Question

Combustion of a `20*g` mass of compound gives `44*g` `CO_2`, and `7.99*g` of water. Given that the molecular mass is `180*g*mol^-1` what are the empirical and molecular formulae?

Answer 1

This question operates under an unspoken assumption. We get (eventually!) a molecular formula of `C_9H_8O_4`.

Explanation:

We can burn organic compounds in a furnace to produce (i) carbon dioxide; and (ii) water. The carbon and hydrogen content of these products COME directly from the hydrocarbon. Sometimes these are expressed as a percentage by mass and SOMETIMES, as here, the measured percentages do not add up to 100%. The missing percentage is presumed to be oxygen, which is NOT measured, but obtained by difference, i.e. `O%=100%-%C-%H`.

Here, the carbon content is measured by `CO_2`, and the hydrogen content is measured by WATER (I think you made a mistake when you posed the question!).

So let's see if we can get something useful.

In `20*g` compound, there are `(44*g)/(44*g*mol^-1)C=1*mol*C`.

In `20*g` compound, there are `(7.99*g)/(18.01*g*mol^-1)xx2H=0.89*mol*H`.

In `20*g` compound, there are presumed to be the following molar quantity of oxygen,

`(20.0*g-12.011*g-0.89*g)/(16*g*mol^-1)O=0.44*mol*O`. (That is the mass balance was due to oxygen!).

We divide thru by the SMALLEST molar quantity, that of oxygen to get a trial formula of `C_(2.27)H_2O`, which we must multiply by #4 # to get whole numbers, i.e. `C_9H_8O_4`, which is the `"empirical formula"`.

But we know that the `"molecular formula"` is a simple whole number multiple of the `"empirical formula"`

And thus............................,

`nxx(9xx12.011+8xx1.00794+4xx15.999)*g*mol^-1`

`=180*g*mol^-1`

Clearly, `n=1`, so here `"molecular formula"` `=` `"empirical formula"` `=` `C_9H_8O_4`

I am sorry for making such a meal of this answer, but if you are an undergrad, you should be aware of the background to this question. If you are at A level, your teacher has no business asking you these sorts of questions.