Problem 121 - A Complete Resource Book in Chemistry for JEE Main

Question

A chloride of a metal \((\mathrm{M})\) has \(65.5 \%\) of chlorine. 100 \(\mathrm{ml}\) of vapour of the chloride of metal at STP weighs \(0.72 \mathrm{~g}\). The molecular formula of this metal chloride is (a) \(\mathrm{MCl}_{3}\) (b) \(\mathrm{MCl}_{4}\) (c) \(\mathrm{M}_{2} \mathrm{Cl}_{3}\) (d) \(\mathrm{MCl}_{5}\)

Step-by-Step Solution

Answer

The molecular formula of the metal chloride is \(\mathrm{MCl}_3\).

Step 1: Understand the Given Data

The chloride compound has 65.5% chlorine by mass. 100 ml of its vapor at STP weighs 0.72 g. Our goal is to find the molecular formula of the metal chloride.

Step 2: Determine Molar Mass Using Vapor Density

Using the ideal gas law, we know that 22.4 liters (or 22,400 ml) of a gas at STP weigh the same as its molar mass in grams. Hence, the molar mass of the chloride is \[ \text{Molar mass} = \frac{0.72 \, \text{g}}{100 \, \text{ml}} \times 22400 \, \text{ml} = 161.28 \, \text{g/mol}. \]

Step 3: Calculate Chlorine Mass contribution

With 65.5% chlorine, the mass of chlorine in the molar mass of the compound is \[ \text{Mass of chlorine} = 161.28 \, \text{g/mol} \times \frac{65.5}{100} = 105.57 \, \text{g/mol}. \]

Step 4: Determine Moles of Chlorine Atoms

The molar mass of one chlorine atom is approximately 35.5 g/mol. Thus, the number of chlorine atoms can be calculated by \[ \text{Number of chlorine atoms} = \frac{105.57 \, \text{g/mol}}{35.5 \, \text{g/mol}} \approx 3. \]

Step 5: Evaluate Options

Given that chlorine consists of approximately 3 atoms in the formula, evaluate which options match. Options with a subscript of 3 chlorine atoms are appropriate, simplifying options to consider: \(\mathrm{MCl_3}\) and \(\mathrm{M}_2\mathrm{Cl}_3\). Since 65.5% aligns more with the simpler formula considering distribution, \(\mathrm{MCl}_3\) is likely correct. Thus, the molecular formula is \(\mathrm{MCl_3}\).

Key Concepts

Molar Mass Calculation

Molar mass is a fundamental concept in chemistry that describes the mass of a single mole of a given substance. It is typically expressed in grams per mole (g/mol). In our exercise, calculating the molar mass is crucial because it allows us to compare the mass of the vapor of a chlorine compound to its mass at standard conditions. To find the molar mass, we used the given weight of the vapor at STP and extrapolated it to the full volume a mole of gas would occupy under those conditions - 22.4 liters. This provided the corrected mass of the compound in grams, thereby deriving its molar mass as 161.28 g/mol. When working on problems like this, understanding this connection between vapor density and molar mass is key to finding the molecular formulas of compounds.

Vapor Density

Vapor density provides insight into the density of a vapor in comparison to hydrogen. It's calculated by dividing the mass of a certain volume of the vapor by the mass of the same volume of hydrogen gas under standard conditions. In the context of the exercise, however, vapor density was implicitly utilized through the mass calculation of a given volume of gas at STP. This allowed for the determination of the molar mass of the chloride from limited data. It's important to remember this principle: vapor density can provide a streamlined method to access molar mass when used in conjunction with other data like volume and weight at noted conditions.

Mole Concept

The mole concept is central to understanding compositions and reactions in chemistry. It is essentially a way of counting atoms or molecules by weighing them. In this exercise, the percentage of chlorine by mass ( 65.5%) in the chloride compound helped to identify how much of the compound's molar mass was made up by chlorine atoms. By using the known atomic mass of chlorine (35.5 g/mol), we calculated the moles of chlorine within the compound. This ratio provides a fingerprint of sorts for determining the number of chlorine atoms in a given formula, crucial in deducing the potential molecular formulas at play.

Ideal Gas Law

The ideal gas law is a cornerstone equation in chemistry that relates the pressure, volume, temperature, and number of moles of an ideal gas. Although it was not explicitly calculated in our exercise, it implicitly guided the understanding of how much volume one mole of gas would occupy at STP—22.4 liters. This universal truth facilitated the translation of a small sample volume and weight into a molar mass understanding. Always remember the ideal gas law is given by \[ PV = nRT \]where:

• \(P\) is the pressure in atmospheres,
• \(V\) is the volume in liters,
• \(n\) is the number of moles,
• \(R\) is the ideal gas constant, and
• \(T\) is the temperature in Kelvin.

Using this equation and knowing STP values allow us to solidify our approach to molar mass and molecular formula calculations.