Question

The percentage composition of acetic acid is found to be 39.9% carbon, 6.7% hydrogen, and 53.4% oxygen, how would you determine the empirical formula of acetic acid?

Answer 1

We assume `100* g` of compound. Therefore the acid contains `39.9* g* C`; `6.7* g* H`; and `53.4*g*O`.

Explanation:

We divide the percentage composition by the atomic mass of each constituent element:

`(39.9*g)/(12.011*g*mol^-1)= 3.32*mol`

`(6.7*g)/(1.0794*g*mol^-1)= 6.65*mol`

`(53.4*g)/(16.00 *g*mol^-1)= 3.34*mol`

We divide thru by the lowest quotient `(3.32)` to get `C:H:O-=1:2:1`.

So the empirical formula is `CH_2O`. And the empirical formula is the simplest whole number ratio that defines constituent atoms in a species.

Now it is a fact that the molecular formula of acetic acid is `H_3C-C(=O)OH = C_2H_4O_2`. Is this a mulitple of the empirical formula; i.e. what is `n` for `(CH_2O)_n = C_2H_4O_2`?