When dealing with molar ratios for liposomal formulation, I have a particular mg of sodium ascorbate and phopshatidylcholine per 5ml as my reference (roughly 1g SA and 500mg PC per 5ml) I want to experiment with 1:1 molar ratio of SA/PC and increase molar ratio of SA ie 1:1, 5:1, 10:1 SA/PC. So I need to calculate molar ratio into weight in mg and then add water.
I would imagine the molar ratios stay the same regardless of the addition of varied amounts of water per mg? I also want to keep between 4 and 6g of sodium ascorbate per 5ml and decrease and want to experiment with molar ratios for optimal EE%.
Please correct and elaborate on any mistakes I have made:
If say if I use a 1:1 ratio sodium ascorbate (198.11g/mol) to phosphatidylcholine (786.11g/mol), the amount required in mg would be 1M x 198.11g/mol = 198.11mg and 1 x 786.129mg = 786.129mg = SA/PC, 1:1, correct? I could add my 5ml to get roughly 0.2g SA per 5ml, my guess is there are too much lipids and not enough encapsulate but I will test the EE%, and move on.
So now I use a 5:1 ratio sodium ascorbate (198.11g/mol) to phosphatidylcholine (786.11g/mol), the amount required in mg would be 5M x 198.11g/mol = 990.55mg and 1 x 786.129mg = 786.129mg, SA/PC, 5:1, correct? I could add my 5ml to get roughly 1g per 5ml SA with 786mg PC per 5ml, correct?
So a 10:1 ratio of SA/PC would simply be, 1981mg SA / 786mg PC? per 5ml?
But what if 5:1 ratio is good but I want to half the amount of PC and maintain the same amount of vitamin C? a 10:1 ratio doubles the amount of SA per 5ml but the amount of PC stays the same. The only solution I see is adding 10ml rather than 5 to 10:1, this then provides roughly 1g SA per 5ml (same as 5:1) but now dilutes the PC content by 50% = 1g SA and 393mg per 5ml.
Its this part that gets confusing for me, because I have other additives I will need to add too at much lower concentrations, for example I have a PC/Sterol ratio of 10:3 and a PC to vitamin E ratio of 10:0.1,
But what if I want to half the amount of lipids
I din't major in Chemistry so any help would be appreciated.
What you propose is summarized in the following molar ratios:
SA:PC = 10:1
PC:S = 10:3
PC:E = 10:0.1
All these binary ratios can be expressed in a single ratio that includes all substances. The key is that in all these molar ratio there is a compound in common with the same relative number. That compound is the PC. In two ratios PC has the relative number 10 and in SA:PC, the number of PC is 1, if you multiply this last ratio by 10 you get an adequate amplified version:
SA:PC = 100:10 (is equivalent to 10:1)
Now all the binary ratio, which have in common the relative amount of PC, can be connected in a global molar ratio:
SA:PC:S:E = 100:10:3:0.1
You can leave it like this or present a more elegant version without decimals (whenever possible) by dividing all the numbers by the smallest number (0,1):
SA:PC:S:E = (100/0.1):(10/0.1):(3/0.1):(0.1/0.1)
SA:PC:S:E = 1000:100:30:1
If you want 1 g of SA in 5 mL of final preparation, the weight (w, in milligrams) of the rest of the compounds is predetermined by the global ratio thus:
wPC = (1000 mg/MMSA) x (100/1000) x MMPC
wS = (1000 mg/MMSA) x (30/1000) x MMS
wE = (1000 mg/MMSA) x (1/1000) x MME
Note: MM is molar mass (molecular weight). On the other hand, observe that in these equations the factors 100/1000, 30/1000 and 1/1000 come from the global molar ratio.
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