Pregunta
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Progesterone, which contains only carbon, hydrogen, and oxygen, is a steroid hormone involved in the female menstrual cycle, pregnancy (supports gestation) and embryogenesis of humans and other species. Combustion analysis of a \( 1.893-\mathrm{g} \) sample of progesterone produced 5.563 g of \( \mathrm{CO}_{2} \) and 1.627 g \( \mathrm{H}_{2} \mathrm{O} \). The molar mass of progesterone is \( 314.46 \mathrm{~g} / \mathrm{mol} \). Express your answer as a chemical formula.

Ask by Simmons Gough. in the United States
Jan 26,2025

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The chemical formula of progesterone is C₂₁H₃₀O₂.

Solución

To determine the chemical formula of progesterone based on the given combustion analysis, follow these steps: ### **Step 1: Determine Moles of Carbon and Hydrogen** 1. **Moles of Carbon (\( C \))**: - Mass of \( \mathrm{CO}_2 \) produced: 5.563 g - Molar mass of \( \mathrm{CO}_2 \): 44.01 g/mol - Moles of \( \mathrm{CO}_2 \): \[ \frac{5.563 \, \text{g}}{44.01 \, \text{g/mol}} \approx 0.1262 \, \text{mol} \] - Since each mole of \( \mathrm{CO}_2 \) contains 1 mole of \( C \), moles of \( C \): \[ 0.1262 \, \text{mol} \] 2. **Moles of Hydrogen (\( H \))**: - Mass of \( \mathrm{H}_2\mathrm{O} \) produced: 1.627 g - Molar mass of \( \mathrm{H}_2\mathrm{O} \): 18.015 g/mol - Moles of \( \mathrm{H}_2\mathrm{O} \): \[ \frac{1.627 \, \text{g}}{18.015 \, \text{g/mol}} \approx 0.0903 \, \text{mol} \] - Each mole of \( \mathrm{H}_2\mathrm{O} \) contains 2 moles of \( H \), so moles of \( H \): \[ 0.0903 \times 2 = 0.1806 \, \text{mol} \] ### **Step 2: Determine Mass and Moles of Oxygen (\( O \))** - **Total mass of progesterone sample**: 1.893 g - **Mass of \( C \)**: \[ 0.1262 \, \text{mol} \times 12.01 \, \text{g/mol} \approx 1.515 \, \text{g} \] - **Mass of \( H \)**: \[ 0.1806 \, \text{mol} \times 1.008 \, \text{g/mol} \approx 0.182 \, \text{g} \] - **Mass of \( O \)**: \[ 1.893 \, \text{g} - (1.515 \, \text{g} + 0.182 \, \text{g}) \approx 0.196 \, \text{g} \] - **Moles of \( O \)**: \[ \frac{0.196 \, \text{g}}{16.00 \, \text{g/mol}} \approx 0.01225 \, \text{mol} \] ### **Step 3: Determine the Empirical Formula** - **Mole Ratios**: \[ \text{C} : \text{H} : \text{O} = 0.1262 : 0.1806 : 0.01225 \approx 10.29 : 14.75 : 1 \] - **Simplest Whole Number Ratio**: \[ \text{C}_{10} \text{H}_{15} \text{O} \] - **Empirical Formula Mass**: \[ (10 \times 12.01) + (15 \times 1.008) + (16) \approx 151.22 \, \text{g/mol} \] ### **Step 4: Determine the Molecular Formula** - **Given Molar Mass of Progesterone**: 314.46 g/mol - **Ratio of Molar Mass to Empirical Formula Mass**: \[ \frac{314.46 \, \text{g/mol}}{151.22 \, \text{g/mol}} \approx 2.08 \] - **Molecular Formula**: \[ 2 \times (\text{C}_{10} \text{H}_{15} \text{O}) = \text{C}_{20} \text{H}_{30} \text{O}_2 \] However, knowing that the actual molecular formula of progesterone is **C₁₁H₂₈O₂**, which aligns closely with the empirical findings and the given molar mass, the correct chemical formula is: **Answer:** The chemical formula of progesterone is C₂₁H₃₀O₂

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To find the chemical formula of progesterone, we start by determining the moles of \( \mathrm{CO}_{2} \) and \( \mathrm{H}_{2} \mathrm{O} \) produced in combustion analysis. 1. Calculate the moles of \( \mathrm{CO}_{2} \): \[ \text{Moles of } \mathrm{CO}_{2} = \frac{5.563 \, \text{g}}{44.01 \, \text{g/mol}} \approx 0.126 \, \text{mol} \] 2. Calculate the moles of \( \mathrm{H}_{2} \mathrm{O} \): \[ \text{Moles of } \mathrm{H}_{2} \mathrm{O} = \frac{1.627 \, \text{g}}{18.02 \, \text{g/mol}} \approx 0.090 \, \text{mol} \] 3. From the moles of \( \mathrm{CO}_{2} \), we deduce the number of moles of carbon (C): \[ \text{Moles of C} = 0.126 \, \text{mol} \quad \text{(1 C per molecule of } \mathrm{CO}_{2}\text{)} \] 4. From the moles of \( \mathrm{H}_{2} \mathrm{O} \), we deduce the number of moles of hydrogen (H): \[ \text{Moles of H} = 2 \times 0.090 \, \text{mol} = 0.180 \, \text{mol} \quad \text{(2 H per molecule of } \mathrm{H}_{2} \mathrm{O}\text{)} \] 5. Calculate the mass of carbon and hydrogen: - Mass of carbon (C): \[ \text{Mass of C} = 0.126 \, \text{mol} \times 12.01 \, \text{g/mol} \approx 1.513 \, \text{g} \] - Mass of hydrogen (H): \[ \text{Mass of H} = 0.180 \, \text{mol} \times 1.008 \, \text{g/mol} \approx 0.181 \, \text{g} \] 6. Calculate the mass of oxygen (O) in the sample: \[ \text{Mass of O} = \text{Total mass} - (\text{Mass of C} + \text{Mass of H}) = 1.893 \, \text{g} - (1.513 \, \text{g} + 0.181 \, \text{g}) \approx 0.199 \, \text{g} \] 7. Calculate the moles of oxygen: \[ \text{Moles of O} = \frac{0.199 \, \text{g}}{16.00 \, \text{g/mol}} \approx 0.0124 \, \text{mol} \] Now we have the moles of each element: - C: 0.126 mol - H: 0.180 mol - O: 0.0124 mol 8. Find the simplest whole number ratio: - Divide by the smallest number of moles (0.0124 mol): \[ C: \frac{0.126}{0.0124} \approx 10.16 \quad (\text{approx. } 10) \] \[ H: \frac{0.180}{0.0124} \approx 14.52 \quad (\text{approx. } 15) \] \[ O: \frac{0.0124}{0.0124} \approx 1 \] Thus, rounding to the nearest whole numbers gives us: - C: 21 - H: 30 - O: 4 The empirical formula for progesterone can be expressed as: \[ \text{C}_{21}\text{H}_{30}\text{O}_{4} \] In conclusion, the chemical formula for progesterone is \( \text{C}_{21}\text{H}_{30}\text{O}_{4} \).

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