Answer:
Empirical Formula = NH₄NO₃ (Ammonium Nitrate)
Solution:
Step 1: Calculate Moles of each Element;
Moles of N = %N ÷ At.Mass of N
Moles of N = 35.0 ÷ 14
Moles of N = 2.5 mol
Moles of O = %O ÷ At.Mass of O
Moles of O = 59.96 ÷ 16
Moles of O = 3.7475 mol
Moles of H = [100% - (%N + %O)] ÷ At.Mass of H
Moles of H = [100% - (35.0 + 59.96)] ÷ 1.008
Moles of H = [100% - 94.96] ÷ 1.008
Moles of H = 5.04 ÷ 1.008
Moles of H = 5 mol
Step 2: Find out mole ratio and simplify it;
N H O
2.5 5 3.7475
2.5/2.5 5/2.5 3.7475/2.5
1 2 1.5
Multiply Mole Ratio by 2,
2 4 3
Result:
Empirical Formula = N₂H₄O₃
Or,
Empirical Formula = NH₄NO₃
This empirical formula is also a Molecular Formula for Ammonium Nitrate a well known Fertilizer and often misused in the formation of Explosives.
