Chemistry
Chemistry
10th Edition
ISBN: 9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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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
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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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