A compound that is usually used as a fertilizer can also be used as a powerful explosive. the compound has the composition 35.00% nitrogen, 59.96% oxygen and the remainder being hydrogen. what is its empirical formula? given it is ionic, suggest a name for the compound.

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.