Question

The percentage of C, H and N in an organic compound are 40%, 13.3% and 46.7% respectively then emprirical formula is 

• A. C₃H₁₃N₃
• B. CH₂N
• C. CH₄N
• D. CH₆N

Answer 1

The Correct Option is C

To determine the empirical formula of an organic compound from the given percentage composition of elements, we follow these steps:

1. Identify the percentages of each element present:
   • Carbon (C) = 40%
   • Hydrogen (H) = 13.3%
   • Nitrogen (N) = 46.7%
2. Assume a 100-gram sample of the compound, so:
   • Mass of C = 40 g
   • Mass of H = 13.3 g
   • Mass of N = 46.7 g
3. Convert these masses to moles using the atomic masses:
   • Moles of C: \frac{40}{12} = 3.33
   • Moles of H: \frac{13.3}{1} = 13.3
   • Moles of N: \frac{46.7}{14} = 3.34
4. Determine the simplest mole ratio by dividing each by the smallest number of moles:
   • Mole ratio of C: \frac{3.33}{3.33} = 1
   • Mole ratio of H: \frac{13.3}{3.33} \approx 4
   • Mole ratio of N: \frac{3.34}{3.33} \approx 1
5. Write the empirical formula using these ratios:
   • This gives the empirical formula as CH₄N

We verify that the calculated empirical formula, CH₄N, matches the given correct answer: CH₄N. Other options do not provide the correct mole ratios, and hence are incorrect.