# 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?

Given that the question mentioned decabrom, the unknown sample is possibly decabromodiphenyl ether, i.e. ${\left({C}_{6} B {r}_{5}\right)}_{2} O$, 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.