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) \(\mathrm{X}_{2}\left(\mathrm{SO}_{4}\right)_{3}-18 \mathrm{H}_{2} \mathrm{O}\) (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

The simplest formula of the compound is \( \mathrm{X}_{2}(\mathrm{SO}_{4})_{3} \cdot 12 \mathrm{H}_{2} \mathrm{O} \).

Step 1: Calculate Moles of Metal X

First, find the moles of the metal X. Given that the mass of X is 5.0 g and its atomic mass is 27, the moles can be calculated using the formula:\[ \text{moles of X} = \frac{\text{mass}}{\text{atomic mass}} = \frac{5.0}{27}. \] Calculating gives approximately 0.185 moles of X.

Step 2: Determine Mass of Hydrated Sulphate

Find the total mass of the crystalline sulphate formed, knowing that it contains 61.7% of the total mass. This is calculated by re-arranging: \[ \frac{\text{mass of sulphate}}{5.0} = 0.617, \] leading to a mass of sulphate as \( \text{mass of sulphate} = \frac{5.0}{0.617} \approx 8.109 \text{ g}.\)

Step 3: Calculate Mass of Water in Sulphate

Since the sulphate contains 48.6% water by mass, calculate the mass of water: \[ \text{mass of water} = 8.109 \times 0.486 = 3.94 \text{ g}. \]

Step 4: Calculate Moles of Water

Convert the mass of water into moles: \[ \text{moles of } \text{H}_2\text{O} = \frac{\text{mass of water}}{\text{molecular mass of water}} = \frac{3.94}{18} \approx 0.219 \text{ moles}. \]

Step 5: Determine the Hydrate Formula

Use the ratio of moles of metal X to moles of water to determine the simplest formula: since the moles of metal X is approximately 0.185 and moles of water is 0.219, let's consider water molecules correspond as nearly 12 water molecules fitting each molecule part of sulphate based on the ratio obtained. Therefore, the compound is \( \mathrm{X}_{2}\left(\mathrm{SO}_{4}\right)_{3} \cdot 12 \mathrm{H}_{2} \mathrm{O} \).

Key Concepts

Moles Calculation

The fundamental concept of chemistry begins with understanding moles, a core measurement unit. Calculating moles involves determining the amount of a substance. In the problem, we are given a mass of metal X, 5 grams. By using its atomic mass, 27, we can calculate the number of moles present. This is calculated by dividing the mass by the atomic mass: \[\text{moles of X} = \frac{5.0}{27} \approx 0.185\]% of metal X. This calculation allows us to bridge the gap between the macroscopic world (what we can measure) and the microscopic world (atoms and molecules). Understanding moles is crucial as it sets the stage for deeper exploration into chemical reactions and formulas.

Hydrate Chemistry

Hydrates are compounds that include water molecules within their crystalline structure. In this exercise, the crystalline sulphate contains 48.6% water by mass. The challenge is to determine how water incorporates into the sulfate structure. Hydrate chemistry is crucial as it impacts everything from the structural integrity of minerals to specific reactions in a chemical process.The mass of water is calculated by knowing the total mass of the hydrated sulphate. By multiplying the percentage of water by the total mass, 8.109 grams in this case, we obtain the mass of the water: \[\text{mass of water} = 8.109 \times 0.486 = 3.94 \text{ g}. \] Next, we convert this mass into moles using the molecular weight of water: \[\frac{3.94}{18} \approx 0.219 \text{ moles}. \] These calculations are essential for forming the correct chemical formula of the hydrate.

Stoichiometry

Stoichiometry is the area of chemistry that deals with the quantitative aspects of chemical reactions. It helps chemists understand the relationships between reactants and products. In our case, stoichiometry allows us to balance and predict the interactions in a hydrated compound and how much of each component is effectively part of the reaction.For the crystalline sulphate, we determine the mass and moles of each element. Dimensions of stoichiometry are employed here to calculate the simple {\( ratio \)} between water and metal X. Using moles of metal X (0.185 moles) and moles of water (0.219 moles), allows us to establish a proportional relationship which hints that for each molecule of sulphate, there is essentially a representation of 12 water molecules. Such essential calculations and balancing acts are a backbone to forming accurate chemical formulas and understanding the behavior of compounds in reactions.

Crystalline Compound

Crystalline compounds like hydrates are unique because they have a defined arrangement and structure. The crystalline nature and associated water in the compound uphold this distinctive form and behavior. In metallurgy and mineral chemistry, shapes, structures, and the presence of water play significant roles in determining properties and reactivity.In this scenario, we are given the compound with a formula to find: including metal X alongside sulphate groups and water. Each crystal lattice or framework adsorbing water molecules defines the hydrate's integrity and overall mass. The arrangement allows compounds to stabilize costs by structuring water molecules within their framework. The eventual determination of \( \mathrm{X}_{2}(\mathrm{SO}_{4})_{3} \cdot 12 \mathrm{H}_{2} \mathrm{O} \)was based on calculated ratios indicating how metal and water cohabitate within the chemical matrix of the sulfate.