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7. An unknown solvent DHB, contains 49.02% C, 2.74% H, and 48.24 % chlorine. If the molar mass is 293.87 g/mol, what is the empirical formula and molecular formula of this compound? vol t

7. An unknown solvent DHB, contains 49.02% C, 2.74% H, and 48.24 % chlorine. If the molar mass is 293.87 g/mol, what is the empirical formula and molecular formula of this compound? vol t

Chemistry: Principles and Practice
Chemistry: Principles and Practice
3rd Edition
ISBN:9780534420123
Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Chapter3: Equation, The Mole, And Chemical Formulas
Section: Chapter Questions
Problem 3.150QE
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**Text from Educational Image:** **Problem 7: Empirical and Molecular Formula Calculation** An unknown solvent (DHB) contains 40.02% carbon (C), 2.74% hydrogen (H), and 48.24% chlorine (Cl). If the molar mass of the compound is 293.87 g/mol, what are the empirical formula and molecular formula of this compound? **Explanation:** To find the empirical formula of a compound, we follow these steps: 1. Convert the percentage composition to grams. Assume 100 grams of the compound, so the composition in grams will be the same as the percentage composition. - Carbon: 40.02 grams - Hydrogen: 2.74 grams - Chlorine: 48.24 grams 2. Convert the grams to moles by dividing by the atomic mass of each element: - Moles of Carbon (C): 40.02 g / 12.01 g/mol = 3.33 moles - Moles of Hydrogen (H): 2.74 g / 1.01 g/mol = 2.71 moles - Moles of Chlorine (Cl): 48.24 g / 35.45 g/mol = 1.36 moles 3. Find the simplest whole number ratio by dividing the moles of each element by the smallest number of moles calculated: - Carbon: 3.33 / 1.36 = 2.45 ≈ 2 - Hydrogen: 2.71 / 1.36 = 2.00 ≈ 2 - Chlorine: 1.36 / 1.36 = 1.00 ≈ 1 Therefore, the empirical formula is \( C_2H_2Cl \). 4. Determine the molecular formula using the molar mass given (293.87 g/mol). First, calculate the molar mass of the empirical formula: - Molar mass of \( C_2H_2Cl \): - Carbon (C): 2 × 12.01 = 24.02 g/mol - Hydrogen (H): 2 × 1.01 = 2.02 g/mol - Chlorine (Cl): 1 × 35.45 = 35.45 g/mol - Total mol
Transcribed Image Text:**Text from Educational Image:** **Problem 7: Empirical and Molecular Formula Calculation** An unknown solvent (DHB) contains 40.02% carbon (C), 2.74% hydrogen (H), and 48.24% chlorine (Cl). If the molar mass of the compound is 293.87 g/mol, what are the empirical formula and molecular formula of this compound? **Explanation:** To find the empirical formula of a compound, we follow these steps: 1. Convert the percentage composition to grams. Assume 100 grams of the compound, so the composition in grams will be the same as the percentage composition. - Carbon: 40.02 grams - Hydrogen: 2.74 grams - Chlorine: 48.24 grams 2. Convert the grams to moles by dividing by the atomic mass of each element: - Moles of Carbon (C): 40.02 g / 12.01 g/mol = 3.33 moles - Moles of Hydrogen (H): 2.74 g / 1.01 g/mol = 2.71 moles - Moles of Chlorine (Cl): 48.24 g / 35.45 g/mol = 1.36 moles 3. Find the simplest whole number ratio by dividing the moles of each element by the smallest number of moles calculated: - Carbon: 3.33 / 1.36 = 2.45 ≈ 2 - Hydrogen: 2.71 / 1.36 = 2.00 ≈ 2 - Chlorine: 1.36 / 1.36 = 1.00 ≈ 1 Therefore, the empirical formula is \( C_2H_2Cl \). 4. Determine the molecular formula using the molar mass given (293.87 g/mol). First, calculate the molar mass of the empirical formula: - Molar mass of \( C_2H_2Cl \): - Carbon (C): 2 × 12.01 = 24.02 g/mol - Hydrogen (H): 2 × 1.01 = 2.02 g/mol - Chlorine (Cl): 1 × 35.45 = 35.45 g/mol - Total mol
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