Burning hydrocarbons: the gas-volume detective

Squirt a mystery hydrocarbon into a tube of oxygen. Spark it. Bang. The hydrocarbon is gone, but it left evidence: how much oxygen vanished and how much carbon dioxide appeared.

From just those volumes you can name the suspect. No mass spectrometer needed, just moles in disguise.

Complete combustion of any hydrocarbon $\text{C}_x\text{H}_y$:

$\text{C}_x\text{H}_y + \left(x + \frac{y}{4}\right)\text{O}_2 \rightarrow x\,\text{CO}_2 + \frac{y}{2}\,\text{H}_2\text{O}$

• Each C atom becomes one $\text{CO}_2$, needing one $\text{O}_2$: that's the $x$.
• $y$ H atoms make $\frac{y}{2}$ water molecules. Those contain $\frac{y}{2}$ O atoms, which come from $\frac{y}{4}$ molecules of $\text{O}_2$.

Gases at the same temperature and pressure react in the ratio of their coefficients, so for propane ($x = 3$, $y = 8$): 10 cm³ of $\text{C}_3\text{H}_8$ needs 50 cm³ of $\text{O}_2$ and makes 30 cm³ of $\text{CO}_2$.