent is less than or equal to M0/c - Va/k, simply do a single saturated pull, you are done.

If the total volume of starting solvent is larger tham this, things are more interesting.

First, simply do a saturated pull if possible.

After this, you would max out the drug by doing infinite pulls with an infinitesimal small volume. At one point this backfires in practice if you lose some drug at each pull and there is an optimal small volume as you say.

However, we can calculate the maximum theoretical limit. This would be what we could get out mathematically with infinite infinitely small pulls.

Alright, let's have some fun!

So, we have (if needed) done one pull to be out of the saturated regime. If we do that pull, let's set that aside and restart the counters once M0/c < Va/k.

Let's call the total amount of solvent available Vst.

If we divide into n smaller volumes, then,

Vs = Vst/n
p = 1 / [1+(Va/Vs)*(1/k)] = 1 / [1+n*(Va/Vst)*(1/k)]

Now we can write the extracted amount as

Rn = M0 * [1-(1-p)^n] = M0 * [1-(1-1 / [1+n*(Va/Vst)*(1/k)])^n]

And now all we need to do is tale the limit for n-> infinity. This is possible, just recognize the form,

lim n-> infinity [1/(1+a*n)]^n = e^(-1/a)

Where e is Euler's number 2.71828... This can be proved her upon request.

Armed with this knowledge,

R∞ = M0 [1 - e^(-k*Vst/Va)]

This gives the maximum drug available for a fixed solvent volume Vst when pulling in the non-saturated regime. The more small pulls done the closer one can get to this in theory. If one is familiar with the TEK being used and knows p for a certain Vs, then this can also be written as,

R∞ = M0 * [1 - e^(-p/(1-p)*Vst/Vs)]

Which is a cool upper limit to be aware of. Since in this situation n = Vst/Vs

R∞ = M0 * [1 - e^(-p*n/(1-p))]

Which can be compared to

Rn = M0 * [1-(1-p)^n]

To understand the loss of doing n pulls instead of infinite pulls for the solvent volume Vst.

Note: typing all this on my phone, there could me mistakes, will double check and update if needed.

Seems like the formulas hold up, would be great if someone could double check them.

Below is a table showing the possible upside to doing infinite pulls (instead of n pulls) for different p and n situations (R∞/Rn - 1). Note that the total solvent volume is fixed at Vst=n*VS and the infinite pulls have infinetly small volumes that add up to Vst). The colors come from the first table in this post. A quick examination shows little upside to doing a very large amount of pulls. Instead, it seems better to simply shoot for the green area once p is known after measuring the mass from the first two pulls in a non-saturated regime.