Question

a) calculate the total number of hydrogen atoms (the actual number, not the number of moles) in 1.78 g NH3.
b) determine the mass percent of hydrogen in 1.78 g NH3.

Answer 1

1 mol NH3 = 1 mol N + 3 mol H

1 mol NH3 = 3 x 6.023 x 10²³ H atoms

17 g NH3 = 3 x 6.023 x 10²³ H atoms

1.78 g = ? H atoms

? = (1.78 g/ 17 g) x 3 x 6.023 x 10²³ H atoms

   = 1.89 x 10²³ H atoms

Therefore, 1.78 g NH₃ has 1.89 x 10²³ Hydrogen atoms.

b)

mass percentage of hydrogen in NH3 = molar mass of 3 hydrogens / molar mass of NH₃ = (3 /17) x 100 = 17.64

Hence,

    17 g NH3 = 17.64 % H

   1.78 g NH3 = ? % H

   ? = (1.78/ 17) x 17.64 %

       = 1.847 %

Therefore,

   mass percent of hydrogen in 1.78 g NH3 = 1.847 %