An unknown compound has the formula C HyO₂. You burn 0.2072 g of the compound and isolate 0.5059 g of CO2 and 0.2071 g of H₂O. What is the empirical formula of the compound? If the molar mass is 72.1 g/mol, what is the molecular formula? (Enter the elements in the order: C, H, O.) Empirical formula: Molecular formula:
An unknown compound has the formula C HyO₂. You burn 0.2072 g of the compound and isolate 0.5059 g of CO2 and 0.2071 g of H₂O. What is the empirical formula of the compound? If the molar mass is 72.1 g/mol, what is the molecular formula? (Enter the elements in the order: C, H, O.) Empirical formula: Molecular formula:
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![### Determining Empirical and Molecular Formulas of a Compound
An unknown compound has the formula \( C_xH_yO_z \). You burn 0.2072 g of the compound and isolate 0.5059 g of \( CO_2 \) and 0.2071 g of \( H_2O \). What is the empirical formula of the compound? If the molar mass is 72.1 g/mol, what is the molecular formula?
**Problem:**
(Enter the elements in the order: C, H, O.)
| Empirical formula: | [---Text Box for Answer---] |
| Molecular formula: | [---Text Box for Answer---] |
In this problem, the goal is to find both the empirical and molecular formulas of the unknown compound. Let's break down the steps necessary to solve this problem.
#### Step-by-Step Method:
1. **Determine the amount of carbon, hydrogen, and oxygen in the compound:**
- **Carbon Analysis:**
- Mass of \( CO_2 \) produced: 0.5059 g
- Since each mole of \( CO_2 \) contains one mole of carbon, find the moles of carbon:
\[
\text{Moles of C} = \frac{\text{Mass of } CO_2}{\text{Molar Mass of } CO_2} = \frac{0.5059 \, \text{g}}{44.01 \, \text{g/mol}} = 0.011497 \, \text{mol}
\]
- **Hydrogen Analysis:**
- Mass of \( H_2O \) produced: 0.2071 g
- Since each mole of \( H_2O \) contains two moles of hydrogen, find the moles of hydrogen:
\[
\text{Moles of H} = 2 \times \frac{\text{Mass of } H_2O}{\text{Molar Mass of } H_2O} = 2 \times \frac{0.2071 \, \text{g}}{18.02 \, \text{g/mol}} = 0.023003 \, \text{mol}
\]
- **Oxygen Analysis:**
- Use the mass of the compound burned and the](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7a354f1c-cdfd-41e5-a863-9a86a504af1c%2F4a91e65d-9f31-46fb-ab07-d71bf76244d0%2Ftsox799_processed.png&w=3840&q=75)
Transcribed Image Text:### Determining Empirical and Molecular Formulas of a Compound
An unknown compound has the formula \( C_xH_yO_z \). You burn 0.2072 g of the compound and isolate 0.5059 g of \( CO_2 \) and 0.2071 g of \( H_2O \). What is the empirical formula of the compound? If the molar mass is 72.1 g/mol, what is the molecular formula?
**Problem:**
(Enter the elements in the order: C, H, O.)
| Empirical formula: | [---Text Box for Answer---] |
| Molecular formula: | [---Text Box for Answer---] |
In this problem, the goal is to find both the empirical and molecular formulas of the unknown compound. Let's break down the steps necessary to solve this problem.
#### Step-by-Step Method:
1. **Determine the amount of carbon, hydrogen, and oxygen in the compound:**
- **Carbon Analysis:**
- Mass of \( CO_2 \) produced: 0.5059 g
- Since each mole of \( CO_2 \) contains one mole of carbon, find the moles of carbon:
\[
\text{Moles of C} = \frac{\text{Mass of } CO_2}{\text{Molar Mass of } CO_2} = \frac{0.5059 \, \text{g}}{44.01 \, \text{g/mol}} = 0.011497 \, \text{mol}
\]
- **Hydrogen Analysis:**
- Mass of \( H_2O \) produced: 0.2071 g
- Since each mole of \( H_2O \) contains two moles of hydrogen, find the moles of hydrogen:
\[
\text{Moles of H} = 2 \times \frac{\text{Mass of } H_2O}{\text{Molar Mass of } H_2O} = 2 \times \frac{0.2071 \, \text{g}}{18.02 \, \text{g/mol}} = 0.023003 \, \text{mol}
\]
- **Oxygen Analysis:**
- Use the mass of the compound burned and the
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