How many mg of THC can be dissolved in 1 ml of ethanol?

ballin jack

ballin jack

Well-Known Member
I can't seem to find a hard number for this anywhere on the internet. I'm trying to find out the extraction efficiency of THC in ethanol.

Just to give a little information, I'm soaking decarbed flower in 151 vodka in the freezer to make a tincture. I'm trying to figure out how much of the THC in the flower I can expect to be absorbed by the alchohol, because obviously it's not going to be 100%. I have heard everything from 20mg/ml to 1g/ml. I would love to know the real answer, so that I can guesstimate how strong the tincture will be.

Here's my math:

8g decarbed flower at 17% THC = 1360 mg THC

I can't really go past this lol, because I don't know how much of the 1360 mg I can expect to be absorbed by the 150 ml of 151 alcohol I am soaking it in. (Is it 90%? 70%? etc...)

I assumed it was 90% just to finish the math, so 1360 mg X .90 = 1224 mg THC

Then 1224mg/150ml = 8.16 mgTHC/ml 151 vodka

The big question though is what percent of the THC will actually be absorbed into the alcohol. Anyone know the answer to this?

Thanks!
 
Hugo Phurst

Hugo Phurst

Well-Known Member
90% sounds good to me. I did that once and it really fucked me up, that and the taste convinced me to switch to edibles.
If you want to be more accurate, you could make hash & then dissolve it in the 151, it would taste a lot better too.

Good luck
 
ballin jack

ballin jack

Well-Known Member
I plan on mixing it 50/50 with CBD oil to dilute the alcohol and add more CBD to the mixture. Then taking it under tongue medicinally. I'm just trying to figure out how many ml will give me 1mg of THC so I can decide how much I want to dose myself then.

I'm kind nervous getting into making hash, or evaporating the alcohol off or anything. I'm hoping to keep it as simple as soaking it, and then diluting if possible.
 
ballin jack

ballin jack

Well-Known Member
Coloradoclear said:
Your extraction process is inefficient. So be realistic in your expectations so you don't make it ridiculously weak .
Click to expand...
No no, that number was just the THC content of the dry bud, before I do anything to it. Like, if I pick this off of the plant, or buy it in a store, 17% of the weight is THC.

What I'm trying to figure out is how efficient the extraction process is lol.

From reading other recipes, I think 90% is about the right number. So 90% of the THC in the bud I can expect to be absorbed into the alcohol. Seems like if there's enough alcohol to cover the top of the bud, it should be more than enough to hold 90% of the THC. I know there will be tons of other things that get absorbed along with it too.

I'll read through that thread Renfro. If anyone else has a hard number for this, feel free to post it. Thanks guys.
 
Fadedawg

Fadedawg

Well-Known Member
Renfro said:
If anyone knows an exact number it would probably be @Fadedawg
Click to expand...
Solly, there's not enough process information provided for exact numbers. I was waiting for more to appear before trying to take a shot at it.

What temperature, how long the soak, any agitation?

How was material prepared? Chunks or ground?

How decarbed? Melted and smeared trichomes are not as easily extracted as intact trichome heads.

I can tell you that using my QWET process and 190 proof, the followup gleaning run typically average typically averages around another 13% or so. A longer soak and more agitation would yield more, but we are trying to avoid the chlorophyll.

https://web.archive.org/web/20150707173953/http://skunkpharmresearch.com/qwet-extraction/

Using 151 proof vis a vis 190 or 200 proof adds complexity to just counting mols. It also raises the dielectric index (polarity). Are you not able to get a higher grade?

Might I suggest a simple minded approach of using a microscope to view the material as you are processing a sample to maximize extraction rate?
 
Thundercat

Thundercat

Well-Known Member
It is pretty easy to make qwet, or qwiso. I would have to think starting with a gram of that and dissolving it into a set amount of alcohol for the tincture might be an easier way to control the dosage. Or if you have the ability to go buy a finished concentrate(i saw you mentioned buying bud at the store) you would know the potancy and should be easier to control dosages.
 
ballin jack

ballin jack

Well-Known Member
Fadedawg said:
Solly, there's not enough process information provided for exact numbers. I was waiting for more to appear before trying to take a shot at it.

What temperature, how long the soak, any agitation?

How was material prepared? Chunks or ground?

How decarbed? Melted and smeared trichomes are not as easily extracted as intact trichome heads.

I can tell you that using my QWET process and 190 proof, the followup gleaning run typically average typically averages around another 13% or so. A longer soak and more agitation would yield more, but we are trying to avoid the chlorophyll.

https://web.archive.org/web/20150707173953/http://skunkpharmresearch.com/qwet-extraction/

Using 151 proof vis a vis 190 or 200 proof adds complexity to just counting mols. It also raises the dielectric index (polarity). Are you not able to get a higher grade?

Might I suggest a simple minded approach of using a microscope to view the material as you are processing a sample to maximize extraction rate?
Click to expand...
I was hoping it would be easier to figure out than this lol. I was going to take all the variables into account by finding the "theoretical" extraction rate in perfect conditions. i.e. 190 proof, i guess absorbing pure THC. Then I was just going to assume I'll reach about 60-70% of that number because of all my factors - Using whole bud instead of just trichomes, using 151 instead of 190 proof, etc.

The way I was thinking about was akin to, "How much salt can be dissolved in 1 cup of water." Should be a pretty easy number to get, right? But I can't even find anything saying like, "in 190 proof ethanol, at 70 degrees F, with pure THC concentrate, xxxx mg of THC can be absorbed per 1ml of ethanol. So I don't even have a starting point.

Sounds like the only way to get a reliable number, from what everyone is saying, is to start with a concentrate, where I know the exact amount of THC I have. Then dissolve that in alcohol. But then it kind of gets me back to the same question. Say I have 1 gram of concentrate - how much ethanol will I need to be able to dissolve all of that? Or is the number so tiny, that the amount of alcohol it takes to cover the concentrate is more than enough to dissolve all of it? :wall:

I fully plan on just buying tinctures at the dispensaries when they are available, but I live in Ohio and the processing plants haven't opened yet. So we're looking at a good 6 months before anything but bud is available. I can't wait to not have to think about this haha.
 
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