A quantity of \(5 \mathrm{~g}\) of a crystalline salt when rendered anhydrous lost \(1.8 \mathrm{~g}\) of water. The formula mass of the anhydrous salt is 160 . The number of molecules of water of crystallization in the salt is (a) 3 (b) 5 (c) 2 (d) 1

Understanding the molar mass calculation is essential for solving many problems in chemistry, including determining the number of molecules of water of crystallization in a hydrate. The molar mass, often measured in grams per mole (\text{g/mol}), represents how much one mole of a substance weighs. A mole is a unit that corresponds to Avogadro's number, which is approximately 6.022\(\times\)10²³ entities (usually atoms or molecules).

Calculating the molar mass involves summing the atomic masses of all atoms within a single molecule of a substance. The atomic masses are provided in the periodic table and the lookup involves simple multiplication and addition. For instance, the molar mass of water (\text{H}₂O) would be calculated as follows:

• Hydrogen (H) has an atomic mass of approximately 1 g/mol. Since there are two hydrogens, we multiply by 2, getting 2 g/mol for hydrogen.
• Oxygen (O) has an atomic mass of approximately 16 g/mol.
• The total molar mass of water is then 2 g/mol + 16 g/mol = 18 g/mol.

In the provided exercise, knowing how to calculate the molar mass is crucial both for the water lost during the process of dehydration and for the anhydrous salt itself. This foundational concept allows students to progress to the next steps and find the ratio of moles of water to moles of anhydrous salt.

Hydrates are compounds that include water molecules physically trapped within their crystalline structure. These water molecules are known as water of crystallization. During the heating process, hydrates can lose these water molecules, transforming into anhydrous salts, which are the same compounds without the water of crystallization.

Identifying Hydrates and Anhydrous Forms

A common way to represent hydrates in chemical formulas is by using a dot to separate the formula of the anhydrous salt from the number of water molecules. For example, Copper(II) sulfate pentahydrate is written as \text{CuSO}₄\(\cdot\)5\text{H}₂O, indicating it contains five moles of water for every mole of Copper(II) sulfate. When this compound is dehydrated, it loses those water molecules and becomes anhydrous \text{CuSO}₄.

To fully grasp the topic, students need to recognize that mass is conserved during the dehydration process and that the loss in mass corresponds exactly to the mass of water removed. In the exercise, the mass of the anhydrous salt is found by subtracting the mass of the water of crystallization, essential for the subsequent stoichiometric calculations.

Stoichiometry is an area of chemistry that involves the calculation of reactants and products in chemical reactions. It is based on the conservation of mass and the concept of the mole. Understanding stoichiometry allows us to relate the quantities of substances involved in a chemical reaction.

In the context of the exercise, stoichiometry is used to determine the number of molecules of water of crystallization by finding the ratio of moles of water to moles of anhydrous salt. Since stoichiometry is founded on the principle that matter is neither created nor destroyed in a chemical reaction, we can analyze mass and mole relationships between reactants and products.

Applying Stoichiometry to the Problem

In the provided solution, stoichiometry is applied through a series of methodical steps: First, the mass of water and anhydrous salt is measured. Then, their respective molar masses are used to find the number of moles of each. Finally, the molar ratio of water to salt tells us the formula for the hydrate. The mole-to-mole comparison intrinsic to stoichiometry is what reveals there are five moles of water for every mole of anhydrous salt in the compound. Remember, stoichiometry is not just about the numbers; it requires understanding the mole concept and how to manipulate molar relationships in chemical substances.