Comments on: Molecules of Caffeine http://www.rough-equivalents.com/2008/05/molecules-of-caffeine/ Making Numbers Fun Fri, 10 Oct 2008 21:18:34 +0000 http://wordpress.org/?v=2.5.1 By: Rough Equivalents » Blog Archive » Parts Per Billion http://www.rough-equivalents.com/2008/05/molecules-of-caffeine/#comment-72 Rough Equivalents » Blog Archive » Parts Per Billion Mon, 19 May 2008 07:33:35 +0000 http://www.rough-equivalents.com/?p=43#comment-72 [...] the "Molecules of Caffeine" post, I calculated the number of caffeine molecules in a can of Mountain Dew. This lead reader [...] [...] the "Molecules of Caffeine" post, I calculated the number of caffeine molecules in a can of Mountain Dew. This lead reader [...]

]]> By: Greg Bulmash http://www.rough-equivalents.com/2008/05/molecules-of-caffeine/#comment-56 Greg Bulmash Sat, 10 May 2008 07:07:44 +0000 http://www.rough-equivalents.com/?p=43#comment-56 @Peter, Thanks for the heads up. I've made corrections above. @Peter,

Thanks for the heads up. I've made corrections above.

]]> By: Peter Chastain http://www.rough-equivalents.com/2008/05/molecules-of-caffeine/#comment-55 Peter Chastain Sat, 10 May 2008 06:01:17 +0000 http://www.rough-equivalents.com/?p=43#comment-55 A caffeine molecule actually has 8 carbon atoms, not 3. Your structural diagram is correct, but the unlabeled atoms (e.g., the node at the far left of the "pentagon," between the two nitrogen atoms) of all such diagrams are assumed to be be carbon. Also, individual carbon-hydrogen bonds are not usually shown, so any node with fewer than four bonds (e.g., the same carbon atom as in the previous example) is assumed to have enough hydrogen atoms to make up the difference. The empirical formula for caffeine is actually C8 H10 N4 O2. (Sorry, I cannot figure out how to enter subscripts in your blog.) So, unless I have made an arithmetic mistake, caffeine weighs 194.1918 g/mol. I leave the rest of the calculation as an exercise for the student. A caffeine molecule actually has 8 carbon atoms, not 3. Your structural diagram is correct, but the unlabeled atoms (e.g., the node at the far left of the "pentagon," between the two nitrogen atoms) of all such diagrams are assumed to be be carbon. Also, individual carbon-hydrogen bonds are not usually shown, so any node with fewer than four bonds (e.g., the same carbon atom as in the previous example) is assumed to have enough hydrogen atoms to make up the difference.

The empirical formula for caffeine is actually C8 H10 N4 O2. (Sorry, I cannot figure out how to enter subscripts in your blog.) So, unless I have made an arithmetic mistake, caffeine weighs 194.1918 g/mol. I leave the rest of the calculation as an exercise for the student.

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