Acompound 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.

fe 55.85 g/mol and h2o 18.02 g/mol.