Combustion analysis of a 31.472 $\mathrm{mg}$ sample of the widely used flame retardant Decabrom gave 1.444 $\mathrm{mg}$ of $\mathrm{CO}_{2}$ . Is the molecular formula of Decabrom $\mathrm{C}_{12} \mathrm{Br}_{10}$ or $\mathrm{C}_{12} \mathrm{Br}_{10} \mathrm{O}$ ?