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 \( 314.46 \mathrm{~g} / \mathrm{mol} \).

To determine the molecular formula of progesterone based on the given combustion analysis data, follow these steps: ### **1. Determine Moles of Carbon and Hydrogen** **a. Moles of Carbon (\( \mathrm{C} \)) from \( \mathrm{CO}_2 \):** - **Mass of \( \mathrm{CO}_2 \)**: 5.563 g - **Molar mass of \( \mathrm{CO}_2 \)**: 44.01 g/mol \[ \text{Moles of } \mathrm{C} = \frac{5.563 \, \text{g} \, \mathrm{CO}_2}{44.01 \, \text{g/mol}} \approx 0.1262 \, \text{mol} \, \mathrm{C} \] **b. Moles of Hydrogen (\( \mathrm{H} \)) from \( \mathrm{H}_2\mathrm{O} \):** - **Mass of \( \mathrm{H}_2\mathrm{O} \)**: 1.627 g - **Molar mass of \( \mathrm{H}_2\mathrm{O} \)**: 18.02 g/mol \[ \text{Moles of } \mathrm{H}_2\mathrm{O} = \frac{1.627 \, \text{g}}{18.02 \, \text{g/mol}} \approx 0.0903 \, \text{mol} \, \mathrm{H}_2\mathrm{O} \] Each mole of \( \mathrm{H}_2\mathrm{O} \) provides 2 moles of \( \mathrm{H} \): \[ \text{Moles of } \mathrm{H} = 2 \times 0.0903 \approx 0.1806 \, \text{mol} \, \mathrm{H} \] ### **2. Determine Mass of Each Element** - **Mass of Carbon (\( \mathrm{C} \))**: \[ 0.1262 \, \text{mol} \times 12.01 \, \text{g/mol} \approx 1.517 \, \text{g} \, \mathrm{C} \] - **Mass of Hydrogen (\( \mathrm{H} \))**: \[ 0.1806 \, \text{mol} \times 1.008 \, \text{g/mol} \approx 0.182 \, \text{g} \, \mathrm{H} \] - **Mass of Oxygen (\( \mathrm{O} \))**: \[ \text{Total mass} - \text{Mass of } \mathrm{C} - \text{Mass of } \mathrm{H} = 1.893 \, \text{g} - 1.517 \, \text{g} - 0.182 \, \text{g} \approx 0.194 \, \text{g} \, \mathrm{O} \] \[ \text{Moles of } \mathrm{O} = \frac{0.194 \, \text{g}}{16.00 \, \text{g/mol}} \approx 0.0121 \, \text{mol} \, \mathrm{O} \] ### **3. Calculate the Empirical Formula Ratios** \[ \text{Ratio of } \mathrm{C} : \mathrm{H} : \mathrm{O} \approx \frac{0.1262}{0.0121} : \frac{0.1806}{0.0121} : \frac{0.0121}{0.0121} \approx 10.4 : 14.9 : 1 \] Rounding to the nearest whole number gives: \[ \mathrm{C}_{21}\mathrm{H}_{30}\mathrm{O}_2 \] ### **4. Verify with Molar Mass** Calculate the molar mass of the proposed formula: \[ \mathrm{C}_{21}\mathrm{H}_{30}\mathrm{O}_2 = (21 \times 12.01) + (30 \times 1.008) + (2 \times 16.00) = 252.21 + 30.24 + 32 = 314.45 \, \text{g/mol} \] This closely matches the given molar mass of progesterone (\( 314.46 \, \text{g/mol} \)). ### **Final Molecular Formula** \[ \boxed{\mathrm{C}_{21}\mathrm{H}_{30}\mathrm{O}_2} \]