Question

What are the empirical and molecular formulae of an organic species, whose percentage composition is: `C:38.7%;H:9.7%;O,51.6%`? A `0.93*g` mass of this material occupies `0.336*L` at `"STP"`.

Answer 1

We get, eventually, `"a molecular formula of"` `C_2H_6O_2`...........

Explanation:

As with all these problems, we assume a mass of `100*g`, and from this mass we calculate the molar quantity of each element.

`"Moles of carbon"=(38.7*g)/(12.011*g*mol^-1)=3.22*mol`.

`"Moles of hydrogen"=(9.7*g)/(1.00794*g*mol^-1)=9.62*mol`.

`"Moles of oxygen"=(51.6*g)/(15.999*g*mol^-1)=3.22*mol`.

(Note that an analyst normally does not give you `%O`, you are given `%C,H,N`, and if these percentages do not sum to `100%`, the balance is ASSUMED to be `%O`.)

And now we divide thru by the smallest atomic quantity to get an empirical formula of `CH_3O`.

And now we use the gas data to estimate a molecular mass. The molar volume an Ideal Gas at `STP` is `22.4*L`.

And thus the molar mass of our vapour is `(0.93*g)/((0.336*L)/(22.4*L*mol^-1))=62.0*g*mol^-1`.

Now it is FACT, the molecular formula is ALWAYS a whole number multiple of the empirical formula, i.e.

`"Molecular formula"=nxx"Empirical formula"`

And thus................................

`nxx(12.011+3xx1.00794+16.00)*g*mol^-1=62.0*g*mol^-1`

We solves for `n`, to get `n=2`, and thus `"molecular formula"` `=` `C_2H_6O_2`.

`"Phew!"`