Mole Concept & Stoichiometry Miscellaneous Exercises — Question 4

(i) Molar mass of NaNO₃ = 23 + 14 + 3(16) = 23 + 14 + 48 = 85 g

Nitrogen in 85 g NaNO₃ = 14 g

∴ Percentage of Nitrogen in NaNO₃ = $\dfrac{14}{85}$ x 100 = 16.47%

(ii) Molar mass of (NH₄)₂SO₄ = 2[14 + 4(1)] + 32 + 4(16) = 36 + 32 + 64 = 132 g

Nitrogen in 132 g of (NH₄)₂SO₄ = 28 g

∴ Percentage of Nitrogen in (NH₄)₂SO₄ = $\dfrac{28}{132}$ x 100 = 21.21%

(iii) Molar mass of CO(NH₂)₂ = 12 + 16 + 2[14 + (2 x 1)]

= 28 + 2(16)

= 28 + 32

= 60 g

Nitrogen in 60 g of CO(NH₂)₂ = 2 x 14 = 28 g

∴ Percentage of Nitrogen in CO(NH₂)₂ = $\dfrac{28}{60}$ x 100 = 46.66%

So, the highest percentage of Nitrogen is in Urea.