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6. Determine the empirical and molecular formula of the following: caffeine, 49.48%C; 5.19%H, 28.85%N and 16.48%O; Molar mass = 194.2. a. b. octene, 85.7%C and 14.3%H; Molar mass = 112. an oxide of phosphorous, 43.7%P and 56.3%O; Molar mass = 284. c.

6. Determine the empirical and molecular formula of the following: caffeine, 49.48%C; 5.19%H, 28.85%N and 16.48%O; Molar mass = 194.2. a. b. octene, 85.7%C and 14.3%H; Molar mass = 112. an oxide of phosphorous, 43.7%P and 56.3%O; Molar mass = 284. c.

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
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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**Chemistry Problem Set** **7. Stoichiometry and Moles of Oxygen** Determine the number of moles of O₂ that react with or are produced by 6.75 moles of the first reactant in each balanced equation: a. \( 2C_2H_6 + 7O_2 \rightarrow 4CO_2 + 6H_2O \) b. \( 2KClO_3 \rightarrow 2KCl + 3O_2 \) c. \( 4Fe + 3O_2 \rightarrow 2Fe_2O_3 \) **8. Mass of Second Reactant** Calculate the mass of the second reactant needed to react with 20.0 grams of the first reactant in each balanced equation: a. \( P_4 + 6Cl_2 \rightarrow 4PCl_3 \) b. \( P_4O_{10} + 6H_2O \rightarrow 4H_3PO_4 \) c. \( Fe_2O_3 + 3CO \rightarrow 2Fe + 3CO_2 \) **9. Reaction Analysis** Using the three balanced equations below, answer the following for each: a. How many grams of the first product can be produced from 20.0 grams of the first reactant and 20.0 grams of the second reactant? b. Identify the limiting reactant. c. Identify the excess reactant. d. Calculate the grams of the excess reactant that reacted. e. Calculate the grams of the excess reactant that remained unreacted. - Equation: \( Fe_2O_3 + 3CO \rightarrow 2Fe + 3CO_2 \) **10. Reaction Yield** A student conducted the reaction shown by reacting 10.0g of \( P_4 \) with excess \( Cl_2 \) and obtained 35.0g of \( PCl_3 \). a. Determine the theoretical yield of \( PCl_3 \). b. Calculate the percent yield of \( PCl_3 \). --- These exercises involve calculations based on stoichiometry, requiring a solid understanding of molecular weights, mole-to-mole relationships in balanced equations, and the concept of limiting and excess reactants.
Transcribed Image Text:**Chemistry Problem Set** **7. Stoichiometry and Moles of Oxygen** Determine the number of moles of O₂ that react with or are produced by 6.75 moles of the first reactant in each balanced equation: a. \( 2C_2H_6 + 7O_2 \rightarrow 4CO_2 + 6H_2O \) b. \( 2KClO_3 \rightarrow 2KCl + 3O_2 \) c. \( 4Fe + 3O_2 \rightarrow 2Fe_2O_3 \) **8. Mass of Second Reactant** Calculate the mass of the second reactant needed to react with 20.0 grams of the first reactant in each balanced equation: a. \( P_4 + 6Cl_2 \rightarrow 4PCl_3 \) b. \( P_4O_{10} + 6H_2O \rightarrow 4H_3PO_4 \) c. \( Fe_2O_3 + 3CO \rightarrow 2Fe + 3CO_2 \) **9. Reaction Analysis** Using the three balanced equations below, answer the following for each: a. How many grams of the first product can be produced from 20.0 grams of the first reactant and 20.0 grams of the second reactant? b. Identify the limiting reactant. c. Identify the excess reactant. d. Calculate the grams of the excess reactant that reacted. e. Calculate the grams of the excess reactant that remained unreacted. - Equation: \( Fe_2O_3 + 3CO \rightarrow 2Fe + 3CO_2 \) **10. Reaction Yield** A student conducted the reaction shown by reacting 10.0g of \( P_4 \) with excess \( Cl_2 \) and obtained 35.0g of \( PCl_3 \). a. Determine the theoretical yield of \( PCl_3 \). b. Calculate the percent yield of \( PCl_3 \). --- These exercises involve calculations based on stoichiometry, requiring a solid understanding of molecular weights, mole-to-mole relationships in balanced equations, and the concept of limiting and excess reactants.
### Determining Empirical and Molecular Formulas **Problem 6:** Find the empirical and molecular formulas for the following compounds: a. **Caffeine** - Composition: 49.48% Carbon (C), 5.19% Hydrogen (H), 28.85% Nitrogen (N), and 16.48% Oxygen (O) - Molar Mass: 194.2 g/mol **Empirical Formula:** [Space for answer] **Molecular Formula:** [Space for answer] b. **Octene** - Composition: 85.7% Carbon (C) and 14.3% Hydrogen (H) - Molar Mass: 112 g/mol **Empirical Formula:** [Space for answer] **Molecular Formula:** [Space for answer] c. **An Oxide of Phosphorus** - Composition: 43.7% Phosphorus (P) and 56.3% Oxygen (O) - Molar Mass: 284 g/mol **Empirical Formula:** [Space for answer] **Molecular Formula:** [Space for answer] ### Notes: - The empirical formula represents the simplest whole-number ratio of elements in the compound. - The molecular formula is a multiple of the empirical formula and indicates the actual number of atoms of each element in a molecule. **Instructions:** Calculate the empirical formula based on the given percent compositions. Then use the molar mass to determine the molecular formula.
Transcribed Image Text:### Determining Empirical and Molecular Formulas **Problem 6:** Find the empirical and molecular formulas for the following compounds: a. **Caffeine** - Composition: 49.48% Carbon (C), 5.19% Hydrogen (H), 28.85% Nitrogen (N), and 16.48% Oxygen (O) - Molar Mass: 194.2 g/mol **Empirical Formula:** [Space for answer] **Molecular Formula:** [Space for answer] b. **Octene** - Composition: 85.7% Carbon (C) and 14.3% Hydrogen (H) - Molar Mass: 112 g/mol **Empirical Formula:** [Space for answer] **Molecular Formula:** [Space for answer] c. **An Oxide of Phosphorus** - Composition: 43.7% Phosphorus (P) and 56.3% Oxygen (O) - Molar Mass: 284 g/mol **Empirical Formula:** [Space for answer] **Molecular Formula:** [Space for answer] ### Notes: - The empirical formula represents the simplest whole-number ratio of elements in the compound. - The molecular formula is a multiple of the empirical formula and indicates the actual number of atoms of each element in a molecule. **Instructions:** Calculate the empirical formula based on the given percent compositions. Then use the molar mass to determine the molecular formula.
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