Posted by Jimmy on Sunday, September 27, 2009 at 4:52am.
I think you are correct. To have the identical mass percentage of C the "other" compound must have a multiple of C with the same multiple of H and none of the answers qualify.
A trick to solving this type of problem is to reduce all of the ratios to either their simplest form or a percentage.
Ex. 3C/6H => 2C/4H => 1C/2H = 50% C
1/2 is the simplest form, and the molecule is 50% Carbon.
With this in mind, you should look over your options once more.
all of above is wrong.
i have the same question for school.
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