 Caproic acid, responsible for the odor of dirty socks, is composed of C, H, and O atoms.
Combustion of a 0.225-g sample of this compound produces 0.512 g CO2 and 0.209 g H2O.
What is the empirical formula of caproic acid? 

To find the empirical formula of caproic acid, we need to determine the moles of carbon, hydrogen, and oxygen in the compound based on the data from the combustion analysis. Here’s the step-by-step process:

### 1. Determine the moles of carbon:
- Given that \( 0.512 \, \text{g} \) of \( \text{CO}_2 \) is produced.
- The molar mass of \( \text{CO}_2 \) is \( 44.01 \, \text{g/mol} \).

\[
\text{Moles of } \text{C} = \frac{0.512 \, \text{g} \, \text{CO}_2}{44.01 \, \text{g/mol}} = 0.01163 \, \text{mol of } \text{CO}_2
\]

Since each mole of \( \text{CO}_2 \) contains 1 mole of carbon:

\[
\text{Moles of } \text{C} = 0.01163 \, \text{mol}
\]

### 2. Determine the moles of hydrogen:
- Given that \( 0.209 \, \text{g} \) of \( \text{H}_2\text{O} \) is produced.
- The molar mass of \( \text{H}_2\text{O} \) is \( 18.02 \, \text{g/mol} \).

\[
\text{Moles of } \text{H}_2\text{O} = \frac{0.209 \, \text{g} \, \text{H}_2\text{O}}{18.02 \, \text{g/mol}} = 0.0116 \, \text{mol of } \text{H}_2\text{O}
\]

Since each mole of \( \text{H}_2\text{O} \) contains 2 moles of hydrogen:

\[
\text{Moles of } \text{H} = 2 \times 0.0116 \, \text{mol} = 0.0232 \, \text{mol}
\]

### 3. Determine the moles of oxygen:
To find the moles of oxygen in caproic acid, subtract the mass of carbon and hydrogen (from the masses of \( \text{CO}_2 \) and \( \text{H}_2\text{O} \)) from the total mass of the sample.

- Mass of carbon (in 0.225 g sample): 
\[
\text{Mass of } \text{C} = 0.01163 \, \text{mol} \times 12.01 \, \text{g/mol} = 0.1397 \, \text{g}
\]

- Mass of hydrogen (in 0.225 g sample): 
\[
\text{Mass of } \text{H} = 0.0232 \, \text{mol} \times 1.008 \, \text{g/mol} = 0.0234 \, \text{g}
\]

- Mass of oxygen:
\[
\text{Mass of } \text{O} = 0.225 \, \text{g} - (0.1397 \, \text{g} + 0.0234 \, \text{g}) = 0.0619 \, \text{g}
\]

- Moles of oxygen:
\[
\text{Moles of } \text{O} = \frac{0.0619 \, \text{g}}{16.00 \, \text{g/mol}} = 0.00387 \, \text{mol}
\]

### 4. Determine the simplest whole number ratio:
To find the empirical formula, divide the moles of each element by the smallest number of moles calculated:

\[
\text{C}: \frac{0.01163}{0.00387} = 3, \quad \text{H}: \frac{0.0232}{0.00387} = 6, \quad \text{O}: \frac{0.00387}{0.00387} = 1
\]

### 5. Conclusion:
The empirical formula of caproic acid is \( \text{C}_3\text{H}_6\text{O} \).

Would you like more details on this process or have any questions?

Here are some related questions you might find useful:
1. How do you find the molecular formula if the molar mass is known?
2. What is the significance of an empirical formula in chemistry?
3. Can you have an empirical formula with non-integer subscripts?
4. How does combustion analysis work in determining empirical formulas?
5. What are common errors to avoid when performing combustion analysis?

**Tip:** Always ensure that your final empirical formula is in the simplest whole-number ratio by dividing each element's mole count by the smallest value calculated.

Caproic acid, responsible for the odor of dirty socks, is composed of C, H, and O atoms. Combustion of a 0.225-g sample of this compound produces 0.512 g CO2 and 0.209 g H2O. What is the empirical formula of caproic acid?

Learn how to calculate the empirical formula of caproic acid from combustion data. This problem covers key stoichiometric concepts, including combustion analysis and empirical formula calculation. Suitable for students in Grades 10-12.

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