Atom Economy MCQs PDF

1. A compound contains 40% carbon, 6.7% hydrogen, and 53.3% oxygen. What is its empirical formula? a) CHO b) CH₂O c) CH₄O d) CH₂O₂ Answer: b) CH₂O Explanation: Convert each percentage to moles by dividing by the atomic masses: C: 40/12 = 3.33, H: 6.7/1 = 6.7, O: 53.3/16 = 3.33. Th...

1. A compound contains 40% carbon, 6.7% hydrogen, and 53.3% oxygen. What is its empirical formula? a) CHO b) CH₂O c) CH₄O d) CH₂O₂ Answer: b) CH₂O Explanation: Convert each percentage to moles by dividing by the atomic masses: C: 40/12 = 3.33, H: 6.7/1 = 6.7, O: 53.3/16 = 3.33. Then, simplify the ratio: 3.33/3.33 : 6.7/3.33 : 3.33/3.33 = 1 : 2 : 1, so the empirical formula is CH₂O. 2. Given the molecular formula C₆H₁₂O₆, what is the empirical formula? a) C₃H₆O₃ b) C₆H₁₂O₆ c) CH₃O d) CH₂O Answer: d) CH₂O Explanation: The empirical formula is the simplest whole number ratio of atoms, which in this case is CH₂O. 3. A hydrocarbon contains 85.7% carbon and 14.3% hydrogen. What is the empirical formula? a) CH₂ b) CH₃ c) CH₄ d) CH Answer: b) CH₃ Explanation: Convert each percentage to moles: C: 85.7/12 = 7.14, H: 14.3/1 = 14.3. Simplify the ratio: 7.14/7.14 : 14.3/7.14 = 1 : 2, so the empirical formula is CH₃. 4. Calculate the atom economy for the following reaction: N₂ + 3H₂ → 2NH₃. (Molar Masses: N₂ = 28 g/mol, H₂ = 2 g/mol, NH₃ = 17 g/mol) a) 24.5% b) 50% c) 85% d) 100% Answer: c) 85% Explanation: Desired product is NH₃ (2 × 17 = 34 g). Total mass of reactants = 28 + 3 × 2 = 34 g. Atom economy = (34/34) × 100 = 100%. 6. Find the atom economy of a reaction producing water as the desired product: 2H₂ + O₂ → 2H₂O. a) 50% b) 67% c) 89% d) 100% Answer: d) 100% Explanation: All the reactants convert to the desired product, water. Therefore, the atom economy is 100%. 7. In the reaction: 2HgO → 2Hg + O₂, calculate the atom economy for the production of mercury. (Hg = 200, O = 16) a) 75.2% b) 92.6%  c) 84.5% d) 94.2% Answer: b) 92.6% Explanation: Desired product is Hg: 2 × 200 = 400 g. Total mass of reactants = (2 × 216) = 432 g. Atom economy = (\(\frac{400}{432}\) × 100)% = 92.6%. 8. A compound with a molecular weight of 84 g/mol contains 71.4% carbon, 14.3% hydrogen, and 14.3% oxygen. What is its molecular formula? a) C₂H₄O b) C₆H₁₂O₆ c) C₆H₁₂ d) C₅H₁₀O Answer: d) C₅H₁₀O Explanation: First, nd the empirical formula: 71.4/12 ≈ 5.95, 14.3/1= 14.3, 14.3/16 ≈ 0.89. Simplify ratio: ~1: 2.4: 0.15. Approximate to nearest integer: C₅H₁₀O. Next, con rm molar mass alignment with provided molar mass, con rm as correct molecular formula. 9. Find the mass of desired product in a reaction with 80% atom economy which used 50g of reactants. a) 50g b) 40g c) 80g d) 30g Answer: b) 40g Explanation: Mass of desired product = Atom economy percentage applied to total reactants: 80% × 50g = 0.80 × 50g = 40g. 10. Calculate the nal mass of non-desired product in a reaction with 75% atom economy producing 55g of desired product. a) 18.3g b) 12.6g c) 8.2g d) 5.6g Answer: c) 18.3g Explanation: At 75% atom economy, 55g desired product corresponds to 75% total used mass; thus total used mass is 55/0.75= 73.3g; non-desired mass = 73.3 - 55 = 18.3g. 11. If the actual yield is 75g and the theoretical yield is 100g, what is the percentage yield? a) 75% b) 80% c) 85% d) 90% Answer: a) 75% Explanation: Percentage yield = (\(\frac{\text{Actual Yield}}{\text{Theoretical Yield}}\) × 100) = (\ (\frac{75}{100}\) × 100)% = 75%. 12. A certain reaction delivers 1.2 moles of product. If the theoretical yield is 2 moles, what is the percentage yield? a) 40% b) 50% c) 60% d) 70% fi fi fi fi  Answer: c) 60% Explanation: Percentage yield = (\(\frac{\text{Actual Yield}}{\text{Theoretical Yield}}\) × 100)% = (\(\frac{1.2}{2}\)× 100)% = 60%. 13. The empirical formula of a compound is CH₂. The molecular mass is 28 g/mol. What is the molecular formula? a) C₂H₄ b) CH₂ c) C₄H₈ d) C₆H₁₂ Answer: a) C₂H₄ Explanation: The empirical formula mass of CH₂ is 14 g/mol. The molar mass is 28 g/mol. The ratio of the molar mass to the empirical formula mass is 2, thus the molecular formula is C₂H₄. 14. If a compound contains 40% sulfur and 60% oxygen by mass, nd its empirical formula. a) SO₂ b) S₂O₃ c) S₂O d) SO₃ Answer: d) SO₃ Explanation: Convert each percentage into moles: S: 40/32 = 1.25, O: 60/16 = 3.75. Normalize ratio: divide by 1.25 → S = 1, O= 3. Thus, SO₃. 15. An organic compound has an empirical formula of CH₂O and a molecular mass of 150g/mol. What is the molecular formula? a) CH₂O b) C₅H₁₀O₅ c) C₆H₁₂O₆ d) C₃H₃O₃ Answer: b) C₅H₁₀O₅ Explanation: Empirical mass of CH₂O is 30g/mol, shows molecular structure by 30× = 150; x = 5 indicates replication of empirical formula into molecular formula C₅H₁₀O₅. 16. What is the atom economy if only 40g out of 100g reactants are desired products? a) 60% b) 40% c) 20% d) 80% Answer: b) 40% Explanation: Atom economy = (\(\frac{\text{Mass of desired product}}{\text{Total mass of reactants}}\)× 100)% = (\(\frac{40}{100}\)× 100)% = 40%. 17. What is the empirical formula of a compound containing 30.5% nitrogen and 69.5% oxygen by mass? a) NO b) NO₂ c) N₂O d) N₂O₃ Answer: b) NO₂ Explanation: Convert percentage into moles: N: 30.5/14 ≈ 2.18, O: 69.5/16 ≈ 4.34. Simpli ed by smaller number: 2.18:4.34 ≈ 1:2. Thus, empirical formula is NO₂. 18. What is the empirical formula of a compound with 40% sulfur and 60% oxygen by mass? a) SO₃ fi fi  b) SO₂ c) S₂O₃ d) S₂O Answer: b) SO₃ Explanation: Ratio by mass to S:O = 40/32 : 60/16 = 1.25:3.75 = 1:3, empirical formula thus SO₃. 19. In a reaction producing H₂O, if 50g H₂ reacts fully with O₂ to create water, how much water will preferably be produced (H₂ = 2g/mol, O₂ = 32 g/mol)? a) 225g b) 450g c) 56g d) 25g Answer: b) 225g Explanation: First establish moles: H₂ position: 50g H₂ 1 mol/2g = 25 mol; utilizing 2 H₂ + O₂ = 2 H₂O means 25g H₂ translates to H₂O yielding 1:1 ratio forming 225g as e ectively mass desired. 20. Calculate number of Oxygen moles in decomposing H₂: Besides 3O₂ into 2H₂O if 72g involved in reaction equiequivalent volume consistent (O₂ = 32g/mol). a) 6 moles b) 8 moles c) 4.5 moles d) 2 moles Answer: a) 6 moles Explanation: O₂ equally arrives H₂ in reaction formation: molar equivalence O₂ operational in forming structure for H₂ at empirical equivalent for creating e ciency. ffi ff