Basic properties and water loss
The behavior of crystalline hydrates is largely determined by how strongly water molecules are held in their structure. There are two main ways this moisture is lost:
- Efflorescence. This is a spontaneous process in which a substance loses its water of crystallization simply by being in dry air. Additional heating is not required for this.
- Thermal treatment. Upon heating, most hydrates decompose, turning into ordinary anhydrous salts. However, in some cases, the water is bound so firmly that extreme conditions are required. For example, the compound $CaCl_2 \cdot 6H_2O$ completely releases all its moisture only upon reaching a temperature of $300^\circ \text{C}$.
It is precisely due to their ability to firmly bind water that the resulting anhydrous salts (such as calcium chloride $CaCl_2$) are actively used in chemistry as reliable desiccants for drying other substances.
Interestingly, the composition of the resulting crystal is formed already at the crystallization stage and directly depends on the temperature regime. A classic example is sodium carbonate:
- If the temperature of the solution is below $32^\circ \text{C}$, a crystal with ten water molecules is formed — $Na_2CO_3 \cdot 10H_2O$.
- If the process proceeds at a temperature above $32^\circ \text{C}$, a compound with only one water molecule is formed — $Na_2CO_3 \cdot H_2O$.
Formula determination (calculation example)
To establish the exact chemical formula, chemists measure the mass loss of a substance upon complete drying. Let us consider this using a classic sodium carbonate problem.
Given: Crystals with a mass of $14.3$ g were isolated from a 40% solution. They were then dried in a desiccator over phosphorus pentoxide ($P_2O_5$) to a constant mass. As a result, an anhydrous powder with a mass of $5.3$ g remained.
The process is described by a simple dehydration scheme: $Na_2CO_3 \cdot xH_2O \rightarrow Na_2CO_3 + xH_2O$
For the calculation, we will need the molar masses: $M(Na_2CO_3) = 106$ g/mol and $M(H_2O) = 18$ g/mol. The mass of the evaporated water will be the difference between the initial and final mass: $14.3$ g $-$ $5.3$ g $=$ $9.0$ g.
Solution steps: We use the stoichiometric rule. The ratio of the mass of the initial wet substance to its total molar mass is equated to the ratio of the mass of the evaporated water to the molar mass of this water (necessarily taking into account the unknown coefficient $x$).
The equation looks like this: $14.3 / (106 + 18x) = 9.0 / 18x$
Solving this mathematical proportion, we find that $x = 10$. Conclusion: The sought formula of the studied substance is $Na_2CO_3 \cdot 10H_2O$.
Nomenclature
The names of crystalline hydrates are built according to a strict but simple logic. Naming scheme: [Greek numeral] + "hydrate" + [name of the salt].
Below is a table of the main prefixes and their corresponding examples:
| Number of molecules | Prefix | Formula example | Name |
|---|---|---|---|
| 1 | Mono- | $LiCl \cdot H_2O$ | Lithium chloride monohydrate |
| 2 | Di- | $CaSO_4 \cdot 2H_2O$ | Calcium sulfate dihydrate |
| 3 | Tri- | $CH_3COONa \cdot 3H_2O$ | Sodium acetate trihydrate |
| 4 | Tetra- | $CaBr_2 \cdot 4H_2O$ | Calcium bromide tetrahydrate |
| 5 | Penta- | $Na_2S_2O_3 \cdot 5H_2O$ | Sodium thiosulfate pentahydrate |
| 6 | Hexa- | $AlCl_3 \cdot 6H_2O$ | Aluminum chloride hexahydrate |
| 7 | Hepta- | $FeSO_4 \cdot 7H_2O$ | Iron(II) sulfate heptahydrate |
| 8 | Octa- | $MgI_2 \cdot 8H_2O$ | Magnesium iodide octahydrate |
| 9 | Nona- | $Al(NO_3)_3 \cdot 9H_2O$ | Aluminum nitrate nonahydrate |
| 10 | Deca- | $Na_2SO_4 \cdot 10H_2O$ | Sodium sulfate decahydrate |
| 12 | Dodeca- | $KAl(SO_4)_2 \cdot 12H_2O$ | Potassium aluminum sulfate dodecahydrate |
Class and exam questions
What is the difference between thermal processing and weathering?
Efflorescence is the loss of water in a dry air environment without heating. Thermal treatment implies targeted temperature elevation, which is necessary to break strong bonds with water (as in the case of calcium chloride hexahydrate, requiring 300 degrees).
Why can a single salt have different hydrate forms?
The number of water molecules embedded into the crystal structure depends on the temperature conditions at the moment of crystallization. For example, sodium carbonate forms a decahydrate at low temperatures and a monohydrate at temperatures above 32 degrees.
How is the molar mass of an unknown hydrate calculated?
It consists of the molar mass of the salt itself and the molar mass of water multiplied by the unknown coefficient 'x' (e.g., 106 + 18x). This coefficient is then found through a proportion based on mass loss.
What else to study on this topic
- Dependence of the composition of crystalline hydrates on the temperature conditions of crystallization.
- Use of anhydrous calcium chloride as a laboratory desiccant.
- Stoichiometric calculations of salt dehydration processes.
- Mechanisms of natural crystal efflorescence in dry air.
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