“The numbers in my recrystallization answer never match.” “I can’t tell which mass belongs in the proportion.” — The precipitated mass comes from three quantities, in order:

  1. how much solute was dissolved while hot,
  2. how much can stay dissolved after cooling,
  3. the difference between the two (the crystals that come out).

This article works through the proportional calculation from solubility per 100 g of water, with a checked example and a free tool.

Solubility and saturated solutions

Solubility is the maximum mass of solute that dissolves in 100 g of water at a given temperature. A solution holding that maximum is saturated, with the ratio:

  • solute: S g
  • water: 100 g
  • saturated solution: 100 + S g

If the solubility is 32, 100 g of water holds 32 g of solute and the saturated solution weighs 132 g. The substance that dissolves is the solute, the liquid that dissolves it (water) is the solvent, and together they form the solution.

Check the basis: solubility may be per 100 g of water or per 100 g of solution. This article and the tool use per 100 g of water.

Precipitation on cooling: the four steps

Potassium nitrate: solubility 109 at 60 °C, 32 at 20 °C (per 100 g of water). What mass of crystals forms when 200 g of saturated solution at 60 °C is cooled to 20 °C?

  1. Solute dissolved while hot, from the saturated ratio: 200 × 109 ÷ (100 + 109) = 200 × 109 ÷ 209 = 104.306… g
  2. The water mass does not change on cooling: 200 − 104.306… = 95.694… g
  3. Apply the 20 °C solubility to that water mass: 32 × 95.694… ÷ 100 = 30.622… g
  4. The difference is the precipitate: 104.306… − 30.622… = 73.684… g

The answer is about 73.7 g. Check: 30.622 ÷ 95.694 = 0.32, the 20 °C ratio.

One-step proportion: from a saturated solution of mass M, the precipitate is M × (S₁ − S₂) ÷ (100 + S₁). Here, 200 × 77 ÷ 209 = 73.684… g — the four steps compressed.

When the water mass is given

If the problem gives the water mass, the maximum dissolved mass in W g of water is S × W ÷ 100, so:

precipitate = (S₁ − S₂) × W ÷ 100

Example: solubility 110 at 60 °C and 32 at 20 °C; 200 g of water saturated at 60 °C, cooled:

(110 − 32) × 200 ÷ 100 = 78 × 2 = 156 g

From a solution mass, convert with “100 g water : (100 + S₁) g solution”.

When water evaporates

Evaporation removes water but leaves the solute; apply the cold solubility to the remaining water.

Example: solubility 64 at 40 °C and 32 at 20 °C; 200 g of water saturated at 40 °C, 100 g evaporated, then cooled:

  1. Solute dissolved at 40 °C: 64 × 200 ÷ 100 = 128 g
  2. Water left after evaporation: 200 − 100 = 100 g
  3. Amount still soluble at 20 °C: 32 × 100 ÷ 100 = 32 g
  4. Precipitate: 128 − 32 = 96 g

For evaporation without cooling, set both solubilities equal.

Common mistakes

  • Using water mass in the denominator. Converting a solution mass to solute mass uses (100 + S₁).
  • Reporting the cold soluble amount as the precipitate. It is still dissolved; subtract the hot amount.
  • Confusing water and solution mass. Whether “200 g” means water or total solution changes the formula.
  • Applying the subtraction to hydrates. CuSO₄·5H₂O crystals carry water away, so the remaining water changes.
  • Expecting precipitation without a solubility drop. No drop means 0 g.
  • Rounding too early. Round once at the end (check digit counts with the significant figures tool).

Verify with the tool

The Tools Hub solubility calculator reproduces the example above.

1

Choose "Precipitation on cooling"

Open the tool and select "Precipitation on cooling".

2

Enter the solubility and the mass

Enter 109 for S₁ and 32 for S₂; choose "From the saturated-solution mass" and enter 200 for M.

3

Read the working and the result

Press Calculate to see "Precipitate = 73.6842 g" and the steps (solute 104.306 g, cold amount 30.622 g, difference). For a water-mass problem, choose "From the water mass".

Tool mentioned in this article

Solubility & Recrystallization Calculator

Precipitation on cooling or evaporation, with the mass-balance steps shown. Free, in your browser, nothing uploaded.

Try it now

For problems that use molarity instead of mass ratios, see the molarity calculator and how to calculate molarity.

Summary

  • The precipitate is solute dissolved while hot minus solute soluble when cold
  • Base the proportion on 100 g of water; convert a solution mass with the (100 + S) ratio
  • With a known water mass, precipitate = (S₁ − S₂) × W ÷ 100
  • With evaporation, apply the cold solubility to the remaining water
  • Hydrates carry water away, so the simple subtraction does not apply
  • Round once, at the end

Write the three lines — hot composition, cold soluble amount, difference — before calculating.

FAQ

Should I start from the water mass or the saturated-solution mass?

Either works: (S₁ − S₂) × W ÷ 100 from W g of water, or M × (S₁ − S₂) ÷ (100 + S₁) from M g of solution.

Does the same formula work when a hydrate precipitates?

No. A hydrate takes water into the crystals, so the remaining water decreases; use separate balances for solute and water. The linked tool assumes anhydrous crystals.

How do I handle evaporation without cooling?

Use the evaporation-and-cooling mode with both solubilities equal; the tool applies the solubility to the water left.

Why do different sources list slightly different solubilities?

Solubility is measured, so values vary slightly by source and temperature. Use the value given in the question or textbook.

References