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Calculation using molecular weight...


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#1 chrisbelle

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Posted 18 April 2009 - 03:21 AM

Dear all, I have a calculation problem...

My liquid has this info on its bottle: has a molecular weight 200, weight 5g.
How come the company gives the unit as 5g since this is a liquid?

I need to make 50mM stock solution OR
100ug/ml solution.

How do I calculate this????

Thanks,
Chris
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#2 pito

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Posted 18 April 2009 - 09:12 AM

Dear all, I have a calculation problem...

My liquid has this info on its bottle: has a molecular weight 200, weight 5g.
How come the company gives the unit as 5g since this is a liquid?

I need to make 50mM stock solution OR
100ug/ml solution.

How do I calculate this????

Thanks,
Chris


what is the molecular weight? 200 ? or 5g? and I suppose they mean molecular mass or even better molar mass , in g/mol? Meaning 5gram/mol.


And it doenst matter whether its a liquid or solid or gas.
Its just the mass of the chemical compounds in your liquid per 1 mol liquid.
I do not see the link with the stock solution you need to make.
The only important thing you need to know is whether the liquid in the bottle you have is allready diluted or not (is it 100% or?)

if you need 100g per ml, then simply make 1 liter with 1000x100g ==> meaning you need 0.1 gram for the liter (if I calculated it correclty)

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


#3 HomeBrew

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Posted 18 April 2009 - 12:42 PM

You need to know the density of the liquid. This will give you grams per ml.

#4 pito

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Posted 19 April 2009 - 02:06 AM

You need to know the density of the liquid. This will give you grams per ml.


My bad,
Homebrew is right, I forgot the density part.

Here is an example calculation:


You have product B with a density of 10g/l , molecular weight of 100g/mol

and you need to make a solution with 5gram per 4liters

then you simply need to take 0.5 liter of your product ( 0.5 l* 10 g/l = 5gram) and add 3.5 liter of water so that you have 3.5+0.5liters= 4 liters in total, containing 5 grams of your product.

(I do assume that your product B is 100% and not diluted yet, if this is the case then you need to recalculate that too.)

However I still do not understand why you would need the molecular weight.

Edited by pito, 19 April 2009 - 02:53 AM.

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





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