Some Basic Concepts of Chemistry

1Some Basic Concepts of Chemistry

MHT - CET Chemistry - XI

01. Some Basic Concepts of Chemistry

1.0 : Prominent Scientists :

Note: In order to give a tribute to Avogadro's contributions related to molecular theory, the number of elementary

entities (atoms, molecules, ions or other particles) in 1 mole of a substance, 6.022 x 1023 is known as

Avogadro number.

1.1 : Introduction :

Q.1. Chemistry has played an important role in the fulfillment of basic needs of man. Explain.

Ans: Increasing population has led to an increase in the demands of basic needs of man (food, clothing and

shelter). Developments in the field of chemistry have helped to cope up with these necessities as follows:

i. Food:

a. The population of a country requires nutritious and hygienic food in sufficient quantity. To achieve

the same, there is a need to manufacture good quality fertilizers and insecticides.

b. The advancement of chemistry has helped many countries to become not only self sufficient but

also an exporter of food commodities.

ii. Clothing:

a. Good quality clothes are required for every individual to adjust with changing environmental conditions.

b. Because of the production of synthetic fibres like nylon, rayon, etc. in factories, this need has

been fulfilled.

iii. Shelter:

a. The human population needs comfortable and well-built houses. Iron, cement and steel are

required in large quantities for construction of such houses.

b. Chemistry has played an important role in the extraction of these metals from their respective

ores.

Q.2. Define chemistry.

Ans: Chemistry is defined as the study of the composition, structure and properties of matter and the reactions

by which one form of matter may be converted into another form.

Q.3. Give reason: Chemistry is called as a central science.

Ans: i. Chemistry is an active evolving science and is of vital importance to the entire world. Although the

Joseph Louis Gay-Lussac (1778 – 1850) i. Formulated the gas law.

(French chemist and physicist) ii. Collected samples of air at different heights and

recorded temperatures and moisture contents.

iii. Discovered that the composition of atmosphere

does not change with increasing altitude.

Amedeo Avogadro (1776 – 1856) i. Published article in French journal on determining

(Italian scholar) the relative masses of elementary particles of

bodies and proportions by which they enter

combinations.

ii. Published a research paper titled ''New

considerations on the theory of proportions and

on determination of the masses of atoms."

Scientists Contributions

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subject has very ancient roots, it is a modern science.

ii. The basic knowledge of chemistry is essential for development of subjects like physics, biology, geology,

engineering, environmental science and many others.

Therefore, it is called as a central science.

Q.4. What are the various branches of chemistry?

Ans: The several branches of chemistry are as follows:

i. Physical chemistry:

It is the branch of chemistry that deals with the structure of matter, the energy changes and the

theories, laws and principles that explain the transformation of matter from one form to another.

ii. Inorganic chemistry:

It is the branch of chemistry that deals with chemistry of elements other than carbon and their

compounds.

iii. Organic chemistry:

This branch of chemistry deals with reactions of the compounds of carbon.

iv. Analytical chemistry:

This is the branch of chemistry which deals with the separation, identification and quantitative

determination of the compositions of different substances.

v. Biochemistry:

This is the branch of chemistry that deals with substances which are constituents of living organisms.

Note:

Pharmaceutical, environmental and nuclear chemistry are also branches of chemistry.

1.2 : Importance and scope of chemistry :

Q.5. Explain the importance and scope of chemistry.

Ans: Chemistry has a wide scope and importance in various fields.

i. Processes based on chemical technology help to extract, purify, synthesize and analyse materials like

iron, steel, aluminium, zinc, alloys like brass, amalgams as well as precious metals like silver, gold,

platinum.

ii. All drugs are synthesized in chemical / pharmaceutical laboratories.

eg.

iii. Photosynthesis is the process through which trees and plants prepare their food using chlorophyll

(green pigment) in presence of sunlight. The process is a simple chemical reaction and takes place

naturally.

CO2 + H

2O

sunlight

chlorophyll food grains/fruits/flowers/cotton/medicine etc.

iv. Fossil fuels like coal, petroleum, natural gas, etc. are combustible chemicals which are used to produce

energy which is used to drive trains, trucks, buses and all automobiles. The energy is also used to

generate electricity. Several electrochemical cells like Daniel 'cell, lead storage cell, dry cell, nickel

cadmium cell, lithium ion cell, fuel cell, etc., are used as a source of energy. These cells are less'

polluting and more efficient. There are attempts being made to convert solar energy into electrical

energy using photovoltaic cells, the solar cells. Attempt is also made to obtain hydrogen from water,

which is used in fuel cells to generate electricity.

i. L-dopa For treatment of Parkinson's disease.

ii. Human insulin For treatment of diabetes.

iii. Cisplatin and taxol Life saving drugs to give relief to cancer patients.

iv. Azidothymidine (AZT) For treatment of AIDS.

v. Tamiflue For treatment of swine flu.

No. Drugs Treatment of diseases





v. With the help of chemistry it is easy to design and generate large number of materials like polymers,

plastic, liquid crystals, adhesives and surface coating materials like latex paints. Knowledge of chemistry

can also be used to synthesize new materials that can act as super conductors at or near room

temperature due to which loss of electricity will get reduced by almost 20%. Microprocessors used in

computers are silicon chips formulated and developed by chemists.

Q.6. Give five applications of subject chemistry which are not mentioned in the book.

Ans: i. Warfare:

With the knowledge of chemistry, various destructive gases and bombs have been invented which are

used during wars.

ii. Cosmetics:

Chemistry has helped to produce good quality cosmetics.

iii. Health:

Chemistry plays an important role in maintaining one's good health by providing knowledge about

proper intake of proteins, carbohydrates, fats, minerals, vitamins, etc.

iv. Education:

Chemistry provides inter-relationship to study the para-chemistry subjects such as Bio-chemistry,

Pharmacy, Herbal Science, Toxicology, Archaeology, Environmental Science, etc.

v. In recent years, chemistry has given us new materials such as super-conducting ceramics, conducting

polymers, optical fibres, micro alloys, carbon fibres, etc. which are used for various purposes.

1.3 : Historical approach to particulate nature are matter :

Q.7. Define matter.

Ans: Matter is anything which has mass and occupies space.

Q.8. Explain the classification of matter on the basis of its chemical composition.

Ans: Matter on the basis of chemical composition can be classified as follows:

i. Pure substances:

Substances which always have afzxed composition are called pure substances.

They are of two types:

a. Elements:

Elements are the pure substances which are made up of only one component.

eg. Gold, silver, copper, mercury, bromine, oxygen etc.

b. Compounds:

Compounds are the pure substances which are made up of two or more components.

eg. Water, ammonia, methane etc.

ii. Mixtures:

A mixture is a simple combination of two or more substances in which the constituent substances

retain their separate identities.

The composition of mixture can be varied to any extent. Therefore, mixtures do not have fixed

composition.

eg. Mixture of ethyl alcohol and water, salt in water, mixture of gases, etc.

Mixtures are of two types:

a. Homogeneous mixture:

A mixture in which the concentration of the constituents remains uniform throughout the mixture

and all the constituents are present in one phase, is called a homogeneous mixture.

eg. Mixture of salt and water.

b. Heterogeneous mixture:

If two or more phases are present in a mixture, it is called a heterogeneous mixture.

eg. Phenol - water system, silver chloride-water system, iron fillings-sand system, etc.

Q.9. What is Phlogiston theory?





Ans: Phlogiston theory:

i. According to this theory, a combustible substance contains phlogiston (a mysterious matter) and some

clax. During combustion, phlogiston evolves and is lost in the atmosphere. Clax remains in the form of

an ash.

ii. Combustion reactions were explained by phlogiston theory.

eg. During the combustion of a candle in a closed container, the air inside the container is saturated

with phlogiston. Since air cannot accommodate more phlogiston, the candle gets extinguished.

Q.10. What is "dephlogisticated air"? Who named it?

Ans: i. Joseph Priestley (a British scientist) focussed sunrays on a substance (mercuric oxide) to heat it.

ii. A gas evolved, in which substances could bum more vigorously than in air.

iii. A burning candle became brighter in this gas.

iv. Priestley was of the view that this gas is the normal air without phlogiston. Hence he named it

"dephlogisticated air".

Q.11. What was the contribution of Sir Henry Cavendish in the phlogiston theory?

Ans: i. Sir Henry Cavendish carried out the reaction of a dilute acid with metals such as zinc, iron, etc. He

named the gas evolved as "flammable air". It was found that this gas burnt in air and in dephlogisticated

air and produced water.

ii. Cavendish suggested that flammable air is water associated with phlogiston. This is in continuation

with the idea of phlogiston.

Q.12. Who ruled out the theory of phlogiston? Why?

Ans: i. The theory of phlogiston was ruled out by Antoine Lavoisier (a French Scientist).

ii. He proved that a part of air is used in chemical reaction during combustion. This part of air was called

oxygen. It means 'acid forming'.

iii. He also proved that oxygen was the gas formed in Joseph Priestley's experiment.

iv. He also proved that the flammable air produced by Cavendish was a new gas, which he named as

hydrogen, meaning 'water forming'.

Q.13. Define and explain the following with the help of examples.

i. Elements ii. Compounds

Ans: i. Elements:

a. An element is defined asa substance which cannot be separated into simpler substances by any

chemical process.

eg. Gold, silver, copper, carbon, etc.

b. Out of 118 elements that have been identified till recently, most are naturally occuring and a few

are prepared in the laboratory (man-made).

c. Chemists use one or two letter symbols to represent elements.

d. All the chemical symbols have first letter capital and second letter small (if present).

e. The symbols of the elements are derived either from English names or from Latin names of the

elements.

eg. Aluminium - Al , Einsteinium - Es, Gold - Au (Aurum)

f. Elements contain only one type of atom. Elements are further classified as metals, non-metals,

metalloids.

ii. Compounds:

a. Compounds are defined as substances of definite compositions which can be decomposed into

two or more substances by a simple chemical process.

eg. Water, sodium chloride, sugar, alcohol, etc.

b. The properties of compounds differ from the properties of the substances and elements obtained

from decomposition of the compounds.

eg. Hydrogen and oxygen are obtained from decomposition of water. Water can be used to

extinguish fire whereas oxygen supports combustion and hydrogen is combustible.





c. Compounds are classified into two subclasses organic compounds and inorganic compounds.

d. Compounds contain two or more components.

Note:

Names and symbols of some elements

Q.14. Classify the following substances into elements, compounds, homogeneous mixtures and

heterogeneous mixtures.

Sand in water, Sodium chloride, Nitrogen, Sodium chloride in water, Pumice stone, Air,

Phenolwater system, Carbon dioxide, Gold

Ans: i. Sand in water : Heterogeneous mixture

ii. Sodium chloride : Compound

iii. Nitrogen : Element

iv. Sodium chloride in water : Homogeneous mixture

v. Pumice stone : Heterogeneous mixture

vi. Air : Homogeneous mixture

vii. Phenol-water system : Heterogeneous mixture

viii. Carbon dioxide : Compound

ix. Gold : Element

Q.15. Distinguish between

i. Mixtures and compounds. ii. Compounds and elements.

Ans: i. Mixtures and compounds:

Aluminium Al Cadmium Cd Mercury Hg Neobium Nb Rhodium Rh

Argon Ar Caesium Cs Holmium Ho Neodymium Nd Rhenium Re

Silver Ag Cerium Ce Iodine I Neptunium Np Sulphur S

Gold Au Curium Cm Irridium Ir Oxygen O Scandium Sc

Actinium Ac Calefomium Cf Krypton Kr Osmium Os Selenium Se

Americium Am Erbium Er Lithium Li Potassium K Strontium Sr

Beryllium Be Einsteinium Es Lanthanum La Phosphorous P Sodium Na

Boron B Fluorine F Lutetium Lu Lead Pb Technicium Tc

Barium Ba Francium Fr Lawrencium Lr Palladium Pd Uranium U

Bismuth Bi Iron Fe Magnesium Mg Platinum Pt Tungsten W

Carbon C Galium Ga Manganese Mn Promethium Pm Vanadium V

Chlorine Cl Germanium Ge Molybdenum Mo Protoactinium Pa Xenon Xe

Calcium Ca Gadolinium Gd Mendelivium Md Plutonium Pu Ytterbium Yb

Chromium Cr Hydrogen H Nitrogen N Radium Ra Zirconium Zr

Cobalt Co Helium He Neon Ne Rubidium Rb

Copper Cu Hafnium Hf Nickel Ni Ruthenium Rn

Element Sym- Element Sym- Element Sym- Element Sym- Element Sym-

bol bol bol bol bol

i. The constituents of a mixture may be present The constituents of a compound are always present

in any ratio. in a fixed ratio.

ii. Mixtures may or may not be homogeneous Compounds are always homogeneous in nature.

in nature.

iii. 'The properties of a mixture are in between The properties of a compound are entirely different

those of its constituents. from those of its constituent elements.

iv. The constituents of a mixture can be easily The constituents of a compound cannot be easily

separated by simple physical means. separated by simple physical means but can be

separated by chemical processes.

v. The melting and boiling points of Chemical compounds possess sharp melting and

mixtures are usually not sharp. boiling points.

No. Mixtures Compounds





ii. Compounds and elements:

Q.16. Define a unit.

Ans: The arbitrarily decided and universally accepted standards used in the measurement of physical quantities

are called units. eg. meter (m), kilogram (kg).

Q.17. State the need of units.

Ans: i. In chemistry, several experiments are carried out which involve observation and collection of both

qualitative and quantitative data. .

ii. Measurement of physical properties such as mass, length, volume, temperature, pressure, time, etc.,

comprise of the quantitative data.

iii. For this purpose, the magnitude or size of physical quantity is compared with a suitable standard.

These units are arbitrarily chosen on the basis of universally accepted standards.

iv. To express any measured property, a number and an appropriate unit has to be used. Only number

does not give any idea about the property.

Q.18. What are the various systems in which units are expressed?

Ans: Units are expressed in various systems like CGS (centimetre for length, gram for mass and second for

time), FPS (foot, pound, second) and MKS (metre, kilogram, second) systems, etc.

Note:

i. During calculations, confinement to one single system of unit is advisable.

ii. NASA's Mars climate orbiter (first weather satellite for mars) was destroyed due to heat. The mission failed

as' there was a confusion while estimating the distance between earth and mars in miles and kilometers.

Q.19. What are st units? Name the fundamental SI units.

Ans: SI Units:

In 1960, the general conference of weights and measures proposed revised metric system, called International

system of Units i.e. SI system abbreviated from its F nch name Systeme Internationale d' Units.

The seven fundamental SI units are as given below:

i. A compound is a substance composed of An element is a pure chemical substance made of

two or more different chemical elements. same type of atoms.

ii. A compound can be separated into simpler Elements cannot be broken down into simpler

substances by chemical reactions. substances.

iii. Compounds contain different elements in a Elements are distinguished by their atomic' number

fixed ratio arranged in a defined manner (number of protons in their nucleus).

through chemical bonds.

iv. A compound is represented using a formula. An element is represented using symbols.

v. The list of compounds is endless but can There are about 118 elements that have been

broadly be classified as ionic and covalent. identified and can be classified as metals, nonmetals

or metalloids.

eg. Sodium chloride (NaCl) , Sodium eg. Iron, copper, silver, gold, etc.

bicarbonate (NaHCO3), etc.

No. Compounds Elements

i. Length Metre m

ii. Mass Kilogram kg

iii. Time Second s

iv. Temperature Kelvin K

v. Amount of substance Mole mol

vi. Electric current Ampere A

vii. Luminous intensity Candela cd

No. Fundamental quantity SI unit Symbol





Q.20. What are derived units?

Ans: The units of all physical quantities can be derived from the seven fundamental S1 units. These units are

known as derived units.

eg. i. Area = Length squared (m2)

ii. Concentration = mole per cubic metre (mol m–3)

Note: The table given below shows some common derived units.

Some common SI prefixes used for expressing big and small numbers:

i. Area Length squared m2

ii. Volume Length cubed m3

iii. Density Mass per unit volume kg m–3

iv. Velocity Distance travelled in unit time ms–1

v. Acceleration Velocity change per unit time ms–2

vi. Force Mass × Acceleration kg m s–2(newton, N)

vii. Pressure Force per unit area kg m–1 s–2

viii. Electric charge Current × Time As (coulomb, C)

ix. Electric potential or Potential Energy per unit charge kg m2 s–2A–1

difference (J A-1 s–1 or

Volt,V or JC–1)

x. Energy Force × distance travelled kg m2 s–2

(work or heat) (Joule s–1)

xi. Concentration Mole per cubic metre mol m–3

xii. Heat capacity Cp = dH / dT JK–1 mol–1

Cv = dE / dT

xiii. Electrochemical equivalent Z = E/F kg C–1 (kg/Coulomb)

No. Physical Quantity Relationship with fundamental unit Unit

Tera- T 1012 1 000 000 000 000

Giga- G 109 1000000000

Mega- M 106 1000000

myria- my 104 1 0000 (this is now obsolete)

kilo- k 103 1 000

hecto- h 102 100

deka- da 10 10

– – – –

deci- d 10–1 0.1

centi- c 10–2 0.01

milli- m 10–3 0.001

micro- µ 10–6 0.000001

nano- n 10–9 0.000000001

picc- p 10–12 0.000000000001

femto- f 10–15 0.000000000000001

Prefix Symbol Magnitude Meaning (multiply by)





1.4 : Laws of chemical combination :

Q.21. What is a chemical combination?

Ans: The process in which the elements combine with each other chemically, to form compounds, is called as

chemical combination.

Q.22. State and explain the laws of chemical combination.

Ans: Laws of chemical combination:

One of the most important aspects of the subject of chemistry is the study of chemical reactions. These

chemical reactions take place according to certain laws called as "Laws of chemical combination". They

are:

i. Law of conservation of mass:

a. The law of conservation of mass states that, "mass is neither created nor destroyed during

chemical combination of matter ".

b. It was first stated by Russian scientist Lomonosove (1765) and later independently stated by

French chemist Antoine Lavoisier (1783) who performed careful experimental studies for various

combustion reactions.

c. Lavoisier observed that the total masses of the reactants (before the reaction) were in agreement

with the total masses of the products (after the reaction).

eg. Consider the reaction for the burning of carbon to form carbon dioxide.

Carbon + Oxygen Carbon dioxide

The sum of the masses of carbon and oxygen (reactants) is always equal to the mass of the

carbon dioxide (product). This is in accordance with the law of conservation of mass.

ii. Law of defmite composition / proportions:

a. The law of definite composition! proportions was stated by French chemist Joseph Proust.

b. The law states that "any pure compound always contains the same elements in a definite proportion

by weight irrespective of its source or method of preparation ".

c. Proust worked with two samples of cupric carbonate; one of which was naturally occuring

cupric carbonate and other was prepared in the laboratory. He found that the composition of

elements present in both the samples of cupric carbonate was same as shown below:

OCu C

Naturally occurring cupric carbonate 51.35 39.91 9.74

Cupric carbonate prepared in the laboratory 51.35 38391 9.74

Percentage

d. The law was further supported by different samples of pure water which contained same amount

of oxygen (88.81 % by weight) and hydrogen (11.19% by weight) and different samples of pure

sugar which contained same amount of carbon (42.1% by weight), hydrogen (6.5% by weight)

and oxygen (51.4% by weight). This was irrespective of the source.

iii. Law of multiple proportions:

a. John Dalton (British scientist) proposed the law of multiple proportions in 1803.

b. The law states that, "if two elements chemically combine with each other forming two or more

compounds with different compositions by mass, then the ratios of masses of the two interacting

elements in the two compounds are small whole numbers ".

eg.

Chemical reaction of carbon with oxygen gives two compounds carbon monoxide and carbon

dioxide. Carbon monoxide is a poisonous, combustible gas. However carbon dioxide is a non-

poisonous, non-combustible gas.

I g of carbon reacts with 1.33 g of oxygen to form carbon monoxide. Ig of carbon reacts with

2.66 g of oxygen to form carbon dioxide. The ratio of weights of oxygen to that of carbon for





carbon dioxide is

2.66 g of oxygen

1.00 g of carbon = 2.66

And the ratio of weights of oxygen to that of carbon for carbon monoxide is

1.33g of oxygen

1.00 g carbon = 1.33

The two ratios are in the proportion 2.66

1.33 = 2 i..e., 2 : 1

Therefore, the ratio of the masses of oxygen that combine with the same mass of carbon is 2: 1

i.e., a simple ratio.

iv. Gay Lussac's law of combining volumes of gases:

a. Joseph Louis Gay Lussac (a French chemist) observed that there exists a definite relationship

between volumes of gaseous reactants and the volumes of gaseous products. He generalized his

observations in the form of a law of combining volumes of gases.

b. Gay Lussac's law states that, when gases react together to produce gaseous products, the

volumes of reactants and products bear a simple whole number ratio with each other, provided

volumes are measured at same temperature and pressure.

c. eg. Under identical conditions of temperature and pressure, lL of hydrogen gas reacts with 1 L of

chlorine gas to produce 2 L of hydrogen chloride gas i.e.

Hydrogen + Chlorine Hydrogen chloride

[1 L] [1 L] [2 L]

[ 1 vol] [ 1 vol] [ 2 vol]

Thus, the ratio of volumes is 1 : 1 : 2

d. Volumes may be measured in any convenient unit such as L, mL, cm3, dm3, etc.

eg. Consider the reaction for the conversion of sulphur dioxide to sulphur trioxide.

Sulphur dioxide + Oxygen Sulphur trioxide

[ 200 mL ] [ 100 mL] [ 200 mL ]

[ 1 vol] [ 12 vol] [ 1 vol]

The ratio of volumes is 2 : 1 : 2.

Note:

The first three laws deal with the mass relationships whereas the fourth law deals with the volume of the

reacting gases.

Q.23. Who opposed the law of definite proportion? How were the objections ruled out?

Ans: i. Berthollet (French scientist) opposed Proust's law of definite proportion.

ii. He gave examples of the substances that contained different proportions of elements.

iii. However, the experimental work of analysis explained by Berthollet was based on impure samples or

incomplete reactions.

Hence, Berthollet's objections were ruled out.

Q.24. Is the law of constant composition true for all types of compounds? Justify your answer.

Ans: No, law of constant composition is not true for all types of compounds. It is true for only those compounds

which are obtained from one isotope.

eg. Carbon exists in two common isotopes: 12C and 14C. When it forms 12CO2, the ratio of masses is 12 :

32 or 3 : 8. However, when it is formed from 14C i.e., 14CO2, the ratio will be 14 : 32 i.e., 7 : 16, which

is not same as in the first case.

Q.25. Verify the law of multiple proportions for the chemical reaction between hydrogen and oxygen.





Ans: The chemical reactio;n of hydrogen with oxygen gives two compounds, water and hydrogen peroxide.

Water contains 88.89% by weight of oxygen and 11.11% by weight of hydrogen. The ratio of the percentages

by weight of oxygen to that of hydrogen is equal to

88.89% by weight of oxygen

11.11%by weig.ht of hydrogen = 8 = 8 : 1

And hydrogen peroxide contains 94.12% by weight of oxygen and 5.88% by weight of hydrogen. The ratio

of the percentages by weights is equal to

94.12%byweight of oxygen

5.88% by weight of hydrogen = 16 = 16 : 1

The ratios of the two compoun dss aarree iIn the proportion 16:1

8:1 = 2 : 1

Therefore, the ratio of weights of oxygen that combine with the same weight of hydrogen in the two

compounds is equal to 2 : 1.

This is in accordance with the law of multiple proportions.

Q.26. Give two examples which support the Gay Lussac's law of combining volumes of gases.

Ans: i. Under identical conditions of temperature and pressure, 2 L of hydrogen gas reacts with lL of oxygen

gas to produce 2 L of steam (water vapour). i.e.,

Hydrogen + Oxygen Steam (water vapour)

2 L 1 L 2 L

(2 vol) (1 vol) (2 vol)

The ratio of volumes, is 2 : 1 : 2.

ii. Under identical conditions of temperature and pressure, 1 L of nitrogen gas reacts with 3 L of hydrogen

gas to produce 2 L of ammonia gas, i.e.

Nitrogen + Hydrogen Ammonia

1 L 3 L 2 L

(1 vol) (3 vol) (2 vol)

Here the ratio of volumes is 1 : 3 : 2

From these two examples, it can be concluded that there exists a simple ratio of whole numbers of

volumes of the gaseous reactants with gaseous products.

Hence, these examples support the Gay Lussac's law.

Note:

i. Gay Lussac's law of combining volumes is applicable only to reactions involving gases and not to

solids and liquids.

ii. The volumes of gases in the chemical reaction are not additive. For example, in case of reaction

between hydrogen and chlorine gases it appears to be additive. However in case of reaction between

hydrogen and oxygen, 2 volumes of hydrogen and 1 volume of oxygen, equal to 3 volumes of reactants

get converted into 2 volumes of the product, steam.

iii. Similarly, in case of formation of ammonia, 1 volume of nitrogen and three volumes of hydrogen, equal

to 4 volumes of reactants, react to get converted into 2 volumes of the product, ammonia.

1.5 : Dalton's atomic theory :

Q.27. State and explain Dalton's atomic theory.

Ans: To provide theoretical justification to the laws of chemical combination, John Dalton postulated a simple

atomic theory of matter (1808).

According to him, "Atom is the smallest indivisible particle of a substance". The basic assumptions of the

theory are ,given below:

i. All matters are made up of tiny, indestructible, indivisible unit particles called atoms. Atoms are the





smallest particles of the element and molecules are the smallest particles of a compound.

ii. All atoms of the same element have same size, shape and mass and all other properties. Atoms of

different elements have different properties.

iii. Compounds are formed when atoms of different elements combine. The atoms in a compound unite

in small whole number ratios like 1 : 1, 1 : 2, 1 : 3, 2 : 1, 2 : 3, etc.

iv. A chemical reaction involves only the separation, combination or rearrangement of integer number of

atoms. During a chemical reaction, atoms are neither created nor destroyed.

Q.28. How was the law of conservation of mass explained by Dalton on the basis of his atomic theory?

Ans: i. Based on the assumptions of atomic theory, Dalton explained the law of conservation of mass.

ii. He stated that, during a chemical reaction, atoms can rearrange and change their partners. Molecules

could be decomposed into atoms.

iii. However, total number of atoms in the reactants and products should be same. Thus, mass is conserved

during a reaction. .

Q.29. How was the law of multiple proportion explained by Dalton on the basis of his atomic theory?

Ans: i. The law of multiple proportion states that, "if two elements chemically combine with each other

forming two or more compounds with different compositions by mass, then the ratios of masses of

two interacting elements in the two compounds are small whole numbers ".

ii. According to Dalton's atomic theory, compounds are formed when atoms of different elements combine.

iii. Dalton explained that under certain conditions, atoms of two types combine in the ratio 1:1 to form a

molecule. Under some other conditions, they may combine in the ratio 1:2 or 1:3 or 2:3, etc.

Note:

Dalton proposed some symbols for some common atoms and molecules. They are shown in the following

figure:

1.6 : Concepts of elements atoms and molecules :

Q.30. Explain the following terms:

i. Atoms ii. Molecules

Ans: i. Atoms:

a. The smallest indivisible particle of an element is called an atom. Thus, it has properties similar to

that of the element.

b. Atoms mayor may not exist freely. Atoms of almost all the elements can react with one another





to form compounds.

c. Every atom of an element has definite mass of the order of 10–27 kg and has a spherical shape of

radius of the order of 10–15 m.

d. The smallest atom of the element is that of hydrogen with mass of 1.667 × 10–27 kg.

eg. Water consists of 3-atoms; 2-hydrogen atoms and l-oxygen atom.

Note:

The properties of the constituent atoms are different from those of their respective compounds.

eg. Carbon dioxide gas is used as a fire extinguisher although carbon is combustible and oxygen supports

combustion.

ii. Molecules:

a. A molecule is an aggregate of two or more atoms. of definite composition which are held together

by chemical bonds.

b. A molecule may contain atoms of same elements or different elements.

c. It is the smallest independent unit of a compound which exists freely. It has all the properties of

the original compound.

d. It cannot be divided into constituent atoms with the help of simple methods but decomposition

into constituent atoms can be achieved under drastic conditions. .

eg. Water molecule (H2O).

1.7 : Atomic and molecular masses :

Q.31. Why do different elements have different atomic masses?

OR

Justify: "It is not possible to weigh a single atom".

Ans: i. Each and every atom of an element consists of fixed number of protons, neutrons and electrons,

which are the subatomic particles.

ii. The number of protons differs from atom to atom.

iii. As the number of subatomic particle changes, the mass of the atom changes.

iv. The size of one atom is of the order of 10–15 m. The mass is of the order of 10–27 kg.

Thus it is not possible to weigh a single atom.

Q.32. How is atomic mass measured?

OR

Why is atomic mass called relative mass?

Ans: i. The size ofa single atom is of the order of 10–15 m and the mass is of the order of 10–27 kg. Thus it is

not possible to weigh a single atom.

ii. The masses of the atoms of the elements can be determined experimentally by using mass spectrograph.

iii. For this purpose, a standard element is chosen and assigned appropriate value of mass of an atom.

iv. By international agreement in 1961, for determining atomic masses, carbon-12, a distinct atom of

carbon, was chosen as the standard with its atomic mass as 12.000 atomic mass unit (abbreviated as

amu).

v. Atomic mass unit is defined as 1/12 of the mass of an atom of carbon-12 equal to 1.6605 × 10–24 g.

vi. The masses of all other atoms are determined relative to the mass of an atom of carbon-12. Thus,

atomic mass is called as relative mass.

Note:

Recently, the unit of atomic mass, amu is replaced by 'u' which means 'unified mass' .

Q.33. Define isotopes.

Ans: Isotopes are the atoms of the same element having same atomic number but different mass number.

Therefore isotopes of an element contain same number of protons and electrons but different number of

neutrons.

Q.34. Explain the need of the term average atomic mass?





Ans: i. Several naturally occurring elements exist in the form of two or more isotopes. They have different

isotopic masses.

ii. In any sample of an element, the isotopes are present in different quantities.

iii. the observed atomic mass of the atom of the element is the average atomic mass of the element. The

natural abundances of the isotopes are taken into consideration for this purpose. The natural abundance

is the percentage occurrence.

iv. Thus, it is the average weight of an atom of the element which is used in calculating the atomic weight

of the element.

v. Chemistry is a macroscopic science and involves a large number of atoms (or molecules). Hence, an

average mass of an atom or average atomic mass is an inevitable term.

eg.

Chlorine has two isotopes, CI–35 and CI–37, present in 75% and 25% proportion respectively. Hence,

the atomic weight of chlorine is the weighed average of these two isotopic weights i.e., (35.0 × 0.75) +

(37.0 × 0.25) = 35.5.

Note:

i. The variation of the exact atomic mass of isotope of elements with its relative abundance is obtained

by using the mass spectrometer with higher precision.

ii. By using Aston's mass spectrometer, it was shown that Neon exists in nature in the form of a mixture

of the following three isotopes:

a. Neon-20 with atomic mass 19.9924 u with natural abundance 90.92%·

b. Neon - 22 with atomic mass 21.9914 u with natural abundance 8.82%.

c. Neon - 21 with atomic mass 20.9940 u with natural abundance 0.26%.

The observed atomic mass of Neon is its average atomic mass which is calculated as shown below:

Average atomic mass of Ne

=

(Atomic mass of Ne-20 × % of Ne-20)+Atomicmassof Ne-22 × % of Ne-22 100+ Atomic mass of Ne-21 × % of Ne-21

= [19.9924 u × 90.92 + 21.9914 u × 8.82 + 20.994 u × 0.26] /100 = 20.1713 u

Q.35. Define the term "Molecular Mass".

Ans: Molecular mass of a substance is defined as the ratio of mass of one molecule of a substance to

th1

12of the

mass of one atom of Carbon- 12.





It is also the alge.braic sum of atomic masses of constituent atoms which con.stitute the molecule.

Q.36. What are the 'characteristics of molecular mass (molar mass)?

Ans: i. Molecular mass is also expressed in amu.

ii. It indicates comparative mass of a molecule of a compound with respect to

th1

12 of the mass of one

atom of Carbon-12.

iii. The molar mass expressed in gram is known as gram molar mass. 1 gram molar mass is also known

as 1 gram molecule or 1 gram mole or 1 mole.

iv. 1 mole of the element is the amount of the element equal to its atomic mass in gram. It is also called

1 gram atom of the element.

1.8 : Avogadro's law :

Q.37. State and explain Avogadro's law.

Ans: i. In the year 1811, Avogadro combined Gay Lussac's law and Dalton's theory and thereby proposed

Avogadro's law.

ii. Avogadro's law states that, "equal volumes of all gases, under identical conditions of temperature and

pressure, contain equal number of molecules ".

iii. Mathematically, Avogadro's law is stated as, "at constant pressure and temperature, volume of a gas

is directly proportional to the number of molecules".

iv. V number of molecules (P, T constant)

Since number of molecules is proportional to the number of moles of gas n,

V n (P, T constant)

or V

n = constant, where n =

mass of gas

molar mass of gas

v. At standard temperature and pressure (STP condition), i.e., at 273.15 K and 1 atmosphere, the volume

of 1 mole of a gas i.e., molar volume of a gas can be calculated by gas equation,

PV = nRT

Volume per mol (molar volume) = V

n

V

n=

RT

P, (R = 0.08206 L atm mol–1 K–1)

= 0.08206 273.15

1atm

= 22.414 L mol–1.

vi. Thus 1 mole of any pure gas occupies a volume of 22.414 L (or 0.022414 m3) at standard temperature

and pressure.

The value 22.414 L mol–1 is called Avogadro's molar volume or molar gas volume at STP.

Q.38. Describe the significance of Avogadro's number.

Ans: The significance of Avogadro's number is as follows:

i. Avogadro's number is equal to the number of molecules present in one gram mole or one gram

molecular weight of any compound. Gram molecular weight of any substance is the weight in grams

of Avogadro's number, i.e., 6.022 × 1023 molecules.

ii. It is equal to the number of molecules in one mole or number of atoms in one gram atomic weight of

an element. Gram atomic weight of an element is the weight of Avogadro number of atoms.

iii. It is equal to the number of molecules in 22.414 dm3 of any gas at STP.

iv. The actual weight of a molecule of a compound or an atom of an element can be calculated using this

number.





Q.39. What is Atomicity?

Ans: Atomicity of a molecule is the total number of atoms of constituent elements combined toform a molecule.

eg. Atomicity of Oxygen (O2) is two, while that of Ozone (O

3) is three.

Note:

He and Ne are monoatomic

O2, N

2 and H

2 are diatomic

O3 is triatomic

P4 is tetraatomic

S8 is Polyatomic.

Q.40. Explain how an atom of an element is distinguished from its molecule.

Ans: i. Consider the formation of hydrogen chloride from hydrogen and chlorine.

ii. Hydrogen and chlorine do not exist in free atomic state but exist in molecular state and molecules take

part in the reaction to form hydrogen chloride.

iii. The reaction may be written as

Hydrogen + Chlorine Hydrogen chloride

According to

Gay Lussac's law: [1 volume] [1 volume] [2 volumes]

According to

Avogadro's law: (V n) [n molecules] [n molecules] [2n molecules]

Dividing by 2n1

molecule2

1molecule

2

[1 molecule]

iv. This shows that 1

2 molecule of hydrogen and

1

2 molecule of chlorine give 1 molecule of HCl. If

molecules of hydrogen and chlorine are considered to be monoatomic, it indicates that atoms are So,

the molecule of hydrogen and chlorine are considered to be diatomic.

Q.41. Explain mole concept.

Ans: i. The mass of one atom of an element or one molecule of a compound is negligibly small and is difficult

to-weigh.

ii. The amount of a substance equal to its atomic mass or molar mass in grams is 1 mole of a substance.

iii. Thus, one mole of a substance is defined as the amount of the substance that contains the number of

particles (atoms, molecules, ions or electrons, etc.) as present in 0.012 kg of carbon–12.

iv. This number of particles is determined to be equal to 6.022 × 1023 particles.

Q.42. What is Avogadro Number (NA)?

Ans: The number of atoms, molecules, ions, or electrons, etc. present in 1 mole of a substance is found to be

equal to 6.022 × 1023, which is called Avogadro Number (NA). Thus, N

A = 6.022 × 1023 mol–1





Note:

i. Number of electrons equal to NA make an electrical charge of one Faraday.

ii. One mole of various gases (of different molar masses) occupy 22.414 L at STP and contain avogadro's

number of molecules.

Q.43. Give the relationship between mass of a gas, number of moles and volume of the gas at STP,

with the help of mole triangle.

Ans: The mole triangle representing the relationship between the mass of gas, number of moles, volume of gas

at STP and the number of molecules.

Note:

i. Number of moles (n) = mass of a substance

molar mass of the substance =

W

M

where, W is mass and M is molar mass of the gas.

ii. One mole ofa gas contains 6.022 × 1023 molecules.

Number of molecules = n × Avogadro number = n × NA= n × 6.022 × 1023

iii. One mole of a gas at STP (273.15 K and 1 atmosphere) occupies 22.414 L





Volume of a gas at STP = n × 22.414 L.

1.9 : Percentage composition and molecular formula :

Q.44. Explain the terms

i. Percentage composition

ii. Empirical formula

iii. Molecular formula

Ans: i. Percentage composition:

The percentage composition of a compound is defined as the percentage by weight of each element

present in the compound.

Percentage (by weight) = Mass of the element in 1 mole of the compound

Gram molecular weight of the compound × 100 %

eg.

One mole of ethanol C2H

5OH (molecular mass 46) contains,

2 moles of carbon atom = 12 × 2 = 24 g

1 mole of oxygen atom = 16 × 1 = 16 g

6 moles of hydrogen atom = 1 × 6 = 6 g

Thus 46 g of ethanol contains 24 g, 16 g and 6 g of carbon, oxygen and hydrogen respectively. Hence,

the percentages of constituent elements are:

Percentage of carbon = 24

46 × 100 = 52.17 %

Percentage of oxygen = 16

46 × 100 = 34.78 %

Percentage of hydrogen = 6

46 × 100 = 13.04 %

Each 100 g sample of ethanol contains 52.17 g carbon, 34.78 g oxygen and 13.04 g hydrogen.

ii. Empirical formula:

The empirical formula of a compound is defined as a chemical formula indicating the relative number

of constituent atoms in a molecule in the simplest ratio.

eg.

The empirical formula of ethanol (C2H

5OH) is C

2H

6O and that of benzene (C

6H

6) is CH.

iii. Molecular formula:

The formula which gives the actual number of each kind of constituent atoms in one molecule of the

compound'is called the molecular formula of the compound.

It is an integral multiple of empirical formula.

eg.

The molecular formula of ethanol is C2H

5OH and that of benzene is C

6H

6.

Q.45. Write the steps for determination of empirical and molecular formula of a compound.

Ans: Steps for determination of empirical and molecular formula of a compound:

i. First the percentage of all the elements present in the compound are experimetally determined. If the

sum of the percentages of the constituent elements is less than 100%, then oxygen is present. The

difference between 100 and sum of the percentages of the constituent elements is the percentage of

oxygen.

ii. The number of moles of each constituent element present in 100g of the substance is obtained by

dividing percentage of the element by its atomic mass.

iii. The ratio of number of moles of constituent elements is determined. It is then converted into smallest

simple whole number ratio.





eg. For ethanol, C2H

5OH, the smallest simple whole number ratio in the order of C : H : O is 2 : 6 : 1.

iv. For fractional ratios, the whole number ratio is obtained by multiplying it with suitable integer. For this,

minor fractions may be neglected.

v. The empirical formula of the compound is obtained by writing the whole numbers of the ratio of

number of moles of elements as the subscripts to the right side of the corresponding atoms of the

elements.

vi. Molecular mass of the substance is experimentally determined. The ratio (r) of molar mass of the

substance to the empirical formula mass of the substance is determined, then

Molecular formula = r × empirical formula of compound.

Q.46. Write empirical and molecular formulae of following compounds:

i. water ii. carbon

iii. methane iv. hydrogen peroxide

v. acetylene vi. glucose

vii. diborane viii. tetraphosphorous decoxide.

Ans:

1.10 : Chemical reactions and stoichiometry :

Q.47. What is a chemical reaction?

Ans: Chemical reaction is a process in which a single substance or many substances interact with each other to

produce one or more substances. They are represented in terms of chemical equation.

eg.

The chemical reaction for the formation of water (product) from gaseous hydrogen and oxygen (reactants)

may be written as,

H2(g)

+ O2(g)

H2O

(g)

This is as shown below:

Q.48. Describe-in brief representation of chemical symbols in a chemical equation.

Ans: Chemical equation use chemical symbols called formulae of reactants and products. For example, symbol

of atomic hydrogen is H. It exists in gaseous state as a diatomic molecule. Therefore, chemical formula of

hydrogen molecule is H2, its gaseous state is indicated by the suffix (g) and written completely as H

2(g).

Similarly other gases are represented as, oxygen O2(g)

, nitrogen N2(g)

, fluorine F2(g)

, chlorine Ch(g)

, bromine

i. Water H2O H

2O

ii. Carbon C C

iii. Methane CH4

CH4

iv. Hydrogen peroxide HO H2O

2

v. Acetylene CH C2H

2

vi. Glucose CH2O C

6H

12O

6

vii. Diborane BH3

B2H

6

viii. Tetraphosphorous decoxide P2O

5P

4O

10

No. Compound Empirical formula Molecular formula





Br2(g)

and Iodine I2(g)

.

At room temperature, bromine exists in liquid state, hence, bromine in liquid state is represented as Br2(7)

while iodine exists in solid state and hence represented as I2(s)

.

Q.49. What are reactants and products in a chemical reaction?

Ans: The starting material which takes part in chemical reaction is called as reactant, the substance formed

after the chemical reaction is called product.

eg. Gaseous hydrogen and oxygen react to form water.

H2(g)

+ O2(g) H2O

(g)

Here, H2(g)

and O2(g)

are reactants and H2O

(g) is product obtained in the reaction.

Q.50. Explain the balancing of mass in the following reaction: H2(g)

+ O2(g)

H2O

(g)

Ans: i. In the given reaction, H2(g)

+ O2(g)

H2O

(g) conservation of mass in not observed.

ii. Hence, balancing of mass is essential.

iii. In order to conserve the mass of oxygen the reaction is balanced as 2H2(g)

+ O2(g)

2H2O

(g)

iv. This balanced chemical equation suggests that two molecules of hydrogen react with one molecule of

oxygen gas to produce two water molecules.

Q.51. Write a note on chemical stoichiometry.

Ans: i. Chemical stoichiometry is a process of making calculations based on formulae and balanced chemical

equations.

ii. Consider the balanced chemical reaction, IN2(g)

+ 3H2(g)

2NH3(g)

Numbers 1, 3 and 2 are coefficients representing the number of moles of N2(g)

, H2(g)

and NH3(g)

respectively.

iii. From this chemical equation, it can be seen that 1 mole of N2(g)

(28 grams) react with 3 moles H2(g)

(6

grams) giving 2 moles of NH3(g)

(34 grams).

Hence from the amount of the reactants, the amount of products that would be formed can be

calculated.

iv. From stoichiometry, the amounts of reactants required to obtain definite amount of products can be

estimated.

Q.52. What are the steps involved in writing a balanced chemical equation by stoichiometry?

OR

What are the steps involved in stoichiometric calculation?

Ans: Following are the steps involved in the stoichiometric calculations:

i. The correct formula of reactants and products must be written. The valencies of the atoms of

compounds should be satisfied.

ii. In order to write balanced chemical equation, following three steps must be followed:

Step - I:

The names of the reactants are written on the left: hand side. A '+' sign is used to separate the reactants.

An arrow is drawn from left: to right and on the right side of the arrow, the names of the products are

written. A '+' sign is used to separate the products.

eg. The reaction for the combustion of methane to form carbon dioxide and water is written as,

methane + oxygen carbon dioxide + water

Step - II:

The chemical equation is rewritten in terms of chemical formula of each substance.

CH4(g)

+ O2(g)

CO2(g)

+ H2O

(g)

Step - III:

To balance the mass of the chemical reaction, proper whole number coefficients are selected for each

reactant and product.

CH4(g)

+ 2O2(g) CO

2(g) + 2H

2O

(g)

This is the balanced chemical equation.





Note :

Some Common compounds :

Q.53. Show that "Law of conservation of mass is fully justified in a balanced chemical equation".

Ans: i. A balanced chemical reaction can be used to establish the weight relationships of reactants and

products.

ii. This is in accordance with the law of conservation of mass, which states that, total mass of reactants

is always equal to total mass of the products.

eg.

The mass relationship is given in following balanced reaction.

(Atomic masses: Fe = 55.85 u, CI = 35.45 u)

2Fe(s)

+ 3Cl2(g) 2FeCl

3(s)

(2 atoms) (3 molecules) (2 molecules)

[2 × 55.85 g] [3 × 35.45 × 2 g] [2 (55.85 + 3 × 35.45) g]

[111.7 g] [212.7 g] [324.4 g]

324.4 g 324.4 g

Hence, the mass is conserved during the reaction.

Q.54. What are limiting and excess reactants?

Ans: i. The stoichiometric coefficients of reactants and products in the balanced chemical equation determines

the amounts of reactants required and products formed in a chemical reaction .

ii. Usually, to save cost cheaper reactant is taken in excess while the costlier reactant is used in lesser

amount.

iii. As and when the reactant, which is taken in lesser amount, gets consumed the reaction stops. For the

reactant, which is taken in excess, only a part of it is consumed while the rest is left behind with the

products.

iv. Thus, the reactant taken in inadequate quantity is called as limiting reactant. It is the reactant that

reacts completely, but limits further progress of the reaction.

v. The excess reactant is the reactant which is taken in excess compared to the limiting reactant. Hence,

some amount of it remains unreacted.

Q.55. In the combustion of methane in air, what is the limiting reactant and why?

Ans: In the combustion of methane in air, methane is the limiting reactant because the other reactant is oxygen

of the air which is always present in excess. Thus, the amounts of carbon dioxide and water formed will

depend upon the amount of CH4 burnt.

i. Phosphoric acid H3PO

4

ii. Sodium phosphate Na3PO

4

iii. Ferric phosphate FePO4

iv. Aluminium phosphate AlPO4

v. Copper phosphate Cu3(PO

4)

2

vi. Ferrous phosphate Fe3(PO

4)

2

vii. Hydrogen chloride HCl

Viii. Sodium chloride NaCl

ix. Cuprous chloride Cu2Cl

2

x. Cupric chloride CuCl2

No. Compound Formula

xi. Ferrous chloride FeCl2

xii. Ferric chloride FeCl3

xiii. Stannous chloride SnCl2

xiv. Stannic chloride SnC4

xv. Sulphuric acid H2SO

4

xvi. Sodium sulphate Na2SO

4

xvii. Copper sulphate CuSO4

xviii. Ferrous sulphate FeSO4

xix. Ferric sulphate Fe2(SO

4)

3

No. Compound Formula





Quick Review :

The classification of matter (On basis of chemical composition):

Different laws and the contributing scientists :

Important Formulae :

1. One moIe of atoms = Mass of element

Atomic mass

2. Mass of one atom = 23

Atomic mass

6.022×10

3. Mass of one molecule = 23

Molecular mass

6.022×10

4. Nurnber of moles (n) = Mass of substance

Molar mass of substance5. Number of molecules = n × Avogadro number6. Volume of gas at STP = n × 22.414 L.

7. Percentage (by weight) = Mass of the element in 1 mole of the compound

Gram molecular weight of the compound × 100

8. Molecular formula = r × empirical formula (where r is ratio of molecular mass to empirical mass).

i. Law of conservation of mass i. Lomonosove (Russian scientist)

ii. Antoine Laviosier (French chemist)

ii. Law of definite composition Joseph Proust (French chemist)

iii. Law of multiple proportions John Dalton (British scientist)

iv. Law of combining volumes of gases Joseph Louis Gav-Lussac (French chemist)

No. Laws Contributinz Scientists





Solved Examples :

Type I : Problems based on average atomic mass

Example 1

Calculate the average atomic mass of naturally occuring argon from the following data:

-1

36

38

40

Isotope Isotopic mass (g mol ) abundance

Ar 35.96755 0.337%

Ar 37.96272 0.063%

Ar 39.9624 99.600%

Solution:

To find : Average atomic mass of naturally occuring argon

Formula : Average atomic mass of argon

=

36 36

38 38

40 40

atomic mass of Ar × percentage of Ar

= + atomic mass of Ar × percentage of Ar 100

+ atomic mass of Ar × percentage of Ar

Calculation :

Average atomic mass of argon

= 35.96755 × 0.337 + 37.96272 × 0.063 + 39.9624 × 99.60

100

= 39.947 g mol–1.

Ans: Average atomic mass of argon = 39.947 g mol–1.

Example 2

Calculate the atomic mass (average) of chlorine using the following data:

35

37

Atomic%Natural abundance

mass

Cl 75.77 34.9689

Cl 24.23 36.9659

Solution:

To find : Average atomic mass of chlorine

Formula : Average atomic mass of chlorine

=

35 35

37 37atomic mass of Cl × percentage of Cl= +atomic mass of Cl × percentage of Cl

100

Calculation :

Average atomic mass of chlorine

= 34.9689 × 75.77 + 36.9659 × 24.23

100

= 35.4528 g mol–1

Ans: Average atomic mass of chlorine = 35.4528 g mol–1.





Example 3

Boron occurs in nature in the form of two isotopes having atomic mass 10 and 11. What are the

percentage abundances of two isotopes in the sample of boron having average atomic mass

10.8?

Solution:

Given : Atomic masses of two isotopes of boron = 10 and 11, Average atomic mass of boron = 10.8

To find : The percentage abundances of two isotopes in the sample of boron having average atomic

mass 10.8

Formula : Average atomic mass = atomic mass of 10 B × percentage

100atomic mass of 10 B × percentage

Calculation : Let the % abundance of 10B isotope = x.

% abundance of 11B isotope = 100 – x.

Average atomic mass = 10.8

From formula, Average atomic mass = x 10 (100 x) 11

100

= 10.8

10 x + 1100 – 11 x = 10.8 × 100

– x = – 1100 + 1080

x = 20

Percentage abundance: 10B = 20%, 11B = (100 – 20) = 80%

Ans: Percentage abundance: 10B = 20%, 11B = 80%.

Type II: Problems based on Avogadro number and mole concept

Example 4

Calculate the number of moles and molecules of ammonia present in 5.6 dm3 of its volume.

Solution:

Let number of moles present in 5.6 dm3 of ammonia = x

Number of moles present in 22.414 dm3 of ammonia = 1

x = 5.6

22.414 = 0.25 mole

1 mole of NH3 (ammonia) = 6.022 × 1023 molecules

Number of molecules in 0.25 mole of NH3 (ammonia)

= 0.25 × 6.022 × 1023

= 1.5055 × 1023 molecules

Ans: The number of moles and molecules in 5.6 dm3 of NH3 are 0.25 mole and 1.5055 × 1023 molecules,

respectively.

Example 5

3.49 g of ammonia/at STP occupies volume of 4.48 dm3, Calculate molar mass of ammonia.

Solution:

Let 'x' grams be the molar mass of NH3.

Volume occupied by 3.49 g of NH3 at S.T.P = 4.48 dm3

Volume occupied by' x' g of NH3 at S.T.P = 22.414 dm3 .... [ 1 mole of any gas occupies 22.414 dm3 at

S.T.P]

x = 22.414 × 3.49

4.48 = 17.46 g mol–1.

Ans: Molar mass of ammonia is 17.46 g mol–1.

Example 6

Calculate the mass of potassium chlorate required to liberate 6.72 dm3 of oxygen at STP, molar





mass of potassium chlorate is 112.5 g mol–1.

Solution:

The molecular formula of potassium chlorate is KClO3.

Required chemical equation:

2KClO3 2KCl + 3O

2

[2 × 112.5 = 225 g] [3 × 22.414 = 67.242 dm3]

Thus, 225 g of potassium chlorate will liberate 67.242 dm3 of oxygen gas.

Let 'x' gram of KClO3 liberate 6.72 dm3 at S.T.P.

x = 225 × 6.72

67.242 = 22.49 g

Ans: Mass of potassium chlorate required is 22.49 g.

Example 7

Calculate the volume of oxygen required for complete combustion of 0.25 mole of methane at

STP.

Solution:


CH4

+ 2O2 CO

2+ 2H

2O

[1 mole] [2 moles]

Thus, 1 mole of CH4 requires 2 moles or 2 × 22.414 dm3 of O

2 for complete combustion.

0.25 mole of CH4 requires 'x' dm3 of O

2 for complete combustion.

x = 0.25 × 2 × 22.414

= 0.5 × 22.4 = 11.207 dm3

Ans: The volume of O2 required is 11.207 dm3.

Example 8

Calculate the volume of hydrogen required for complete hydrogenation of 0.25 dm3 of ethyne at

STP.

Solution:


C2H

2+ 2H

2 C2H

6

[1 mole] [2 moles]

1 mole of C2H

2 occupies 22.414 dm3

2 moles of H2 occupies 44.828 dm3

Thus, 22.414 dm3 of C2H

2 requires 44.828 dm3 of H

2 for complete hydrogenation.

Let 0.25 dm3 of C2H

2 require 'x' dm3 of Hyfor complete hydrogenation.

x = 0.25 × 44.828

22.414 = 0.5 dm3

Ans: The volume of hydrogen required is 0.5 dm3.

Example 9

Calculate the number of atoms of hydrogen present in 5.6 g of urea (molar mass of urea = 60 g

mol–1). Also calculate the number of atoms of N, C and O.

Solution:

Given : Mass of urea = 5.6 g

Molar mass of urea = 60 g mol–1

To find : The number of atoms of hydrogen, nitrogen, carbon and oxygen

Calculation:

Molecular formula of urea: CO(NH2)

2

1 molecule of urea has total 8 atoms, out of which 4 atoms are of H, 2 atoms are of N, 1 of





C and 1 of O.

1 mole or 60 g of urea contains 6.022 × 1023 molecules

60 g of urea has 4 × 6.022 × 1023 atoms of hydrogen.

Number of 'H' atoms in 5.6 g of urea = 5.6 × 4 × 6.022 × 1023

60 = 2.248 × 1023 atoms of

hydrogen.

Similarly, 60 g of urea has 2 × 6.022 × 1023 atoms of nitrogen

Number of N atoms in 5.6 g of urea =

235.6 × 2 × 6.022 × 10

60 = 1.124 × 1023 atoms of

nitrogen.

Similarly,

60 g of urea has 1 × 6.022 × 1023 atoms of carbon

Number of C atoms in 5.6 g of urea =

235.6 × 1 × 6.022 × 10

60 = 0.562 × 1023 atoms of

carbon.

Similarly,

60 g of urea has 1 × 6.022 × 1023 atoms of oxygen

Number of O atoms in 5.6 g of urea =

235.6 × 1 × 6.022 × 10

60 = 0.562 × 10 atoms of

oxygen.

Ans: 5.6 g of urea contains 2.248 × 1023 atoms of H,

1.124 × 1023 atoms of N,

0.562 × 1023 atoms of C and

0.562 × 1023 atoms of O.

Example 10

Calculate the number of atoms of 'C', 'H' and 'O' in 72.5 g of isopropanol, C3H

7OH (molar mass

60)

Solution:

Given : Mass of isopropanol (C3H

7OH) = 72.5 g

Molar mass of isopropanol = 60 g

Total number of atoms = 12, out of which, there are 3 carbon atoms, 8 hydrogen atoms and

1 oxygen atom.

To find : The number of atoms of C, H, and O

Calculation : In 60 g of C3H

7OH, there are 3 × 6.022 × 1023 atoms of carbon.

In 72.5 g of C3H

7OH, let there be 'x' atoms of carbon

x = 72.5 × 3 × 6.022 × 1023

60

= 21.829 × 1023

= 2.183 × 1024 atoms of carbon.

Similarly,

In 60 g of C3H

7OH, there are 8 × 6.022 × 1023 atoms of hydrogen.

In 72.5 g of C3H

7OH, let there be 'y' atoms of hydrogen.

Y =

2372.5 × 8 × 6.022 × 10

60= 58.21 × 1023 atoms of hydrogen

= 5.821 × 1024 atoms of hydrogen.

Similarly,





In 60 g of C3H

7OH, there are 1 × 6.022 × 1023 atoms of oxygen

In 72.5 g of C3H

7OH, let there be 'z' atoms of oxygen.

z =

2372.5 × 1 × 6.022 × 10

60 = 7.277 × 1023 atoms of oxygen.

Ans: 72.5 g of isopropanol contains 2.183 × 1024 atoms of C,

5.821 × 1024 atoms of H and

7.277 × 1023 atoms of O.

Example 11

Calculate the number of moles and the volume in litres of the following gases at STP:

i. 1.6 g of oxygen

ii. 3. 5 × 10–3 kg of nitrogen

iii. 85 × 10–3 kg of hydrogen sulphide

Solution:

i. For Oxygen :

One mole of O2

= 2 × 16 = 32 g (molar mass)

= 22.414 dm3 at STP

Number of moles of O2

= weight(w) 1.6

molar mass(M) 32

= 0.05

Volume of oxygen (at STP)

= moles × 22.414 L

= 0.05 × 22.414 L

= 1.12 L or dm3

ii. For Nitrogen:

One mole of N2

= 2 × 14

= 28 g = 28 × 10–3 kg (molar mass)

= 22.414 dm3 at STP

Number of moles of N2

= w

M =

3

3

3.5 × 10

28 × 10

= 0.125

Volume of N2 (at STP) = moles × 22.414 L

= 0.125 × 22.414

= 2.8 L or dm3

iii. For Hydrogen sulphide:

One mole of H2S = (2 × 1) + (1 × 32) = 34 g i.e., 34 × 10–3 kg (molar mass) = 22.414 dm3 at STP

Number of moles of H2S =

w

M =

3

3

85 × 10

34 10

= 2.5

Volume of H2S (at STP) = moles × 22.414 L

= 2.5 × 22.414 = 56 L or dm3

Ans: i. Number of moles and volume of oxygen in 1.6 g of oxygen are 0.05 and 1.12 L respectively.

ii. Number of moles and volume of nitrogen in 3.5 × 10–3 kg of nitrogen are 0.125 and 2.8 L respectively.

iii. Number of moles and volume of hydrogen sulphide in 85 × 10–3 kg of hydrogen sulphide are 2.5 and

56 L respectively.

Example 12

Calculate the number of moles and number of molecules of 12 g of carbon, 64 g of oxygen and 72





g of water.

(Atomic masses: C = 12, O = 16, H = 1)

Solution:

Given : Atomic masses of C = 12, O = 16 and H = 1.

To find : The number of moles and number of molecules of 12 g of carbon, 64 of oxygen and 72 g of

water.

Formula :

1. Number of moles (n) = Mass of the substance(w)

Atomicmass or molecular massof substance(M)

2. Number of molecules = n × Avogadro number

Calculation :

a. For Carbon:

i. Number of moles = w

M =

12

12 ( Atomic mass of C= 12)

ii. Number of molecules = 1 × 6.022 × 1023 = 6.022 × 1023

b. For Oxygen:

i. Number of moles = 64

32 ( Molecular mass of O

2 = 2 × 16 = 32)

= 2

ii. Number of molecules = 2 × 6.022 × 1023

= 12.044 × 1023 = 1.2044 × 1024

c. For Water:

i. Number of moles = 72

18 ( Molecular mass of H

2O = (2 × 1) + (1 × 16) = 18)

= 4

ii. Number of molecules = 4 × 6.022 × 1023

= 24.088 × 1023 = 2.4088 × 1024

Ans: i. Number of moles and number of molecules in 12 g of carbon are 1 and 6.022 × 1023 respectively.

ii. Number of moles and number of molecules in 64 g of oxygen are 2 and 1.2044 × 1024 respectively.

iii. Number of moles and number of molecules in 72 g of water are 4 and 2.4088 × 1024 respectively.

Example 13

Calculate the mass of the following

i. 0.25 mole of iron ii. 2.5 moles of ammonia

iii. 250 molecules of sodium chloride iv. 1.2 moles of methane

Solution:

i. 0.25 mole of iron (Fe):

Atomic mass of iron = 56

1 mole of Fe 56 g of Fe 56 × 10–3 kg of Fe

0.25 mole of Fe 56 × 10–3 × 0.25 14 × 10–3 kg of Fe

Mass of 0.25 mole of Fe = 1.4 × 10–2 kg.

ii. 2.5 moles of ammonia (NH3):

Molecular mass of ammonia = (1 × 14) + (3 × 1) = 17

(N) (3H)

1 mole of ammonia 17 g of ammonia 17.0 × 10–3 kg of ammonia

2.5 moles of ammonia 17.0 × 10–3 × 2.5 42.5 × 10–3 kg of ammonia

Mass of 2.5 moles of ammonia = 4.25 × 10–2 kg.

iii. 250 molecules of sodium chloride (NaCl):





Molecular mass of NaCl = 23 + 35.5 = 58.5

(Na) (CI)

1 mole of NaCl 58.5 g of NaCl

58.5 × 10–3 kg of NaCl

6.022 × 1023 molecules of NaCl 58.5 × 10–3 kg of NaCl

250 molecules of NaCl 3

23

58.5 10 250

0.022 10

2.429 × 10–23 kg

Mass of 250 molecules of NaCl = 2.429 × 10–23 kg.

iv. 1.2 moles of methane (CH4) :

Molecular mass of CH4 = 12 + (4 × 1) = 16

(C) (4H)

1 mole of methane = 16 g of methane = 16.0 × 10–3 kg of methane

1.2 moles of methane = 16.0 × 10–3 × 1.2 = 19.2 × 10–3 kg.

Mass of 1.2 moles of methane = 1.92 × 10–2 kg.

Ans: i. Mass of 0.25 moles of iron is 1.4 × 10–2 kg.

ii. Mass of 2.5 moles of ammonia is 4.25 × 10–2 kg.

iii. Mass of 250 molecules of sodium chloride is 2.429 × 10–23 kg.

iv. Mass of 1.2 moles of methane is 1.92 × 10–2 kg.

Example 14

What will be the mass of one 12C atom in g?

Solution:

1 mole of 12C atom 6.022 × 1023 atom 12 g

Mass of 6.022 × 1023 atoms of 12C = 12 g

Mass of 1 atom of 12C = 23

12

6.022 10 g = 1.9927 × 10–23 g

Ans: Mass of one 12C atom is 1.9927 × 10–23 g.

Type III : Problems based on percentage composition, molecular formula and empirical formula

Example 15

Phosphoric acid is widely used in carbonated beverages, detergents, toothpastes and fertilizers.

Calculate the mass percentages of H, P and O in phosphoric acid if atomic masses are H = 1,

P = 31 and O = 16.

Solution:

Given : Atomic mass of H = 1, P = 31 and O = 16

To find : The mass percentage of hydrogen, phosphorous, oxygen in H3PO

4

Formula : %(by weight) = Mass of element in one mole of compound

Gram molecular weight of compound × 100

Calculation : Molecular formula of phosphoric acid : H3PO

4

Molar mass of H3PO

4 = 3 × (1) + 1 × (31) + 4 × (16)

= 3 + 31 + 64

= 98 g mol–1

Percentage of Hydrogen = 3

98 × 100

= 3.06 %

Percentage of phosphorus = 31

98 × 100

= 31.63 %






98 × 100

= 65.31 %

Ans: Mass percentage of hydrogen, phosphorous and oxygen in phosphoric acid are 3.06%, 31.63% and

65.31 % respectively.

Example 16

Calculate the mass percentage composition ofthe elements in nitric acid (H = 1, N = 14, 0 = 16).

Solution:

Given : Atomic mass of H = 1, N = 14 and O = 16

To find : The mass percentage of H, N and O in HNO3



Calculation : Molecular formula of nitric acid : HNO3

Molar mass = 1 × (1) + 1 × (14) + 3 × (16) = 1 + 14 + 48 = 63 g mol–1

Percentage of hydrogen = 1

63 × 100

= 1.59 %

Percentage of nitrogen = 14

63 × 100

= 22.22%


63 × 100

= 76.19 %

Ans: Mass percentage of hydrogen, nitrogen and oxygen in nitric acid are 1.59%, 22.22% and 76.19%

respectively.

Example 17

Calculate the mass percentage of different elements present in sodium sulphate (Na2SO

4).

Solution:

Given : Molecular formula of sodium sulphate = Na2SO

4

To find : The mass percentage of Na, S and O in sodium sulphate



Calculation : Atomic mass of Na = 23, S = 32, O = 16

Molar mass of Na2SO

4= 2 × (23) + 1 × (32) + 4 × (16)

= 46 + 32 + 64

= 142 g mol–1

Percentage of sodium = 46

142 × 100 = 32.39 %

Percentage of sulphur = 32

142 × 100 = 22. 54 %


142 × 100 = 45.07 %

Ans: Mass percentage of sodium, sulphur and oxygen m sodium sulphate are 32.39, 22.54 and 45.07





respectively.

Example 18

Analysis of vitamin C shows that it contains 40.92% carbon by mass, 4.58% hydrogen and 54.50%

oxygen. Determine the empirical formula of vitamin C.

Solution:

Given : Analysis of vitamin C shows,

Percentage mass of carbon = 40.92%

Percentage mass of hydrogen = 4.58%

Percentage mass of oxygen = 54.50%

To find : The empirical formula of vitamin C

Calculation :

Moles of carbon = % of carbon

Atomic mass of carbon =

40.92

12 = 3.41

Moles of hydrogen = % of hydrogen

Atomic mass of hydrogen =

4.58

1 = 4.58

Moles of oxygen = % of oxygen

Atomic mass of oxygen =

54.50

16 = 3.406 = 3.41

Ratio of number of moles of C : H : O = 3.41

3.41 = 1 :

4.58

3.41 = 1 34 =

3.41

3.41 = 1

Ratio = 1 : 1.34 : 1

Multiply by 3 to get whole number

Ratio = 3 : 4.02 : 3 = 3 : 4 : 3

The empirical formula of compound Vitamin C is C3H

4O

3.

Ans: Empirical formula of Vitamin C is C3H

4O

3.

Example 19

Determine the empirical formula of an oxide of iron which contains 69.9% iron and 30.1 %

oxygen by mass. (NCERT)

Solution:

Given : Percentage mass of iron = 69.9 %

Percentage mass of oxygen = 30.1 %

To find : The empirical formula of an oxide of iron

Calculation :

Moles of iron = % of iron

Atomic mass of iron =

69.9

55.85 = 1.25

Moles of oxygen = % of oxygen


30.1

16 = 1.88

Ratio of number of moles of Fe : O = 1.25

1.25 = 1 :

1.88

1.25 = 1.50

Ratio = 1 : 1.50

Multiply by 2 to get whole number

Ratio = 2 : 3

The empirical formula is Fe2O

3

Ans: The empirical formula of an oxide of iron is Fe2O

3.




MH - CET Chemistry - XI

Example 20

Calculate the percentage of water of crystallization in the sample pf blue •vitrlol (CuS04.5H:zO)

Solution:

To find : The percentage of water of crystallization in the sample of blue vitriol

Formula :

% of H2O =

Weight of water in one mole of vitriol

Molar mass of vitriol

Calculation : Molecular mass of CuSO4.5H

2O = 63.5 + 32 + (4 × 16) + (5 × 18) = 249.5

For formula,

% of H2O =

90

249.5 × 100 = 36.07%

Ans: The percentage of water of crystallization in blue vitriol is 36.07%.

Example 21

An inorganic compound contained 24.75% (w/w) potassium and 34.75% (w/w) manganese and

some other common elements. Give the empirical formula of the compound (K = 39 u, Mn = 59

u, O = 16 u)

Solution:

Given : Atomic mass of K = 39 u, Mn = 59 u, and O = 16 u.

Percentage of potassium and manganese = 24.75 % and 34.75 % respectively.

To find : The empirical formula of the given inorganic compound

Calculation : Percentage of potassium = 24.75 %

Percentage of manganese = 34.75 %

Total percentage = 59.50 %

Remaining must be that of oxygen

Percentage of oxygen = 100 – 59.50 = 40.50 %

Moles of potassium = %of potassium

Atomic mass of potassium =

24.75

39 = 0.635

Moles of manganese = % of manganese

Atomic mass of manganese =

34.75

59 = 0.589

Mole of oxygen = % of oxygen


40.50

16 = 2.53

Ratio of K : Mn : O = 0.635

0.589 :

0.589

0.589 :

2.53

0.589 = 1.08 : 1 : 4.29 = 1 : 1 : 4

Ans: The empirical formula of given inorganic compound is KMnO4.

Example 22

Phosphoric acid used in carbonated beverages contain 3.086% (w/w) hydrogen and 31.61% (w/

w) phosphorous and remaining oxygen. If the atomic masses of hydrogen, phosphorous and

oxygen are 1.01 u, 31.0 u and 16 u respectively and if the molar mass of phosphoric acid is 98.03

g mol–1. What is the molecular formula of phosphoric acid?

Solution:

Given : Atomic mass of H = 1.01 u, P = 31.0 u and O = 16 u

The molar mass of phosphoric acid = 98.03 g mol–1

Percentage of hydrogen and phosphorus = 3.086 % and 31.61 % respectively.

To find : The molecular formula of phosphoric acid

Calculation : Percentage of hydrogen = 3.086 %





Percentage of phosphorous = 31.61 %

Total percentage = 34.696 %

Remaining amount is oxygen

Percentage of oxygen = 100 – 34.696 = 65.304 %

Moles of hydrogen = % mass of hydrogen

Atomic mass of hydrogen =

3.086

1.01 = 3.05

Moles of phosphorous = % mass of phosphorous

Atomic mass of phosphorous =

31.61

31 = 1.019

Moles of Oxygen = % mass of oxygen


65.304

16 = 4.08

Ratio of R : P : O = 3.05

1.019 :

1.019

1.019 :

4.08

1.019 = 3 : 1 : 4

Hence, the empirical formula of phosphoric acid = H3P04

Empirical formula mass = 3 (1.01) + 1 (31) + 4(16)

= 3.03 + 31 + 64

= 98.03 g mol–1

Thee ratio r = Molar mass

Empirical formula mass =

98.03

98.03 = 1

Molecular formula = r × Empirical formula

= 1 × H3PO

4

= H3PO

4

Ans: The molecular formula of phosphoric acid is H3PO

4.

Type IV : Problems based on stoichiometry

Example 23

Calculate the mass of iron which will be converted into oxide (Fe3O

4) by the action of 18 g of

steam on it.

Solution:

The chemical equation representing the reaction is:

3Fe + 4H2O Fe

3O

4 + 4H

2

[3 × 56] [4 × 18]

[168 g] [72 g]

Thus, 72 g of steam reacts with 168 g of iron.

18 g of steam will react with 168

72 × 18 = 42 g of iron

Ans: Mass of iron which will be converted into oxide by action of 18 g of steam is 42 g.

Example 24

How much copper can be obtained from 100 g of copper sulphate (CuSO4)?

Solution:

1 mole of copper can be obtained from 1 mole of copper sulphate

Atomic mass of copper = 63.5 u

Molar mass of copper sulphate= 1 × (63.5) + 1 × (32) + 4 × (16)

= 63.5 + 32 + 64

= 159.5 g mol–1

63.5 g of copper canbe obtained from 159.5 g of copper sulphate





Amount of copper that can be obtained from 100 g of copper sulphate is 63.5

159.5 × 100 = 39.81 g

Ans: 39.81 g of copper can be obtained from 100 g of copper sulphate.

Example 25

1.5 g of an impure sample of sodium sulphate dissolved in water was treated with excess of

barium chloride solution when 1.74 g of BaSO4 was obtained as dry precipitate. Calculate the

percentage purity of the sample.

Solution:

1.5 g of impure Na2SO

4 2Treated with BaCl 1.74 g of BaSO

4

The chemical equation representing the reaction is:

Na2SO

4+ BaCl

2 BaSO4

+ 2NaCl

[(2 × 23) + 32 + (4 × 16)] [137 + 32 + (4 × 16)]

[46 + 32 + 64] [137 + 32 + 64]

[142 g] [233 g]

To calculate the mass of Na2SO

4 from which 1.74 g of BaSO

4 is obtained:

233 g of BaSO4 is produced from 142 g of Na

2SO

4.

Mass of Na2SO

4 from which 1.74 g of BaSO

4 would be obtained =

142

233 × 1.74 = 1.06 g

The mass of pure Na2SO

4 present in 1.5 g of impure sample = 1.06 g

To calculate the percentage purity of the impure sample:

1.5 g of impure sample contains 1.06 g of pure Na2SO

4

100 g of the impure sample will contain 1.06

1.5 × 100 = 70.67 g of pure Na

2SO

4

Ans: Percentage purity of the sample is 70.67 %.

Example 26

Calculate the amount of lime Ca(OH)2 , required to remove hardness of 50,000 L of well water

which has been found to contain 1.62 g of calcium bicarbonate per 10 L.

Solution:

Calculation of total Ca(HCO3)

2 present:

10 L of water contains 1.62 g of Ca(HCO3)

2

50,000 L of water will contain 1.62

10 × 50,000 = 8100 g of Ca(HCO

3)

2

Calculation of lime required:

The balanced equation for the reaction:

Ca(HCO3)

2+ Ca(OH)

2 2CaCO3 + 2 H

2O

[1 mole] [1 mole]

[40 + (1 + 12 + 48) × 2] [40 + (16 + 1) × 2]

[40 + l22] [40 + 34]

[162] [74]

162 g of Ca(HCO3)

2 requires 74 g of lime.

Mass of lime required by 8100 g of Ca(HCO3)

2 =

74

162 × 8100 g = 3700 g = 3.7 kg

Ans: The amount of lime required to remove hardness of 50,000 L of well water, with 1.62 g of calcium

bicarbonate per 10 L is 3.7 kg.

Type V : Problems based on limiting and excess reactants





Example 27

Gold reacts with chlorine at 150 °C as per balanced chemical reaction, 2Au + 3Ch 2AuCh.

10 g each of gold and chlorine are sealed in a container and heated at 150 °C till the reaction is

complete. Name the limiting and excess, reactants. Also calculate the amount of AuCh formed

and the mass of the excess reactants left behind.

(Atomic masses: Au = 196.97 u, Cl = 35.45 u)

Solution:

The reaction of gold, with chlorine to produce AuCh is 2Au + 3Cl2 2AuCl

3. 2 moles of gold reacts

with 3 moles of Cl2 to given 2 moles of AuCl

3 will be obtained.

2Au + 3Cl2 2AuCl

3

[2 × 196.97 g] + [3 × 35.45 × 2] [2(196.97 + 3 × 35.45)]

[393.94 g] + [212.7 g] [606.64]

[606.64 g] [606.64 g]

Theoretical ratio of moles of gold to chlorine = 2

3 = 0.667

Actual amount is 10 g of Au and 10 g Cl2

Number of moles of Au in 10 g = 10

196.97 = 0.0508

Number of moles of Cl2 in its 10 g =

10

70.90 = 0.141 mol of Cl

2

Actual ratio of moles of gold to chlorine = 0.0508

0.141 = 0.36

Theoretical ratio of moles of Au to Cl2 is 0.667 and actual ratio of moles of Au to Cl

2 is 0.36

Thus actual ratio is smaller than theoretical ratio

Au is the limiting reactant, and Cl2 is the excess reactant.

2 moles of Au react with 3 moles of Cl2

0.050 8 moles of Au will react with 0.0508 × 3

2 = 0.7672 mo 1es of Cl

2

Moles of Cl2 remaining = 0.141 – 0.0762

= 0.0648 moles of Cl2

Mass of Cl2 reacted = 0.0762 × 70.90 = 5.40 g (70.90 g is molar mass of Cl

2)

Mass of Cl2 left behind = 10 – 5.40

= 4.6 g

2 moles of Au give 2 moles of AuCl3

0.0508 mole of Au will give = 0.0508 mole of AuCl3

Mass of AuCl3, formed = 0.0508 × 303.32 (Molar mass of AuCl

3 = 303.32 g)

= 15.409 g AuCl3

Ans: Au is the limiting reactant, while Cl2 is the excess reactant. The amount of AuCl

3 formed is 15.409 g and

the mass of the excess reactant (Chlorine) left behind is 4.6 g.

Additional Theory Questions :

Q.1. Define:

i. physical chemistry. Refer Q.4.i.

ii. inorganic chemistry. Refer Q.4.ii.

iii. organic chemistry. Refer Q.4.iii.





iv. analytical chemistry. Refer Q.4.iv.

v. biochemistry. Refer Q.4. v.

Q.2. Explain the following:

i. Pure substance. Refer Q.8.i.

ii. Mixtures. Refer Q.8.ii.

Q.3. Define the following with suitable examples:

i. Homogeneous mixtures. Refer Q.8. ii.a.

ii. Heterogeneous mixtures. Refer Q.8.ii.b.

Q.4. State and explain the law of conservation of mass. Refer Q.22.i.

Q.5. State and explain the law of definite composition / proportions. Refer Q.22.ii.

Q.6. State and explain the law of multiple proportions. Refer Q.22.iii.

Q.7. State and explain Gay Lussac's law of combining volumes of gases. Refer Q.22.iv.

Q.8. Define elements, compounds and molecules. Refer Q.13.i.a, Q.13.ii.a. and Q.30.ii.a.

Q.9. Show that molar gas volume at STP is 22.414 L or 0.022414 m'.

OR

Prove that 1 mole of any gas at STP always has a volume of 22.414 L. Refer Q.37.(iii to vi)

Practice Problems :

Type I

1. The relative abundance of various isotopes of

silicon is: Si (28) = 92.25%, Si (29) = 4.65% and

Si (30) = 3.10%. Calculate the average atomic

mass of silicon.

Type II

2. How many molecules and atoms of sulphur are

present in 0.1 mole of Sgmolecules?

3. Calculate the number of moles of iodine in a

sample containing 1.0 × 1022 molecules.

4. Calculate moles of Ammonia in 0.1284 kg of

Ammonia.

5. O2 is present in one litre flask at a pressure of

7.6 × 10–10 mm of Hg. Calculate no. of O2

molecules at 0 °C.

6. Calculate the number of moles In the following:

i. 9.0 × 10–2 kg water

ii. 89.6 dm3 nitrogen at STP

iii. 0.5 g hydrogen

iv. 7.1 × 10–2 kg chloride ions.

7. Calculate the number of moles In the following:

i. 0.108 kg of water

ii. 0.277 kg of carbon

iii. 1.058 × 10–5 kg of sodium ions

iv. 1.7 × 10–2 kg of hydrogen sulphide

8. Calculate the weights of the following:

i. 0.3 mole of carbon

ii. 0.1 mole of hydrogen chloride

iii. 2.4 moles of bromide .ions

9. Calculate the number of molecules In the following

i. 4.5 × 10–5 kg of chlorine

ii. 0.28 dm3 of oxygen at STP

Type III

10. Calculate the percentage composition of the

following compound:

i. Urea CO(NH2)

2ii. CuSO

4.5H

2O

11. Calculate the mass percentage composition of

copper pyrites 9uFeS2

12. Calculate the percentage composition of glucose

(C6H

12O

6).

13. A compound containing sodium, sulphur and

oxygen has the following percentage composition

Na = 29.11%, S = 40.51% and O = 30.38% and

molecular weight 158. Find the empirical formula.

Type IV

14. 1.0 g of a mixture of carbonates of calcium and

magnesium gave 240 cm3 of CO2 at STP calculate

the percentage composition of the mixture.

Type V

15. 50.0 kg of N2(g)

and 10.0 kg of H2(g)

are mixed to

produce NH3(g)

i. Calculate the NH3(g)

formed.

ii. Identify the limiting reagent In this reaction

if any.





Multipal Choice Questions

1.0 Prominent scientists

1. The law of combining volumes of gases was

formulated by ______

a) Avogadro b) Gay-Lussac

c) Aristotle d) Joseph Priestley

1.1 Introduction

2. The branch of chemistry which deals with carbon

compounds is called ______ chemistry.

a) organic b) inorganic

c) carbon d) bio

1.2 Importance and scope of chemistry

3. _____ is used for the treatment of parkinson's

disease.

a) L-dopa b) Taxol

c) Cisplatin d) Tamiflue

1.3 Historical approach to particulate nature of

matter.

4. The phlogiston theory was suggested for

a) neutralisation reaction

b) oxidation reaction

c) reduction reaction

d) combustion reaction

5. Electrochemical equivalent has unit _____ .

a) kg m s–1 b) kg m2 s–1

c) kg C–1 d) kg m–1 s–2

6. A is a simple combination of two or more

substances in which the constituent substances

retain their separate identities.

a) compound b) mixture

c) alloy d) amalgam

7. Which one of the following is NOT a mixture?

a) Iodized table salt

b) Gasoline

c) Liquefied Petroleum Gas (LPG)

d) Distilled water

8. Magnitude of 'pico–' is _____

a) 10–12 b) 10–15

c) 1012 d) 1015

1.4 Laws of chemical combination

9. The sum of the masses of reactants and products

is equal in any physical or chemical reaction. This

is in accordance with _______

a) law of multiple proportion

b) law of definite composition

c) law of conservation of mass

d) law of reciprocal proportion

10. A sample of calcium carbonate (CaCO3) has the

following percentage composition :

Ca = 40 %; C = 12 %; O = 48 %

If the law of constant proportions is true, then

the weight of calcium in 4 g of a sample of calcium

carbonate from another source will be

a) 0.016 g b) 0.16 g

c) 1.6 g d) 16 g

11. Two elements, A and B, combine to form two

compounds in which 'a' g of A combines with 'b.'

and 'b2' g. of B respectively. According to law of

multiple proportion _____ .

a) b1 = b

2

b) b1 and b

2 bear a simple whole number ratio

c) a and b1 bear a whole number ratio

d) no relation exists between b1 and b

2

12. At constant temperature and pressure, one litre

of nitrogen gas reacts with three litres of hydrogen

gas to produce two litres of ammonia gas. This is

in accordance with

a) law of multiple proportion

b) law of definite composition

c) law of Conservation of mass

d) law of combining volumes of gases

1.5 Dalton's atomic theory

13. ______ proposed some symbols for some

common atoms and molecules.

a) Democritus b) Newton

c) Thomson d) Dalton

1.6 Concepts of elements, atoms and molecules

14. A/An _____ is an aggregate of two or more

atoms of definite composition which are held

together by chemical bonds.

a) ion b) molecule

c) compound d) mixture

1.7 Atomic and molecular masses

15. One mole of oxygen weighs _____

a) 8 g b) 32 g

c) 16 g d) 6.023 × 1023 g

16. The number of atoms present in a molecule of a





substance is called _____

a) atomicity b) volume

c) density d) mass

1.8 Avogadro's law

17. The number of molecules in 22.4 cm3 of nitrogen

gas at STP is ______

a) 6.023 × 1020 b) 6.023 × 1023

c) 22.4 × 1020 d) 22.4 × 1023

18. The number of molecules present in 8 g of oxygen

gas are ______

a) 6.022 × 1023 b) 3.011 × 1023

c) 12.044 × 1023 d) 1.505 × 1023

19. 11.2 cm3 of hydrogen gas at STP, contains

______ moles

a) 0.0005 b) 0.01

c) 0.029 d) 0.5

20. 4.4 g of an unknown gas occupies 2.24 L of

volume under STP conditions. The gas may be

_____

a) CO2

b) CO

c) O2

d) SO2

21. Under similar conditions, oxygen and nitrogen are

taken in the same mass. The ratio of their volumes

will be _

a) 7 : 8 b) 3 : 5

c) 6 : 5 d) 9 : 2

22. Atomicity of ozone is _____ .

a) 1 b) 2

c) 3 d) 4

23. 19.7 kg of gold was recovered from a smuggler.

How many atoms of gold were recovered?

(Au = 197)

a) 197 b) 6.02 × 1023

c) 6.02 × 1024 d) 6.02 × 1025

24. The number of atoms in 4.25 g of NH3 is

approximately

a) 1 × 1023 b) 1.5 × 1023

c) 2 × 1023 d) 6.022 × 1023

25. What is the mass of 0.5 mole of ozone molecule?

a) 8g b) 16 g

c) 24 g d) 48 g

26. Mole triangle is the relationship between the mass

of a gas, the number of moles, the volume of gas

at S.T.P. and the ______

a) number of electrons

b) number of molecules

c) pressure at S.T.P

d) temperature at S.T.P

1.9 Percentage composition and molecular

formula

27. The percentage of oxygen in NaOH is

a) 40 b) 60

c) 8 d) 10

28. Which of the following has same molecular

formula and empirical formula?

a) CO2

b) C6H

12O

6

c) C2H

4d) all of these

29. The empirical formula of C2H

2 is ______ .

a) C2H

4b) CH

c) CH4

d) C4H

10

1.10 Chemical reactions and stoichiometry

30. _____ reactant is the reactant that reacts

completely but limits further progress of the

reaction.

a) Oxidizing b) Reducing

c) Limiting d) Excess

31. If 0.5 mol of BaCl2 is mixed with 0.2 mol of

Na3PO

4, the maximum number of moles of

Ba3(PO

4)

2 that can be formed is ______

a) 0.7 b) 0.5

c) 0.3 d) 0.1

32. Two elements X (atomic mass = 75) and Y

(atomic mass = 16) combine to give a compound

having 75.8% of X . The formula of the compound

is ______

a) XY b) X2Y

c) XY2

d) X2Y

3

Answer Keys

1. b) 2. a) 3. a) 4. d) 5. c) 6. b) 7. d) 8. a) 9. c) 10. c)

11. b) 12. d) 13. d) 14. b) 15. b) 16. a) 17. b) 18. d) 19. a) 20. a)

21. a) 22. c) 23. d) 24. d) 25. c) 26. b) 27. a) 28. a) 29. b) 30. c)

31. d) 32. d)





Answers to Practice Problems :

1. 28.11 u.

2. 6.022 × 1022 molecules, 4.82 × 1023 atoms of

sulphur

3. 0.0166 mol

4. 7.55 moles

5. 2.68 × 1010

6. i. 5 moles

ii. 4.0 moles

iii. 0.25 moles

iv. 2 moles

7. i. 6 moles

ii. 23.08 moles

iii. 4.6 × 10–4 mole

iv. 0.5 mole

8. i. 3.6 × 10–3 kg

ii. 3.65 × 10–3 kg

iii. 0.192 kg

9. i. 3.82 × 1020

ii. 7.52 × 1021

10. i. % of carbon = 20%

% of oxygen = 26.67%

% of nitrogen = 46.67%

% of hydrogen = 6.67%

ii. % of Cu = 25.45%

% of S = 12.83%

% of O = 57.72%

% of H = 4.00

11. Cu = 34.64%, Fe = 30.44%, S = 34.9%

12. C = 40.0%, H = 6.67%, O = 53.33%

13. Na2S

2O

3

14. CaCO3 : 57.5%, MgCO

3 : 42.5%

15. i. 56.67 kg

ii. Hydrogen



Some Basic Concepts of Chemistry

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