Combustion analysis of 63.8 mg of a compound containing carbon, hydrogen, and oxygen produced 145.0 mg of CO2 and 59.38 mg of H2O. What is the empirical formula for the compound?

• A. C4H10O
• B. C3H8O
• C. C2H4O
• D. C3H6O2

Answer: (C)

Verified step by step guidance

1. Start by determining the amount of carbon in the compound. Use the mass of CO2 produced to find the mass of carbon. The molar mass of CO2 is 44.01 g/mol, and the molar mass of carbon is 12.01 g/mol. Calculate the moles of CO2 using the formula: $\frac{145.0}{44.01}$. Then, use the ratio of carbon in CO2 to find the mass of carbon: $\frac{12.01}{44.01}$.
2. Next, determine the amount of hydrogen in the compound using the mass of H2O produced. The molar mass of H2O is 18.02 g/mol, and the molar mass of hydrogen is 1.01 g/mol. Calculate the moles of H2O using the formula: $\frac{59.38}{18.02}$. Then, use the ratio of hydrogen in H2O to find the mass of hydrogen: $\frac{2.02}{18.02}$.
3. Calculate the mass of oxygen in the compound. Subtract the mass of carbon and hydrogen from the total mass of the compound to find the mass of oxygen: $63.8-\mathrm{mass\; of\; carbon}-\mathrm{mass\; of\; hydrogen}$.
4. Convert the masses of carbon, hydrogen, and oxygen to moles using their respective molar masses: Carbon: $\frac{\mathrm{mass\; of\; carbon}}{12.01}$, Hydrogen: $\frac{\mathrm{mass\; of\; hydrogen}}{1.01}$, Oxygen: $\frac{\mathrm{mass\; of\; oxygen}}{16.00}$.
5. Determine the empirical formula by finding the simplest whole number ratio of moles of carbon, hydrogen, and oxygen. Divide each mole value by the smallest number of moles calculated in the previous step to get the ratio. Adjust the ratio to the nearest whole number to find the empirical formula.