The percentages of C, H, and N in an organic compound are 40%, 13.3%, and 46.7% respectively. The empirical fo

Question

The percentages of C, H, and N in an organic compound are 40%, 13.3%, and 46.7% respectively. The empirical formula of the compound is:

Accepted Answer

1. **Determine the number of moles of each element:** Assume a $100	ext{ g}$ sample of the compound to directly convert percentages to mass.
   - Moles of Carbon (C): $\frac{40	ext{ g}}{12	ext{ g/mol}} = 3.33	ext{ mol}$
   - Moles of Hydrogen (H): $\frac{13.3	ext{ g}}{1	ext{ g/mol}} = 13.3	ext{ mol}$
   - Moles of Nitrogen (N): $\frac{46.7	ext{ g}}{14	ext{ g/mol}} \approx 3.335	ext{ mol}$
2. **Determine the simplest molar ratio:** Divide the mole values by the smallest number of moles calculated ($3.33	ext{ mol}$).
   - Ratio for C: $\frac{3.33}{3.33} = 1$
   - Ratio for H: $\frac{13.3}{3.33} \approx 4$
   - Ratio for N: $\frac{3.335}{3.33} \approx 1$
3. **Construct the Formula:** The simplest whole number ratio of C:H:N is $1:4:1$. Therefore, the empirical formula is $	ext{CH}_4	ext{N}$.

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