7 replies to this topic

[chandch]

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

Edited by chandch, 30 November 2012 - 10:35 AM.

[pito]

Do you know what it mean? mM and molecular weight?

[bob1]

You need the density of the solution to work this one out...

[chandch]

Hi.. It's millimolar (mM). Can you give me an example calculation please. The density is 1.002g/ml.

Edited by chandch, 30 November 2012 - 03:32 PM.

[John Forsberg]

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).

Edited by John Forsberg, 30 November 2012 - 05:23 PM.

[casandra]

John Forsberg, on 30 November 2012 - 05:22 PM, 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?

Edited by casandra, 30 November 2012 - 09:01 PM.

[John Forsberg]

casandra, on 30 November 2012 - 08:58 PM, 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.

[chandch]

Thank you all. Now I understand how to calculate moles from liquid conc.