The Vespiary
The Hive => Stimulants => Topic started by: 12cheman12 on September 07, 2004, 10:07:00 AM
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HCL mole weight is 36.47
and one mole of pseudoephedrine weighs 165.23 (not to sure if this weight is the freebase or hcl form though)
Does this mean if i had 165.23 grams of pseudo FB i would need 36.47 grams of HCL to turn the pseudo back to hydrochloride form in a perfect ratio?
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pseudo freebase=165.232 and is ~80% weight of the hcl salt
does seem a little mind boggling though doesn't it
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165(pseudo FB)/ 16.5 = 10gs
36(hcl)/16.5 = 2.8gs
So does this mean if i had 10Gs of pseudo FB i would need 2.8 Gs of HCL?
And if my HCL is 30% i would need 9.3 mls?
its not really the maths im worried about i just want to know if you need equal molar ratios when titrating so that there wont be any excess hcl.
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yes
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EPHEDRINE HCl, C10H15NO.HCl molecular weight MW = 201.73 grams/mole
EPHEDRINE FREEBASE, C10H15NO molecular weight MW = 165.23 grams/mole
The freebase has a melting point 40 deg C with a ½ H2O. The boiling point is 225 deg C.
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(165.23/201.73) x 100 = 81.9% is EPHEDRINE of the EPHEDRINE HCl
(36.5/201.73) x 100 = 18.1% is HCl of the EPHEDRINE HCl
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81.9% + 18.1% = 100% or 165.23 + 36.5 = 201.73 for EPHEDRINE HCl, C10H15NO.HCl
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Compensate in your E's weight to maintain the ratios !
" Pseudo HCl is about 202gr per mole and psuedo free base is about 166gr per mole.
So 166 divided by 202 is a ratio of 0.82 FB E
( because less weight is more E with the fb )
FBE = .82
I = 1.2
LGRP = .33
which equals
FBE = 1
I = 1.45
LGRP = .4
Hydrochloride E
HCL E = 1
I = 1.2
LGRP = .33