## Making a 100mM solution...

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Melissah
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### Making a 100mM solution...

Hey everyone,

It's been a long time since Chem 100 indeed...

I'm trying to make a 100mM solution of ammonium bicarbonate (NH4HCO3) and am a little stuck on how to do it. I was getting some help from a colleague and when he was instructing me via email, he said to use "04 grams in 50mL water" ... unfortunately he's gone for the week now, and I'm not sure if he meant 4 grams or 0.4 grams...

Any suggestions?

canalon
Inland Taipan
Posts: 3909
Joined: Thu Feb 03, 2005 2:46 pm
easy for a 100mM solution you want:
MW x 0.1 x V

with V the volume of the solution you want in liter
and MW the molecular weight of your compound.
Patrick

Science has proof without any certainty. Creationists have certainty without
any proof. (Ashley Montague)

canan5000
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Posts: 17
Joined: Tue Dec 04, 2007 6:17 am
Location: Corvallis OR
If you need to make a solution with the concentration of 100mM you need the molecular weight of 79g/mol of substance and dilute to a liter you need 7.9grams of substance mixed with 1L of water to make a 100mM solution or as your collegue said 4grams in a 50ml of water will work as well
you have come here because society has no further use for you

Canan

blcr11
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Joined: Fri Mar 30, 2007 4:23 am
Just don't forget the implied units. Writing 100 mM as 0.1 means 0.1 mole/L. So V has to be in L (50 ml = 0.05 L) or you won't get the correct answer--those pesky orders of magnitude again, but for different reasons. For the anhydrous salt it works out to 0.4 g rounded off to the nearest tenth of a gram.

blcr11
Viper
Posts: 672
Joined: Fri Mar 30, 2007 4:23 am
Hmm. I get 79 g/mole x 0.1 mole/L x 0.05 L = 0.395 g or 0.4 g of anyhdrous ammonium bicarbonate--unless I'm doing something I ortent.
Last edited by blcr11 on Thu Dec 06, 2007 7:13 pm, edited 1 time in total.

canalon
Inland Taipan
Posts: 3909
Joined: Thu Feb 03, 2005 2:46 pm
blcr11 wrote:Just don't forget the implied units. Writing 100 mM as 0.1 means 0.1 mole/L. So V has to be in L (50 ml = 0.05 L) or you won't get the correct answer--those pesky orders of magnitude again, but for different reasons. For the anhydrous salt it works out to 0.4 g rounded off to the nearest tenth of a gram.

I said to put the volume in liters.
And I really suggest the original poster to have a look back at those Chemistry 101 lessons on the dilutions etc. Because that is something that will be useful all the time in biology, and you cannot always rely on other people for such basic calculations.
Patrick

Science has proof without any certainty. Creationists have certainty without
any proof. (Ashley Montague)

Melissah
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Posts: 5
Joined: Mon Dec 03, 2007 7:37 pm
Thanks for the all help, looks like he meant 0.4 grams as opposed to 4 g.

I appreciate your concern, Patrick... however, I'm not overly concerned about it, because this is not something I generally need to do, and really, I just couldn't remember the formula for figuring out the calculation.

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