Chemistry Basics and Applications

CHAPTER 1
SOME BASIC CONCEPTS OF
CHEMISTRY
BY : VEENU GUPTA
(PGT CHEMISTRY)
BY: VEENU MAHAJAN FOR CLASS XITH STUDENTS 1
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LEARNING OBJECTIVES:
By the end of this part of the chapter ,Learner will be able to :
• Explain importance and scope of Chemistry
• Explain matter and its classification
• Define and describe the different laws of chemical combinations like, law of conservation of
mass, law of definite proportion, law of multiple proportion ,Gay Lussac’s law of gaseous
volume and Avogadro’s law by giving suitable examples.
• Define atomic mass, relative atomic mass, molecular mass and formula mass
• Define and calculate average atomic mass.
• Understand the postulates of Dalton’s atomic theory and relate them with the laws of chemical
combinations.
• Solve numerical based on the concentration terms
• Calculate Empirical & molecular formula
• Solve numerical based on stoichiometry and limiting reagents
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IMPORTANCE OF CHEMISTRY
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CHEMISTRY HAS A DIRECT IMPACT ON OUR LIFE AND HAS WIDE
RANGE OF APPLICATIONS IN DIFFERENT FIELDS.
THESE ARE GIVEN BELOW:
• 1. IN AGRICULTURE AND FOOD:
(i) It has provided chemical fertilizers such as urea, calcium phosphate,
sodium nitrate, ammonium phosphate etc.
(ii) It has helped to protect the crops from insects and harmful bacteria,
by the use ‘ of certain effective insecticides, fungicides and pesticides.
(iii) The use of preservatives has helped to preserve food products like
jam, butter, squashes etc. for longer periods.
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2. IN HEALTH AND SANITATION:
(i) Chemistry has provided mankind with a large number of life-
saving drugs. Today, dysentery and pneumonia are curable due to
discovery of sulpha drugs and penicillin life-saving drugs.
Cisplatin and taxol have been found to be very effective for cancer
therapy and AZT (Azidothymidine) is used for AIDS victims.
(ii) Disinfectants such as phenol are used to kill the micro-organisms
present in drains, toilet, floors etc.
(iii) A low concentration of chlorine i.e., 0.2 to 0.4 parts per million
(ppm) is used ’ for sterilization of water to make it fit for drinking
purposes.
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MEDICINES OR DRUGS
• Chemicals which may be used for the treatment of diseases and
for reducing the suffering from pain are called medicines or drugs.
• The branch of science which makes use of chemicals for the
treatment of diseases [therapeutic effect] is called Chemotherapy.
• Some important classes of drugs are:
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A. ANTACIDS
The chemical substances which neutralize the excess acid in gastric
juice and raise the pH to an appropriate level in stomach are called
antacids.
The most commonly used antacids are weak bases such as sodium
bicarbonate [sodium hydrogen carbonate, ( NaHCO3)], magnesium
hydroxide [Mg(OH)2] and aluminium hydroxide [Al(OH)3].Generally
liquid antacids are more effective than tablets because they have
more surface area available for interaction and neutralisation acid.
Milk is a weak antacid.
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Histamine stimulates the secretion of pepsin and hydrochloric acid.
The drug cimetidine [Tegamet] was designed to prevent the
interaction of histamine with the receptors present in the stomach.
Cimetidine binds to the receptors that triggers the release of acid
the stomach. This results in release of lesser amount of acid.
Now ranitidine (zantac), omeprazole and lansoprazole are used
for hyperacidity
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2. TRANQUILIZERS (PSYCHOTHERAPEUTIC DRUGS)
• Chemical substances used for the treatment of stress, anxiety,
irritability and mild or even severe mental diseases, are known as
tranquilizers. These affect the central nervous system and induce
sleep for the patients as well as eliminate the symptoms of
emotional distress. They are the common constituents of sleeping
pills.
• Noradrenaline is one of the neurotransmitter that plays a role in
mood change. If the level of noradrenaline is low, the signal sending
activity becomes low, and the person suffers from depression. In such
situations antidepressant drugs are required.
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• These drugs inhibit the enzymes which catalyse the
degradation of noradrenaline. If the enzyme is inhibited, this
important neurotransmitter is slowly metabolized and can
activate its receptor for longer periods of time, thus
counteracting the effect of depression.
• Iproniazid and phenelzine are two such drugs
• Barbituric acid and its derivatives viz. veronal, amytal, nembutal,
luminal, seconal are known as barbiturates. Barbiturates are
hypnotic, i.e., sleep producing agents.
• Equanil is used to control depression and hypertension.
• Non-hypnotic chlorodiazepoxide and meprobamate are relatively
mild tranquilizers suitable for relieving tension
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• 3. Analgesics
• Medicines used for getting relief from pain are called analgesics. These are of
two types :
• 1. Narcotics: Drugs which produce sleep and unconsciousness are called
narcotics. These are habit forming drugs. For example, morphine and codeine.
Morphine diacetate is commonly known as heroin.
• 2. Non-narcotics: These are non-habit forming chemicals which reduce mild to
moderate pain such as headache, toothache, muscle and joint pain, etc. These
are also termed as non-addictive. These drugs do not produce sleep
unconsciousness. Aspirin (2-acetoxybenzoic acid) is most commonly used
analgesic with antipyretic properties. Now these days because its anti-blood
clotting action, aspirin is widely used to treat heart-attacks.Aspirin is toxic for
liver and sometimes also causes bleeding from- stomach. So, naproxen,
ibuprofen, paracetamol,dichlorofenac sodium are other widely used analgesics.
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• Antipyretics
• These are the chemical substance which reduce body temperature during high
fever. Paracetamol, aspirin, phenacetin (4-hydroxy acetanilide), analgin and
novalgin, etc., are common antipyretics. Out of these, paracetamol (4-
acetamidophenol) is most common.
• Antimicrobials
• An antimicrobial tends to kill or prevent development of microbes .It inhibit the
pathogenic action of microbes such as bacteria, fungi and virus selectively.
• [Sulpha drugs constitute a group of drugs which are derivatives of
sulphanilamide and have great antimicrobial capacity, thus, these are widely
used against diseases such as dyptheria, dysentry, tuberculosis, etc.]
• In structure these drugs are analogues of p-amino benzoic acid.
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• Penicillin was the first antibiotic discovered (by Alexander Fleming) in
1929. It is a narrow-spectrum antibiotic.
• Ampicillin and amoxicillin are semi-synthetic modifications of penicillin.
Penicillin is not suitable to all persons and some Persons are allergic to it.
Consequently, it is essential to test the patients for sensitivity (or allergy)
to penicillin, before it is administered.
• In India, penicillin is manufactured at Pimpri and Rishikesh (Uttarakhand).
• Broad-spectrum antibiotics are effective against different types of
harmful microorganisms. e.g., Tetracycline and chloramphenicol is given
in case of typhoid, dysentery, fever etc
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4. ANTISEPTICS
• These are the chemicals which either kill or prevent the growth of microorganisms.
Antiseptics are applied to the living tissues such wounds, cuts, ulcers and skin diseases in
the form of antiseptic creams like furacin and soframycin. e.g., Some important examples
of antiseptics are
• (i) Dettol is a mixture of chloroxylenol and terpineol.
• (ii) Bithional is added to soaps to impart antiseptic properties to reduce the odours
produced by bacterial organic matter on the skin.
• (iii) Tincture of iodine is a 2-3% solution of iodine in alcohol, which is a powerful
antiseptic for wounds.
• (iv) Iodoform (CHI3) is also used as an antiseptic for wounds.
• (v) Boric acid is used in dilute aqueous solution as a treatment for diaper rash, sun burn,
insect bites and sting.
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5. DISINFECTANTS
• These are the chemical substances which kill microorganisms not safe to be
applied to the living tissues. They are generally kill the microorganisms
present on inanimate objects such as drainage systems, instruments, etc.
• Some common examples of disinfectants are as follows :
• (i) 1% phenol solution is disinfectant while in lower concentration 0.2%
solution of phenol is antiseptic.
• (ii) 0.2-0.4 ppm aqueous solution of chlorine is used for sterilisation of water
to make it fit for drinking purpose.
• (iii) SO2 at very low concentrations behaves like disinfectant.
• (iv) Formaldehyde (HCHO) in the disinfecting rooms and operation gaseous
theatres forms is used in hospitals.
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6. ANTIFERTILITY DRUGS
• These are the chemical substances used to control the
pregnancy. These are also called oral contraceptives. They
belong to the class of natural products, known as steroids.
• Birth control pills essentially contain a mixture of synthetic
estrogen and progesterone derivatives.
• Norethindrone is widely used as antifertility drug.
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F. FOOD PRESERVATIVES
• These are the chemical substances added to food to prevent their spoilage due
to microbial growth (bacteria, yeasts and moulds) and to retain their nutritive
value for longer periods .
• The most commonly used preservatives include table salt, sugar, vegetable oil,
vinegar, citric acid. spices and sodium benzonate (C6H5COONa). Salts of sorbic
acid and propanoic acid are also used” preservatives for cheese, baked food,
pickles, meat and fish products.
• 1. Sodium benzoate is metabolised by conversion into hippuric acid
(C6H5CONHCH2COOH), which is ultimately urine. It is used in soft drinks and
acidic foods.
• 2. Antioxidants like BHT(butylated hydroxytoluene) and BHA (butylated
hydroxyanisole) retard the action of oxygen on the food and help in the
preservation of food materials.
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3. SAVING THE ENVIRONMENT:
The rapid industrialization all over the world has resulted in lot of pollution.
Poisonous gases and chemicals are being constantly released in the atmosphere.
They are polluting environment at an alarming rate. Scientists are working day and
night to develop substitutes which may cause lower pollution.
• For example, CNG (Compressed Natural Gas), a substitute of petrol, is very effective
in checking pollution caused by automobiles.
4. APPLICATION IN INDUSTRY:
Chemistry has played an important role in developing many industrially ^
manufactured fertilizers, alkalis, acids, salts, dyes, polymers, drugs, soaps,
detergents, metal alloys and other inorganic and organic chemicals including new
materials contribute in a big way to the national economy.
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9. CHEMICALS IN FOOD
1. Artificial Sweetening Agents
• Sucrose (table sugar) and fructose are the most widely used
natural sweeteners. But they add to our calorie intake and
promote tooth decay. To avoid these problems many people take
artificial sweeteners.
• Organic substances which have been synthesized in lab are
known to be many times sweeter than cane sugar. Such
compounds are known as artificial sweetening agents or artificial
sweeteners
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SOME IMPORTANT ARTIFICIAL SWEETENERS ARE GIVEN
BELOW:
• (i) Saccharin (o-sulphobenzimide)
• Discovered by Johns- Hopkins in 1879 (University of USA).
• It is the most popular artificial sweetener. It is 550 times as sweet as cane
Sugar, since it is insoluble in water, so it is sold in the market as its soluble
or calcium salt.
• It is non-biodegradable so excreted from the body in urine (unchanged). Its
use is of great value for diabetic persons and people who need to control
intake of calories.
• acid. Accumulation of phenylpyruvic acid is harmful especially to infants
due to brain damage and mental retardation.
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(II) ASPARTAME
• It is the methyl ester of the dipeptide derived from phenylalanine
aspartic acid. It is also known as ‘Nutra sweet’.
• It decomposes at baking or cooking temperatures and hence, can used
only in cold food and soft drinks.
• Aspartame has the same amount of calories as sugar (4 cal per gm).
• Aspartame should not be used by people suffering from the genetic
disease known as PKU (phenyl ketone urea). Because in such people
decomposition of aspartame gives phenylpyruvic
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• (iii) Alitame
• It is quite similar to aspartame but more stable than aspartame. It is 2000
times as sweet as sucrose. The main problem for such sweetener is the control
of sweetness of the substance to which it is added because it is high potency
sweetener.
• (iv) Sucralose
• It is a trichloro derivative of sucrose. It’s appearance and taste are like sugar. It
is stable at cooking temperature. It is almost 600 times 88 sweet as sucrose.
However, it neither provides calories nor causes toot.1i decay.
• (v) Cyclamate
• It is N-cyclohexylsulphamate. It is only 20 times sweeter than cane sugar.
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10. CLEANSING AGENTS
• The word detergent means cleansing agent. Actually detergent word is derived from
Latin word ‘detergere’ means “to wipe off”, Cleansing agents are the substance which
remove dirt and have cleansing action in water. These are also called surfactants.
• Detergents can be classified into two types: Soapy detergents or soaps, and Non-soapy
detergents or soap less soap.
• 1. Soaps: Soaps are sodium or potassium salts of higher fatty acids (containing 15-18
carbon atoms) e.g., stearic acid, oleic acid and palmitic acid. Sodium salts of fatty acids
are known as hard soaps while the potassium salts of fatty acids are known as soft
soaps.
• Hard soaps are prepared by cheaper oil and NaOH while soft soaps are prepared by oil
of good quality and KOH. The soft soaps do not contain free alkali, produce more lather
and are used as toilet soaps, shaving Soaps and shampoos.
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PREPARATION OF SOAPS
• Soaps containing ‘Sodium salts are formed by heating fat
(glyceryl ester fatty acid) with aqueous sodium hydroxide
solution. This reaction is known as saponification.
• The solution left after removing the soap contains glycerol,
which can be recovered by fractional distillation. To improve
the quality of soaps desired colours, perfumes and medicinal
chemical substances, added.
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TYPES OF SOAPS
• Different kind of soaps are made by using different raw materials.
• Toilet soaps These are prepared by using better grade of fat or oil
and care is taken to remove excess alkali. Colour and perfumes are
added to make these more attractive.
• Floating soaps These can be prepared by beating tiny bubbles into
the product before it hardens.
• Transparent soaps These are made by dissolving the in ethanol and
then evaporating the excess solvent.
• Medicated soaps Medicated soaps are prepared by some
antiseptics like dettol or bithional.
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• Shaving soaps These contain glycerol to prevent drying. A gum called rosin
is added while making them. It forms sodium rosinate which lather well.
• Laundry soaps These sodium silicate, borax and contain sodium fillers like
carbonate. sodium rosins
• Soap Chips These are made by running a thin sheet of melted soap on a
cool cylinder and scraping off the soaps in small broken pieces.
• Soap grannules These are dried miniature soap bubbles.
• Soap powder and scouring soaps These contain a scouring agent (abrasive)
such as powdered pumice or finely divided sand and builders like sodium
carbonate and trisodium phosphate. Builders make the soaps act more
quickly.
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DISADVANTAGES OF SOAPS
• Soap is good cleansing agent and is 100% biodegradable microorganisms present
in sewage water can completely oxidise soap to CO2, As a result, it does not create
any pollution problem. However soaps have two disadvantages:
(i) Soaps cannot be used in hard water since calcium magnesium ions present in hard
water produce curdy precipitates of calcium and magnesium soaps.
• These insoluble soaps separate as scum in water and causes hinderance to
washing because the precipitate adheres onto the fibre of the cloth as gummy
mass. Thus, a lot of soap is wasted if water. is hard.
(ii) Soaps cannot be used in acidic solutions since acids precipitate the insoluble free
fatty acids which adhere to the fabrics and thus, reduce the ability of soaps to
remove oil and grease from fabrics.
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SOAPLESS SOAP/SYNTHETIC DETERGENTS
• Synthetic detergents have all the properties of soaps but actually does not contain any soap,
so they are known as ‘soapless soaps’. Straight chain alkyl group containing detergents are
biodegradable whereas branched chain alkyl group containing detergents are non-
biodegradable.
• Unlike soaps, synthetic detergents can be used in both soft and hard water. This is due to the
reason that calcium and magnesium salts of detergents like their sodium salts are also soluble
in water. Synthetic detergents are mainly classified into three categories:
• 1. Anionic Detergents: These are sodium salts of sulphonated long chain alcohols
or hydrocarbons.
• (i) Alkyl hydrogen sulphates formed by treating long chain alcohols with concentrated
sulphuric acids are neutralised with alkali to form anionic detergents.
(ii) Alkyl benzene sulphonates are obtained by neutralising alkyl benzene sulphonic acids
with alkali. In such detergents, the anionic part of the molecule is involved in the cleansing
action. They are mostly used for household work and in tooth paste
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• 2. Cationic Detergents
• These are quaternary ammonium salts of arnines with acetates, chlorides or
bromides as anion. For example,
• Cationic detergents are used in hair conditioner. They have germicidal
properties but are expensive therefore, these are of limited use.
• 3. Non-ionic Detergents
• Such detergents does not contain any ion in their constitution. One such
detergent can be obtained by reaction of stearic acid and polyethylene glycol.
• Liquid dish washing detergents are non-ionic type, Mechanism of cleansing
action of this type of detergents is the same as that of soaps.
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ADVANTAGES OF SYNTHETIC DETERGENTS OVER SOAPS
• 1. Synthetic detergents can be used even in case of hard water whereas
soaps fail to do so.
2. Synthetic detergents can be used in the acidic medium while soaps
cannot because of their hydrolysis to free acids.
3. Synthetic detergents are more soluble in water and hence,
form better lather than soaps.
4. Synthetic detergents have a stronger cleansing action than soaps.
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CHEMISTRY IN COLOURING MATTER
• The natural or synthetic colouring matter which are used in solution to stain
materials especially fabrics are called dyes.
• All colouring substances are not dyes, e.g., azobenzene, a coloured substance
does ‘not act as dye.
• A dye have following characteristics :.
• It must have a suitable colour.
• It can be fixed on the fabric either directly or with the help of mordant.
• It must be resistant to the action of water, acid and alkalies. The groups,
responsible for colour, are called chromophore, e.g.,
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CLASSIFICATION OF DYES ON THE BASIS OF CONSTITUTION
• (i) Nitro or nitroso dye Chromophore NO2 or NO group, Auxochrome = -OH
group, e.g., picric acid, martius yellow, Gambine, naphthol yellow-S.
(ii) Azo dye, e.g., bismark brown, methyl orange, methyl red, congo red, etc.
(iii) Anthraquinone dye e.g., alizarin
(iv) Indigo is the oldest known dye. Other examples are tyrian purple,
indigosol.
(v) Phthalein dye e.g., phenolphthalein, fluorescein, eosin, mercurochrome.
(vi) Triarybnethane dye, e.g., malachite green, rosaniline.
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CLASSIFICATION OF DYES ON THE BASIS OF APPLICATION
• (i) Direct dyes These dyes applied directly to fibre and are more useful to the
fabrics containing H-bonding like cotton, rayon, wool, silk and nylon, e.g., martius
yellow, congo reu/etc.
• (ii) Acid dyes These are water soluble and contain porar/acidic groups which
interact with the basic group of e.g., Orange-I, congo red, methyl orange, etc.
These dyes does not have affinity for cotton but are used for silk, wool, etc.
• (iii) Basic dyes These dyes contain basic group (like NHz group) and react with
anionic sites present on the fabric. These are used to dye nylons and polyester,
e.g., butter yellow, magenta (rosaniline), aniline yellow, etc.
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• (iv) Vat dyes Being water insoluble, these cannot be applied directly. These
are first reduced to a colourless soluble form by a reducing agent in large vats
and then, applied to fabrics. After applying, these are oxidised to insoluble
coloured form by exposure. to air or some oxidising agents, e.g., Indigo, tyrian
purple, etc.
• (v) Mordant dyes These are applied with the help of a binding material (e.g.,
metal ion, tannic acid or metal hydroxide) called mordant. Depending upon
the metal ion used, the same dye can give different colours. Alizarin is an
important example of such dyes.
• (vi) Ingrain dye These dyes are synthesised directly on the fabric. These are
water insoluble and particularly suitable for cotton fibres. Azo dyes belong to
this group of dye.
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CHEMISTRY IN COSMETICS
• Cosmetics are used for decorating, beautifying or improving complexion of skin. Some
of the cosmetics of daily use are as
• 1. Creams: These are stable emulsions of oils or fats in water and contain emmollients
(to prevent water loss) and humectants (to attract water) as two fundamental
components.
• 2. Perfumes: These solutions have pleasent odour and invariably consist of three
ingredients: a vehicle (ethanol + H20), fixative e.g., sandalwood oil, benzoin, glyceryl
diacetate etc.) and odour producing substance (e.g., terpenoids like linalool,
anisaldehyde (p-methoxy- benzaldehyde etc.)
• 3. Talcum Powder: It is used to reduce irritation of skin. Talc (Mg3(OH)2Si4O10), chalk,
zinc sterate and a suitable perfume are the constituents of talcum powder.
• 4. Deodorants: These are applied to mask the body odour. These possess antibacterial
properties. Aluminium salts, ZnO, Zn02, (C17H35COO)2Zn can be used in deodorant
preparation.
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MATTER
• Anything which has mass and occupies space is called matter.
For example, book, pencil, water, air are composed of matter
as we know that they have
mass and they occupy space.
• Classification of Matter
There are two ways of classifying the matter:
(A) Physical classification
(B) Chemical classification
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(A) PHYSICAL CLASSIFICATION:
• Matter can exist in three physical states:
1. Solids 2. Liquids 3. Gases
1. Solids: The particles are held very close to each other in an orderly fashion and
there is not much freedom of movement.
Characteristics of solids: Solids have definite volume and definite shape.
2. Liquids: In liquids, the particles are close to each other but can move around.
Characteristics of liquids: Liquids have definite volume but not definite shape.
3. Gases: In gases, the particles are far apart as compared to those present in solid
or liquid states. Their movement is easy and fast
• Characteristics of Gases: Gases have neither definite volume nor definite shape.
They completely occupy the container in which they are placed.
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PROPERTIES
Solid Liquid Gas
1. VOLUME Definite Definite Indefinite
2. SHAPE Definite Indefinite Indefinite
3. INTER MOLECULAR
FORCE OF ATTRACTION
Very high Moderate Negligible / Very
low
4. ARRANGEMENT OF
MOLECULES
Orderly arranged Free to move
within the volume
Free to move every
where
5. INTER MOLECULAR
SPACE
Very small Slightly greater Very great
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Properties
Solid Liquid Gas
6 . Compressibility Not compressible Not compressible Highly
compressible
7. Diffusion They can diffuse
due to kinetic energy of
liquid/gases
Can diffuse And
rate of diffusion is very fast
Can diffuse And
rate of diffusion is very fast
8. Expansion on
heating
Very little Very little Highly expand
9. Rigidity Very rigid Not rigid knownas
fluid
Not rigid and
known as fluid
9. Fluidity Can’t flow Can flow Can flow
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(B) CHEMICAL CLASSIFICATION:
• Based upon the composition, matter can be divided into two main types:
• 1. Pure Substances 2. Mixtures.
1. Pure substances: A pure substance may be defined as a single substance (or
matter) which cannot be separated by simple physical methods.
Pure substances can be further classified as (i) Elements (ii) Compounds
(i) Elements: An element consists of only one type of particles. These particles may
be atoms or molecules.For example, sodium, copper, silver, hydrogen, oxygen etc. are
some examples of elements. They all contain atoms of one type. However, atoms of
different elements are different in nature. Some elements such as sodium . or copper
contain single atoms held together as their constituent particles whereas in some
others two or more atoms combine to give molecules of the element. Thus, hydrogen,
nitrogen and oxygen gases consist of molecules in which two atoms combine to give
the respective molecules of the element.
(ii) Compounds: It may be defined as a pure substance containing two or more
elements combined together in a fixed proportion by weight and can be decomposed
into these elements by suitable chemical methods. Moreover, the properties of a
compound are altogether different from the constituting elements.
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TYPES OF COMPOUNDS
• The compounds have been classified into two types. These are:
(i) Inorganic Compounds: These are compounds which are obtained
from non-living sources such as rocks and minerals. A few
examples are: Common salt, marble, gypsum, washing soda etc.
•
(ii) Organic Compounds are the compounds which are present in plants
and animals. All the organic compounds have been found to contain
carbon as their essential constituent. For example, carbohydrates,
proteins, oils, fats etc.
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2. MIXTURES:
• The combination of two or more elements or compounds which are not chemically combined
together and may also be present in any proportion, is called mixture. A few examples of
mixtures are: milk, sea water, petrol, lime water, paint glass, cement, wood etc.
Types of mixtures: Mixtures are of two types:
• (i) Homogeneous mixtures: A mixture is said to be homogeneous if it has a uniform composition
throughout and there are no visible boundaries of separation between the constituents.For
example: A mixture of sugar solution in water has the same sugar water composition
throughout and all portions have the same sweetness.
(ii) Heterogeneous mixtures: A mixture is said to be heterogeneous if it does not have uniform
composition throughout and has visible boundaries of separation between the various
constituents. The different constituents of a heterogeneous mixture can be seen even with
naked eye.For example: When iron filings and sulphur powder are mixed together, the mixture
formed is heterogeneous. It has greyish-yellow appearance and the two constituents, iron and
sulphur, can be easily identified with naked eye
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DIFFERENCES BETWEEN COMPOUNDS AND MIXTURES COMPOUNDS
• Compounds
• 1. In a compound, two or more elements are
combined chemically.
2. In a compound, the elements are present in
the fixed ratio by mass. This ratio cannot
change.
3. Compounds are always homogeneous i.e.,
they have the same composition
throughout.
4 In a compound, constituents cannot be
separated by physical methods
5. In a compound, the constituents lose their
identities i.e., i compound does not show the
characteristics of the constituting elements.
• Mixtures
1. In a mixture, or more elements or
compounds are simply mixed and not
combined chemically.
2. In a mixture the constituents are not
present in fixed ratio. It can vary
3. Mixtures may be either homogeneous or
heterogeneous in nature.
4. Constituents of mixtures can be
separated by physical methods.
5, In a mixture, the constituents do not lose
their identities i.e., a mixture shows the
characteristics of all the constituents
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LAWS
OF
CHEMICAL COMBINATION
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LAWOFCONSERVATIONOFMASS:
• This law was proposed by Antoine Lavoisier in 1744.
• According to this law, “Matter can neither be created nor destroyed in a
chemical reaction” ie Matter is indestructible.
• In all the physical changes and chemical reactions, total mass of the products is the same
as the total mass of the reactants.
• It is a derivation of Dalton’s atomic theory ‘atoms neither created nor destroyed’.
• Example: N2 + 3H2 → 2NH3
• 28g + 3x2g = 2x17g
34g = 34g
• Total masses of reactants = Total masses of products + Masses of un reacted reactants.
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WHAT ARE LIMITATIONS OF LAW OF
CONSERVATION OF MASS?
• We know from the definition of the law of conservation of mass that the
total mass of the products must be equal to the total mass of the reactants
in any physical or chemical change. But in a nuclear reaction, some the
mass gets converted into energy creating an imbalance between the mass
of reactants and the products. Therefore, the total mass is not conserved.
This is the limitation of the conservation of mass.
• The other limitation of conservation of mass is according to Einstein’s theory,
the relation between two quantities is given by E = mc2, which means that
energy and mass are interconvertible. Therefore, for the law of
conservation of mass to be valid mass and energy of the system must be
conserved.
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Numerical: What weight of NaCl is decomposed by 4.9g of H2SO4 if 6g of NaHSO4 and
1.825g of HCl are produced in the same reaction?
Solution:
NaCl + H2SO4 → NaHSO4 + HCl
x + 4.9 = 6 + 1.825
x + 4.9 = 7.825
x = 7.825- 4.9
x = 2.925g
Practice questions:
1. wWhen 4.2g Of NaHCO3 is added to a solution of CH3COOH weighing 10g, it is
observed that 2.2 g of CO2 is released. The residue is found to weigh 12.0g. Show that
these observations are in agreement ith the law of conservation of mass.
2. 1.7 g Of AgNO3 dissolved in 100g of water were mixed with 0.585 g of NaCl dissolved in
100g of water when 1.435 gram of AgCl and 0.85 gram of NaNO3 were formed. Justify
that the data illustrates law of conservation of mass.
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• A pure chemical compound always consist of same elements combined together in a
fixed (definite) proportion by mass.
• Example: CO2 can be prepared in number of ways
(i) by burning Coke in air C + O2 → CO2
(ii) by the decomposition of lime stone (CaCO3) on heating:
CaCO3 → CaO + CO2
(iii) By the action of dilute HCl on washing soda or baking soda:
Na2CO3 + 2 HCl → 2NaCl + CO2+ H2O
in all the samples of CO2, carbon and oxygen are in the ratio (By mass) :
C : O
12 : 16 X2
12 : 32
3 : 8
5/27/2021

Example: The mass of copper oxide obtained by heating 2.16 g of metallic copper with nitric acid
and subsequent ignition was found to be 2.7 g. In another experiment 1.15 gram of copper oxide
on reduction yielded 0.92 gram of copper. Show that the data illustrates the law of constant
composition.
In the first experiment
Weight of copper = 2.16 g Weight of copper oxide = 2.70 g
Weight of oxygen =2.70 – 2.16 = 0.5 4g
Percentage of copper = weight of copper/ weight of copper oxide X100= 2.16/ 2.70 x100 = 80%
Percentage of oxygen= weight of Oxygen / weight of copper oxide X100= 0.54/ 2.70 x100= 20%
In the second experiment
Weight of copper = 0.92 g Weight of copper oxide = 1.15g
Weight of oxygen = 1.15 –0 .92= 0.23g
Percentage of copper =weight of copper/ weight of copper oxideX100= 0.92/1.15 x 100 = 80%
Percentage of oxygen= weight of Oxygen/weight of copper oxide X100= 0.23/ 1.15 x100 = 20%
Since the ratio of weights of copper and oxygen in the two compounds remains the same, the
law of definite proportion is illustrated.
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5/27/2021

LAW OF MULTIPLE PROPORTIONS:
• The law was given by Dalton in
1803.
• If two elements combine to
form two or more compounds,
the weights of one of the
elements which combine with a
fixed weight of the other in
these compounds bear a simple
whole number ratio by weight.
5/27/2021

5/27/2021

Example 2: COMPOUNDS OF CARBON AND OXYGEN
The number of parts by weight of oxygen which combined with 12 parts by weight of carbon in
the two oxides are 16 and 32 respectively. i.e. they are in the ratio of 1:2 which is a simple
whole number ratio.
Numerical:A metal forms two oxides. The higher oxide contains 80% metal. 0.72 g of the lower
oxide give 0.80g of higher oxide when oxidized. Show that the data illustrates the law of
multiple proportions.
Solution: In The Higher Oxide
Weight of metal = 0.8x80/100= 0.64 g
Weight of oxygen= 0.8- 0.64 = 0.16g
In Lower Oxide:
Weight of metal will remains same as the higher oxide = 0.64 g
Weight of oxygen = 0.72 – 0.64 = 0.08 g
The ratio of weights of oxygen which combines with a fixed weight of metal in the II oxide is:
0.16 : 0.08 = 2 : 1.
As the ratio is simple whole number in nature, the law of multiple proportion is illustrated.
5/27/2021

5/27/2021

LAW OF RECIPROCAL PROPORTION
• This law was proposed by “Richer” in 1794
• This law states that ‘when two elements combine separately with third
element and form different types of molecules, their combining ratio is
directly reciprocated if they combine directly.”
• The above law is the basis of law of equivalent masses.
• For example: ratio of masses of carbon and sulphur which combine with
fixed mass (32 parts) of oxygen is 12:32 or 3:8 …(1)
• In ratio of masses of carbon and sulphur is 12:64 or 3:16 …(2)
• The two ratios (1) and (2) are related to each other by
2:1 i.e. first ratio is integral multiple of second.
• Limitations of Law of Reciprocal Proportion
• Law of reciprocal proportions is not obeyed by- SO3, SO2, O2.
5/27/2021

5/27/2021

GAY LUSSAC’S LAW OF COMBINING VOLUMES
• Examples:
(i) Combination of hydrogen and chlorine
H2 + Cl2 → 2HCl
1vol 1 vol. 2 volumes. (Ratio1:1:2)
(ii) Combination between nitrogen and hydrogen:
N2 + 3H2 → 2NH3
1vol. 3 vol. 2 vol. ( Ratio1:3:2)
Note: volume should be in same units.
5/27/2021

AVOGADRO’S HYPOTHESIS
• Under similar conditions of temperature and pressure equal volume of all
gases contain equal number of molecules or moles.
• Limitations of Avogadro's Law
• Avogadro's Law is applicable only to gases. Avogadro stated that for any
ideal gas, 1 mole of particles will occupy a specific volume at a specific ratio
of pressure and temperature. So it does not apply to a liquid or solid.
5/27/2021

5/27/2021

DALTON’S ATOMIC THEORY.
POSTULATES OF DALTON’S ATOMIC THEORY:
1. Matter is made up of extremely small particles called atoms.
2. Atoms of same elements are identical in all respect that is in size, mass and also in
properties. [law of definite proportion].
3. Atoms of different elements differ in all respects and have also different properties.
4. Atom is the smallest portion of matter, which can take part in chemical combination.
5. Atoms of the same or different elements combine together to form compound atoms
(molecules).
6. When atoms combine with one another to form compounds atoms or molecules, they do so
in simple whole number ratios. [law of multiple proportion].
7. Atoms can neither be created nor destroyed in a physical change or chemical reaction.[ Law
of conservation of mass]
5/27/2021

LIMITATIONS OF DALTON’S ATOMIC THEORY
• It explain the laws based on weight but not on volume i.e. it could give no
explanation for the Gay Lussac’s Law.
• It fail to explain why atoms of different elements differ in size and properties.
• It could not explain why atoms combine to form molecules.
• It fail to explain the bonding between the atoms to form molecules.
• He said atom is indivisible, but we know that atom is divisible and the subatomic
particles are electrons protons and neutrons.
• He said that atoms of the same elements have same mass, but isotopes are the
atoms of same elements having different masses.
• He said that atoms of different elements have different masses but isobars are
the atoms of different elements having same masses.
5/27/2021

ATOMIC AND MOLECULAR MASSES.
• Atom is very small, so gram is a very big unit to express the mass of the atom.
• So it was suggested that the mass of atom should be expressed as relative mass
which is known as atomic mass unit (amu) or unified mass (u) .
• Isotope of carbon 12 was selected for expressing the relative atomic mass.
• Relative atomic mass of = mass of one atom of an element.
1/12 X mass of one atom of carbon (C-12)
For example:
(i) if the relative atomic mass of an element is 24u it means that this atom is twice
the mass of C 12 atom.
(ii) If the relative atomic mass of an element is 4 u, it means this element has one
third the mass of C-12 atom. [ u= unified mass].
5/27/2021

AVERAGE ATOMIC MASS
• We know that atomic mass is equal to the sum of number of neutron and
number of protons.
• Number of neutrons and protons are always in whole number, so their sum
i.e. atomic mass should also be in whole numbers.
• But atomic mass of some elements are found to be in fractions. For example
chlorine has atomic mass of 35.5 u. It can only be explained by the concept
of average atomic mass.
• The average atomic mass of an element is actually the average mass of all
the isotopes of that element present in nature.
5/27/2021

CALCULATIONS OF AVERAGE ATOMIC MASS
• If an element has two isotopes having masses ‘a’ and ‘b’ and their abundance
ratio in nature is ‘m’ and ‘n’ respectively, then their average atomic mass can
be given by the formula:
average atomic mass = a X m + b X n
m + n
For example: The element chlorine has two isotopes Cl-35 and Cl-37 and their
relative proportions are 3 :1. What is the average atomic mass of chlorine.
Average atomic mass of chlorine is = 35 X 3 + 37 X 1 = 105 + 37 = 142
3+1. 4 4
= 35.5u
5/27/2021

MOLECULAR MASS AND GRAM MOLECULAR MASS
• Molecular mass: It is the sum of atomic masses of all the atoms
which constitutes a molecule of that substance.
• For example: molecular mass of CaCO3
= 40 + 12 + (3 x 16) = 100 u
Gram molecular mass: The quantity of a substance whose mass in
grams is numerical equal to its relative molecular mass. It is also
called gram molecule.
For Example: gram molecular mass of CaCO3 = 100g.
5/27/2021

CALCULATE THE MOLECULAR MASS OF THE FOLLOWING:
(I) H20 (II) C02 (III) CH4
• Answer:
• (i) Molecular mass of H2O =2(1.008 amu)+16.00 amu=18.016 amu
(ii) Molecular mass of CO2= 12.01 amu + 2 x 16.00 amu = 44.01 amu
(iii) Molecular mass of CH4= 12.01 amu + 4 (1.008 amu) = 16.042 amu
5/27/2021

FORMULA MASS
• For ionic compounds the molecular mass of one
formula unit is called formula mass.
• Formula mass of sodium chloride (NaCl)
= 23+35.5= 58.5 u
5/27/2021

MOLE CONCEPT AND MOLAR MASSES
•
Term mole was suggested by Ostwald (Latin word mole = heap)
A mole is defined as the amount of substance which contains same number of
elementary particles (atoms, molecules or ions) as the number of atoms
present in 12 g of carbon (C-12).
•It may be emphasised that the mole of a substance always contain the same
number of entities, no matter what the substance may be. In order to determine
this number precisely, the mass of a carbon– 12 atom was determined by a mass
spectrometer and found to be equal to 1.992648 × 10–23 g. Knowing that one
mole of carbon weighs 12 g, the number of atoms in it is equal to 6.0221367
x1023atoms/mol This number of entities in 1 mol is so important that it is given
a separate name and symbol. It is known as ‘Avogadro constant’, denoted by NA
in honour of Amedeo Avogadro.
•
5/27/2021

• To really appreciate largeness of this number, let us write it with all the zeroes
without using any powers of ten. 602213670000000000000000 Hence, so
many entities (atoms, molecules or any other particle) constitute one mole of
a particular substance.
• We can, therefore, say that 1 mol of hydrogen atoms = 6.022×1023 atoms
• The mass of one mole of a substance in grams is called its molar mass
• mol = 6.023 * 1023 atoms = one gram-atom = gram atomic mass
• 1 mol = 6.023 * 1023 molecules = gram molecular mass
• In gaseous state at STP (T = 273 K, p = 1 atm)
• Gram molecular mass = 1 mol= 22.4 L = 6.022 * 1023 molecules
• The volume occupied by one mole molecules of a gaseous substance is called
molar volume or gram molecular volume
5/27/2021

• Number of moles = amount of substance (in gram) / molar mass
• Number of moles =Number of molecule / NA
• Number of molecules in I g compound = NA / g-molar mass
• Number of molecules in 1 cm3 (1 mL) of an ideal gas at STP is
called Loschmidt number (2.69x 1019).
• [One amu or u (unified mass) is equal to exactly the 1 / 12th of
the mass of 12C atom, i.e., 1 amu or u = 1 / 12 * mass of one
carbon (C12) atom
• 1 amu = 1 / NA= 1 Dalton = 1.66x 10-24 g
• One mole of electrons weighs 0.55 mg (5.5x 10-4 g).
5/27/2021

5/27/2021

Question: In three moles of ethane (C2H6), calculate the following:
(i) Number of moles of carbon atoms.
(ii) Number of moles of hydrogen atoms.
(iii) Number of molecules of ethane.
Solution.
(i) 1 mole of C2H6 contains 2 moles of carbon atoms.
Number of moles of carbon atoms in 3 moles of C2H6 = 3 x 2 = 6
(ii) 1 mole of C2H6 contains 6 moles of hydrogen atoms.
Number of moles of hydrogen atoms in 3 moles of C2H6 = 3 x 6 = 18
(iii) 1 mole of C2H6 = 6.022 x 1023 molecules
Number of molecules in 3 moles of C2H6 = 3 x 6.022 x 1023
= 1.807 x 1024 molecules
5/27/2021

Which one of the following will have largest number of atoms?
(I) 1 g Au(s)
(Ii) 1 g Na(s)
(Iii) 1 g Li(s)
(Iv) 1 g Of Cl2(g)
5/27/2021

• Calculate the number of atoms
in each of the following:
(i) 52 moles of Ar
(ii) 52 u of He
(iii) 52 g of He.
5/27/2021

EMPIRICAL FORMULA & MOLECULAR FORMULA
• An empirical formula represents the simplest whole number ratio
of various atoms present in a compound whereas the molecular
formula shows the exact number of different types of atoms
present in a molecule of a compound.
• Molecular Formula =Empirical Formula x n
• If the mass per cent of various elements present in a compound is
known, its empirical formula can be determined. Molecular
formula can further be obtained if the molar mass is known.
• The following example illustrates this sequence.
5/27/2021

PROBLEM 1.2 A COMPOUND CONTAINS 4.07 % HYDROGEN, 24.27 % CARBON AND
71.65 % CHLORINE. ITS MOLAR MASS IS 98.96 G. WHAT ARE ITS EMPIRICAL AND
MOLECULAR FORMULAS ?
• Step 1. Conversion of mass per cent to grams. Since we are having mass per cent, it is
convenient to use 100 g of the compound as the starting material. Thus, in the 100 g sample of
the above compound, 4.07g hydrogen is present, 24.27g carbon is present and 71.65 g
chlorine is present.
• Step 2. Convert into number moles of each element :Divide the masses obtained
above by respective atomic masses of various elements.
Moles of hydrogen = 4.07 g/ 1.008 g = 4.04
Moles of carbon = 24.27 g/ 12.01g = 2.021
Moles of chlorine = 71.65g 35.453 g = 2.021
• Step 3. Divide the mole value obtained above by the smallest number :Since
2.021 is smallest value, division by it gives a ratio of 2:1:1 for H:C:Cl . In case the ratios are not
whole numbers, then they may be converted into whole number by multiplying by the suitable
coefficient.
5/27/2021

• Step 4. Write empirical formula by mentioning the numbers after writing
the symbols of respective elements. CH2Cl is, thus, the empirical formula of
the above compound.
• Step 5. Writing molecular formula
• (a) Determine empirical formula mass Add the atomic masses of various
atoms present in the empirical formula. For CH2 Cl, empirical formula mass
is 12.01 + 2 × 1.008 + 35.453 = 49.48 g
• (b) Divide Molar mass by empirical formula mass
• Molar mass /Empirical formula mass = 98.96 g =/49.48g = 2 = (n)
• (c) Multiply empirical formula by n obtained above to get the molecular
formula
• Empirical formula = CH2Cl, n = 2. Hence molecular formula is C2H4Cl2 .
5/27/2021

QUESTION: Determine the empirical formula of an oxide of
iron which has 69.9% iron and 30.1% oxygen by mass.
Solution:
Hence, the empirical formula is Fe2CO3
5/27/2021

• Question: Determine the molecular formula of an oxide of iron in
which the mass per cent of iron and oxygen are 69.9 and 30.1
respectively.
Solution.
For empirical formula, Fe2O3.
Molecular mass of Fe2O3 x 1 = 2 x 56 + 3 x 16 = 112 + 48 = 160
Molecular formula = n X (Empirical formula)
∴ n= Molecular formula /Empirical formula =160/160=1
∴ Molecular formula = (Fe2O3) x 1 = (Fe2O3)
5/27/2021

• Question 34.
A welding fuel gas contains carbon
and hydrogen only. Burning a small
sample of it in oxygen gives 3.38 g
carbon dioxide, 0.690 g of water
and no other products. A volume of
10.0 L (measured at STP) of this
welding gas is found to weigh 11.6
g. Calculate
(i) empirical formula,
(ii) molar mass of the gas, and
(iii) molecular formula.
5/27/2021

5/27/2021

STOICHIOMETRY AND STOICHIOMETRIC CALCULATIONS
• The word ‘stoichiometry’ is derived from two Greek words - stoicheion
(meaning element) and metron (meaning measure).
• Stoichiometry, thus, deals with the calculation of masses (sometimes
volumes also) of the reactants and the products involved in a chemical
reaction.
• Before understanding how to calculate the amounts of reactants required
or the products produced in a chemical reaction, let us study what
information is available from the balanced chemical equation of a given
reaction.
• Let us consider the combustion of methane. A balanced equation for this
reaction is as given below :
5/27/2021

• CH4 (g) + 2O2 (g) → CO2 (g) + 2 H2O (g)
• Here, methane and dioxygen are called reactants and carbon dioxide and
water are called products. Note that all the reactants and the products are
gases in the above reaction and this has been indicated by letter (g) in the
brackets next to its formula. Similarly, in the case of solids and liquids, (s) and
(l) are written respectively.
• The coefficients 2 for O2 and H2O are called stoichiometric coefficients.
Similarly the coefficient for CH4 and CO2 is one in each case. They represent
the number of molecules (and moles as well) taking part in the reaction or
formed in the reaction. Thus, according to the above chemical reaction,
5/27/2021

CH4 (g) + 2O2 (g) → CO2 (g) + 2 H2O (g)
o One mole of CH4 (g) reacts with two moles of O2 (g) to give one mole of CO2 (g)
and two moles of H2O(g)
• One molecule of CH4 (g) reacts with 2 molecules of O2 (g) to give one molecule of
CO2 (g) and 2 molecules of H2O(g)
• 22.7 L of CH4 (g) reacts with 45.4 L of O2 (g) to give 22.7 L of CO2 (g) and 45.4 L of
H2O(g)
• 16 g of CH4 (g) reacts with 2×32 g of O2 (g) to give 44 g of CO2 (g) and 2×18 g of
H2O (g).
From these relationships, the given data can be interconverted as follows :
mass ⇌ moles ⇌ no.of molecules
Mass/ Volume = Density
5/27/2021

Calculate the amount of water (g) produced by the
combustion of 16 g of methane.
• Solution The balanced equation for combustion of methane is :
• CH4 (g) + 2O2 (g) → CO2 (g) + 2 H2O (g)
• (i)16 g of CH4 corresponds to one mole.
• (ii) From the above equation, 1 mol of CH4 (g) gives 2 mol of H2O (g).
• 1 mol H2O = 18 g H2O
• Hence 2 mol H2O = 2 × 18 g of H2O = 36 g H2O
5/27/2021

How many moles of methane are required to
produce 22 g CO2 (g) after combustion?
According to the chemical equation,
CH4 (g) + 2O2 (g) → CO2 (g) + 2 H2O (g)
 44g CO2 (g) is obtained from 16 g CH4 (g).
 [ ∵1 mol CO2 (g) is obtained from 1 mol of CH4 (g)]
 mole of CO2 (g) = 22 g CO2 (g) × 1 mol CO2 (g) /44 g CO2 (g) = 0.5 mol CO2 (g)
 Hence, 0.5 mol CO2 (g) would be obtained from 0.5 mol CH4 (g) or 0.5 mol of
CH4 (g) would be required to produce 22 g of CO2 (g).
5/27/2021

• Question 4.
Calculate the amount of carbon dioxide that could be produced when
(i) 1 mole of carbon is burnt in air.
(ii) 1 mole of carbon is burnt in 16 g of dioxygen.
(iii) 2 moles of carbon are burnt in 16 g of dioxygen.
Solution:
(i) Hence, 1 mole of C produces 44 g of CO2
(ii)32 g O2 reacts with 12g C to produce 44 g of CO2
∵ 16 g O2 reacts with C to produce=(44/32)x 16
=22g of CO2
Hence, O2 is the limiting reagent.
(iii) 64 g O2 reacts with C to produce 88 g of CO2
∵ 16 g O2 reacts with C to produce = (88/64)X16=
• =22g of CO2
5/27/2021

Calcium carbonate reacts with aqueous HCI to give CaCi2 and
CO2 according to the reaction,
CaC03(S) + 2HCI(Aq) —> CaCI2(Aq)+ CO2(g) + H2O(I).
What mass of CaC03 is required to react completely with 25 ml of
0.75 m HCl?
5/27/2021

5/27/2021

LIMITING REAGENT
• Many a time, the reactions are carried out when the reactants are not
present in the amounts as required by a balanced chemical reaction. In such
situations, one reactant is in excess over the other.
• The reactant which is present in the lesser amount gets consumed after
sometime and after that no further reaction takes place whatever be the
amount of the other reactant present.
• Hence, the reactant which gets consumed, limits the amount of product
formed and is, therefore, called the limiting reagent.
• In performing stoichiometric calculations, this aspect is also to be kept in
mind.
5/27/2021

• Solution.
According to the equation, one mole of
A reacts with one mole of B and one
atom of A reacts with one molecule of B.
(i) B is limiting reagent because 200
molecules of B will react with 200 atoms
of A and 100 atoms of A will be left in
excess.
(ii) A
(iii) Both will react completely because it
is stoichiometric mixture. No limiting
reagent.
(iv) B
(v) A
• Question 23.
In a reaction :A + B2 → AB2
Identify the limiting reagent, if any,
in the following reaction mixtures.
(i) 300 atoms of A + 200 molecules
of B
(ii) 2 mol A + 3 mol B
(iii) 100 atoms of A + 100 molecules
of B
(iv) 5 mol A + 2.5 mol B
(v) 2.5 mol A + 5 mol B
5/27/2021

Question: Dinitrogen and dihydrogen
react with each other to produce
ammonia according to the following
chemical equation:
N2(G) + H2(G) → 2NH3(G)
•
(i) Calculate the mass of ammonia
produced if 2.00 x 103 g dinitrogen
reacts with 1.00 x 103 g of dihydrogen.
(ii) Will any of the two reactants remain
unreacted?
(iii) If yes, which one and what would
be its mass?
Solution.
The balanced chemical equation is N3 +
3H3 → 2NH3
5/27/2021

MASS PER CENT
It is obtained by using the following
relation:
Mass per cent =
(Mass of solute /Mass of solution) x100
5/27/2021

Calculate the mass percent of different elements
present in sodium sulphate (Na2 SO4).
•
5/27/2021

Molarity: It is defined as
the number of moles of
solute in 1 litre of the
solution. It is denoted by
“M”
. Molality: It is defined as
the number of moles of
solute present in 1 kg of
solvent. It is denoted by
“m”.
5/27/2021

Question 12.
If the density of methanol is 0.793 kg L-1 what is its volume needed
for making 2.5 L of its 0.25 M solution?
• Solution. Moles of methanol present in 2.5 L of 0.25 M solution
5/27/2021

QUESTION:
Calculate the
molarity of a
solution of ethanol
in water in which the
mole fraction of
ethanol is 0.040.
5/27/2021

How are 0.50 Mol Na2CO3 And 0.50 M Na2CO3 different?
Solution.
1 mol Na2CO3 = 2 x 23 + 12 + 3 x 16 = 106 g mol-1
0.50 mol Na2CO3 0.50 x 106= 53 g
0.50 M Na2CO3 solution means that 0.50 moles or 53 g of
Na2CO3 are dissolved in 1000 ml of solution.
5/27/2021

5/27/2021

MOLE FRACTION: It is the ratio of number of moles of a particular
component to the total number of moles of the solution. For a solution
containing n2 moles of the solute dissolved in n1 moles of the solvent.
5/27/2021

Thank you
5/27/2021

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