An organic compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen. What is the empirical formula of the compound? [O = 16.0, C = 12.0, H = 1.0]

Step 1: Analyze the question We are given the percentage composition by mass and must find the simplest whole-number ratio of atoms (the empirical formula).

Step 2: The concept Empirical formula steps:

1. Assume 100 g, so the percentages become masses in grams.
2. Divide each mass by its atomic mass to get moles.
3. Divide by the smallest number of moles.
4. Round to the simplest whole-number ratio.

Step 3: Solution

| Element | Mass (g) | Moles = mass ÷ atomic mass | |---|---|---| | C | 40.0 | 40.0 ÷ 12.0 = 3.33 | | H | 6.7 | 6.7 ÷ 1.0 = 6.7 | | O | 53.3 | 53.3 ÷ 16.0 = 3.33 |

Divide by the smallest (3.33):

• C: 3.33 ÷ 3.33 = 1
• H: 6.7 ÷ 3.33 ≈ 2
• O: 3.33 ÷ 3.33 = 1

The ratio is C : H : O = 1 : 2 : 1, so the empirical formula is CH₂O.

Step 4: Answer The correct option is C. CH₂O.

Tip: If the percentages for C and O give equal mole values and H is about double, think CH₂O (the formula pattern of carbohydrates and methanal). Always divide by the smallest mole value, and don't round too early. Values like 1.5 mean you should multiply everything by 2.