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7.) 0.446 grams of an unknown alcohol (composed of C, O, and H only) was combusted (using combustion analysis), producing 0.982 g CO2 and 0.534 g H20. What is the empirical formula of this alcohol?

7.) 0.446 grams of an unknown alcohol (composed of C, O, and H only) was combusted (using combustion analysis), producing 0.982 g CO2 and 0.534 g H20. What is the empirical formula of this alcohol?

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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### Chemistry Empirical Formula Exercises #### Exercise 7 **Problem Statement:** 0.446 grams of an unknown alcohol (composed of Carbon (C), Oxygen (O), and Hydrogen (H) only) was combusted using combustion analysis, producing 0.982 grams of Carbon Dioxide (CO₂) and 0.534 grams of Water (H₂O). What is the empirical formula of this alcohol? **Solution Approach:** 1. **Find moles of CO₂ produced:** - Molecular weight of CO₂ = 12.01 (C) + 2 * 16.00 (O) = 44.01 g/mol - Moles of CO₂ = 0.982 grams / 44.01 g/mol 2. **Find moles of H₂O produced:** - Molecular weight of H₂O = 2 * 1.008 (H) + 16.00 (O) = 18.016 g/mol - Moles of H₂O = 0.534 grams / 18.016 g/mol 3. **Determine the moles of C from CO₂:** - Each mole of CO₂ contains 1 mole of C. 4. **Determine the moles of H from H₂O:** - Each mole of H₂O contains 2 moles of H. 5. **Find the mass of C and H in the original compound:** 6. **Determine the mass of O by difference:** - Total mass = mass of C + mass of H + mass of O - Mass of O = total mass - (mass of C + mass of H) 7. **Find moles of O.** 8. **Determine the simplest whole number ratio of moles of C, H, and O.** #### Exercise 8 **Problem Statement:** In the course of the combustion analysis of an unknown compound containing only Carbon (C), Hydrogen (H), and Nitrogen (N), 12.923 grams of Carbon Dioxide (CO₂) and 6.608 grams of Water (H₂O) were measured. Treatment of the residual nitrogen with H₂ gas resulted in 2.501 grams of Ammonia (NH₃). What is the empirical formula? **Solution Approach:** 1. **Find moles of CO₂ produced:** - Molecular weight
Transcribed Image Text:### Chemistry Empirical Formula Exercises #### Exercise 7 **Problem Statement:** 0.446 grams of an unknown alcohol (composed of Carbon (C), Oxygen (O), and Hydrogen (H) only) was combusted using combustion analysis, producing 0.982 grams of Carbon Dioxide (CO₂) and 0.534 grams of Water (H₂O). What is the empirical formula of this alcohol? **Solution Approach:** 1. **Find moles of CO₂ produced:** - Molecular weight of CO₂ = 12.01 (C) + 2 * 16.00 (O) = 44.01 g/mol - Moles of CO₂ = 0.982 grams / 44.01 g/mol 2. **Find moles of H₂O produced:** - Molecular weight of H₂O = 2 * 1.008 (H) + 16.00 (O) = 18.016 g/mol - Moles of H₂O = 0.534 grams / 18.016 g/mol 3. **Determine the moles of C from CO₂:** - Each mole of CO₂ contains 1 mole of C. 4. **Determine the moles of H from H₂O:** - Each mole of H₂O contains 2 moles of H. 5. **Find the mass of C and H in the original compound:** 6. **Determine the mass of O by difference:** - Total mass = mass of C + mass of H + mass of O - Mass of O = total mass - (mass of C + mass of H) 7. **Find moles of O.** 8. **Determine the simplest whole number ratio of moles of C, H, and O.** #### Exercise 8 **Problem Statement:** In the course of the combustion analysis of an unknown compound containing only Carbon (C), Hydrogen (H), and Nitrogen (N), 12.923 grams of Carbon Dioxide (CO₂) and 6.608 grams of Water (H₂O) were measured. Treatment of the residual nitrogen with H₂ gas resulted in 2.501 grams of Ammonia (NH₃). What is the empirical formula? **Solution Approach:** 1. **Find moles of CO₂ produced:** - Molecular weight
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