If the relative molecular mass of ammonium nitrate is 80, calculate the percentage of nitrogen and oxygen in ammonium nitrate.

[N = 14, H = 1, O = 16]

Given,

molecular mass of ammonium nitrate = 80

mass of nitrogen in 1 mole of ammonium nitrate (NH₄NO₃) is 2(14) = 28 g

percentage of nitrogen in ammonium nitrate = $\dfrac{28}{80}$ x 100 = 35%

mass of oxygen in 1 mole of ammonium nitrate (NH₄NO₃) is 3(16) = 48 g

percentage of oxygen in ammonium nitrate = $\dfrac{48}{80}$ x 100 = 60%

Hence, percentage of nitrogen is 35% and percentage of oxygen is 60%