Elemental analysis of a pure compound indicated that the compound had 72.2% C, 8.50% H and the remainder as O. If 0.250 moles of the compound weighs 41.55 g, what is the molecular formula of the compound?

• A. C₅H₇O
• B. C₁₀H₁₄O₂
• C. C₆H₈O₂
• D. C₁₂H₁₆O₄

Answer: (B)

Verified step by step guidance

1. Determine the empirical formula by converting the percentage composition to moles. Assume 100 g of the compound, which gives 72.2 g of C, 8.50 g of H, and the remainder as O. Calculate the moles of each element using their atomic masses: C (12.01 g/mol), H (1.008 g/mol), and O (16.00 g/mol).
2. Calculate the moles of each element: \( \text{moles of C} = \frac{72.2}{12.01} \), \( \text{moles of H} = \frac{8.50}{1.008} \), \( \text{moles of O} = \frac{100 - 72.2 - 8.50}{16.00} \).
3. Determine the simplest whole number ratio of moles of C, H, and O by dividing each by the smallest number of moles calculated in the previous step.
4. Use the empirical formula to find the empirical formula mass. Then, calculate the molar mass of the compound using the given mass and moles: \( \text{molar mass} = \frac{41.55 \text{ g}}{0.250 \text{ moles}} \).
5. Determine the molecular formula by dividing the molar mass by the empirical formula mass to find the multiplier. Multiply the subscripts in the empirical formula by this number to get the molecular formula.