What is the empirical formula of a compound that is 64.8% C, 13.6% H, and 21.6% O by mass?

To determine the empirical formula of a compound, we start by converting the percentages of each element into grams. This means we can assume we have 100 grams of the compound, which gives us:

• Carbon (C): 64.8 grams
• Hydrogen (H): 13.6 grams
• Oxygen (O): 21.6 grams

Next, we convert the masses of each element to moles by dividing by their atomic masses:

• Carbon: 64.8 g / 12.01 g/mol ≈ 5.4 moles
• Hydrogen: 13.6 g / 1.008 g/mol ≈ 13.49 moles
• Oxygen: 21.6 g / 16.00 g/mol ≈ 1.35 moles

Now, we divide each of the mole values by the smallest number of moles calculated:

• Carbon: 5.4 moles / 1.35 moles ≈ 4
• Hydrogen: 13.49 moles / 1.35 moles ≈ 10
• Oxygen: 1.35 moles / 1.35 moles ≈ 1

This gives us a ratio of approximately C₄H₁₀O₁.

Thus, the empirical formula of the compound is C₄H₁₀O.