Concept:To identify the compound with the highest elemental percentage, calculate the ratio of the mass of Nitrogen within one mole of the compound to the total molar mass of that compound.
Formula:$$ \%\text{N} = \left( \frac{\text{Mass of Nitrogen}}{\text{Molar mass of compound}} \right) \times 100 $$
Solution:- A) NO: Molar mass = \( 14 + 16 = 30 \). \( \%\text{N} = (14 / 30) \times 100 = 46.6\% \).
- B) \( \text{NO}_2 \): Molar mass = \( 14 + 32 = 46 \). \( \%\text{N} = (14 / 46) \times 100 = 30.4\% \).
- C) \( \text{N}_2\text{O} \): Molar mass = \( 28 + 16 = 44 \). \( \%\text{N} = (28 / 44) \times 100 = 63.6\% \).
- D) \( \text{N}_2\text{O}_5 \): Molar mass = \( 28 + 80 = 108 \). \( \%\text{N} = (28 / 108) \times 100 = 25.9\% \).
- By direct comparison, \( \text{N}_2\text{O} \) (Nitrous oxide) contains the highest mass percentage of Nitrogen (63.6%).
Why other options are incorrect:- NO, \( \text{NO}_2 \), \( \text{N}_2\text{O}_5 \): These all feature higher ratios of Oxygen atoms relative to Nitrogen atoms compared to \( \text{N}_2\text{O} \), thereby heavily diluting the percentage of Nitrogen by mass.
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