Menthol, from oil of mint, has a characteristic odor. The compound contains only C, H, and O. If 89.6 mg of menthol burns completely in O2, and gives 252 mg of CO2 and 103 mg of H2O, what is the empirical formula of menthol? (Enter the elements in the order: C, H, O.) Empirical formula:
The Correct Answer and Explanation is :
To determine the empirical formula of menthol, let’s break down the problem step by step.
Step 1: Determine the moles of carbon and hydrogen
When menthol burns, it reacts with oxygen to produce carbon dioxide (CO2) and water (H2O). The masses of CO2 and H2O produced can be used to find the moles of carbon and hydrogen in the menthol sample.
- Moles of carbon: Each mole of CO2 contains 1 mole of carbon (C). Therefore, we can calculate the moles of carbon from the mass of CO2. Moles of carbon=Mass of CO2Molar mass of CO2=252 mg44.01 g/mol=252×10−3 g44.01 g/mol=5.72×10−3 mol\text{Moles of carbon} = \frac{\text{Mass of CO2}}{\text{Molar mass of CO2}} = \frac{252 \, \text{mg}}{44.01 \, \text{g/mol}} = \frac{252 \times 10^{-3} \, \text{g}}{44.01 \, \text{g/mol}} = 5.72 \times 10^{-3} \, \text{mol} Since each mole of CO2 contains 1 mole of carbon, the moles of carbon in the sample are 5.72 × 10⁻³ mol.
- Moles of hydrogen: Each mole of H2O contains 2 moles of hydrogen. Therefore, the moles of hydrogen can be calculated from the mass of H2O. Moles of hydrogen=Mass of H2OMolar mass of H2O=103 mg18.015 g/mol=103×10−3 g18.015 g/mol=5.72×10−3 mol\text{Moles of hydrogen} = \frac{\text{Mass of H2O}}{\text{Molar mass of H2O}} = \frac{103 \, \text{mg}}{18.015 \, \text{g/mol}} = \frac{103 \times 10^{-3} \, \text{g}}{18.015 \, \text{g/mol}} = 5.72 \times 10^{-3} \, \text{mol} Since each mole of H2O contains 2 moles of hydrogen, the moles of hydrogen are 2×5.72×10−3=11.44×10−3 mol2 \times 5.72 \times 10^{-3} = 11.44 \times 10^{-3} \, \text{mol}.
Step 2: Calculate the mass of oxygen
The remaining mass in the menthol sample is oxygen, which comes from both the menthol and the oxygen gas that reacts with it. To find the mass of oxygen in the sample:
- The total mass of menthol is 89.6 mg.
- The mass of carbon and hydrogen combined is: Mass of carbon=5.72×10−3 mol×12.01 g/mol=0.0687 g=68.7 mg\text{Mass of carbon} = 5.72 \times 10^{-3} \, \text{mol} \times 12.01 \, \text{g/mol} = 0.0687 \, \text{g} = 68.7 \, \text{mg} Mass of hydrogen=11.44×10−3 mol×1.008 g/mol=0.0115 g=11.5 mg\text{Mass of hydrogen} = 11.44 \times 10^{-3} \, \text{mol} \times 1.008 \, \text{g/mol} = 0.0115 \, \text{g} = 11.5 \, \text{mg} The mass of oxygen is then: Mass of oxygen=89.6 mg−68.7 mg−11.5 mg=9.4 mg\text{Mass of oxygen} = 89.6 \, \text{mg} – 68.7 \, \text{mg} – 11.5 \, \text{mg} = 9.4 \, \text{mg} Convert this to grams: 9.4 mg=9.4×10−3 g9.4 \, \text{mg} = 9.4 \times 10^{-3} \, \text{g}
- Moles of oxygen: Using the molar mass of oxygen (16.00 g/mol), we calculate the moles of oxygen: Moles of oxygen=9.4×10−3 g16.00 g/mol=5.88×10−4 mol\text{Moles of oxygen} = \frac{9.4 \times 10^{-3} \, \text{g}}{16.00 \, \text{g/mol}} = 5.88 \times 10^{-4} \, \text{mol}
Step 3: Calculate the empirical formula
Now that we have the moles of carbon, hydrogen, and oxygen, we can determine the empirical formula by dividing each by the smallest number of moles.
- Moles of carbon: 5.72×10−3 mol5.72 \times 10^{-3} \, \text{mol}
- Moles of hydrogen: 11.44×10−3 mol11.44 \times 10^{-3} \, \text{mol}
- Moles of oxygen: 5.88×10−4 mol5.88 \times 10^{-4} \, \text{mol}
The smallest number of moles is 5.88×10−4 mol5.88 \times 10^{-4} \, \text{mol}. Dividing each by this number:
- Carbon: 5.72×10−35.88×10−4=9.7\frac{5.72 \times 10^{-3}}{5.88 \times 10^{-4}} = 9.7 (approximately 10)
- Hydrogen: 11.44×10−35.88×10−4=19.5\frac{11.44 \times 10^{-3}}{5.88 \times 10^{-4}} = 19.5 (approximately 20)
- Oxygen: 5.88×10−45.88×10−4=1\frac{5.88 \times 10^{-4}}{5.88 \times 10^{-4}} = 1
Thus, the empirical formula of menthol is approximately: C10H20O\text{C}_{10}\text{H}_{20}\text{O}
Conclusion:
The empirical formula of menthol is C10H20O\text{C}_{10}\text{H}_{20}\text{O}. This indicates that menthol contains 10 carbon atoms, 20 hydrogen atoms, and 1 oxygen atom in its simplest form.