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 1.893 - g sample of progesterone produced 5.563 g of co2 and 1.627 g h2o. the molar mass of progesterone is 314.46 g/mol. part a find the molecular formula for progesterone. express your answer as a chemical formula.

Explanation:

Step1: Calculate moles of carbon

The mass of $CO_2$ is $m_{CO_2}=5.563\ g$. The molar mass of $CO_2$ is $M_{CO_2}=44.01\ g/mol$. The number of moles of $CO_2$, $n_{CO_2}=\frac{m_{CO_2}}{M_{CO_2}}=\frac{5.563\ g}{44.01\ g/mol}= 0.1264\ mol$. Since 1 mole of $CO_2$ contains 1 mole of carbon, the moles of carbon in the progesterone sample, $n_C = 0.1264\ mol$.

Step2: Calculate moles of hydrogen

The mass of $H_2O$ is $m_{H_2O}=1.627\ g$. The molar mass of $H_2O$ is $M_{H_2O}=18.02\ g/mol$. The number of moles of $H_2O$, $n_{H_2O}=\frac{m_{H_2O}}{M_{H_2O}}=\frac{1.627\ g}{18.02\ g/mol}=0.0903\ mol$. Since 1 mole of $H_2O$ contains 2 moles of hydrogen, the moles of hydrogen in the progesterone sample, $n_H=2\times n_{H_2O}=2\times0.0903\ mol = 0.1806\ mol$.

Step3: Calculate mass of carbon and hydrogen

The mass of carbon, $m_C=n_C\times12.01\ g/mol=0.1264\ mol\times12.01\ g/mol = 1.518\ g$. The mass of hydrogen, $m_H=n_H\times1.008\ g/mol=0.1806\ mol\times1.008\ g/mol=0.1821\ g$.

Step4: Calculate mass of oxygen

The mass of the progesterone sample is $m = 1.893\ g$. The mass of oxygen, $m_O=m-(m_C + m_H)=1.893\ g-(1.518\ g + 0.1821\ g)=0.1929\ g$.

Step5: Calculate moles of oxygen

The molar mass of oxygen is $M_O = 16.00\ g/mol$. The moles of oxygen, $n_O=\frac{m_O}{M_O}=\frac{0.1929\ g}{16.00\ g/mol}=0.01206\ mol$.

Step6: Find the empirical - formula ratio

Divide the number of moles of each element by the smallest number of moles ($n_O = 0.01206\ mol$).
For carbon: $\frac{n_C}{n_O}=\frac{0.1264\ mol}{0.01206\ mol}\approx10.5$.
For hydrogen: $\frac{n_H}{n_O}=\frac{0.1806\ mol}{0.01206\ mol}\approx15$.
For oxygen: $\frac{n_O}{n_O}=1$. Multiply each ratio by 2 to get whole - numbers. The empirical formula is $C_{21}H_{30}O_2$.

Step7: Calculate the empirical - formula mass

The empirical - formula mass of $C_{21}H_{30}O_2$ is $M_{empirical}=21\times12.01\ g/mol+30\times1.008\ g/mol + 2\times16.00\ g/mol=252.21\ g/mol+30.24\ g/mol+32.00\ g/mol = 314.45\ g/mol$.

Step8: Determine the molecular formula

The molar mass of progesterone is $M = 314.46\ g/mol$. Since $\frac{M}{M_{empirical}}=\frac{314.46\ g/mol}{314.45\ g/mol}\approx1$, the molecular formula is the same as the empirical formula, $C_{21}H_{30}O_2$.

Answer:

$C_{21}H_{30}O_2$