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) \(\mathrm{X}_{2}(\mathrm{SO}_{4})_{3} \cdot 18\, \mathrm{H}_{2}\mathrm{O}\) is the compound formula.

Step 1: Determine Sulphate Compound Mass

First, find the mass of the entire sulphate compound. We know it makes up 61.7% of the 5.0 g sample. Calculate this by: \[ \text{Mass of sulphate compound} = 0.617 \times 5.0\, \text{g} = 3.085\, \text{g} \]

Step 2: Calculate Crystalline Water Mass

Now find the mass of the water of crystallization, which makes up 48.6% of the sulphate compound mass. Calculate this: \[ \text{Mass of water} = 0.486 \times 3.085\, \text{g} = 1.4993\, \text{g} \]

Step 3: Determine Anhydrous Sulphate Mass

Subtract the mass of water from the total mass of the sulphate compound to find the anhydrous sulphate mass: \[ \text{Mass of } X_2(SO_4)_3 = 3.085\, \text{g} - 1.4993\, \text{g} = 1.5857\, \text{g} \]

Step 4: Calculate Moles of Metal X

Calculate the moles of Metal \(X\) using its atomic mass (27 g/mol). The entire sample is 5.0 g of \(X\). \[ \text{Moles of X} = \frac{5.0\, \text{g}}{27\, \text{g/mol}} = 0.1852\, \text{mol} \]

Step 5: Calculate Moles of Anhydrous Sulphate

Using the mass of \(X_2(SO_4)_3\) (1.5857 g), find its moles. The molar mass of \(X_2(SO_4)_3\) is \(2 \times 27 + 3 \times 96 = 342\, \text{g/mol}\).\[ \text{Moles of } X_2(SO_4)_3 = \frac{1.5857\, \text{g}}{342\, \text{g/mol}} = 0.00464\, \text{mol} \]

Step 6: Verify Number of Waters in Hydrate

Calculate the moles of water of crystallization: \[ \text{Moles of water} = \frac{1.4993\, \text{g}}{18\, \text{g/mol}} = 0.0833\, \text{mol} \] Ratios: - For \(X_2(SO_4)_3\): \(0.00464\, \text{mol}\)- For \(H_2O\): \(0.0833\, \text{mol}\)Calculate the ratio: \[ \frac{0.0833}{0.00464} \approx 18 \] The number of water molecules matches the multiple of 18 in one formula unit.

Key Concepts

Crystalline Hydrates

Crystalline hydrates are fascinating structures in chemistry where water molecules are included in the crystal lattice of a salt. It's a common phenomenon among certain ionic compounds. These hydrates are formed when the water molecules are weakly bound to the ions in the crystal.

One way to express the composition of crystalline hydrates is through their chemical formulas. For example, in the chemical formula \( X_2(SO_4)_3 \cdot 18 \text{H}_2\text{O} \), the dot indicates that each formula unit of the anhydrous compound \( X_2(SO_4)_3 \) is associated with 18 water molecules.

**Why is this important?**

• Hydrates can affect the compound's physical properties such as solubility, color, and stability.
• They are important for calculating the correct formula weight, which you'll use in chemical equations and stoichiometry.

Recognizing and calculating the amount of water in such a compound helps in understanding the compound's complete composition. It also plays a role in determining the simplest or empirical formula of the hydrate.

Stoichiometry

Stoichiometry is essentially the arithmetic of chemistry. It's the calculation of reactants and products in chemical reactions. It involves the use of balanced chemical equations to determine the quantities of substances involved.

**Why is stoichiometry important in this exercise?**

• Stoichiometry allows us to calculate how much of a certain component is present in a compound based on its percentage composition by mass.
• In the given exercise, stoichiometry helps us determine the mass of the sulphate compound and the water of crystallization in the crystalline hydrate.

By understanding stoichiometry, you can convert between mass, moles, and molecules. It's fundamental when dealing with chemical reactions and helps determine empirical formulas based on experimental data.

Mole Concept

The mole concept is a cornerstone of chemistry that provides a bridge between the atomic and macroscopic worlds. The mole allows chemists to count particles like atoms and molecules in a given mass of a substance.

**Essential aspects of the mole concept:**

• The mole is defined as exactly \(6.022 \times 10^{23}\) entities, known as Avogadro's number.
• The molar mass of a substance is the mass of one mole of that substance, measured in grams per mole.

In the exercise, the mole concept is crucial for finding out how many moles of metal \(X\), sulphate, and water of crystallization are present.

For instance, knowing that the moles of \(X\) are calculated by dividing the mass of \(X\) by its atomic mass helps in verifying the composition of the hydrate. Just as the mole helps in basic terms to assess how much of the ingredient is in your chemical 'recipe', it maintains a standard measure that ensures accuracy in chemical analysis and synthesis.