|MadSci Network: Chemistry|
Hello Wim. This is a question I had several years ago. There are published techniques in the literature (one such technique resides in an older edition of Skoog and West's Analytical Chemistry text...maybe 2nd edition?). The most available (and probably simplest) method for performing this experiment will be with a GC. If you make dilute solutions of beer (I think that there are too many dissolved solids in undiluted beer, which will coat the GC column and may cause future problems)that has been left out for several hours to allow carbonation to disperse (I recall that dissolved carbonation is an interference), and then run against dilute ethanol solutions (to generate a calibration curve), you should be able to determine what the alcohol % is in the original beer. As a side note: % alcohol is typically expressed as % weight/weight (called percent by weight) or % volume/volume (called percent by volume). the % by weight is the weight of alcohol divided by the weight of the whole solution. The percent by volume is the volume of alcohol divided by the total volume. If the concentration is X% volume/volume, then the concentration % weight/weight is about 0.8 * % volume/volume since the density of alcohol is about 0.8 g/mL and that of water is about 1 g/mL, assuming dilute solution so that the solution is effectively all water. As such, % by volume will always be larger. I hope this helps. If you are looking for a procedure, try a few analytical chemistry texts. Please feel free to email me if you have further questions. Best Regards, Mike
Try the links in the MadSci Library for more information on Chemistry.