Author Topic: Pseudo FB and HCL mole ratios  (Read 2002 times)

0 Members and 1 Guest are viewing this topic.

12cheman12

  • Guest
Pseudo FB and HCL mole ratios
« on: September 07, 2004, 10:07:00 AM »
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?


ozmosis

  • Guest
"Pseudo FB and HCL mole ratios"
« Reply #1 on: September 07, 2004, 10:37:00 AM »
pseudo freebase=165.232 and is ~80% weight of the hcl salt

does seem a little mind boggling though doesn't it

12cheman12

  • Guest
165(pseudo FB)/ 16.5 = 10gs 36(hcl)/16.5 =...
« Reply #2 on: September 07, 2004, 11:04:00 AM »
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.


placebo

  • Guest
yes
« Reply #3 on: September 07, 2004, 11:17:00 AM »
yes


WizardX

  • Guest
EPHEDRINE HCl, C10H15NO.HCl
« Reply #4 on: September 07, 2004, 12:34:00 PM »
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.

-----------------------------------------------------------------------------------------
(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
-----------------------------------------------------------------------------------------

81.9% + 18.1% = 100% or 165.23 + 36.5 = 201.73 for EPHEDRINE HCl, C10H15NO.HCl

BullwinkleMoose

  • Guest
Compensating to maintain ratios for FB Pseudo E
« Reply #5 on: September 08, 2004, 07:50:00 AM »
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