# Given combustion data, how would we find the empirical formula of C_7H_5N_3O_6?

Aug 26, 2017

${C}_{7} {H}_{6} {N}_{3}$........is what we would say on the given data, and we would be DEAD WRONG.

#### Explanation: The $\text{empirical formula}$ is the simplest whole number ratio defining constituent atoms in a species, and I wish I had a fiver for every time I've written or said that......

And you were given the elemental composition that IGNORED the oxygen atoms, and got the number of hydrogen atoms wrong.

We gots ${C}_{6} {H}_{2} \left(C {H}_{3}\right) {\left(N {O}_{2}\right)}_{3} \equiv {C}_{7} {H}_{5} {N}_{3} {O}_{6}$, and here the empirical formula corresponds to the molecular formula.

TNT is used as an explosive because of its ease of manufacture, and its resistance to detonation under standard conditions. Of course, when properly primed, and detonated, it is a highly potent explosive.

Note that when microanalysis is performed upon an organic sample, we measure %C, %H, %N, by combustion analysis. The missing percentage (if the measured percentages do not sum to 100%) is ASSUMED to be that of oxygen. Oxygen is a difficult element to measure microanalytically. And often in combustion, we add EXTRA oxidant, for instance, $S {b}_{2} {O}_{5}$, to get the sample to combust completely.