ICSE Class 10 Chemistry Chapter 05 Mole Concept and Stoichimetry

Mole Concept and Stoichiometry

Syllabus

Mole Concept and Stoichiometry

(i) Gay-Lussac's Law of Combining Volumes; Avogadro's Law. Idea of mole - a number just as dozen, a gross; Avogadro's Law - statement and explanation; Gay-Lussac's Law of Combining Volumes - statement and explanation; "the mass of 22-4 litres of any gas at S.T.P. is equal to its molar mass" (Questions will not be set on formal proof but may be taught for clear understanding) - simple calculations based on the molar volume.

(ii) Refer to the atomicity of hydrogen, oxygen, nitrogen and chlorine (proof not required). The explanation can be given using equations for the formation of HCl, NH3 and H2O.

(iii) Relative atomic masses (atomic weight) and relative molecular mass (molecular weights) : either H = 1 or 12C = 12 will be accepted; molecular mass = 2 × vapour density (formal proof not required). Deduction of simple (empirical) and molecular formula from the percentage composition of a compound; the molar volume of a gas at S.T.P.; simple calculations based on chemical equations; both reacting weight and volumes. Idea of relative atomic mass and relative molecular mass - standard H atom, or 1/12th of carbon 12 atom. Relating mole and atomic mass; arriving at gram atomic mass and then gram atom; atomic mass is a number dealing with one atom; gram atomic mass is the mass of one mole of atoms. Relating mole and molecular mass arriving at gram molecular mass and gram molecule - molecular mass is a number dealing with a molecule, gram molecular mass is the mass of one mole of molecules. Molecular mass = 2 × vapour density (questions will not be set on formal proof but may be taught for clear understanding); simple calculations based on the formula. Deduction of simple (empirical) and molecular formula from the percentage composition of a compound.

5A. Gay-Lussac's Law and Avogadro's Law

5.1 Introduction

You have learnt in Class IX that all gases behave similarly under similar conditions of temperature and pressure as expressed by gas laws.

Gas laws are certain rules applicable to a gas in respect of change in temperature (T) or pressure (P) or volume (V). The change in any one of these affects the other two.

Pressure-Volume Relationship or Boyle's Law

It states that the volume of a given mass of dry gas is inversely proportional to its pressure at a constant temperature.

\(P_1V_1 = P_2V_2 = k\) at constant temperature

Temperature-Volume Relationship or Charles's Law

It states that volume of a given mass of a dry gas is directly proportional to its absolute (kelvin) temperature, if the pressure is kept constant.

OR

The pressure remaining constant, the volume of a given mass of a dry gas increases or decreases by 1/273 of its volume for each 1°C increase or decrease in temperature respectively.

\(\frac{V_1}{T_1} = \frac{V_2}{T_2} = k\) at constant pressure

Gas Equation

On combining both the laws we conclude that the volume of a given mass of a dry gas varies inversely as the pressure and directly as the absolute temperature.

\(V \propto \frac{1}{P} \times T\)

or

\(V = \frac{T}{P} \times \text{constant}\)

or

\(\frac{PV}{T} = k\) (constant)

If the volume of a given mass of a gas changes from \(V_1\) to \(V_2\), pressure from \(P_1\) to \(P_2\) and temperature from \(T_1\) to \(T_2\) then,

\(\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} = k\) (constant)

Standard Temperature Pressure (STP)

Since the volume of a gas changes remarkably with change of temperature and pressure, it becomes necessary to choose standard values of temperature and pressure to which gas volumes can be referred.

The standard values chosen are : 0°C or 273 K for temperature and 1 atmospheric pressure or 760 mm or 76 cm of Hg for pressure.

Temperature Scales

Absolute Scale or Kelvin Scale : A temperature scale with absolute zero (zero kelvin) as the starting point is called the absolute scale or the kelvin scale.

Absolute zero = 0 K = - 273°C

5.2 Gay-Lussac's Law of Combining Volumes

Gay-Lussac (1805) observed that a simple relation exists between the volumes of hydrogen and oxygen, which react together to form water. Thus, one litre of oxygen requires two litres of hydrogen to form two litres of water vapour. So, he found that oxygen and hydrogen react in the ratio 1 : 2 (by volume). When he extended his study to the volumes of other reacting gases, he noted similar simple relationships. Consequently, he generalised these observations as the law of combining volumes of gases.

Gay-Lussac's Law of Combining Volumes

When gases react, they do so in volumes which bear a simple ratio to one another, and to the volume of the gaseous product, provided that all the volumes are measured at the same temperature and pressure.

Note : Gay-Lussac's Law is valid only for gases. The volumes of solids and liquids are considered to be zero.

The law may be illustrated by the following examples involving gases or vapours :

1. Hydrogen chloride

One volume of hydrogen when mixed with one volume of chlorine, gives two volumes of hydrogen chloride gas at the same temperature and pressure.

\(H_2 + Cl_2 \to 2HCl\)

Thus, the ratio 1 : 1 : 2 is simple.

2. Ammonia

One volume of nitrogen combined with three volumes of hydrogen gives two volumes of ammonia at the same temperature and pressure.

\(N_2 + 3H_2 \to 2NH_3\)

Thus, the ratio 1 : 3 : 2 is simple.

3. Carbon dioxide

Two volumes of carbon monoxide on combustion with one volume of oxygen gives two volumes of carbon dioxide at the same temperature and pressure.

\(2CO + O_2 \to 2CO_2\)

Thus, the ratio 2 : 1 : 2 is simple.

Numericals based on Gay-Lussac's Law

Example 1 : What volume of oxygen would be required to burn completely 200 mL of acetylene (C2H2) and what would be the volume of carbon dioxide formed ?

Solution :

\(C_2H_2 + O_2 \to CO_2 + H_2O\)

On balancing,

\(2C_2H_2 + 5O_2 \to 4CO_2 + 2H_2O\) 2 vols 5 vols 4 vols Nil

According to the above equation, 2 volumes of acetylene will require 5 volumes of oxygen for complete combustion to produce 4 volumes of carbon dioxide.

- 200 mL of acetylene will require

\(= \frac{5}{2} \times 200 \text{ mL} = 500 \text{ mL of oxygen}\), i.e., 500 mL of oxygen is required to produce 200 × 2 = 400 mL of carbon dioxide.

Answer : 500 mL of oxygen is required to burn completely 200 mL of acetylene thereby producing 400 mL of carbon dioxide.

Example 2 : If 6 litres of hydrogen and 5-6 litres of chlorine are mixed and exploded, what will be the composition by volume of the resulting gaseous mixture ?

Solution :

\(H_2 + Cl_2 = 2HCl\) 1 vol 1 vol 2 vols

Since 1 volume of chlorine reacts with 1 volume of hydrogen,

- 5-6 litres of chlorine will react with only 5-6 litres of hydrogen.

- (6 - 5.6) i.e., 0-4 litres of hydrogen will remain unreacted.

Since the volume of HCl gas formed is twice that of chlorine used,

- volume of HCl formed will be 5-6 × 2 = 11-2 litres.

Answer : 11.2 litres of HCl and 0-4 litres of residual hydrogen.

Note : The reactant which is completely used up in a reaction is known as Limiting reagent or Limiting reactant. In example 2 chlorine is the limiting reagent.

Example 3 : What volume of propane is burnt for every 100 cm³ of oxygen in the reaction.

\(C_3H_8 + 5O_2 \to 3CO_2 + 4H_2O\)

(Gas volumes measured under the same conditions).

Solution :

From the above reaction, it is clear that for every 5 volumes of oxygen, 1 volume of propane is burnt.

- Volume of propane burnt for every 100 cm³ of oxygen = \(\frac{1}{5}\) × 100 = 20 cm³

Answer : 20 cm³ of propane is burnt.

Example 4 : 80 cm³ of methane is mixed with 200 cm³ of pure oxygen at room temperature and pressure. The mixture is then ignited when it burns as illustrated by the equation :

\(CH_4(g) + 2O_2(g) \to CO_2(g) + 2H_2O(l)\)

Calculate the composition of the resulting mixture if it is cooled to initial room temperature and pressure.

Solution :

\(CH_4(g) + 2O_2(g) \to CO_2(g) + 2H_2O(l)\) 1 vol 2 vols 1 vol Nil

By Gay-Lussac's law,

(i) 1 vol of methane requires 2 vols of oxygen.

- 80 cm³ of methane requires

\(= 2 \times 80 = 160 \text{ cm}^3\) oxygen

(ii) 1 vol of methane produces 1 vol of carbon dioxide.

- 80 cm³ of methane produces 80 cm³ of carbon dioxide.

Hence, the composition of gaseous mixture after reaction is :

Methane = (80 - 80) = 0 Carbon dioxide = 80 cm³ Oxygen = (200 - 160) = 40 cm³ Water = Negligible.

Example 5 : 40 cm³ of methane reacts with chlorine according to the following equation.

Equation :

\(CH_4 [g] + 2Cl_2 [g] \to CH_2Cl_2 [g] + 2HCl [g]\)

Calculate the volume of HCl gas formed and chlorine gas required.

Solution :

By Gay-Lussac's law :

1 Vol of CH4 produces 2 Vols of HCl

- 40 cm³ of CH4 will produce

\(2 \times 40 = 80 \text{ cm}^3 \text{ of HCl}\).

Since 1 Vol of CH4 require 2 Vols of chlorine,

- 40 cm³ of methane will require

\(2 \times 40 = 80 \text{ cm}^3 \text{ of chlorine}\).

Answer : 80 cm³ of HCl is formed. 80 cm³ of chlorine is required.

Example 6 : 200 cm³ of ethylene [C2H4] is burnt in just sufficient air (containing 20% oxygen to form carbon dioxide gas and steam. If all measurements are made at constant pressure and 100°C, find the composition of the resulting mixture.

Solution :

The reaction involved is given by :

\(C_2H_4(g) + 3O_2(g) \xrightarrow{100°C} 2CO_2(g) + 2H_2O(g)\) 1 vol. 3 vols. 2 vols. 2 vols.

By Gay-Lussac's Law;

(i) 1 vol. of ethylene requires oxygen = 3 vols

- 200 cm³ of ethylene will require oxygen

\(= 3 \times 200 = 600 \text{ cm}^3\).

(ii) 1 vol. of ethylene produces carbon dioxide = 2 vols

- 200 cm³ of ethylene will produce carbon dioxide = 2 × 200 = 400 cm³.

(iii) 1 vol. of ethylene produces steam = 2 vols

- 200 cm³ of ethylene will produce steam

\(= 2 \times 200 = 400 \text{ cm}^3\).

When oxygen is 20%, unreacted air is = 80%

When oxygen is 600 cm³, then unreacted air

is = \(\frac{80 \times 600}{20} = 2400 \text{ cm}^3\).

Hence, composition of mixture after reaction :

(i) Carbon dioxide = 400 cm³

(ii) Steam = 400 cm³

(iii) Unreacted air = 2400 cm³.

Example 7 : 20 mL of hydrogen, 10 mL of carbon monoxide and 20 mL of oxygen are exploded in an eudiometer. What will be the volume and composition of the mixture of gases, after cooling to room temperature ?

Write each reaction separately.

Solution :

\(2H_2 + O_2 = 2H_2O\) 2 vols 1 vol. Nil 20 mL 10 mL

\(2CO + O_2 = 2CO_2\) 2 vols 1 vol. 2 vols 10 mL 5 mL 10 mL

Total volume of oxygen used = (10 + 5) i.e. 15 mL.

Volume of oxygen left over = 20 - 15 = 5 mL

Volume of carbon dioxide formed = 10 mL

- Total volume of gases = 15 mL

Answer : Total volume of gaseous mixture after cooling to room temperature would be 15 mL, containing 5 mL oxygen and 10 mL carbon dioxide.

Example 8 : A sample of coal gas contained 45% H2, 30% CH4, 20% CO, and 5% C2H2 by volume. 100 mL of gaseous mixture was mixed with 160 mL of oxygen and exploded. Calculate the volume and the composition of the resulting mixture, when cooled to room temperature and pressure.

Solution :

100 mL of gaseous mixture will contain 45 mL H2, 30 mL CH4, 20 mL CO and 5 mL C2H2.

Equations representing the involved combustion are :

(a) Hydrogen

\(2H_2 + O_2 = 2H_2O\) 2 vols 1 vol Nil 45 mL 22.5 mL

(b) Methane

\(CH_4 + 2O_2 = CO_2 + 2H_2O\) 1 vol 2 vol 1 vol. Nil 30 mL 60 mL 30 mL

(c) Carbon monoxide

\(2CO + O_2 = 2CO_2\) 2 vols 1 vol. 2 vols 20 mL 10 mL 20 mL

(d) Acetylene

\(2C_2H_2 + 5O_2 = 4CO_2 + 2H_2O\) 2 vols 5 vols 4 vols Nil 5 mL 12.5 mL 10 mL

Total volume of oxygen used = (22.5 + 60 + 10 + 12.5) = 105 mL

- Volume of oxygen left = (160 - 105) = 55 mL

Total volume of carbon dioxide formed = 30 + 20 + 10 = 60 mL

- Total volume of gases left = 55 mL

Answer : 55 mL residual oxygen and 60 mL carbon dioxide and 115 mL total volume.

5.3 Avogadro's Law

Amedo Avogadro in 1811 put forward a hypothesis based on the relationship between the number of molecules in equal volumes of different gases under similar conditions.

Avogadro's law states that "equal volumes of all gases under similar conditions of temperature and pressure contain the same number of molecules."

This means that one litre of hydrogen contain the same number of molecules as are present in one litre of oxygen or in one litre of chlorine or of any other gas, provided the volumes of all gases are measured at the same temperature and pressure.

Avogadro further suggested that the smallest particle of a gaseous element is the molecule and not the atom though it may contain one, two or more atoms.

Example : A molecule of N2 is made of two atoms of nitrogen, a molecule of NH3 is made of one atom of nitrogen and three atoms of hydrogen.

He made the following distinction between atoms and molecules of a gaseous element.

An atom is the smallest particle of an element that can take part in a chemical reaction; however, it may or may not exist independently.

A molecule is the smallest particle of an element or a compound that can exist by itself; it never breaks up except for taking part in a chemical reaction.

Atomicity

The number of atoms in a molecule of an element is called its atomicity.

(a) Monoatomic

Monoatomic molecule is composed of only one atom.

Examples : Inert gases like Helium, Neon, Argon, etc.

(b) Diatomic

Diatomic molecule is composed of two similar atoms.

Examples : H2, O2, Cl2, N2, etc.

(c) Triatomic

Triatomic molecule is composed of three similar atoms.

Example : Ozone gas (O3).

(d) Tetratomic

Tetratomic molecule is composed of four similar atoms.

Example : Phosphorus (P4).

(e) Octatomic

Octatomic molecule is composed of eight similar atoms.

Example : Sulphur (S8).

Molecules made up of same type of atoms are homoatomic molecules, e.g. phosphorus (P4), ozone (O3) etc. while molecules made up of different types of atoms are hetero-atomic molecules, e.g. HCl, NH3 etc.

Examples based on Avogadro's Law

Example 9 : Under same conditions of temperature and pressure, you collect 2 litres of carbon dioxide, 3 litres of chlorine, 5 litres of hydrogen, 4 litres of nitrogen and 1 litre of sulphur dioxide. In which gas sample will there be (i) the greatest number of molecules ? (ii) the least number of molecules ? Justify your answer.

Solution :

Equal volumes of all gases under similar conditions of temperature and pressure contain the same number of molecules. So, under the same conditions of temperature and pressure, if volume of the gas is decreased, the number of molecules will also decrease.

Hence,

(i) 5 litres of hydrogen contain the greatest number of molecules.

(ii) 1 litre of sulphur dioxide contains the least number of molecules.

Example 10 : If 50 cc of a gas A contains y molecules, how many molecules of gas B will be present in 25 cc of B under same conditions ?

Solution :

50 cc contains y molecules.

- 25 cc will contain \(\frac{y}{2}\) molecule.

According to Avogadro's law, 50 cc of A will contain equal molecules as 50 cc of B.

- 25 cc will contain half of what 50 cc contains.

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