Question

combustion analysis of a 13.42 - g sample of estradiol (which contains only carbon, hydrogen, and oxygen) produced 39.01 g co₂ and 10.65 g h₂o. the molar mass of estradiol is 272.38 g/mol. part a find the molecular formula for estradiol. express your answer as a chemical formula.

Explanation:

Step1: Calculate moles of carbon

The molar mass of $CO_2$ is $44.01\ g/mol$. Moles of $CO_2=\frac{39.01\ g}{44.01\ g/mol}=0.8864\ mol$. Since 1 mole of $CO_2$ contains 1 mole of carbon, moles of carbon in estradiol is also $0.8864\ mol$.

Step2: Calculate moles of hydrogen

The molar mass of $H_2O$ is $18.02\ g/mol$. Moles of $H_2O=\frac{10.65\ g}{18.02\ g/mol}=0.591\ mol$. Since 1 mole of $H_2O$ contains 2 moles of hydrogen, moles of hydrogen in estradiol is $2\times0.591\ mol = 1.182\ mol$.

Step3: Calculate mass of carbon and hydrogen

Mass of carbon = $0.8864\ mol\times12.01\ g/mol = 10.646\ g$. Mass of hydrogen = $1.182\ mol\times1.008\ g/mol = 1.192\ g$.

Step4: Calculate mass of oxygen

Mass of oxygen = $13.42\ g-(10.646\ g + 1.192\ g)=1.582\ g$.

Step5: Calculate moles of oxygen

The molar mass of oxygen is $16.00\ g/mol$. Moles of oxygen = $\frac{1.582\ g}{16.00\ g/mol}=0.0989\ mol$.

Step6: Find the empirical - formula ratio

Divide the number of moles of each element by the smallest number of moles (moles of oxygen). For carbon: $\frac{0.8864\ mol}{0.0989\ mol}\approx9$. For hydrogen: $\frac{1.182\ mol}{0.0989\ mol}\approx12$. For oxygen: $\frac{0.0989\ mol}{0.0989\ mol}=1$. The empirical formula is $C_9H_{12}O$.

Step7: Calculate the empirical - formula mass

Empirical - formula mass = $(9\times12.01\ g/mol)+(12\times1.008\ g/mol)+(1\times16.00\ g/mol)=128.196\ g/mol$.

Step8: Find the ratio of molar mass to empirical - formula mass

$n=\frac{272.38\ g/mol}{128.196\ g/mol}\approx2$.

Step9: Determine the molecular formula

Multiply the sub - scripts in the empirical formula by $n$. The molecular formula is $C_{18}H_{24}O_2$.

Answer:

$C_{18}H_{24}O_2$