Crystallisation on Cooling

Introduction

Cooling can reduce a solute's capacity in a solvent and force some dissolved material to crystallize. The simplest calculation subtracts cool solubility from hot solubility on the same solvent basis. That arithmetic works only when the starting solution, solvent amount and crystal composition are understood.

Core explanation

Suppose a salt dissolves at 60 g per 100 g water when hot and 25 g per 100 g water when cool. A hot saturated solution prepared with 200 g water contains 120 g dissolved salt. After cooling to equilibrium, that same 200 g water can hold 50 g dissolved. If no solvent is lost and the crystals are the same anhydrous salt, 120 − 50 = 70 g crystals can form. The liquid left behind is called mother liquor; its mass is 200 + 50 = 250 g under this simplified account.

The calculation is a solute mass balance. Initial dissolved solute equals solute in final mother liquor plus solute in crystals. If the hot solution was not saturated, use its actual starting dissolved mass rather than the hot solubility limit. For example, if only 80 g had been dissolved in the 200 g water, cooling to the 50 g capacity would form 30 g crystals, not 70 g. The hot graph value tells maximum possible starting content, not what was necessarily present.

Cooling without reaching equilibrium may produce less crystal mass than predicted. Supersaturation can persist; crystals may remain tiny or dissolved material may stay above the final equilibrium limit for a time. Actual collected mass may also be lower if some crystals remain in the mother liquor or are lost during filtration and transfer. A theoretical mass balance and experimental recovery are different quantities.

Hydrated crystals require special treatment. If a dissolved anhydrous salt forms a crystal containing water, each gram of crystal removes both solute and water from the mother liquor. The remaining solvent mass is not the same as the initial water mass, so simply multiplying the cool solubility by the starting water mass can be wrong. A problem involving CuSO₄·5H₂O, for example, must keep track of the formula water and specify whether the solubility is reported as anhydrous CuSO₄ or hydrate. Similar care is required if water evaporates during heating or cooling.

Choosing a recrystallization solvent involves more than a large solubility difference. The compound should be stable, impurities should be separable, and the solvent should be handled appropriately. A steep solubility curve can help theoretical recovery, but crystal purity and practical recovery depend on the full procedure.

Step-by-step reasoning

1. Determine the actual starting dissolved solute mass. 2. Record the starting solvent mass and whether it changes. 3. Read cool solubility for the same chemical basis and scale it to the remaining solvent. 4. Subtract final dissolved solute from initial dissolved solute to find ideal crystal mass. 5. Check crystal hydration, solvent evaporation and incomplete equilibration before interpreting recovery.

Visual explanation

Draw a hot beaker labeled “200 g water + 120 g dissolved salt.” An arrow marked cooling leads to a cold beaker with “200 g water + 50 g dissolved salt” in the liquid and “70 g crystals” at the bottom. A box around all three salt numbers illustrates 120 = 50 + 70.

Real-world analogy

A warehouse can hold 120 boxes during one season but only 50 under a later capacity rule. Seventy boxes must leave if it began full. The analogy mirrors subtraction, though real crystals form through thermodynamic and kinetic processes rather than a manager's decision.

Real-world example

Growing crystals from a warm saturated solution can make solid material visible during cooling. A student may weigh fewer crystals than the simple prediction because some remain dissolved, are trapped on glassware or are lost during handling. The theoretical mass is a benchmark for interpreting recovery.

Why?

Why compare both solubilities using the same solvent mass? The hot and cold values are often tabulated per 100 g solvent. Scaling both to the actual water mass makes the difference a real solute mass that can be crystallized.

Common misconception

“Hot solubility minus cold solubility always equals collected crystal mass.” It equals an ideal mass only if the hot solution starts saturated, solvent amount remains fixed, crystals have the assumed composition, and final equilibrium is reached.

Worked example

A substance has solubility 48 g per 100 g water at 70 °C and 12 g at 20 °C. A hot saturated sample uses 75 g water. Initially dissolved mass is 48 × 0.75 = 36 g. At 20 °C, the mother liquor holds 12 × 0.75 = 9 g. Ideal anhydrous crystal mass is 36 − 9 = 27 g. Salt accounting checks: 9 + 27 = 36 g. The final liquid mass is 75 + 9 = 84 g, excluding crystals.

Quick check

1. If the hot sample is unsaturated, should you always use the hot solubility as its starting dissolved mass? Answer: No. Use the actual dissolved amount; the hot solubility is only an upper equilibrium limit for that solvent mass.

Exam focus

Write the solute mass-balance equation and state assumptions. Check whether crystals are hydrates and whether heating has changed the water mass.

Advanced insight

Crystal size and purity depend on nucleation and growth, while the equilibrium solubility difference sets an ideal amount. Slow cooling may produce different crystals from rapid cooling even when the final equilibrium mass balance is the same.

Summary

Cooling-induced crystallization is calculated from initial dissolved solute minus final equilibrium dissolved solute. Scale values to the actual solvent mass and use the actual initial amount. Hydration, evaporation and incomplete equilibration can change the simple result or its experimental recovery.

Practice questions

1. Hot and cold solubilities are 40 and 10 g per 100 g water. Find ideal crystals from a hot saturated solution with 150 g water. Answer: The difference is 30 g per 100 g water, so 30 × 1.5 = 45 g anhydrous crystals under constant-water conditions. 2. What if the hot sample actually contained only 45 g dissolved in that 150 g water? Answer: Cold capacity is 10 × 1.5 = 15 g, so 45 − 15 = 30 g crystallizes ideally. 3. Why can hydrated crystals require a different calculation? Answer: They remove water along with solute from the mother liquor, so the solvent mass used for final solubility changes.