# 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?

Mar 27, 2017

This question operates under an unspoken assumption. We get (eventually!) a molecular formula of ${C}_{9} {H}_{8} {O}_{4}$.

#### 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 $C {O}_{2}$, 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 $20 \cdot g$ compound, there are $\frac{44 \cdot g}{44 \cdot g \cdot m o {l}^{-} 1} C = 1 \cdot m o l \cdot C$.

In $20 \cdot g$ compound, there are $\frac{7.99 \cdot g}{18.01 \cdot g \cdot m o {l}^{-} 1} \times 2 H = 0.89 \cdot m o l \cdot H$.

In $20 \cdot g$ compound, there are presumed to be the following molar quantity of oxygen,

$\frac{20.0 \cdot g - 12.011 \cdot g - 0.89 \cdot g}{16 \cdot g \cdot m o {l}^{-} 1} O = 0.44 \cdot m o l \cdot O$. (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 ${C}_{2.27} {H}_{2} O$, which we must multiply by 4  to get whole numbers, i.e. ${C}_{9} {H}_{8} {O}_{4}$, which is the $\text{empirical formula}$.

But we know that the $\text{molecular formula}$ is a simple whole number multiple of the $\text{empirical formula}$

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

$n \times \left(9 \times 12.011 + 8 \times 1.00794 + 4 \times 15.999\right) \cdot g \cdot m o {l}^{-} 1$

$= 180 \cdot g \cdot m o {l}^{-} 1$

Clearly, $n = 1$, so here $\text{molecular formula}$ $=$ $\text{empirical formula}$ $=$ ${C}_{9} {H}_{8} {O}_{4}$

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.