Transition-metal complexes: observations and isolated yield

What you’ll learn

• Distinguish hydroxide precipitation from ligand substitution.
• Calculate complex yield using the stated formula and hydration.

Understand the idea

Core practical 14 concerns a transition-metal complex. In solution, ligand substitution changes the species attached to a metal ion and may change its colour. A precipitate is a separate solid phase, so formation of a pale-blue copper(II) hydroxide precipitate and its dissolution in excess ammonia are different observations. A small amount of ammonia can act as a base; excess ammonia can act as a ligand and stabilise a soluble deep-blue copper complex.

Write formulae with the correct metal charge, ligand number and overall charge. In an aqueous description, the deep-blue copper species can be represented as $[\mathrm{Cu(NH_3)_4(H_2O)_2}]^{2+}$. An isolated solid may be specified using a different hydration formula. For yield calculations, use the exact product formula and molar mass supplied; water of crystallisation contributes to the mass but is not extra metal.

The metal-ion amount often determines the maximum amount of complex when other components are in excess. A dry recovered mass, colour and other supplied purity evidence support evaluation of the product, but colour alone does not establish purity. Residual solution can inflate mass, while dissolution and transfer losses can reduce recovery. Ammonia-containing and metal-containing materials require the school's ventilation and waste controls; the task here is to interpret a preparation record rather than reproduce a synthesis.

Key formulae

$$\mathrm{Cu^{2+}}+2\mathrm{OH^-}\rightarrow\mathrm{Cu(OH)_2(s)}$$

$$[\mathrm{Cu(H_2O)_6}]^{2+}+4\mathrm{NH_3}\rightleftharpoons[\mathrm{Cu(NH_3)_4(H_2O)_2}]^{2+}+4\mathrm{H_2O}$$

$$n_{\mathrm{complex,max}}=n_{\mathrm{metal}}\quad\text{for one metal ion per formula unit}$$

$$\mathrm{yield}\ (\%)=\frac{m_{\mathrm{isolated}}}{n_{\mathrm{complex,max}}M_{\mathrm{complex}}}\times100$$

Worked example

A supplied record uses $2.497\ \mathrm{g}$ of $\mathrm{CuSO_4\cdot5H_2O}$, molar mass $249.7\ \mathrm{g\,mol^{-1}}$. Each mole can give one mole of the specified complex salt, molar mass $245.7\ \mathrm{g\,mol^{-1}}$. Other components are in excess and the dry recovered mass is $1.966\ \mathrm{g}$. Calculate the yield.

1. The amount of copper starting material is $2.497/249.7=0.01000\ \mathrm{mol}$.
2. The maximum complex amount is also $0.01000\ \mathrm{mol}$, so its theoretical mass is $0.01000\times245.7=2.457\ \mathrm{g}$.
3. The yield is $100(1.966/2.457)=80.016\%$, which is approximately $80.0\%$.

Answer: Approximately $80.0\%$.