The ratio of the mass percentages of ‘C & H’ and ‘C & O’ of a saturated acyclic organic compound ‘X’ are 4 : 1 and 3 : 4 respectively. Then, the moles of oxygen gas required for complete combustion of two moles of organic compound ‘X’ is ________.

Solution

Chemistry ma’am · 4 steps

Take 12 g of carbon, read off H and O from the ratios, then make the empirical formula a real molecule.

1. Fix the masses. Ratios of mass percentages are the same as ratios of masses. Take 12 g of C. Then C : H = 4 : 1 gives 3 g of H, and C : O = 3 : 4 gives 16 g of O.
2. Empirical formula. In moles this is 12/12 = 1 C, 3/1 = 3 H and 16/16 = 1 O. So the empirical formula is CH₃O.
3. Molecular formula. CH₃O has an odd number of H atoms, so it cannot be a stable molecule. Doubling it gives C₂H₆O₂ (ethylene glycol). It fits CₙH₂ₙ₊₂O_z with n = 2, so it is saturated and acyclic.

   C₂H₆O₂ + 5/2O₂ → 2CO₂ + 3H₂O

4. So. One mole of X needs 5/2 mol of O₂, so two moles need 2 × 5/2 = 5 mol of O₂.

The common slip: 2.5. Balancing combustion for the empirical formula CH₃O instead of C₂H₆O₂ gives 2.5. Working out the oxygen for one mole of X and forgetting to double it gives 2.5 as well.

The idea to keep. Turn mass ratios into moles by fixing one element's mass. Then check the empirical formula: a molecule made of only C, H and O must have an even number of H atoms.

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