An organic compound contains 72% carbon, 12% hydrogen and 16% oxygen by mass. What is the empirical formula of the compound? [ H = 1.0. C = 12.0, O = 16.0 ]

Analyze the question: We're given percentage composition and asked for the empirical formula, which is the simplest whole-number ratio of atoms.

Concept: Divide each percentage (mass in 100 g) by the atomic mass to get moles. Then divide by the smallest number of moles to get the ratio.

Solution:

| Element | Mass (g) | Moles = mass / atomic mass | |---|---|---| | C | 72 | 72 / 12 = 6.0 | | H | 12 | 12 / 1 = 12.0 | | O | 16 | 16 / 16 = 1.0 |

Divide by the smallest (1.0):

• C : H : O = 6 : 12 : 1

So the empirical formula is C₆H₁₂O.

Check against the options:

• A. C₃H₈O gives a ratio of 3 : 8 : 1, which does not match.
• B. C₆H₁₂O gives a ratio of 6 : 12 : 1, which matches.
• C. C₆H₁₂O₂ gives a ratio of 6 : 12 : 2, which does not match.
• D. C₁₂H₁₂O₁₁ does not match.

Tip: When the percentages add up to 100% (72 + 12 + 16 = 100), just treat them as grams. Then follow the steps: percentage → moles → divide by the smallest → whole-number ratio. If you get a decimal like 1.5, multiply all the ratios by 2 to get whole numbers.