I've been stumped by a question in my textbook concerning combustion analysis.

A $\pu{0.5438g}$ sample of a liquid consisting of only $\ce{C}$, $\ce{H}$, and $\ce{O}$ was burned in pure oxygen, and $\pu{1.039g}$ of $\ce{CO2}$ and $\ce{0.6369g}$ of $\ce{H2O}$ were obtained. What is the empirical formula of the compound?

Firstly, I calculated the amount of substance of carbon and hydrogen by dividing the masses of each product by their respective molar masses. \begin{align} (1.0390/44.01) = n(\ce{C}) &= 0.02360\\ 2\cdot(0.6369/18.02) = n(\ce{H}) &= 0.0706\\ \end{align}

My textbook then states that I must find the sum of the masses of carbon and hydrogen, and then deduct this sum from the mass of the liquid compound to get the mass of oxygen: \begin{align} 0.02360 \cdot 12.01 &= 0.283535\\ 0.0706 \cdot 1.01 &= 0.28637\\ 0.5438 - (0.283535 + 0.28637) &= m(\ce{O}) \\ m(\ce{O}) &= \pu{0.1888g}\\ \end{align}

I have no qualms with this approach, but why can't I find the amount of oxygen directly from the products? \begin{align} \text{I.e.}&& 2\cdot(1.0390/44.01) &= n(\ce{O}) &\text{in }\ce{CO2}\\ \text{and} && (0.6369/18.02) &= n(\ce{O}) &\text{in }\ce{H_2O}.\\ \end{align}

$$\text{Amount of oxygen in liquid compound} = n(\ce{O})\text{ in }\ce{CO2} + n(\ce{O})\text{ in }\ce{H2O}.$$

I know this doesn't work, but my question is why? It seems perfectly intuitive to me, and yet it doesn't work.


3 Answers 3


Let's first write the chemical equation for the reaction (I will leave it unbalanced because the stoichiometry is irrelevant): $$\ce{C_xH_yO_z + O2 -> CO2 + H2O}$$ To do this question, we need to find the mole ratio of carbon, hydrogen and oxygen in $\ce{C_xH_yO_z}$, then we can find its empirical formula.

From the above equation, it is clear that all the carbon in $\ce{CO2}$ originated from the compound and that all the hydrogen $\ce{H2O}$ originated from the compound. However, all the oxygen in $\ce{CO2}$ and $\ce{H2O}$ didn't just come from the compound, it came from the compound and gaseous oxygen. Therefore, from your method, you are just finding the total amount of oxygen in the reactants, not the number of moles of oxygen in the compound.

  • $\begingroup$ While it is certainly unnecessary for the argument, I believe a balanced reaction equation would make it a lot stronger. $\endgroup$ Apr 4, 2017 at 11:11

Both (at the time of writing) other answers are absolutely correct, but I'd like to approach it from a slightly different angle. I think it is one of the most important traits of a chemist to analyse the initial situation correctly. Read the question very carefully again. Below (in the spoiler) I highlighted the part which is an important information that you might have overlooked.

A $\pu{0.5438g}$ sample of a liquid consisting of only $\ce{C}$, $\ce{H}$, and $\ce{O}$ was burned in pure oxygen, and $\pu{1.039g}$ of $\ce{CO2}$ and $\ce{0.6369g}$ of $\ce{H2O}$ were obtained. What is the empirical formula of the compound?

In most of these texts you might alternatively read
[...] was burnt in excess oxygen, [...]

I will always suggest writing the balanced equation first, as usually the chemistry becomes more obvious converting a text into a mathematical construct:

$$\ce{C_xH_yO_z + 1/2(y/2 + 2$x$ - $z$) O2 -> x CO2 + y/2 H2O}$$

from there you can see that as long as $z \neq \frac{y}2 + 2x$ you will need more oxygen for a complete combustion. It will later also make it easier to find the actual empirical formula.

One final advice: Always write down the units when calculating values. It'll help you find mistakes easier. For example, it should have rather been: \begin{align} \frac{\pu{1.039g}}{\pu{44.01g/mol}} = n(\ce{C}) &= \pu{0.0236mol}\\ 2\cdot\frac{\pu{0.6369g}}{\pu{18.02g/mol}} = n(\ce{H}) &= \pu{0.0706mol}\\ \end{align}


Oxygen is found in both reactants, and the amount of O2 used for combustion (pure oxygen) is unknown and usually just given as 'excess' in order to have a complete combustion, i.e. the only products are H2O and CO2.

You are interested only in the oxygen that was originally part of the molecule, but you cannot tell which oxygen atom in each molecule of CO2 or H2O comes from the molecule being tested, and which comes from the pure oxygen you used for combustion. Therefore, you have to calculate it indirectly.

(You COULD have determined the molecule's oxygen content directly if the oxygen atoms were marked in some way either in the molecule or the pure oxygen gas. As a theoretic example, you could use a certain isotope of oxygen which can be detected easily in some way.)


General Chemistry: Principles and Modern Applications; Petrucci, Herring, Madura, Bissonnette; Pearson, 2011; ISBN13: 9780136121497 - pp. 82-83 (for owners of older editions, it should be in chapter 3, combustion analysis section)


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