[fledarmus]

so i got the volume of acetyl chloride to be 1.0467 ml.

im lost at figuring out the volume of methanol.
i have 0.0489 mol of tryp (1000ml of solution/0.75 mol of tryptophan) = 6.52ml of tryp of solution

I did a little correcting on your equation. You do have to be careful of your units - M is not moles of your compound/liter of your compound, but moles of your compound/liter of solution. In your case, for 1 gram of tryptophan, you'll need about 6.5 mL methanol.

As Discodermolide says, you probably don't need that much methanol for the reaction, but in this case you're also using it for the solvent and tryptophan isn't particularly soluble in methanol at room temperature. For a reaction I don't know, I usually start somewhere on the order of 0.5M to 1.0M concentration in solvent as a first try, so I wouldn't think 0.75M is unreasonable.

[viet]

thanks for your help guys!
say i have a certain amount(#) of tryptophan-methyl-ester hcl (1eq) and put it in dichlromethane (0.18M)
i would have:
#g tryp-ome·hcl (1mol/204.3g tryp) = ## mol tryp
to find out how much dichlorometh needed:

X ml                 1000 ml
-------------- =   -----------    => X ml = ### ml of ch2cl2
## mol tryp       0.18 mol

is that right?

[Borek]

Double # invoked a LaTeX interpreter and corrupted your post. Please read about post formatting - you can use LaTeX to nicely format your equations instead of using ASCII for that purpose.

[viet]

sorry i couldnt modify the prev post due to different ip.

say i have a certain amount(#) of tryptophan-methyl-ester hcl (1eq) and put it in dichlromethane (0.18M)
i would have:
[tex](an\,amount\,tryptophanmetholester·HCl) \frac {1mol}{204.3g\,tryp}) = X\,mol\,tryp [/tex]       
want to know how much CH2Cl2 needed:
[tex] \frac {unknown\,ml\,CH2Cl2}{X\,mol\,tryp} = \frac {1000ml\,soln}{0.18 mol\,CH2Cl2}[/tex]

[fledarmus]

sorry i couldnt modify the prev post due to different ip.

say i have a certain amount(#) of tryptophan-methyl-ester hcl (1eq) and put it in dichlromethane (0.18M)
i would have:
[tex](an\,amount\,tryptophanmetholester·HCl) \frac {1mol}{204.3g\,tryp}) = X\,mol\,tryp [/tex]       
want to know how much CH2Cl2 needed:
[tex] \frac {unknown\,ml\,CH2Cl2}{X\,mol\,tryp} = \frac {1000ml\,soln}{0.18 mol\,CH2Cl2tryptophan}[/tex]

Once again, your 0.18M is referring to 0.18 moles tryptophan/1000 mL solution.