3 replies to this topic

[Mad Researcher]

Hi, 

 

I need to make drugs at 10 mM, 1 mM, 0.1 mM, 0.01 mM and 0.001 mM

The stock conc of drug A = 3.2 mg/ml which is dissolved in Methanol

 

Drug A = 387.5 g/mol

Drug B = 144.2 g/mol

Drug C =  546.7 g/mol

Drug D = 392.4 g/mol

 

Could you please help with the calculations?

 

Thank you.

 

[bob1]

I'll do A, you would need to have the w/v concentrations for the others too. 

 

3.2 mg/ml = 3.2 g/l (so that your units for molarity are the same as concentration)

 

n=mass/molar mass

n= 3.2 g /387.5 g/mol

n=0.00826 mol

 

It just so happens that because the calculation is done on the per litre amount, you also have this as a concentration (i.e. 0.00826 mol/l)

 

0.00826 = 8.26 mmol/l or 8.26 mM. You don't have a high enough concentration to get your highest dose.

[Mad Researcher]

I'll do A, you would need to have the w/v concentrations for the others too. 
 
3.2 mg/ml = 3.2 g/l (so that your units for molarity are the same as concentration)
 
n=mass/molar mass
n= 3.2 g /387.5 g/mol
n=0.00826 mol
 
It just so happens that because the calculation is done on the per litre amount, you also have this as a concentration (i.e. 0.00826 mol/l)
 
0.00826 = 8.26 mmol/l or 8.26 mM. You don't have a high enough concentration to get your highest dose.

Thanks. How do I make the lower points i.e 0.1 mM and so on?

[bob1]

There are a couple of options:

 

1) Dilute your stock using V1C1 = V2C2 to work out the amounts needed for each of the higher dilutions. This will probably only work feasibly for the 1, 0.1 and maybe 0.01 mM concentrations.

 

or the preferred method for these sorts of things - serial dilution:

 

2) Make your stock to something convenient (use formula above; 5 mM might work or 1 mM) then do the appropriate dilution series to get the final concentrations. Note that you have 1:10 dilutions at each step, so if you start with 1 mM then all you need to do is take 1 part stock and 9 parts  diluent for each of the subsequent dilutions.