4. Find the empirical formula for a compound with 65.4% C. 5.5% H, and 29.1% a a. C_(5)H_(5)O_(1) b. C_(3)H_(3)O C. C_(5)H_(6)O_(2) d. CHO

To find the empirical formula for the compound, we need to determine the simplest whole number ratio of the elements present in the compound.<br /><br />Given information:<br />- 65.4% Carbon (C)<br />- 5.5% Hydrogen (H)<br />- 29.1% Oxygen (O)<br /><br />Step 1: Convert the percentages to grams.<br />- Carbon (C): 65.4 g<br />- Hydrogen (H): 5.5 g<br />- Oxygen (O): 29.1 g<br /><br />Step 2: Convert the grams to moles.<br />- Carbon (C): 65.4 g / 12.01 g/mol = 5.45 mol<br />- Hydrogen (H): 5.5 g / 1.008 g/mol = 5.47 mol<br />- Oxygen (O): 29.1 g / 16.00 g/mol = 1.81 mol<br /><br />Step 3: Determine the simplest whole number ratio of the elements.<br />- Divide each mole value by the smallest mole value (1.81 mol).<br />- Carbon (C): 5.45 mol / 1.81 mol = 3<br />- Hydrogen (H): 5.47 mol / 1.81 mol = 3<br />- Oxygen (O): 1.81 mol / 1.81 mol = 1<br /><br />The empirical formula for the compound is $C_{3}H_{3}O$.<br /><br />Therefore, the correct answer is:<br />b. $C_{3}H_{3}O$