An unknown compound contains only carbon, hydrogen, and oxygen. Using combustion analysis, you burned 19.31 mg of the sample and collected 17.96 mg of H2O and 43.89 mg of CO2. What is the empirical formula of this compound?

• A. C2H6O
• B. C3H6O
• C. C2H4O
• D. C3H8O

Answer: (B)

Verified step by step guidance

1. Start by determining the moles of carbon in the CO2 produced. Use the molar mass of CO2 (44.01 g/mol) to convert the mass of CO2 to moles: \( \text{moles of CO2} = \frac{43.89 \text{ mg}}{44.01 \text{ g/mol}} \). Since each mole of CO2 contains one mole of carbon, the moles of carbon will be the same as the moles of CO2.
2. Next, calculate the moles of hydrogen in the H2O produced. Use the molar mass of H2O (18.02 g/mol) to convert the mass of H2O to moles: \( \text{moles of H2O} = \frac{17.96 \text{ mg}}{18.02 \text{ g/mol}} \). Since each mole of H2O contains two moles of hydrogen, multiply the moles of H2O by 2 to find the moles of hydrogen.
3. Determine the mass of carbon and hydrogen in the original sample using their respective moles and atomic masses (carbon: 12.01 g/mol, hydrogen: 1.008 g/mol). Calculate the mass of carbon: \( \text{mass of carbon} = \text{moles of carbon} \times 12.01 \text{ g/mol} \) and the mass of hydrogen: \( \text{mass of hydrogen} = \text{moles of hydrogen} \times 1.008 \text{ g/mol} \).
4. Find the mass of oxygen in the sample by subtracting the mass of carbon and hydrogen from the total mass of the sample: \( \text{mass of oxygen} = 19.31 \text{ mg} - (\text{mass of carbon} + \text{mass of hydrogen}) \). Convert this mass to moles using the molar mass of oxygen (16.00 g/mol): \( \text{moles of oxygen} = \frac{\text{mass of oxygen}}{16.00 \text{ g/mol}} \).
5. Determine the empirical formula by finding the simplest whole number ratio of moles of carbon, hydrogen, and oxygen. Divide the moles of each element by the smallest number of moles calculated in the previous steps. Adjust the ratios to the nearest whole numbers to find the empirical formula.