Empirical Formula From Combustion Analysis

Combustion analysis determines the empirical formula of organic compounds by measuring the masses of (\text{CO}_2) and (\text{H}_2\text{O}) produced upon complete combustion. All carbon in the compound converts to (\text{CO}_2), and all hydrogen converts to (\text{H}_2\text{O}). Step 1 — mass of C: moles of (\text{CO}_2 = 0.66/44 = 0.015) mol, so moles of C (= 0.015) mol, mass of C (= 0.015 \times 12 = 0.18) g. Step 2 — mass of H: moles of (\text{H}_2\text{O} = 0.27/18 = 0.015) mol, so moles of H (= 0.030) mol, mass of H (= 0.030 \times 1 = 0.030) g. Step 3 — mass of O by difference: mass of O (= 0.45 - 0.18 - 0.030 = 0.24) g, moles of O (= 0.24/16 = 0.015) mol. Step 4 — mole ratio C:H:O (= 0.015:0.030:0.015 = 1:2:1). Empirical formula: (\text{CH}_2\text{O}). Option (\text{C}_2\text{H}_4\text{O}) has the ratio 2:4:1, which cannot be simplified to 1:2:1; it is not the simplest ratio. Option (\text{CH}_3\text{O}) would require a C:H:O ratio of 1:3:1, inconsistent with the calculated 1:2:1. Option (\text{C}_2\text{H}_6\text{O}) implies 2:6:1 ratio and is inconsistent with any combination of the given data. This is a JEE Advanced-level combustion analysis problem following the NCERT approach of determining elemental composition by back-calculation from combustion products. Plausibility check: masses of C (0.18 g) + H (0.03 g) + O (0.24 g) (= 0.45) g, exactly matching the original sample mass.