# How do you add chemical formulas?

Sep 16, 2017

You add them by conserving mass and charge.......

#### Explanation:

And the method of half equations employed in redox chemistry is an excellent example of this process.

For the oxidation of ethanol to acetic acid, we could propose the given oxidation reaction as:

${H}_{3} \stackrel{- I I I}{C} - \stackrel{- I}{C} {H}_{2} O H + {H}_{2} O \rightarrow {H}_{3} \stackrel{- I I I}{C} - \stackrel{+ I I I}{C} {O}_{2} H + 4 {H}^{+} + 4 {e}^{-} \text{ (i)}$

And something, here $M n {O}_{4}^{-}$, is reduced down to $M {n}^{+ 2}$:

$\stackrel{V I I +}{\text{Mn"rarrstackrel(+II)"Mn}}$

$\text{MnO"_4^(-) + "8H"^(+)+ "5"e^(-) rarr"Mn"^(2+)+"4H"_2"O} \left(l\right)$ $\text{(ii)}$

Deep-red purple permanganate ion is reduced down to almost colourless $M {n}^{2 +}$. to give the final redox equation, we cross multiply $\left(i\right)$ and $\left(i i\right)$ so that electrons, conceptual particles that we use for convenience, DO NOT appear in the final equation:

And thus $5 \times \left(i\right) + 4 \times \left(i i\right)$ gives...............

$4 M n {O}_{4}^{-} + 12 {H}^{+} + 5 {H}_{3} C - C {H}_{2} O H \rightarrow 5 {H}_{3} C - C {O}_{2} H + 4 M {n}^{2 +} + 11 {H}_{2} O \left(l\right)$

Which, if I have done my sums right, is balanced with respect to mass and charge; as indeed it must be if we purport to represent chemical reality.