In combustion analysis at room temperature and pressure, the volumetric ratio of gases equals their molar ratio; by dividing all gas volumes by the smallest volume, we can determine the moles of each product, then calculate the carbon and hydrogen atoms in the original hydrocarbon by accounting for oxygen atoms in CO2 and H2O, ultimately deducing the hydrocarbon formula.
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Combustion Analysis A Level Question WalkThrough | OCR A ChemistryHinzugefügt:
Let's solve this A level combustion analysis question. [music] As the reaction is taking place at room temperature and pressure, we can assume the volumetric ratio is equivalent to the molar [music] ratio. So, we can divide all the volumes of gases provided to us in the question by the volume of gas which is the smallest [music] to find that ratio. That tells us we have six moles of oxygen gas molecules in this combustion. In other words, 12 moles of oxygen atoms. Eight moles of those oxygen atoms are accounted for by the [music] four carbon dioxide molecules. That means the other four moles of oxygen atoms must be found in the water molecules. Therefore, we must have four moles of water molecules. Now, all eight moles of hydrogen atoms found in those water molecules must have come from the combustion of the hydrocarbon.
So, it must have the formula [music] H8.
And all four carbon atoms found in the formula of [music] the four moles of carbon dioxide molecules must also have come from the combustion of that hydrocarbon. Bringing us the formula of the hydrocarbon must have also contained C4. [music] So, therefore, taking those two facts together, we can deduce the formula of the hydrocarbon being burnt was C4H8.
[music]
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