how much PMK ethyl glycidate (CAS: 28578-16-7) is required to obtain 1 liter of PMK (CAS: 4676-39-5)?

[antrax]

and with 1 liter of PMK, how much MDMA can be obtained?
Thank you, all!!

 

[Frit Buchner]

The mol.wt ratio is 175/250 I.E. 7/10. My math says that is equal to 1.43L to obtain 1liter
The second question involves a lot of variables. Choose a route and I'll try to look it up bit I'm not prepared to go it 5 times

 

[antrax]

The mol.wt ratio is 175/250 I.E. 7/10. My math says that is equal to 1.43L to obtain 1liter
The second question involves a lot of variables. Choose a route and I'll try to look it up bit I'm not prepared to go it 5 times

Frit BuchnerWould you be so kind as to calculate how much precursor CAS 52190-28-0 to obtain 1kg of bk-MDMA crystals? since I see you very pro
or much P2NP is required to obtain 1kg of speed sulphate.

 

[Frit Buchner]

I'm not even a chemists assistant. I know nothing. Here's what you do;
Search [whatever precursor] mol.wgt ( molecular weight)
2-bromo-3,4-methylproprophenone mol.weight=257.1grams in a mol.

Search for your target chemical mol.wt
Beta-keto methylenedioxymethamphetamine mole weight =227.23g/mol( I rounded)

For every 257.1g of 2-d-3,4-mdp there is (in theory)227.23g of bk-mdma
227.23g/257.1=0.88381, 88%

That's theoretical perfect quantitative yield. If people are getting 80% yield/ you hope to get 80%, then you multiply that by 0.8
88.381x0.8=70.74%
If 1kg is 70.74%, 1/70.74x10=1.292
1.292kg of 2-b-md-p to make 1 kg of bk-mdma

The yield I didn't look up, I just used a number that sounded reasonable (70%) this is for illustration purposes only. Now you don't have to look like an idiot who couldn't pass the 6th grade.

 

[antrax]

I'm not even a chemists assistant. I know nothing. Here's what you do;
Search [whatever precursor] mol.wgt ( molecular weight)
2-bromo-3,4-methylproprophenone mol.weight=257.1grams in a mol.

Search for your target chemical mol.wt
Beta-keto methylenedioxymethamphetamine mole weight =227.23g/mol( I rounded)

For every 257.1g of 2-d-3,4-mdp there is (in theory)227.23g of bk-mdma
227.23g/257.1=0.88381, 88%

That's theoretical perfect quantitative yield. If people are getting 80% yield/ you hope to get 80%, then you multiply that by 0.8
88.381x0.8=70.74%
If 1kg is 70.74%, 1/70.74x10=1.292
1.292kg of 2-b-md-p to make 1 kg of bk-mdma

The yield I didn't look up, I just used a number that sounded reasonable (70%) this is for illustration purposes only. Now you don't have to look like an idiot who couldn't pass the 6th grade.

Frit Buchnerexcuse my ignorance, 1.292kg of 2-b-md-p to make 1 kg of bk-mdma, crystal or powder?

 

[Frit Buchner]

excuse my ignorance, 1.292kg of 2-b-md-p to make 1 kg of bk-mdma, crystal or powder?

antraxYes. Either. This is theoretical chemistry and theoretical chemistry produces pure chemicals. 98% mdma could be a crystal or you could not crystallize it and it would be a powder. You could crush the crystals to powder, and then recrystallize them back to crystals. A chemical doesn't have to be very pure to crystallize either. On the other hand crystallization can purify a chemical if done right

 

[antrax]

Yes. Either. This is theoretical chemistry and theoretical chemistry produces pure chemicals. 98% mdma could be a crystal or you could not crystallize it and it would be a powder. You could crush the crystals to powder, and then recrystallize them back to crystals. A chemical doesn't have to be very pure to crystallize either. On the other hand crystallization can purify a chemical if done right

Frit Buchnerbut can bk-MDMA crystals be orange-brown in color and still have high purity?