- December 6, 2007 at 6:28 pm #8747MelissahParticipant
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…
- December 6, 2007 at 6:41 pm #79080canalonParticipant
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.
- December 6, 2007 at 6:55 pm #79083canan5000Participant
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
- December 6, 2007 at 7:07 pm #79086blcr11Participant
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.
- December 6, 2007 at 7:12 pm #79087blcr11Participant
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.
- December 6, 2007 at 7:12 pm #79088canalonParticipantquote blcr11:
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.
- December 6, 2007 at 7:41 pm #79089MelissahParticipant
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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