cyantific said:
The Cielo Tek states that monomescaline citrate salt is ~57% as strong as Mescaline HCl. How does this correlate to the numbers above?
The ratio of the (milli)molar masses of the respective salts gives their relative 'strengths' so 248/402=~0.617 (61.7%)
By an amusing coincidence, the ratio with freebase
mescaline (211g/mol), 211/402, gives 52.5% but if you take the average of these two, (52.5%+61.7%)/2, you get the 57% value; however, let's take a look at exactly what it says on the wiki:
In CIELO Tek on the wiki be said:
The product is monomescaline citrate salt of 93.1% purity [emphasis added] based on NMR results from endlessness (see fig 6a), which is ~57% as strong as Mescaline HCl.
So, following through with the working: 93.1%*61.7%=57.4%.
Always check for
all the relevant data, and always show your working!
Looking a bit further through the
wiki we find:
Fig. 6a: NMR results. Only mescaline was detected. Monomescaline salt form was confirmed. Preliminary early calculation indicates 93.1% purity with the remaining 6.9% being water, consistent with the salt form (MesH)H2Cit·1.5H2O (see appendix for notation description). This is a tentative result that needs verification. Credit: endlessness.
So, the hydration level of the salt explains both the 93.1% purity result, and the need for residual water in the EA at the crystallisation stage.
This data point, when combined with the
solubility data for water in EA, can also inform us wrt a maximum concentration for
mescaline above which the crystallisation process will possibly begin to break down due to insufficient water. Additionally, it seems
a minimum of about 2% water is required for (anhydrous) citric acid to dissolve successfully in EA. This is a large molar excess compared to the 57mmol of CA that was used, so citric acid remains the limiting factor rather than the water.
If Endlessness' experiment proves to be representative, the theoretical maximum yield of monomescaline citrate sesquihydrate that can be obtained via CIELO in 250mL EA is ~24.8g. The equivalent amount of
mescaline HCl would be 14.3g. However, the 11.1g may well have been an arbitrary amount. We should also note that Endlessness' method deviated from the standard CIELO through the use of CASEA. Moreover, neither the volume of cactus extract EA, its suspected water content, nor the final yield of MMC is stated in that excerpt.
Regardless of that, here are my workings:
11.1g CA anhyd C6H8O7=72+8+112=192g/mol 11.1/192=
0.0578125mol
5ml water H2O=18g/mol 5/18=
0.27777777777777777778mol
250ml EA MW irrelevant
M C11H17NO3=132+17+14+48=211g/mol 211*0.0578125=
12.1984375g
MC1.5H2O=211+192+27=430g/mol 430*0.0578125=
24.859375g
MHCl=211+36.5=247.5g/mol 247.5*0.0578125=
14.30859375g