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
sample of progesterone produced 5.563 g of and 1.627 g
. The molar mass of progesterone is .

Ask by Washington Higgins. in the United States

Jan 26,2025

Question

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
sample of progesterone produced 5.563 g of and 1.627 g
. The molar mass of progesterone is .

Ask by Washington Higgins. in the United States

Jan 26,2025

UpStudy AI · Accepted Answer

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**

$$
\boxed{\mathrm{C}_{21}\mathrm{H}_{30}\mathrm{O}_2}
$$
The molecular formula of progesterone is $$ \mathrm{C}_{21}\mathrm{H}_{30}\mathrm{O}_2 $$.

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