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Molar calculation from liquid

molarity

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7 replies to this topic

#1 chandch

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Posted 30 November 2012 - 10:33 AM

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.


#2 pito

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Posted 30 November 2012 - 11:56 AM

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

If you don't know it, then ask it! Better to ask and look foolish to some than not ask and stay stupid.


#3 bob1

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Posted 30 November 2012 - 01:58 PM

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

#4 chandch

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Posted 30 November 2012 - 03:27 PM

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.


#5 John Forsberg

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Posted 30 November 2012 - 05:22 PM

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.


#6 casandra

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Posted 30 November 2012 - 08:58 PM

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.

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#7 John Forsberg

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Posted 30 November 2012 - 09:11 PM

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.

#8 chandch

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Posted 01 December 2012 - 11:16 AM

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





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