Problem 78 - A Complete Resource Book in Chemistry for JEE Main

Question

\(5.0 \mathrm{~g}\) of a certain metal, \(\mathrm{X}\) (atomic mass \(=27\) ) is converted into \(61.7 \%\) crystalline sulphate containing \(48.6 \%\) by mass of water of crystallization. The simplest formula of the compound is (a) \(\mathrm{X}_{2}\left(\mathrm{SO}_{4}\right)_{3} \cdot 18 \mathrm{H}_{2} \mathrm{O}\) (b) \(\mathrm{X}_{2}\left(\mathrm{SO}_{4}\right)_{3} \cdot 14 \mathrm{H}_{2} \mathrm{O}\) (c) \(\mathrm{X}_{2}\left(\mathrm{SO}_{4}\right)_{3} \cdot 16 \mathrm{H}_{2} \mathrm{O}\) (d) \(\mathrm{X}_{2}\left(\mathrm{SO}_{4}\right)_{3} \cdot 12 \mathrm{H}_{2} \mathrm{O}\)

Step-by-Step Solution

Answer

(a) \(X_2(SO_4)_3 \cdot 18H_2O\)

Step 1: Calculate the mass of crystalline sulphate

Given that 5.0 g of metal X is converted into 61.7% crystalline sulphate, we need to find the mass of the crystalline sulphate. Apply the percentage formula: \[\text{Mass of crystalline sulphate} = \frac{61.7}{100} \times 5.0 \, \text{g} = 3.085 \, \text{g}.\]

Step 2: Calculate mass of anhydrous sulphate

The sulphate contains 48.6% water by mass. First, find mass of water: \[\text{Mass of water} = \frac{48.6}{100} \times 3.085 \, \text{g} = 1.5 \, \text{g}.\]Then calculate mass of anhydrous sulphate:\[\text{Mass of anhydrous sulphate} = 3.085 \, \text{g} - 1.5 \, \text{g} = 1.585 \, \text{g}.\]

Step 3: Determine moles of metal X

The initial mass of metal X is 5.0 g. Convert this mass to moles using its atomic mass, 27:\[\text{Moles of } X = \frac{5.0 \, \text{g}}{27 \, \text{g/mol}} = 0.185 \, \text{mol}.\]

Step 4: Determine moles of anhydrous sulphate

Assume its formula is \(X_2(SO_4)_3\), which means per formula unit there are two moles of X. Calculate the number of moles of anhydrous sulphate from its mass:\[\text{Molar mass of anhydrous } X_2(SO_4)_3 = 2 \times 27 + 3 \times 96 = 342 \, \text{g/mol}.\]Calculate moles of \(X_2(SO_4)_3\):\[\text{Moles of } X_2(SO_4)_3 = \frac{1.585 \, \text{g}}{342 \, \text{g/mol}} = 0.00463 \, \text{mol}.\]

Step 5: Calculate formula of hydrate

Determine moles of water:\[\text{Molar mass of } H_2O = 18 \, \text{g/mol}\]\[\text{Moles of } H_2O = \frac{1.5 \, \text{g}}{18 \, \text{g/mol}} = 0.0833 \, \text{mol}.\]The ratio of moles of water to moles of \(X_2(SO_4)_3\):\[\text{Water to Sulphate Ratio} = \frac{0.0833}{0.00463} \approx 18.\]Thus, the formula of the hydrate is \(X_2(SO_4)_3 \cdot 18H_2O\).

Key Concepts

Stoichiometry

Stoichiometry is a branch of chemistry that helps us quantify the relationships in chemical reactions, expressed in terms of the ratios of different substances involved. It’s like the recipe book for chemical reactions, guiding us on how much of each ingredient to use. In this exercise, stoichiometry helps us understand the transformation of metal X into its sulfate form.

• Mole Concept: In stoichiometry, the mole concept is fundamental. It allows us to relate mass to amount of substance, using the molar mass as a bridge. For instance, in the given problem, the mass of metal X is converted into moles using its atomic mass.
• Mass Ratios: We also employ mass-to-mass calculations, such as transforming the mass of metal X into the corresponding mass of crystalline sulfate and then to anhydrous sulfate. This ensures our chemistry 'recipe' is accurately followed.

Understanding stoichiometry ensures that the balance in chemical equations is maintained and that we can predict the amounts of products and reactants in a chemical reaction.

Empirical Formula

The empirical formula represents the simplest whole-number ratio of the elements within a compound. It acts like a basic representation of how different atoms combine in nature.

• Determination: Calculating an empirical formula involves determining the moles of the individual elements present in a given mass of compound. In this exercise, the empirical formula involves determining the mole ratio of the metal X and sulfate ions.
• Simplification: Once you compute the moles, they are usually simplified to yield the empirical formula. This enables us to identify the potential formulas of the hydrate based on the moles of metal, sulfate, and water.

The empirical formula provides a crucial foundation to understand the composition of a compound before diving into more complex molecules like hydrates.

Crystallization in chemistry

Crystallization is a process through which solids are organized into a structured pattern, forming crystals. In chemistry, it refers to the formation of a crystalline solid from a solution or melt.

• Formation of Hydrates: In the context of this problem, we deal with crystalline hydrates, which are solid crystals containing water molecules within their structured framework.
• Water of Crystallization: Water is integrated into the crystal structure, often as a crucial component. Here, we calculate the 'water of crystallization' to establish the complete formula of the compound, seen when transforming metal X into its crystalline sulfate form.
• Balancing Ratios: Determining the exact number of water molecules involves examining the ratios of the moles of water to the compound, ensuring correct stoichiometry.

Crystallization reveals the importance of structural organization in chemical compounds, illustrating how components come together to form specific crystalline patterns.