Part 6: Core Practical 12: prepare and evaluate a complex

Connect ligand exchange to isolation of tetraamminecopper(II) sulfate monohydrate, then calculate a theoretical yield and explain losses, contamination and uncertainty.

The aqueous complex and isolated salt need different formulae

In Pearson’s CP12 example, hydrated copper(II) sulfate reacts with ammonia and the isolated product is tetraamminecopper(II) sulfate monohydrate, written [Cu(NH₃)₄]SO₄·H₂O. In the deep-blue aqueous solution the complex is represented as [Cu(NH₃)₄(H₂O)₂]²⁺. Do not copy the aqueous ion formula into the solid’s formula-mass calculation: the supplied crystalline formula determines the theoretical mass.

Copper remains +2 throughout. Ammonia replaces water-derived ligands; sulfate acts as the counterion in the product salt. The empirical preparation equation conserves one copper and one sulfate per formula unit, so one mole of CuSO₄·5H₂O can produce one mole of [Cu(NH₃)₄]SO₄·H₂O when ammonia is in excess.

The expected observations are dissolution of blue copper sulfate crystals, possible initial pale-blue hydroxide as ammonia is added, deep-blue solution in excess ammonia and deep-blue crystals on isolation. Colour supports formation of an ammine complex but is not by itself proof of purity or the precise hydration state.

Dissolve, exchange ligands, then lower solubility

Follow the teacher-approved preparation on the stated scale. Pearson’s method uses about 1.4–1.6 g CuSO₄·5H₂O, accurately measured by difference, dissolved in 4 cm³ water using a hot-water bath. Remove from the bath, then add 2 cm³ concentrated ammonia with stirring in a fume cupboard. Mix the solution into 6 cm³ ethanol and cool in ice before filtering the crystals.

A small water volume dissolves the starting material without leaving an unnecessarily large solvent volume in which product remains dissolved. Gentle warming speeds dissolution. Excess ammonia drives ligand exchange towards the ammine complex. Ethanol acts as an antisolvent: the product is less soluble in the mixed solvent, so it crystallises. Cooling further reduces solubility and improves recovery.

These purpose statements matter when adapting a method. Adding much more water may make dissolution easier but reduce the amount of product crystallised. Adding more ammonia than needed can change solvent volume and waste handling rather than guarantee higher purity. The specified recipe balances solubility, conversion and isolation.

Collect and dry crystals without undoing the preparation

Use a Büchner funnel with a correctly fitted filter paper and a thick-walled side-arm flask connected to the vacuum system. Reduced pressure below the paper draws mother liquor through while the crystals remain above. Clamp the flask and follow the laboratory sequence for releasing suction to prevent backflow; an ordinary thin conical flask is not designed for this use.

Transfer crystals quantitatively and use small portions of cold ethanol to rinse the vessel and wash the solid. Washing removes adhering mother liquor and soluble impurities; cold solvent limits product dissolution. Large warm washes can improve removal of impurities at the cost of losing more product. A few crystals left on every surface can significantly lower recovery on this small scale.

Press crystals gently between clean dry filter papers, moving to dry areas as needed, and measure the dry mass. Do not use strong heating to force dryness: the hydrated ammine salt can lose water or ammonia and change composition. Retained solvent raises measured mass; decomposition can lower it. A repeated stable mass under the approved drying conditions is better evidence of dryness than appearance alone.

The method develops accurate weighing, controlled heating, crystallisation and vacuum filtration. Explaining these operations in an exam is different from demonstrating competent practical work for 9CH0/04; the preparation and records must also be completed under supervision.

Worked example: theoretical mass and percentage yield

Original illustrative data: 1.560 g CuSO₄·5H₂O is used with ammonia in excess and 1.210 g of dry [Cu(NH₃)₄]SO₄·H₂O is recovered. Use Cu 63.5, S 32.1, O 16.0, N 14.0 and H 1.0. Mr(CuSO₄·5H₂O) = 63.5 + 32.1 + 4(16.0) + 5(18.0) = 249.6. Mr(product) = 63.5 + 4(17.0) + 32.1 + 4(16.0) + 18.0 = 245.6.

n(starting copper salt) = 1.560/249.6 = 0.006250 mol. The equation gives a 1:1 product ratio, so theoretical product amount = 0.006250 mol. Theoretical product mass = 0.006250 × 245.6 = 1.535 g. Percentage yield = 1.210/1.535 × 100 = 78.8% to three significant figures.

The theoretical mass is slightly less than the starting mass because the solid’s formula mass changes; it is not equal to the copper sulfate mass. The calculation assumes the stated hydrate and excess ammonia. If another reagent were limiting, compare moles divided by their stoichiometric coefficients before identifying the theoretical product amount.

Pearson’s example uses rounded S = 32.0, giving formula masses 249.5 and 245.5. Small differences from an alternative table are not chemical contradictions: state and use one internally consistent set of relative atomic masses, and keep unrounded values until the final answer.

A high yield does not automatically mean a good preparation

Suppose the same preparation gave 1.620 g ‘product’. Apparent yield would be 1.620/1.535 × 100 = 105.5%. The reaction has not made more copper than was present: retained solvent, impurities or an incorrect product formula must be investigated. Even a value below 100% can contain impurities, so yield and purity are independent assessments.

For an illustrative balance uncertainty of ±0.001 g per reading, a mass found by two weighings has a maximum uncertainty of ±0.002 g. The relative contributions for 1.560 g starting material and 1.210 g product are 0.128% and 0.165%; the simple maximum percentage uncertainty in their ratio is about 0.293%. This is much smaller than a 21.2% shortfall in yield, so balance reading uncertainty alone cannot explain that loss.

Use the actual balance specification and actual weighing method when calculating uncertainty. Replication estimates reproducibility, but a shared solvent loss or incomplete reaction can recur in every trial. Evaluate conversion, recovery and purity separately and propose changes that target the observed cause rather than merely ‘use better equipment’.

Quick checks

Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official Edexcel mark allocations.

Sources