Retinoic acid - (Dec/02/2014 )
Hello,
Kind of embarrassing post. I'm trying to make a 10uM retinoic acid (from Sigma) http://www.sigmaaldrich.com/content/dam/sigma-aldrich/docs/Sigma/Product_Information_Sheet/1/r2625pis.pdf
But suck on the calculations. It says "A stock solution of 0.01 M (3 mg/ml) RA in absolute ethanol was stored at -70°C for up to two weeks." I dissolve 3mg of R.A. in 1 mL of 200 proof ethanol, and then take 5 uL of that mixture and add it to 5mL of media. Are my steps correct?
Critical information missing: what final concentration are you aiming for?
Assuming that 3 mg/ml makes 0.01 mol/l (or 10 mmol/l) then adding 5 ul to 5 ml (1:1000 dilution) will give you 0.01 mmol/l.
What "calculation" is necessary? Your stock solution wants to be at 3 mg/ml. You take 3 mg, and add it to 1 ml, and (ta-da), you have 3 mg/ml.
bob1 on Wed Dec 3 08:19:53 2014 said:
Critical information missing: what final concentration are you aiming for?
Assuming that 3 mg/ml makes 0.01 mol/l (or 10 mmol/l) then adding 5 ul to 5 ml (1:1000 dilution) will give you 0.01 mmol/l.
I want to create RPMI media with the following concentrations: 10uM, 15uM and 20uM from the 3 mg or R.A. dissolved in 1mL of 200 proof ethanol.
Ok, that makes it a bit easier.
Again assuming that 3 mg/ml is equivalent to 0.01 mol/l - you should check this, use n=mass/molar mass and c=n/v!
0.01 mol/l = 10 mmol/l or 10,000 umol/l
C1V1=C2V2
10000 umol/l x V1 = 20 umol/l x 5 ml
or 10000 x V1 = 100
Solve for V1
V1 =100/10000 =??? ml (note that this volume is in ml, convert to ul!)
repeat for the rest.
bob1 on Wed Dec 3 19:31:23 2014 said:
Ok, that makes it a bit easier.
Again assuming that 3 mg/ml is equivalent to 0.01 mol/l - you should check this, use n=mass/molar mass and c=n/v!
0.01 mol/l = 10 mmol/l or 10,000 umol/l
C1V1=C2V2
10000 umol/l x V1 = 20 umol/l x 5 ml
or 10000 x V1 = 100
Solve for V1
V1 =100/10000 =??? ml (note that this volume is in ml, convert to ul!)
repeat for the rest.
Thanks. I was going about it the wrong way.