nannah - 12-2-2014 at 11:01
Hey, today on the bus home i came to wonder if it is in any way possible to scale down reactions?
Is it just so simple that you just measure up only half the amount of reagents that it says. So if it say 1 mole, you take ½ mole?
Thanks in advance. 
Zyklon-A - 12-2-2014 at 11:08
Ummm.... yeah, no shit. Of course you can.
Ever heard of stoichiometry?
[Edited on 12-2-2014 by Zyklonb]
woelen - 12-2-2014 at 11:13
For many reactions it indeed is a matter of multiplying all amounts with a certain factor, but for highly exothermic and for highly endothermic
reactions the outcome of smaller (or larger) amounts may be different. Some reactions keep running smoothly to (near) completion when performed
according to the published procedure, while they do not work well when performed on a much smaller scale, because the amount of heat retained in the
mix is lower and additional heating is required. The same reaction might require cooling when scaled up to larger amounts.
So, check whether the reaction is strongly exothermic or endothermic. Changing amounts may require additional heating or cooling in such cases.
Praxichys - 12-2-2014 at 11:51
What woelen says is true. Also consider transfer losses along the way.
For instance, if a filter paper retains 150mg of product after being scraped, and you are making 10g, that is a loss of 1.5% yield to product stuck in
the filter. If you use the same filter paper for a scaled down reaction making 1g of product, that is a loss of 15% yield.
Consider losses of reagents sticking to glass and stirring rods, or filtrates being absorbed into filter paper and evaporating. Also consider solvent
evaporative losses and the surface area to volume ratio of solutions in large containers.
However, in general, going form a 1 molar scale to half molar scale should not usually be a problem since both are fairly large scale. Also, as woelen
says, if a reaction is strongly endothermic or exothermic or must be heated or cooled rapidly, scaling up can be a problem.
nannah - 12-2-2014 at 12:33
Ok, thank you guys. I am new to this, and i really appreciate the help.