Molar calculation from liquid - (Nov/30/2012 )
Hi all
I am trying to calculate 0.1mM, 10mM, 1mM and 1M concentration. The molecular weight is 90.12 (It is in liquid form and 99% pure). How do I bring down to the required concentration, if I need 5ml each.
Please help.
Regards
Chandch
Do you know what it mean? mM and molecular weight?
You need the density of the solution to work this one out...
Hi.. It's millimolar (mM). Can you give me an example calculation please. The density is 1.002g/ml.
1.002g/ml is 1,002 g/L, so divided by 90.12 g/mol = 11.12 mol/L (M) for the undiluted liquid.
To make 5 ml of a 1M solution, you would mix 1M/11.12M * 5 ml = 0.450 ml of your chemical plus water to 5 ml. I would normally just dilute that 1M stock to 10 mM with a 1:100 (50 ul + 4.95 ml water) dilution, then make the 1 mM stock by making a 1:10 dilution of the 10 mM stock (500 ul + 4.5 ml water), then another 1:10 dilution of the 1 mM stock to get the final 0.1 mM solution (again, 500 ul + 4.5 ml water). I don't like pipetting really small volumes (because of poor accuracy of pipettes at those volumes), so I'd avoid doing dilutions straight from your 11.12M solution to get the 1 mM and 0.1 mM solutions (those would be 1:1,112 and 1:11,120 dilutions, with 4.5 ul and 0.45 ul of your undiluted chemical in 5 ml water).
John Forsberg on Sat Dec 1 01:22:23 2012 said:
1.002g/ml is 1,002 g/L, so divided by 90.12 g/mol = 11.12 mol/L (M) for the undiluted liquid.
To make 5 ml of a 1M solution, you would mix 1M/11.12M * 5 ml = 0.450 ml of your chemical plus water to 5 ml. I would normally just dilute that 1M stock to 10 mM with a 1:100 (50 ul + 4.95 ml water) dilution, then make the 1 mM stock by making a 1:10 dilution of the 10 mM stock (500 ul + 4.5 ml water), then another 1:10 dilution of the 1 mM stock to get the final 0.1 mM solution (again, 500 ul + 4.5 ml water). I don't like pipetting really small volumes (because of poor accuracy of pipettes at those volumes), so I'd avoid doing dilutions straight from your 11.12M solution to get the 1 mM and 0.1 mM solutions (those would be 1:1,112 and 1:11,120 dilutions, with 4.5 ul and 0.45 ul of your undiluted chemical in 5 ml water).
This calculation is ok but shouldn't we factor in the 99% "purity" of the solution or is this just splitting hairs?
casandra on Sat Dec 1 04:58:07 2012 said:
This calculation is ok but shouldn't we factor in the 99% "purity" of the solution or is this just splitting hairs?
I suppose we could. I did use all 4 significant figures in the calculations (so the 1% would affect the numbers there certainly), but with the pipetting, it would only change the volumes used by about 1%, which is smaller than the error on many pipettes. If we correct for the 99% (and assume the contaminants also have the same density), then the volume of chemical to use for the 1M stock changes to 0.454 ml plus water to 5 ml. I'm not sure I trust my 1 ml pipette to be accurate to 5 ul, although someone with nicer pipettes might know theirs are that accurate.
Thank you all. Now I understand how to calculate moles from liquid conc.