Combustion analysis of a 31.472 mg sample of the widely used flame retardant Decabrom gave 1.444 mg of #CO_2#. Is the molecular formula of Decabrom #C_12Br_10O# or #C_12H_10O#?

1 Answer
Sep 29, 2015

There are insufficient data to give a molecular formula. The presence of bromine is likely, but at the moment we just don't know.


Bromine is a common substituent in flame retardants. While we can infer the presence of bromine, the balance might as well be oxygen, whose content is generally not reported in microanalysis. The presence of bromine and other halogens can be quantitatively determined by a Mohr titration.

Given that the question mentioned decabrom, the unknown sample is possibly decabromodiphenyl ether, i.e. #(C_6Br_5)_2O#, a common flame retardant. Such a formula may give results consistent with these data. Before we settle on a formula, we need more quantitative data, i.e. halogen content. If you haven't got these, you can't settle on a formula.