Lesson eLearning

In 2.00 g of the complex “[Cu(NH3)4]SO4”, how many grams of ammonia ligands are present?

easy 1 mark Class 11 · Biology · Chemical Coordination and Integration
Answer
0.598 g

The correct answer is 0.598 g — Thus the mass of ammonia ligands in 2.00 g of the complex is 0.598 g.

Step-by-step solution

  1. Determine the molar mass of “[Cu(NH3)4]SO4”: Cu (63.55) + 4×NH3 (4×17.03) + SO4 (96.06) = 227.73 g/mol.
  2. Calculate moles of the complex: 2.00 g ÷ 227.73 g/mol = 8.78×10⁻³ mol.
  3. Each mole of complex contains 4 moles of NH3, so moles of NH3 = 4 × 8.78×10⁻³ = 3.51×10⁻² mol.
  4. Convert moles of NH3 to grams: 3.51×10⁻² mol × 17.03 g/mol = 0.598 g.
  5. Thus the mass of ammonia ligands in 2.00 g of the complex is 0.598 g.
Common mistake: Neglecting the multiple ligands per complex molecule

Concept tested: Stoichiometry of coordination complexes

Published · How this solution was prepared and checked

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