10 mL of gaseous hydrocarbon on combustion gives 40 mL of CO₂( g) and 50 mL of water vapour. Total number of carbon and hydrogen atoms in the hydrocarbon is _________ .

Solution

Chemistry ma’am · 3 steps

For gases, volume ratio is mole ratio, so read the formula straight off the product volumes.

1. Volumes as moles. At the same temperature and pressure, equal volumes of gases hold equal numbers of molecules (Avogadro's law). So 10 mL hydrocarbon, 40 mL CO₂ and 50 mL water vapour are in the mole ratio 1 : 4 : 5.

   CₓH_y + (x + y/4) O₂ → x CO₂ + y/2 H₂O

2. Find x and y. One molecule of hydrocarbon gives 4 molecules of CO₂, so x = 4. It also gives 5 molecules of H₂O, so y/2 = 5 and y = 10.
3. So. The hydrocarbon is C₄H₁₀ (butane), with 4 + 10 = 14 carbon and hydrogen atoms in total.

The common slip: 9. Taking the water ratio of 5 as the hydrogen count gives C₄H₅ and a total of 9. Each H₂O has two H atoms, so y = 10.

The idea to keep. In gas combustion problems, divide each product volume by the fuel volume. CO₂ gives the carbon count directly, but each H₂O carries two hydrogens, so double it.

Answer from the NTA final answer key. Solution by PrepWiser.