Mesitylene is a liquid hydrocarbon. Burning 0.582 g of the compound in oxygen gives 1.92 g of CO2 and 0.524 g of H₂O. What is the empirical formula of mesitylene?
Mesitylene is a liquid hydrocarbon. Burning 0.582 g of the compound in oxygen gives 1.92 g of CO2 and 0.524 g of H₂O. What is the empirical formula of mesitylene?
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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![**Empirical Formula Determination of Mesitylene**
Mesitylene is a liquid hydrocarbon. When 0.582 grams of mesitylene is burned in oxygen, it produces 1.92 grams of carbon dioxide (CO₂) and 0.524 grams of water (H₂O). The goal is to determine the empirical formula of mesitylene.
**Problem Statement:**
When mesitylene is burned in oxygen:
- Mass of mesitylene burned: 0.582 g
- Mass of CO₂ produced: 1.92 g
- Mass of H₂O produced: 0.524 g
**Question:**
What is the empirical formula of mesitylene?
**Step-by-Step Solution:**
1. **Calculate Moles of CO₂ and H₂O:**
- Moles of CO₂ = \(\frac{\text{mass of CO₂}}{\text{molar mass of CO₂}}\)
\[
\frac{1.92 \text{ g}}{44.01 \text{ g/mol}} = 0.0436 \text{ mol CO₂}
\]
- Moles of H₂O = \(\frac{\text{mass of H₂O}}{\text{molar mass of H₂O}}\)
\[
\frac{0.524 \text{ g}}{18.02 \text{ g/mol}} = 0.0291 \text{ mol H₂O}
\]
2. **Determine Moles of Carbon and Hydrogen:**
- Each mole of CO₂ contains 1 mole of Carbon (C).
\[
\text{Moles of C} = 0.0436 \text{ mol}
\]
- Each mole of H₂O contains 2 moles of Hydrogen (H).
\[
\text{Moles of H} = 0.0291 \text{ mol} \times 2 = 0.0582 \text{ mol}
\]
3. **Calculate Mass of Carbon and Hydrogen:**
- Mass of Carbon = moles of Carbon \(\times\) atomic mass of Carbon (12.01 g/mol)
\[
0.0436 \text{ mol} \times 12.01 \text{ g/mol} = 0.523 \text{ g}](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7a354f1c-cdfd-41e5-a863-9a86a504af1c%2F2c0af587-a16e-4a54-87bf-6d737ba465dc%2Fiihkca8_processed.png&w=3840&q=75)
Transcribed Image Text:**Empirical Formula Determination of Mesitylene**
Mesitylene is a liquid hydrocarbon. When 0.582 grams of mesitylene is burned in oxygen, it produces 1.92 grams of carbon dioxide (CO₂) and 0.524 grams of water (H₂O). The goal is to determine the empirical formula of mesitylene.
**Problem Statement:**
When mesitylene is burned in oxygen:
- Mass of mesitylene burned: 0.582 g
- Mass of CO₂ produced: 1.92 g
- Mass of H₂O produced: 0.524 g
**Question:**
What is the empirical formula of mesitylene?
**Step-by-Step Solution:**
1. **Calculate Moles of CO₂ and H₂O:**
- Moles of CO₂ = \(\frac{\text{mass of CO₂}}{\text{molar mass of CO₂}}\)
\[
\frac{1.92 \text{ g}}{44.01 \text{ g/mol}} = 0.0436 \text{ mol CO₂}
\]
- Moles of H₂O = \(\frac{\text{mass of H₂O}}{\text{molar mass of H₂O}}\)
\[
\frac{0.524 \text{ g}}{18.02 \text{ g/mol}} = 0.0291 \text{ mol H₂O}
\]
2. **Determine Moles of Carbon and Hydrogen:**
- Each mole of CO₂ contains 1 mole of Carbon (C).
\[
\text{Moles of C} = 0.0436 \text{ mol}
\]
- Each mole of H₂O contains 2 moles of Hydrogen (H).
\[
\text{Moles of H} = 0.0291 \text{ mol} \times 2 = 0.0582 \text{ mol}
\]
3. **Calculate Mass of Carbon and Hydrogen:**
- Mass of Carbon = moles of Carbon \(\times\) atomic mass of Carbon (12.01 g/mol)
\[
0.0436 \text{ mol} \times 12.01 \text{ g/mol} = 0.523 \text{ g}
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