[Brünnhilde]

Hi,

I'm going to do an experiment in which I need a final concentration of 3 mM of a probe in a 200 μL cell lysate.
If the probe has a molecular mass of, say, 74 g mol^-1 I have derived that I need to add a final volume of

$$V_1\cdot C_1=V_2\cdot C_2\Rightarrow V_1=\frac{V_2\cdot C_2}{C_1}=\frac{V_2C_2MV}{m}$$

 - as $$C=\frac{n}{V}=\frac{m}{MV}$$ .

To attain a mass of my probe I can weight off I decide to dissolve 5 mg of it in 800 μL PBS (buffer), so I get

$$V_1=\frac{200\cdot 10^{-6}L\,3\cdot 10^{-3}mol\,L^{-1}74.085g\,mol^{-1}800\cdot 10^{-6}L}{5\cdot 10^{-3}g}=\sim 6 μL$$

To me this seems like a reasonable volume of my dissolved drug to add to the lysate, but I am uncertain if it will truly give me a final concentration of 3 mM. Maybe I've only complicated matters with the formula as I'm not sure I use it correctly - especially regarding the dilution. Thoughts?

I hope I've made myself somewhat clear
Thanks!

[Hunter2]

3 mM = 3mmol/l = 3*10^-3mol/l

For 200 µl: c =n/V => n =c*V = 3*10^-3 mol/l * 200 * 10^-6 l = 6* 10^-7 mol = 0,6 µmol

m = n *M  m =0,6 µmol *74 µg/µmol = 44,4 µg