You perform a combustion analysis on a 255 mg sample of a substance that contains only C, H, and O, and you find that 561 mg CO2 is produced, along with 306 mg of H20. (Combustion means that you react the molecule with 02. The products of a combustion reaction are only carbon dioxide and water.) If the substance contains only C, H, and O, what is the empirical formula? If the molar mass of the compound is 180 g/mol, what is the molecular formula of the compound?
You perform a combustion analysis on a 255 mg sample of a substance that contains only C, H, and O, and you find that 561 mg CO2 is produced, along with 306 mg of H20. (Combustion means that you react the molecule with 02. The products of a combustion reaction are only carbon dioxide and water.) If the substance contains only C, H, and O, what is the empirical formula? If the molar mass of the compound is 180 g/mol, what is the molecular formula of the compound?
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
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![**Combustion Analysis of a Substance Containing C, H, and O**
**Introduction:**
Combustion analysis is a common laboratory method used to determine the empirical formula of a compound that contains carbon, hydrogen, and oxygen. This method involves burning a known mass of the compound and measuring the quantities of carbon dioxide (CO₂) and water (H₂O) produced.
**Example Analysis:**
- You perform a combustion analysis on a 255 mg sample of a substance that contains only C, H, and O, and you find that 561 mg CO₂ is produced, along with 306 mg of H₂O.
**Explanation of Combustion:**
Combustion means that you react the molecule with O₂. The products of a combustion reaction are only carbon dioxide and water.
**Steps for Analysis:**
1. **If the substance contains only C, H, and O, what is the empirical formula?**
2. **If the molar mass of the compound is 180 g/mol, what is the molecular formula of the compound?**
**Detailed Explanation and Calculations:**
1. **Determine the Moles of Carbon and Hydrogen:**
- Calculate the moles of CO₂ produced:
\[
n_{CO₂} = \frac{561\, \text{mg}}{44.01\, \text{g/mol}} = 0.561\, \text{g} \div 44.01\, \text{g/mol} = 0.01275\, \text{mol}
\]
- Since each mole of CO₂ contains 1 mole of C:
\[
\text{Moles of C} = 0.01275\, \text{mol}
\]
- Calculate the moles of H₂O produced:
\[
n_{H₂O} = \frac{306\, \text{mg}}{18.02\, \text{g/mol}} = 0.306\, \text{g} \div 18.02\, \text{g/mol} = 0.01698\, \text{mol}
\]
- Since each mole of H₂O contains 2 moles of H:
\[
\text{Moles of H} = 2 \times 0.01698\, \text{mol} = 0.03396\,](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7aed86dd-d5c3-48c6-a690-9abed80f4c22%2F3f6fa7ea-7214-4b36-9842-8e7358ed7aef%2Fsgrifv4.jpeg&w=3840&q=75)
Transcribed Image Text:**Combustion Analysis of a Substance Containing C, H, and O**
**Introduction:**
Combustion analysis is a common laboratory method used to determine the empirical formula of a compound that contains carbon, hydrogen, and oxygen. This method involves burning a known mass of the compound and measuring the quantities of carbon dioxide (CO₂) and water (H₂O) produced.
**Example Analysis:**
- You perform a combustion analysis on a 255 mg sample of a substance that contains only C, H, and O, and you find that 561 mg CO₂ is produced, along with 306 mg of H₂O.
**Explanation of Combustion:**
Combustion means that you react the molecule with O₂. The products of a combustion reaction are only carbon dioxide and water.
**Steps for Analysis:**
1. **If the substance contains only C, H, and O, what is the empirical formula?**
2. **If the molar mass of the compound is 180 g/mol, what is the molecular formula of the compound?**
**Detailed Explanation and Calculations:**
1. **Determine the Moles of Carbon and Hydrogen:**
- Calculate the moles of CO₂ produced:
\[
n_{CO₂} = \frac{561\, \text{mg}}{44.01\, \text{g/mol}} = 0.561\, \text{g} \div 44.01\, \text{g/mol} = 0.01275\, \text{mol}
\]
- Since each mole of CO₂ contains 1 mole of C:
\[
\text{Moles of C} = 0.01275\, \text{mol}
\]
- Calculate the moles of H₂O produced:
\[
n_{H₂O} = \frac{306\, \text{mg}}{18.02\, \text{g/mol}} = 0.306\, \text{g} \div 18.02\, \text{g/mol} = 0.01698\, \text{mol}
\]
- Since each mole of H₂O contains 2 moles of H:
\[
\text{Moles of H} = 2 \times 0.01698\, \text{mol} = 0.03396\,
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