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(3.10: Similar to Example 3.21) Upon combustion of 0.250-g sample of a compound containing carbon, hydrogen, and chlorine, 0.451 grams of CO2 and 0.0617 grams of H₂O were produced. What is the empirical formula of the compound? O CH3Cl₂ O C3H₂CI O CH3CI O C₂H3Cl2

(3.10: Similar to Example 3.21) Upon combustion of 0.250-g sample of a compound containing carbon, hydrogen, and chlorine, 0.451 grams of CO2 and 0.0617 grams of H₂O were produced. What is the empirical formula of the compound? O CH3Cl₂ O C3H₂CI O CH3CI O C₂H3Cl2

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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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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 **Problem Statement:** Upon combustion of a 0.250-g sample of a compound containing carbon, hydrogen, and chlorine, 0.451 grams of CO₂ and 0.0617 grams of H₂O were produced. What is the empirical formula of the compound? **Options Provided:** - \( \mathrm{CH_3Cl_2} \) - \( \mathrm{C_3H_2Cl} \) - \( \mathrm{CH_3Cl} \) - \( \mathrm{C_2H_3Cl_2} \) **Steps to Solve:** 1. **Calculate the Moles of Carbon:** - From \( \mathrm{CO_2} \), determine the moles of carbon. - Molar Mass of \( \mathrm{CO_2} \) = 44.01 g/mol. - Moles of \( \mathrm{CO_2} \) = \( \frac{0.451 \text{ g}}{44.01 \text{ g/mol}} = 0.0103 \text{ mol} \). - Since each mole of \( \mathrm{CO_2} \) contains 1 mole of carbon: - Moles of carbon = 0.0103 mol. 2. **Calculate the Moles of Hydrogen:** - From \( \mathrm{H_2O} \), determine the moles of hydrogen. - Molar Mass of \( \mathrm{H_2O} \) = 18.02 g/mol. - Moles of \( \mathrm{H_2O} \) = \( \frac{0.0617 \text{ g}}{18.02 \text{ g/mol}} = 0.00343 \text{ mol} \). - Since each mole of \( \mathrm{H_2O} \) contains 2 moles of hydrogen: - Moles of hydrogen = 2 × 0.00343 mol = 0.00686 mol. 3. **Determine the Mass of Carbon and Hydrogen:** - Mass of carbon = 0.0103 mol × 12.01 g/mol = 0.124 g. - Mass of hydrogen = 0.00686 mol × 1.008 g/mol = 0.00692 g
Transcribed Image Text:### Determining the Empirical Formula **Problem Statement:** Upon combustion of a 0.250-g sample of a compound containing carbon, hydrogen, and chlorine, 0.451 grams of CO₂ and 0.0617 grams of H₂O were produced. What is the empirical formula of the compound? **Options Provided:** - \( \mathrm{CH_3Cl_2} \) - \( \mathrm{C_3H_2Cl} \) - \( \mathrm{CH_3Cl} \) - \( \mathrm{C_2H_3Cl_2} \) **Steps to Solve:** 1. **Calculate the Moles of Carbon:** - From \( \mathrm{CO_2} \), determine the moles of carbon. - Molar Mass of \( \mathrm{CO_2} \) = 44.01 g/mol. - Moles of \( \mathrm{CO_2} \) = \( \frac{0.451 \text{ g}}{44.01 \text{ g/mol}} = 0.0103 \text{ mol} \). - Since each mole of \( \mathrm{CO_2} \) contains 1 mole of carbon: - Moles of carbon = 0.0103 mol. 2. **Calculate the Moles of Hydrogen:** - From \( \mathrm{H_2O} \), determine the moles of hydrogen. - Molar Mass of \( \mathrm{H_2O} \) = 18.02 g/mol. - Moles of \( \mathrm{H_2O} \) = \( \frac{0.0617 \text{ g}}{18.02 \text{ g/mol}} = 0.00343 \text{ mol} \). - Since each mole of \( \mathrm{H_2O} \) contains 2 moles of hydrogen: - Moles of hydrogen = 2 × 0.00343 mol = 0.00686 mol. 3. **Determine the Mass of Carbon and Hydrogen:** - Mass of carbon = 0.0103 mol × 12.01 g/mol = 0.124 g. - Mass of hydrogen = 0.00686 mol × 1.008 g/mol = 0.00692 g
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