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]

  1. A.CH3Br
  2. B.C2H5Br
  3. C.C3H5Br
  4. D.C3H7Br
  5. 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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