# What are the empirical and molecular formulas of benzene?

Jul 4, 2018

$\underbrace{\text{Benzene"_"Empirical formula}} \equiv C H$

#### Explanation:

$\underbrace{\text{Benzene"_"Molecular formula}} \equiv {C}_{6} {H}_{6}$

The $\text{empirical formula}$ is the simplest whole number ratio defining constituent atoms in a species...and thus for benzene, whose $\text{molecular formula}$ is ${C}_{6} {H}_{6}$...the $\text{empirical formula}$ is simply $C H$...

Typically, we interrogate the empirical formula by experimental means (and that is what $\text{empirical}$ means, $\text{by experiment}$), usually combustion analysis, to give the percentage by mass of, carbon, hydrogen, and nitrogen, and by a Mohr titration to give halide content. And then some means is devised to measure the MOLECULAR MASS of the species...for benzene of course this is $78 \cdot g \cdot m o {l}^{-} 1$...and thus with the empirical formula...we calculate that ${\left(C H\right)}_{n} = 78 \cdot g \cdot m o {l}^{-} 1$

${\left(12.011 \cdot g \cdot m o {l}^{-} 1 + 1.00794 \cdot g \cdot m o {l}^{-} 1\right)}_{n} = 78 \cdot g \cdot m o {l}^{-} 1$

And thus $n = 6$...and the molecular formula is ${C}_{6} {H}_{6}$.

While this is fairly trivial in this instance, for more complex organic and inorganic molecules, some means must be employed to measure the molecular/formula mass of the molecule... And this is not always straightforward..