NECO 2003 Chemistry · Question 27
An organic compound gave the following results on analysis: C = 29.3% ; H = 5.7%; Br = 65%. What is the…
Question
An organic compound gave the following results on analysis: C = 29.3% ; H = 5.7%; Br = 65%. What is the empirical formula of the compound? [C = 12, H = 1, Br=80]
- A.CH3Br
- B.C2H5Br
- C.C3H5Br
- D.C3H7Br
- E.C4H9Br
Answer: D. C3H7Br
Explanation written with AI and checked against the answer key
Note: the percentages given (29.3 + 5.7 + 65 = 100.0) add up to 100, so we can use them directly.
Step 1: Analyze the question We need the empirical formula, which is the simplest whole-number ratio of atoms in the compound. The method is to divide each percentage by its atomic mass, then divide by the smallest result.
Step 2: Divide by atomic masses
- C: 29.3 ÷ 12 = 2.44
- H: 5.7 ÷ 1 = 5.7
- Br: 65 ÷ 80 = 0.8125
Step 3: Divide by the smallest value (0.8125)
- C: 2.44 ÷ 0.8125 ≈ 3.0
- H: 5.7 ÷ 0.8125 ≈ 7.0
- Br: 0.8125 ÷ 0.8125 = 1
Step 4: Write the formula The ratio C : H : Br = 3 : 7 : 1, so the empirical formula is C₃H₇Br.
Check: Molar mass = 36 + 7 + 80 = 123.
- %C = 36/123 = 29.3% ✓
- %H = 7/123 = 5.7% ✓
- %Br = 80/123 = 65.0% ✓
Tip: In empirical formula questions, always follow this order: percentage → divide by atomic mass → divide by smallest → round to whole numbers. If you get a ratio like 1.5, multiply everything by 2. Here the ratios came out as clean whole numbers, which confirms the answer.
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