A 13.04 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 19.12 grams of CO2 and 7.828 grams of H2O are produced. In a separate experiment, the molar mass is found to be 60.05 g/mol. Determine the empirical formula and the molecular formula of the organic compound. Enter the elements in the order C, H, O empirical formula = molecular formula =

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### Combustion Analysis of an Organic Compound

**Problem Statement:**

A 13.04 gram sample of an organic compound containing Carbon (C), Hydrogen (H), and Oxygen (O) is analyzed through combustion analysis. The analysis results in the production of 19.12 grams of carbon dioxide (CO₂) and 7.828 grams of water (H₂O).

In a separate experiment, the molar mass is determined to be 60.05 g/mol. The task is to determine the empirical formula and the molecular formula of the organic compound.

**Instructions:**

Enter the elements in the order C, H, O.

- **Empirical Formula:** [TextBox]
- **Molecular Formula:** [TextBox]
Transcribed Image Text:### Combustion Analysis of an Organic Compound **Problem Statement:** A 13.04 gram sample of an organic compound containing Carbon (C), Hydrogen (H), and Oxygen (O) is analyzed through combustion analysis. The analysis results in the production of 19.12 grams of carbon dioxide (CO₂) and 7.828 grams of water (H₂O). In a separate experiment, the molar mass is determined to be 60.05 g/mol. The task is to determine the empirical formula and the molecular formula of the organic compound. **Instructions:** Enter the elements in the order C, H, O. - **Empirical Formula:** [TextBox] - **Molecular Formula:** [TextBox]
Expert Solution
Step 1

Mass of sample = 13.04 g 

Mass of CO2 = 19.12 g 

Mass of H2O = 7.828 g 

Number of moles of C 

= 19.12 g CO2 × (1 mol CO2 / 44.01 g CO2 ) × (1 mol C / 1 mol CO2

= 0.434 mol C 

Number of moles of H 

= 7.828 g H2O × ( 1 mol H2O / 18.02 g H2O) × ( 2 mol H / 1 mol H2O ) 

= 0.8688 mol H 

 

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