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A compound was analyzed and found to contain 52.17% C, 13.13% H and 34.78% O. What is the empirical formula for the compound? O CH3O O C₂H4O O C4H1202 O C₂H₂O

A compound was analyzed and found to contain 52.17% C, 13.13% H and 34.78% O. What is the empirical formula for the compound? O CH3O O C₂H4O O C4H1202 O C₂H₂O

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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**Determining the Empirical Formula of a Compound** **Problem Statement:** A compound was analyzed and found to contain 52.17% Carbon (C), 13.13% Hydrogen (H), and 34.78% Oxygen (O). What is the empirical formula for the compound? **Options:** - A) CH₃O - B) C₂H₄O - C) C₄H₁₂O₂ - D) C₂H₆O **Explanation:** To find the empirical formula, follow these steps: 1. **Convert percentages to grams:** Assume you have 100 grams of the compound. This means you would have: - 52.17 grams of Carbon - 13.13 grams of Hydrogen - 34.78 grams of Oxygen 2. **Convert grams to moles:** Use the atomic masses of the elements (C: 12.01 g/mol, H: 1.01 g/mol, O: 16.00 g/mol) to find the moles of each element: - Moles of Carbon: \( \frac{52.17 \text{ g}}{12.01 \text{ g/mol}} = 4.34 \text{ moles} \) - Moles of Hydrogen: \( \frac{13.13 \text{ g}}{1.01 \text{ g/mol}} = 13.00 \text{ moles} \) - Moles of Oxygen: \( \frac{34.78 \text{ g}}{16.00 \text{ g/mol}} = 2.17 \text{ moles} \) 3. **Divide by the smallest number of moles:** This will give the smallest whole number ratio: - Ratio for Carbon: \( \frac{4.34}{2.17} = 2 \) - Ratio for Hydrogen: \( \frac{13.00}{2.17} = 6 \) - Ratio for Oxygen: \( \frac{2.17}{2.17} = 1 \) 4. **Write the empirical formula:** The empirical formula that corresponds to the smallest whole number ratio of C:H:O is \( \text{C}_2\text{H}_6\text{O} \). Therefore, the correct
Transcribed Image Text:**Determining the Empirical Formula of a Compound** **Problem Statement:** A compound was analyzed and found to contain 52.17% Carbon (C), 13.13% Hydrogen (H), and 34.78% Oxygen (O). What is the empirical formula for the compound? **Options:** - A) CH₃O - B) C₂H₄O - C) C₄H₁₂O₂ - D) C₂H₆O **Explanation:** To find the empirical formula, follow these steps: 1. **Convert percentages to grams:** Assume you have 100 grams of the compound. This means you would have: - 52.17 grams of Carbon - 13.13 grams of Hydrogen - 34.78 grams of Oxygen 2. **Convert grams to moles:** Use the atomic masses of the elements (C: 12.01 g/mol, H: 1.01 g/mol, O: 16.00 g/mol) to find the moles of each element: - Moles of Carbon: \( \frac{52.17 \text{ g}}{12.01 \text{ g/mol}} = 4.34 \text{ moles} \) - Moles of Hydrogen: \( \frac{13.13 \text{ g}}{1.01 \text{ g/mol}} = 13.00 \text{ moles} \) - Moles of Oxygen: \( \frac{34.78 \text{ g}}{16.00 \text{ g/mol}} = 2.17 \text{ moles} \) 3. **Divide by the smallest number of moles:** This will give the smallest whole number ratio: - Ratio for Carbon: \( \frac{4.34}{2.17} = 2 \) - Ratio for Hydrogen: \( \frac{13.00}{2.17} = 6 \) - Ratio for Oxygen: \( \frac{2.17}{2.17} = 1 \) 4. **Write the empirical formula:** The empirical formula that corresponds to the smallest whole number ratio of C:H:O is \( \text{C}_2\text{H}_6\text{O} \). Therefore, the correct
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