Question

There is loss of weight when a mixture of \(\mathrm{Li}_{2} \mathrm{CO}_{3}\) and \(\mathrm{Na}_{2} \mathrm{CO}_{3} .10 \mathrm{H}_{2} \mathrm{O}\) is heated strongly. This loss is due to (1) \(\mathrm{Li}_{2} \mathrm{CO}_{3}\) due to decomposition into \(\mathrm{Li}_{2} \mathrm{O}\) and gascous \(\mathrm{CO}_{2}\). (2) Both \(\mathrm{Li}_{2} \mathrm{CO}_{3}\) and \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) decompose to \(\mathrm{Li}_{2} \mathrm{O}\) and \(\mathrm{Na}_{2} \mathrm{O}\) losing gascous \(\mathrm{CO}_{2}\). (3) Due to loss of \(\mathrm{CO}_{2}\) from \(\mathrm{Li}_{2} \mathrm{CO}_{3}\) and water from \(\mathrm{Na}_{2} \mathrm{CO}_{3} \cdot 10 \mathrm{H}_{2} \mathrm{O}\) (4) Due to sublimation of \(\mathrm{Li}_{2} \mathrm{CO}_{3}\)

Short answer

Due to loss of \(\mathrm{CO_{2}}\) from \(\mathrm{Li_{2}CO_{3}}\) and water from \(\mathrm{Na_{2}CO_{3} \cdot 10 H_{2}O}\).

Step-by-step solution

- Identify the Compounds and Reactions

First, recognize the compounds involved: \ \(\mathrm{Li_{2}CO_{3}}\) and \(\mathrm{Na_{2}CO_{3} \cdot 10 H_{2}O}\). Assess their likely behavior when heated strongly.

- Analyze Decomposition of Lithium Carbonate

When \(\mathrm{Li_{2}CO_{3}}\) is heated, it decomposes into \(\mathrm{Li_{2}O}\) and releases gaseous \(\mathrm{CO_{2}}\) as follows: \ \[ \mathrm{Li_{2}CO_{3} \rightarrow Li_{2}O + CO_{2}} \].

- Analyze Behavior of Sodium Carbonate Decahydrate

When \(\mathrm{Na_{2}CO_{3} \cdot 10 H_{2}O}\) is heated, it loses its water of crystallization and transforms into anhydrous \(\mathrm{Na_{2}CO_{3}}\) as follows: \ \[ \mathrm{Na_{2}CO_{3} \cdot 10 H_{2}O \rightarrow Na_{2}CO_{3} + 10 H_{2}O (g)} \].

- Determine the Reason for Weight Loss

Considering the reactions: (1) \(\mathrm{Li_{2}CO_{3}}\) loses weight by releasing \(\mathrm{CO_{2}}\), and (2) \(\mathrm{Na_{2}CO_{3} \cdot 10 H_{2}O}\) loses weight by releasing water. So, the reason for weight loss is the loss of \(\mathrm{CO_{2}}\) from \(\mathrm{Li_{2}CO_{3}}\) and the loss of water from \(\mathrm{Na_{2}CO_{3} \cdot 10 H_{2}O}\).

Key concepts

Thermal Decomposition

Thermal decomposition is a chemical reaction where a compound is broken down into simpler substances when heated. In the context of the exercise, heating \(\mathrm{Li_{2}CO_{3}}\) results in its decomposition into lithium oxide \(\mathrm{Li_{2}O}}\) and carbon dioxide gas \(\mathrm{CO_{2}}\). This process can be represented by the equation: \[\mathrm{Li_{2}CO_{3}} \rightarrow \mathrm{Li_{2}O}} + \mathrm{CO_{2}}\].
This reaction illustrates a key point about thermal decomposition: the original compound is broken into at least one solid product and one gas. The release of gas (\mathrm{CO_{2}} in this case) contributes significantly to the observed weight loss. Understanding this helps clarify why certain compounds lose mass when heated.

Weight Loss in Chemical Reactions

Weight loss during chemical reactions, especially thermal decompositions, is due to the release of gaseous products. In the given exercise, both \(\mathrm{Li_{2}CO_{3}}\) and \(\mathrm{Na_{2}CO_{3} \cdot 10 \mathrm{H_{2}O}}\) exhibit weight loss upon heating. For \(\mathrm{Li_{2}CO_{3}}\), this loss is explained by the formation and release of \(\mathrm{CO_{2}}\), while for \(\mathrm{Na_{2}CO_{3} \cdot 10 \mathrm{H_{2}O}}\), it is the water of crystallization that is lost as steam:
\[\mathrm{Na_{2}CO_{3} \cdot 10 \mathrm{H_{2}O}} \rightarrow \mathrm{Na_{2}CO_{3}}\ + 10 \mathrm{H_{2}O}} (g)\].
These chemical changes result in the release of light, gaseous molecules, translating to a decrease in the overall weight of the original compounds. Therefore, understanding the formation and release of gaseous products is crucial in predicting and explaining weight loss in such reactions.

Hydrated vs Anhydrous Compounds

Hydrated compounds contain water molecules integrated into their crystalline structure. \(\mathrm{Na_{2}CO_{3} \cdot 10 \mathrm{H_{2}O}}\) (sodium carbonate decahydrate) is a prime example, with ten water molecules per formula unit.
When such hydrated compounds are heated, these water molecules are released, transforming the compound from a hydrated state to an anhydrous (water-free) form. The corresponding reaction for \mathrm{Na_{2}CO_{3} \cdot 10 \mathrm{H_{2}O}} can be written as:
\[\mathrm{Na_{2}CO_{3} \cdot 10 \mathrm{H_{2}O}} \rightarrow \mathrm{Na_{2}CO_{3}}\ + 10 \mathrm{H_{2}O}} (g)\].
This dehydration process is reversible, meaning that anhydrous \(\mathrm{Na_{2}CO_{3}}\) can absorb water from the environment to revert back to its hydrated form. Understanding the difference between hydrated and anhydrous compounds helps in comprehending many thermal decomposition reactions and their associated weight losses.