Given 4.80g of ammonium carbonate, find:

Number of moles of the compound

Number of moles of ammonium ions

Number of moles of carbonate ions

Number of moles of hydrogen atoms

Number of hydrogen atoms

1) 0.05 mol.

2) 0.1 mol.

3) 0.05 mol.

4) 0.4 mol.

5) 2.4 x 10²³ molecules.

Explanation:

1) Number of moles of the compound:

no. of moles of ammonium carbonate = mass/molar mass = (4.80 g) / (96.09 g/mol) = 0.05 mol.

2) Number of moles of ammonium ions:

Ammonium carbonate is dissociated according to the balanced equation:

(NH₄) ₂CO₃ → 2NH₄⁺ + CO₃²⁻.

It is clear that every 1.0 mole of (NH₄) ₂CO₃ is dissociated to produce 2.0 moles of NH₄⁺ ions and 1.0 mole of CO₃²⁻ ions.

∴ The no. of moles of NH₄⁺ ions in 0.05 mol of (NH₄) ₂CO₃ = (2.0) (0.05 mol) = 0.1 mol.

3) Number of moles of carbonate ions:

Ammonium carbonate is dissociated according to the balanced equation:

(NH₄) ₂CO₃ → 2NH₄⁺ + CO₃²⁻.

It is clear that every 1.0 mole of (NH₄) ₂CO₃ is dissociated to produce 2.0 moles of NH₄⁺ ions and 1.0 mole of CO₃²⁻ ions.

∴ The no. of moles of CO₃²⁻ ions in 0.05 mol of (NH₄) ₂CO₃ = (1.0) (0.05 mol) = 0.05 mol.

4) Number of moles of hydrogen atoms:

Every 1.0 mol of (NH₄) ₂CO₃ contains:

2.0 moles of N atoms, 8.0 moles of H atoms, 1.0 mole of C atoms, and 3.0 moles of O atoms.

∴ The no. of moles of H atoms in 0.05 mol of (NH₄) ₂CO₃ = (8.0) (0.05 mol) = 0.4 mol.

5) Number of hydrogen atoms:

It is known that every mole of a molecule or element contains Avogadro's number (6.022 x 10²³) of molecules or atoms.

Using cross multiplication:

1.0 mole of H atoms contains → 6.022 x 10²³ atoms.

0.4 mole of H atoms contains →? atoms.

∴ The no. of atoms in 0.4 mol of H atoms = (6.022 x 10²³ molecules) (0.4 mole) / (1.0 mole) = 2.4 x 10²³ molecules.