Question:medium

The elemental analysis of an organic compound gave:
C: 38.71%, H: 9.67%, O: 51.67%. What is the empirical formula of the compound?

Show Hint

To determine the empirical formula from elemental analysis data, first calculate the moles of each element, then find the smallest mole ratio and simplify it to get the empirical formula.
Updated On: Apr 18, 2026
  • \( \text{CH}_2\text{O} \)
  • \( \text{CH}_3\text{O} \)
  • \( \text{CH}_4\text{O} \)
  • \( \text{CHO} \)
  • \( \text{CH}_5\text{O} \)
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: Understanding the Concept:
The empirical formula shows the simplest whole-number ratio of the different atoms present in a chemical compound.
We determine this ratio by converting mass percentages into relative moles for each element.
Step 2: Key Formula or Approach:
The number of moles for each element is calculated as \(\frac{\text{Percentage by mass}}{\text{Atomic mass}}\).
Then, we divide each mole value by the smallest calculated mole value to find the simplest ratio.
Step 3: Detailed Explanation:
Let us assume we have a 100 g sample, which means the percentages translate directly to grams.
Mass of C = 38.71 g.
Mass of H = 9.67 g.
Mass of O = 51.67 g.
We calculate the moles using standard atomic masses (C = 12 g/mol, H = 1 g/mol, O = 16 g/mol):
Moles of C = \(\frac{38.71}{12} \approx 3.22 \text{ mol}\).
Moles of H = \(\frac{9.67}{1} = 9.67 \text{ mol}\).
Moles of O = \(\frac{51.67}{16} \approx 3.22 \text{ mol}\).
Next, we divide each by the smallest mole value (3.22) to get the atomic ratio:
C ratio = \(\frac{3.22}{3.22} = 1\).
H ratio = \(\frac{9.67}{3.22} \approx 3\).
O ratio = \(\frac{3.22}{3.22} = 1\).
This results in a simple whole-number ratio of 1:3:1 for C:H:O.
Step 4: Final Answer:
The empirical formula of the given compound is CH\(_{3}\)O.
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