matter in our surroundings and it.docx

1

MATTER AND ITS FORMS

Matter can be defined as anything that occupies space, (i.e., has a volume), possesses mass, offers resistance and can be felt through one or more of our senses.

Till very recently, it was assumed that matter can neither be created nor destroyed. Scientists have established that there are two fundamental entities in the universe : Matter and Energy.

Is Matter Around us Pure

Matter can be broadly divided into two major groups, 'Pure' and 'Impure'.

Purity' as a chemical concept signifies something quite different. When we say a substance is pure, it means that the constituent particles that make up the substance are of only one type and have the same chemical nature. For example, in chemical terms, pure water implies that it is made of only one type of molecule i.e., H2O. Accordingly, the chemical classification of matter specifies two main categories of substances, pure substances and mixtures (impure substances).

Matter that is divided into pure and impure substances can be further categorized. Pure substances can be divided into 'elements' and 'compounds';

Impure substances, commonly called 'mixtures' can be further divided into homogeneous' and 'heterogeneous' mixtures.

Pure Substances

A pure substance has the same composition throughout. For example, different samples of water, prepared by different methods, by different people at different places always consist of hydrogen and oxygen in the ratio 1:8 by mass and 2:1 by volume. The distinctive features of pure substances are:

A pure substance is composed of the same kind of particles e.g. hydrogen, oxygen, water, sodium chloride, etc.

A pure substance is homogeneous, irrespective of its origin or method of preparation.

A pure substance has definite properties, characteristic of itself.

A pure substance has the same composition throughout. For example, different samples of water, prepared by different methods, by different people at different places always consist of hydrogen and oxygen in the ratio 1:8 by mass and 2:1 by volume. If any sample has a different ratio of these elements, then it certainly is not water.

I I T - J E E | M E D I C A L | F O U N D A T I O N S

2

Remember :

A solution of salt in water or sugar solution being homogeneous appears to consist of one type of particles. But it is made up of more than one kind of particles. Hence it is not a pure substance. It is a mixture.

Pure substances can be classified into elements and compounds.

Lavoisier, a French chemist, was the first to do this on the basis of quantitative studies. He showed that when we heat mercuric oxide it changes into mercury and oxygen.

Mercuric oxide is a compound because it decomposes into simpler substances, whereas mercury and oxygen cannot be further decomposed into anything simpler as they are elements.

Element

An element is defined as a pure substance as it is made up of only one kind of atoms having the same atomic number.

The smallest particle of an element is the atom, which has all the properties of that element. It cannot be further reduced to simpler substances by ordinary physical or chemical processes. Example : Hydrogen and oxygen

Remember :

Noble gases, some metals, carbon, silicon etc. have only one atom in their molecules. These are called monatomic molecules. Some elements have two atoms in their molecules. These are called diatomic molecules e.g., hydrogen (H2), oxygen (O2), nitrogen (N2), chlorine (Cl2), etc. Those elements that have more than two atoms in their molecules are said to be polyatomic. For example a molecule of phosphorus consist four atoms (P4), and that of sulphur contains eight atoms (S8).

Elements can be broadly divided into four categories: Metals, non-metals, metalloids and noble gases.

Metals : These are generally solids with characteristics such as hardness, malleability, ductility, high tensile strength, lustre and ability to conduct heat and electricity. Example: Copper, iron, zinc etc.

Non-metals : These are generally non-lustrous, brittle, poor conductors of heat and electricity. Example: Sulphur, phosphorus, nitrogen etc.

Metalloids : Metalloids are those elements that have properties, which lie in-between those of metals and non-metals. Example: Arsenic, tin, bismuth etc.

Noble gases : Are a group of six elements that do not combine with other elements and tend to exist by themselves. They are characterized by extreme chemical inactivity.. Example: Neon, helium, argon etc.

Compound

A compound is a pure substance that is formed by the combination of atoms of two or more elements by either transfer or by sharing of electrons.

The atoms of the different elements in a compound are chemically combined in a fixed and constant proportion. If this proportion is different, the same elements produce entirely a different compound.

Example : Water is a compound of hydrogen and oxygen present in the ratio of 2:1 by volume and 1:8 by weight; if the ratio by volume changes to 1:1 and by weight to 1:16, it forms an entirely new compound called hydrogen peroxide. Thus, compounds are represented by a definite formula with its constituents combined in fixed proportions.

While the combination of two or more elements forms a compound, similarly, two or more simple compounds can form a complex compound by a process called synthesis.

A compound as a pure substance can be decomposed into simpler substances by some suitable chemical technique. However, the properties of compounds are absolutely different from the properties of the elements that constitute the compound.

For instance, hydrogen is combustible and oxygen is a supporter of combustion. But their product, water, neither burns, nor helps in burning. It actually extinguishes fire.


3

Another familiar example is that of sodium and chlorine. Sodium is a violently reactive metal while chlorine is a highly poisonous gas with a choking and irritating smell. One cannot even think of consuming a piece of sodium or breathing in chlorine as such. But their product, i.e., sodium chloride, is consumed by all of us daily in the form of common salt.

Formation of a compound is a chemical process and always involves exchange of energy.

For example, hydrogen burns in oxygen to form water, liberating heat. Nitrogen combines with oxygen to form nitric oxide by absorbing heat. During the process of photosynthesis, carbon dioxide and water combine to form carbohydrate with absorption of light energy. The elements present in a compound can be re-obtained only by chemical processes. Water can be decomposed to hydrogen and oxygen only by electrolysis. If we bring a magnet near a sample of iron sulphide, the iron present in the iron sulphide cannot be separated.

We can summarize the properties of compounds as follows :

A compound cannot be separated into its constituents by mechanical or physical means.

Properties of a compound differ entirely from those of its constituent elements.

Energy changes are involved in the formation of a compound.

The constituent elements in a compound are in a fixed proportion by weight.

A compound is a homogenous substance. That is it is same throughout in properties and composition.

A compound has a fixed melting point and boiling point. For example, ice melts at 0oC.

Impure Substances - Mixtures

Impure substances are commonly called mixtures. A mixture is a material containing two or more elements or compounds that are in close contact and are mixed in any proportion. The constituents of mixtures exhibit their individual properties. These constituents can be separated by physical means.

Example : air, gunpowder, etc.

Mixtures can be homogeneous or heterogeneous. A homogeneous mixture has a uniform composition through out its mass. For example, sugar or salt dissolved in water, alcohol in water, etc. While in a heterogeneous mixture the composition is not uniform throughout its mass. Different portions of a heterogeneous mixture show different properties. There are visible sharp boundaries. Example: Oil and water, salt and sand, etc.

We can now summarize the properties of mixtures as follows:

A mixture may be homogenous or heterogeneous.

The constituents of a mixture can be separated by physical means like filtration, evaporation, sublimation and magnetic separation.

In the preparation of a mixture, energy is neither evolved nor absorbed.

A mixture has no definite melting and boiling point.

The constituents of a mixture retain their original set of properties. For example, magnet attracts iron filings in a mixture of sand and iron powder.

Types of Mixtures

Matter MixtureType Example

Solid Solid mixture Iron filings and sulphur

Solid Liquid mixture Common salt and water

Solid Gas mixture Air entrapped in soil


4

Matter MixtureType Example

Liquid Gas mixture Oxygen dissolved in water

Gas Gas mixture Air containing hydrogen, oxygen, nitrogen, carbon dioxide etc.

Liquid Liquid mixture Water and alcohol

Q : Differences Between Mixtures and Compounds

Mixtures Compounds

A mixture c b spdrated into its constituents by physical processes (filtration, evaporation, sublimation, distillation)

A compound cannot be separated into its constituents by physical processes. It can be separated by chemical means.

A mixture shows the properties of its constituents A compound has a new set of properties different from its constituents

Composition of a mixture varies and the constituents are present in any proportion by weight. ft does not have a definite formula.

The composition of a compound is fixed and the constituents are present in fixed proportions by weight. It has a definite formula.

The constituents do not react chemically, thus no energy changes takes place

Chemical reactions take place and energy changes in the form of heat and light are involved

A mixture does not have a fixed melting point and boiling point Examples: air, sand and salt

A compound has a fixed melting point and boiling point Examples: 1.120 (water), FeS (iron sulphide)

Remember :

The components hydrogen and oxygen cannot be separated by physical methods such as filtration or evaporation.

Hydrogen and oxygen are present in a fixed proportion of 1: 8 by weight

Energy changes accompany the formation of this compound i.e., heat and light are given out.

Properties of water are entirely different from the constituents, hydrogen and oxygen.

The boiling point of water is 100oC at 76 cm of Hg i.e., one atmospheric pressure.

Types of Mixtures

Mixtures can be classified based on the composition and can be broadly divided into three groups, depending on whether the constituents are elements or compounds or both.

Element with an Element (forming alloys)

a) Oxygen and nitrogen

b) Sodium and mercury (amalgam)

c) Copper and zinc alloy.

Compound with a Compound

a) Water and salt

b) Water and alcohol

c) Salt and sugar.


5

Element with a Compound

a) Oxygen and water (air dissolved in water)

b) Oxygen, nitrogen, carbon dioxide and water vapour (air).

Mixtures can also be grouped on the basis of their physical states.

Mixtures of Solids, Liquids and Gases

Solid Liquid Gas

Solid salt and sugar salt and water Dust in air

Liquid Mercury and copper Alcohol and water clouds

Gas Hydrogen and palladium Oxygen and water Air

Characteristics of Compounds and Mixtures

Characteristics Compounds Mixtures

Composition Made up of atoms of elements in a fixed proportion

Made up of elements, or compounds, or both in any proportion

Nature Particles are of the same kind Particles are of different kinds

Structure Always homogeneous May or may not be homogeneous

Separation Components can be separated only by chemical means

Components can be separated by physical means

Energy changes

Energy is always evolved or absorbed Generally no energy is evolved or absorbed

Appearance Components cannot be seen separately Components may or may not be seen separately

Preparation Always involves a chemical change Involves only physical change

Properties Entirely different from those of the constituents

No property of their own. Show the average properties of all the constituents

Constituents of mixtures in matter

Separation of the Constituents of Mixtures

The separation of the various constituents of different mixtures depends on the properties of the constituents. Some of these properties are size, density, magnetic nature of particles, solubility, miscibility, differences in boiling and melting points, sublimability, diffusibility and absorbance.


6

Sedimentation and DecantationSedimentation is the process by which insoluble heavy particles in a liquid are allowed to settle down. This is a simple process that most people employ at home. For example, suppose you are making some tea and have boiled the water and added the tealeaves into the water. Then you realize that you cannot find the strainer. You may look for a clean piece of cloth but do not succeed. What would you do? Keep the tea with the leaves, aside for some time. The tealeaves begin to settle down. This settling down of the particles in lower part of the container is called sedimentation. Another example of sedimentation is the settling of mud particles in water.

Decantation is the process by which, a clear liquid obtained after sedimentation, is transferred into another container, without disturbing the settled particles. After the tealeaves have settled down, the clear tea (liquor) from the top can be poured into a cup. This transfer of the clear tea is called decantation.

LAWS OF CHEMICAL COMBINATIONFrench chemist A. Lavoisier laid the foundation to the scientific investigation of matter by describing that substances react by following certain laws. These laws are called the laws of chemical combination. These later on formed the basis of Dalton's Atomic Theory of Matter.

Law of Conservation of Mass

According to this law, during any physical or chemical change, the total mass of the products remains equal to the total mass of the reactants.


7

Law of conservation of mass is also known as 'Law of indestructibility of matter'.

Example 1 : If 10 grams of CaCO3 on heating gave 4.4g of CO2 and 5.6g of CaO, show that these observations are in

agreement with the law of conservation of mass.

Solution : Mass of the reactants = 10gMass of the products = 4.4 + 5.6g = 10gSince the mass of the reactants is equal

to the mass of the products, the observations are in agreement with the law of conservation of mass.

Law of Constant Proportion

Another French chemist, Joseph Proust, stated this law as 'A chemical compound always contains same elements combined together in the same proportion by mass'.

For example, pure water obtained from different sources such as river, well, spring, sea, etc., always contains hydrogen and

oxygen together in the ratio of 1: 8 by mass. Similarly, carbon dioxide CO2 can be obtained by different methods such as,

Burning of carbon

Heating of lime stone

By action of dilute HCl on marble pieces

But the different samples of CO2 always contain carbon and oxygen in the ratio of 3: 8.

Example 2 : When 1.375g of cupric oxide was reduced on heating in a current of hydrogen, the weight of copper that

remained was 1.098g. In another experiment 1.179g of copper was dissolved in nitric acid and resulting copper nitrate converted into cupric oxide by ignition. The weight of cupric oxide formed was 1.476g. Show that these results illustrate the law of constant proportion.

Solution:

First experiment

Copper oxide = 1.375g

Copper left = 1.098g

Oxygen present = 1.375 - 1.098 = 0.277g

Second Experiment

Copper taken = 1.179g

Copper oxide formed = 1.476g

Oxygen present = 1.476 - 1.179 = 0.297g

Percentage of oxygen is approximately the same in both the above cases. So the law of constant composition is illustrated.

Law of Multiple Proportions


8

John Dalton (1803) stated this law as 'when two elements combine with each other to form two or more than two compounds, the mass of the element which combine with the fixed mass of the other bears a simple whole number ratio'.

Carbon monoxide (CO) : 12 parts by mass of carbon combines with 16 parts by mass of oxygen.

Carbon dioxide (CO2) : 12 parts by mass of carbon combines with 32 parts by mass of oxygen.

Ratio of the masses of oxygen that combines with a fixed mass of carbon (12 parts) 16: 32 or 1: 2

Examples 3:

Hydrogen and oxygen are known to form 2 compounds. The hydrogen content in one is 5.93% while in the other it is 11.2%. Show that this data illustrates the law of multiple proportions.

Solution:

In the first compound hydrogen = 5.93%

Oxygen = (100 – 5.93) = 94.07%

In the second compound hydrogen = 11.2%

Oxygen = (100 – 11.2) = 88.88%

Ratio of the masses of oxygen that combine with fixed mass of hydrogen 15.86: 7.9 or 2:1

The ratio illustrates the law of multiple proportions.

Atoms and Atomic Theory

Atom was derived from the Greek word 'atoms', meaning 'indivisible'. Thus, it was considered to be indivisible.

Dalton's Atomic Theory

John Dalton provided a simple theory of matter to provide theoretical justification to the laws of chemical combinations in 1805. The basic postulates of the theory are :

All substances are made up of tiny, indivisible particles called atoms.

Atoms of the same element are identical in shape, size, mass and other properties.

Each element is composed of its own kind of atoms. Atoms of different elements are different in all respects.

Atom is the smallest unit that takes part in chemical combinations.

Atoms combine with each other in simple whole number ratios to form compound atoms called molecules.

Atoms cannot be created, divided or destroyed during any chemical or physical change.

Nature of Atom


9

The atom is the smallest particle of an element, which may or may not have independent existence. It is made up of sub-atomic particles like electrons, protons and neutrons. Atoms of one type of element differ from those of the other due to different number of sub-atomic particles.

Representation of an Atom by a Symbol

Symbols of Common Metals

Sl. No Name in English Symbol

1. Lithium Li

2. Sodium Na

3. Magnesium Mg

4. Aluminium Al

5. Potassium K

6. Calcium Ca

7. Vanadium V

8. Chromium Cr

9. Manganese Mn

10. Iron Fe

11. Cobalt Co

12. Nickel Ni

13. Copper Cu

14. Zinc Zn


10

Sl. No Name in English Symbol

15. Gallium Ga

16. Strontium Sr

17. Molybdenum Mo

18. Silver Ag

19. Cadmium Cd

20. Tin Sn

21. Antimony Sb

22. Barium Ba

23. Tungsten W

24. Platinum Pt

25. Mercury Hg

26. Lead Pb

27. Bismuth Bi

28. Polonium Po

29. Radium Ra

30. Uranium U

Symbols of Chemically Active Non-metals

Sl. No Non - metal Physical State Symbol

1. Hydrogen Gas H

2. Nitrogen Gas N

3. Oxygen Gas O

4. Fluorine Gas F

5. Chlorine Gas Cl

6. Bromine Liquid Br


11

Sl. No Non - metal Physical State Symbol

7. Astatine Solid At

8. Carbon Solid C

9. Iodine Solid I

10. Sulphur Solid S

11. Phosphorus Solid P

12. Silicon Solid Si

Symbols of Chemically Inactive Non-metals or Noble Gases

Sl. No Nobel Gas Symbol

1. Helium He

2. Neon Ne

3. Argon Ar

4. Krypton Kr

5. Xenon Xe

6. Radon Rn

Atomic Mass Unit

The mass of 1/12th 12C isotope of carbon is now the standard used for defining one atomic mass unit (a.m.u). It is equal to

1.66 x 10-24 g.

Atomic Mass of an Element or Relative Atomic Mass

The atomic mass of an element is a relative quantity and is the mass of one atom of the element relative to 1/12 the

mass of one carbon-12 atom.

Thus,Relative atomic mass (RAM) of an element is the number of times one atom of an element is heavier than 1/12 the mass of an atom of carbon [C-12].

For example, if 1 atom of Na weighs as much as 23 parts of 1/12 of 12C isotopes, then the atomic mass of sodium is 23 a.m.u.

Remember:


12

Atomic masses are not expressed as whole numbers because natural elements are mixtures of two or more isotopes. The

atomic weight of chlorine is 35.43 amu. Chlorine exists as two isotopes in the ratio 3:1

Therefore average atomic masses are not whole numbers.

Gram Atomic Mass (GAM)

The atomic mass of an element expressed in grams is called the gram atomic mass of an element. It is also called gram atom.

Example: The gram atoms present in 46 grams of sodium

Gram atomic mass is the relative atomic mass of an element expressed in grams.

Molecule FormationAtoms of most of the elements do not exist independently. They either form molecules or ions.

An atom having a charge on it is called an ion. When an atom loses or gains charge, it gets converted into a positively or negatively charged ion respectively. Ions may consist of single charged atoms or a group of atoms that have a net charge on them.

Molecules are formed by combination of atoms or ions by chemically bonding to each other or held by attractive forces due to the charges.

A molecule is smallest particle of an element or compound, which is capable of independent existence.

If the molecule is made up of atoms of the same element it is said to be homo-atomic. Examples: O2, N2

If the molecule is made up of atoms of the different elements it is said to be hetero-atomic. Examples: CO2, H2O, SO3, HCl


13

Depending upon the number of atoms in one molecule of the element it can be classified as:

The number of atoms, which constitute one molecule of an element, is called its atomicity.

Phosphorus: P4 - 4 atomicity

Sulphur: S8 - 8 atomicity

Oxygen: O2 - 2 atomicity

Recently a different form of carbon - C60 has been discovered. It is named Buckministerfullerene and has an atomicity of sixty.

A molecule of a compound consists of two or more atoms of different elements joined together in a fixed ratio.

Examples: CuSO4 contains Cu - 1 atom, S - 1 atom, O - 4 atoms

CHEMICAL AND MOLECULAR FORMULA

Every chemical substance is known by a specific name. But these names can sometimes be cumbersome, confusing and not provide sufficient information about its chemical composition. To overcome this, each chemical compound is represented by a chemical formula that gives its composition (constituent elements present) and the number of elements of each type present. A formula will represent only the simple ratio of positive and negative ions.

Molecular formula

A chemical formula that represents the composition of a molecule of the substance in terms of the symbols of the elements present in the molecule is also called molecular formula. Discrete molecules can be described by this formula. As it represents


14

one molecule of the substance giving the names and number of atoms of the various elements present, it denotes the molecular mass of the substance.

For example, the molecular formula of water is H2O, which means that one molecule of water contains two atoms of hydrogen and one atom of oxygen. This also represents the molar mass, which is the sum of the gram atomic mass of all the atoms.

Gram atomic mass of 2 hydrogen atoms = 2 x 1.008 g

Gram atomic mass of oxygen atom = 16 g

Total molecular mass = 18 g

Molecular Formulae of Important Acids

AcidMolecular

formulaAcid

Molecular

formula

Hydrochloric acid HCL Acetic acid CH3COOH

Nitric acid HNO3 Carbonic acid H2CO3

sulphuric acid H2SO4 Sulphurous acid H2SO3

phosphoric acid H3PO4 Nitrous acid HNO2

Molecular Formulae of Important Bases

BaseMolecular

formulaBase

Molecular

formula

Ammonium hydroxide NH4OH Aluminium hydroxide Al(OH)3

Sodium hydroxide NaOH Zinc hydroxide Zn(OH)2

Potassium hydroxide KOH Iron(II) hydroxide Fe(OH)2

Calcium hydroxide Ca(OH)2 Iron(II) hydroxide Fe(OH)2

Magnesium hydroxide Mg(OH)2 Copper(II) hydroxide Cu(OH)2

Molecular Formulae of Important salts - Oxides

oxideMolecular

formulaOxide

Molecular

formula

Sodium oxide Na2O Iron(II) oxide FeO

Potassium oxide K2O Iron(II)oxide Fe2O3

Calcium oxide CaO Lead(II)oxide PbO

Magnesium oxide MgO Lead(4)oxide PbO2


15

oxideMolecular

formulaOxide

Molecular

formula

Aluminum oxide Al2O3 copper(II)oxide CuO

Zinc oxide ZnO Mercury(II)oxide HgO

Molecular Formulae of Important salts - Carbonates and Bicarbonates

Carbonate or Hydrogen carbonate Molecular Formula Carbonate or Hydrogen Carbonate Molecular Formula

Sodium carbonate Na2CO3 Magnesium hydrogen carbonate Mg(HCO3)3

Sodium hydrogen carbonate NaHCO3 Aluminum carbonate Al2(C03)3

Potassium carbonate K2C03 Zinc carbonate ZnCO3

Potassium hydrogen carbonate KHCO3 Iron (II) carbonate FeCO3

Calcium carbonate CaCO3 Lead (II) carbonate PbCO3

Calcium hydrogen carbonate Ca(HCO3)2 Tin (II) carbonate SnCO3

Magnesium carbonate MgCO3 Copper (II) carbonate CuCO3

Molecular MassThe molecular mass of a substance is the relative mass of its molecule as compared with the mass of a 12C atom taken as 12-

units. It indicates the number of times; one molecule of the substance is heavier than atom.

Calculation of Molecular Mass

Molecular mass is equal to sum of the atomic masses of all atoms present in one molecule of the substance.

Example: H2O

Mass of H atom = 1

Mass of 2H atoms = 2

Mass of O atom = 16

Molecular mass = 2 + 16 = 18g

Relative Molecular Mass

Relative molecular mass or RMM is the molecular weight of an element or a compound. It is the number of times one molecule

of the substance is heavier than 1/12 the mass of an atom of carbon (12C).


16

Gram Molecular Mass

Molecular mass expressed in grams is numerically equal to gram molecular mass of the substance.

Molecular mass of O2 = 32Gram molecular mass of O2 = 32g

Gram molecular mass is the relative molecular mass expressed in grams.

MOLE CONCEPT

Since it is not possible to calculate the weight of particles individually, a collection of such particles called mole is taken for all

practical purposes. It was discovered that the number of atoms present in 12g of carbon of 12C isotope is 6.023 x 1023 atoms.

This is referred to as Avogadro number after the discoverer Avogadro.

A mole of a gas is the amount of a substance containing 6.023 x 1023 particles. It is a basic unit of the amount or quantity of a

substance. The substance may be atoms, molecules, ions or group of ions.

Avogadro discovered that under standard conditions of temperature and pressure, (1 atm and 273 K) a sample of gas occupies a volume of 22.4 L.


17

Molar Volume

One mole of any gas at STP will have a volume of 22.4 L. This is called molar volume.

The molar volume [22.4 L at STP] plays a vital role in stoichiometric calculations because it is the link between volume and mass in reactions involving gases.

Relationship Between Gram Molecular Weight and Gram Molecular Volume

Gram molecular weight (GMW) or mole is the relative molecular mass of a substance expressed in grams. It is also called gram molecular weight of that element.

Gram molecular volume (GMV) or molar volume is the volume occupied by one-gram molecular weight of a gas at STP (Standard temperature and pressure).

All gases of equal volumes contain same number of molecules under the same conditions of temperature and pressure. Equal number of molecules of different gases will occupy equal volumes under the same conditions of temperature and pressure.

One mole of gas = 6.023 x 1023 molecules


18

1 mole of a gas = 22.4 L at STP

Important Relationships

1 mole of an atom = 1 gram atomic weight of an atom

1 mole of a molecule = 1 gram molecular weight of molecule

1 mole of a gas = 22.4 liters of gas at STP

1 mole of a substance = 6.023 x 1023, atoms, molecules or ions

1 molar volume = 22.4 dm3 /L at STP

Example: 1

Calculate the volume occupied by 2.8 g of N2 at STP.

Solution:

Molecular weight of N2 = 2 x 14 = 28 g

28 g of N2 at STP occupies = 22.4 L

2.8 g of N2 at STP = ?

2.8 g of N2 at STP occupies a volume of 2.24 L.

Example: 2

Calculate gram molecular weights of the following gases:

a. N2 (if 360 cm3 at STP weighs 0.45g)

b. Cl2 (if 308 cm3 at STP weighs 0.97g)

Solution:

Back to Top

a. 360 cm3 of N2 = 0.45g

22.4L of gas = 1 gram molecular weight

22.4L = 22,400 cm3. (1L = 1000 cm3)

360 cm3 of N2 = 0.45g

22,400 cm3 of N2 = ? Gram molecular weight of N2 is 28 g.b. 308 cm3 Cl2 = 0.979 g22.400

cm3 of Cl2 = ?


19

Molecular weight of Cl2 = 71.9 g

Example: 3

What is the volume of 32 g of sulphur dioxide measured at STP?

Solution:

Molecular formula = SO2

Molecular weight = 1 x 32 + 2 x 16 = 64 g

64g of SO2 occupies 22.4 L

32 g of SO2 = ?

Volume of 32 g of SO2 is 11.2 litres.

Example: 4

Back to Top

Calculate the volume at S.T.P. of 7.1g of chlorine.

Solution:

Cl = 35.51

mole of a substance = 22.4 L

1 Mole of a substance = 1 GMM

1 GMM of Cl2 = 71 g

71 g of Cl2 = 22.4 L

7.1 g of Cl2 = ?

7.1 g of Cl2 will occupy a volume of 2.24 litres.

Example: 5

Calculate the number of moles of nitrogen in 7g of nitrogen.

Solution:

1 mole of N2 = 1 GMM

1 mole of N2 = 2 x 14g = 28g

1 mole = 28 g

? = 7 g

7g of nitrogen is equal to 0.25 moles.

Example: 6


20

Calculate the mass of 0.4 moles of water.

Solution:

1 GMM of water (H2O) = 2 x 1 + 16 = 18 g.18 g = 1 moleXg = 0.4 moles 0.4 moles of water

weigh 7.2 g.

Example: 7

Back to Top

Calculate the gram atoms present in 8g of oxygen.

Solution:

Back to Top

Example: 8

Back to Top

Calculate the gram molecules present in 45 g of water.

Solution:

Back to Top

1 molecule = 2 + 16g = 18g of H2O? = 45 g of H2O

Example: 9

Back to Top

Calculate the number of molecules in 500g of sodium chloride.

Solution:

Back to Top

1 GMM = 6.023 x 1023 molecules23 + 35.5 g = 58.5g = 1 GMM of NaCl58.5g = 6.023 x 1023molecules500g = ?

The number of molecules in 500g of sodium chloride= 34.2 x 6.023 x1023 molecules

Example: 10


21

Back to Top

About 0.48 g of a gas forms 100 cm3 of vapours at STP. Calculate the gram molecular weight of the gas.

Solution:

Back to Top

22.4L of a gas = 1 GMM100 cm3 of gas = 0.48 g22.4 x 1000 of gas = ? Gram

molecular weight of the gas is 107.52 g.

Numericals Based on Percentage Composition

Back to Top

All elements are represented by symbols and all compounds represented by chemical formulae indicating the number of atoms of elements and also the proportion of the atoms in the compound.Example: Hydrogen atom is represented as H.Hydrogen molecule is represented as H2.A compound of hydrogen, water is represented as H2O.In H2O - proportion of atoms H : O = 2 : 1

Knowing the proportion of atoms in a compound, the percentage composition can be calculated. Percentage composition of a compound is the percent by weight of each element present in it.

Example: 11

Back to Top

Calculate the percentage by weight of all the elements present in calcium carbonate.

Solution:

Back to Top

Calcium carbonate = CaCO3 Ca = 40, C = 12, O = 16GMM = 1 x 40 + 1 x 12 + 3 x 16= 40 + 12 + 48 = 100 g

Example: 12

Back to Top

Calculate the percentage by weight of potassium in potassium dichromate.


22

Solution:

Back to Top

Potassium dichromate = K2Cr2O7GMM = (2 x 39) + (2 x 52) + (7 x 16)= 78 + 104 + 112 g = 294 g

Empirical and Molecular Formula of a Compound

Back to Top

Numericals Based on Empirical Formula

Back to Top

Empirical formula is the formula of a compound, which shows the simplest whole number ratio between the atoms of the elements in the compound. It does not indicate the actual number of atoms of the elements present but the simplest whole number ratio.

Example:

Back to Top

Organic Compound Empirical Formula Molecular Formula

Benzene C1H1 C6H6

Glucose C1H2O1 C6H12O6

How can we differentiate a molecular formula from an empirical formula? If the subscripts in the formula have a common divisor, it is usually a molecular formula. Generally the empirical formula is multiplied by this common divisor to get the molecular formula.

Example:

Back to Top

Empirical formula of acetic acid is CH2OMolecular formula is CH3COOH = C2H4O2

C1H2O1 x 2 = C2H4O2 [Molecular formula]

Steps for Calculation

Calculate the percentage by weight of each element.

Find out relative number of atoms by dividing percentage of weight by atomic weight.

Choose the simplest and the smallest ratio; divide all the ratios by it

If whole numbers are not obtained, then multiply it by a smallest integer to make it whole.


23

Example: 13

Back to Top

An oxide of iron contains 72.41% of iron. Calculate the empirical formula for the oxide of iron [Fe = 56; O=16].

Solution:

Back to Top

Element % by wt At wt Relative No.of Atoms Simple Ratio

Fe 72.41 56 72.41/56 = 1.29 1.29/1.29 = 1 x 3 = 3

O 27.59 16 27.59/16 = 1.72 1.72/1.29 = 1.33 x 3 = 4

Therefore simple ratio = Fe3O4.Empirical formula = Fe3O4.

Determination of Molecular Formula from Empirical Formula

Back to Top

Molecular formula is the chemical formula, which represents the actual numbers of atoms of each element present in a compound.

Steps for Calculation

Back to Top

Calculate empirical formula

Use vapour density if given

If molecular weight given, calculate 'n' using this formula

Molecular formula = n x empirical formula

Example: 14

Back to Top

Calculate the molecular formula of a compound with vapour density of 30 having 40% carbon; 6.67% of hydrogen and the rest is oxygen.

Solution:

Back to Top

Element % by wt At wt Rlative No.of Atoms Simple Ratio

C 40 12 40/12 = 3.33 3.33/3.33 = 1


24

Element % by wt At wt Rlative No.of Atoms Simple Ratio

H 6.67 1 6.67/1 = 6.67 6.67/3.33 = 2

O 100 - 46.67 16 53.33/16 = 3.33 3.33/3.33 = 1

Empirical formula = C1H2O1Empirical formula weight = 12 x 1 + 2 x 1 + 1 x 16= 12 + 2 + 16= 30 g

Molecular weight = 2 x vapour density= 2 x 30= 60Molecular weight = n x empirical weight60 = n x 30 Molecular

formula = n x empirical formula = 2 x CH2O = C2H4O2

Example: 15

Back to Top

A compound has molecular formula C5H10., what is its empirical formula?

Solution:

Back to Top

Ratio of C atoms to H atoms is 5:10 = 1:2Empirical formula is C1H2

Atoms and Molecules - Test QuestionsQuestion 1

Back to Top

Question: What is a symbol? What information does it convey?

Answer: A short hand representation of an element is called symbol. It represents the following:

Name of the element

One atom of the element

One mole of atoms. It represents 6.023 x 1023 atoms of the element.

A definite mass of the element

Question 2

Back to Top

Question: What is the difference between symbol of an element and formula of an element?

Answer: Symbol of an element represents the name of the element. It also represents one atom of the element. Example: H represents hydrogen and C represents carbon A formula of an element represents the number of atoms in the molecule of the compound. One molecule of hydrogen element contains two atoms of hydrogen; therefore the formula of hydrogen is H2. 2H represents two separate atoms of hydrogen, whereas H2 represents 1 molecule of hydrogen similarly the molecular formula of oxygen element and chlorine element is O2 and Cl2.

Question 3


25

Back to Top

Question: Define atomic mass of an element.

Answer: The atomic mass of an element is the relative mass of its atom as compared to the mass of C - 12 atom taken as 12 units.

Question 4

Back to Top

Question: State the laws of chemical combination.

Answer: Laws of chemical combinations are:

i) Law of Conservation of Mass: The law states that during any physical or chemical change, the total mass of the product remains equal to the total mass of the reactants.

ii) Law of Constant Composition: The law states that a chemical compound always contains same elements combined together in the same proportion by mass.

iii) Law of Multiple Proportions: The law states that when two elements combine with each other to form two or more compounds, the masses of one of the elements, which combine with fixed mass of the other, bear a simple whole number ratio to one another.

Question 5

Back to Top

Question: In an experiment it was found that litharge, red oxide of lead and lead peroxide contained 92.83%, 90.6% and 86.6% of lead respectively. Show that these figures are in agreement with the Law of Multiple Proportions.

Answer: In litharge, the amount of lead = 92.83%

The amount of oxygen = 100 - 92.83 = 7.17% 7.17g of O2 combines with 92.83g of lead

In red oxide of lead, the amount of Pb = 90.6%

The amount of oxygen = 100 - 90.6 = 9.4 %

9.4 g of oxygen combines with 90.6g of lead

In lead peroxide, the amount of lead = 86.6%

The amount of oxygen = 100 - 86.6 = 13.4%

13.4g of oxygen combines with 86.6g of lead


26

Thus, the different weights of lead combining with fixed weight (1g) of oxygen are in the ratio 12.947: 9.638: 6.462 = 2:1.5: 1 or 4: 3: 2

This is in line with the law of multiple proportions.

Question 6

Back to Top

Question: What do you understand by atomicity of an element? Give an example of a polyatomic molecule.

Answer: The number of atoms that constitute one molecule of an element is called its atomicity. Sulphur is a polyatomic molecule: S8 - 8 is the atomicity.

Question 7

Back to Top

Question: What do the following stand for? (i) P4 and 4P (ii) O2 and 2O

Answer:

P4 - 1 molecule of phosphorus 4P - 4 atoms of phosphorus

O2 - 1 molecule of oxygen 2O - 2 atoms of oxygen

Question 8

Back to Top

Question: How many gram atoms are present in 69 grams of sodium?

Answer:

The number of gram atoms present in 69 grams of sodium is 3.

Question 9

Back to Top

Question: The mass of a single atom of an element Z is 2.65x10-23g. What is its gram atomic mass?

Answer: 1 atom of element Z has mass = 2.65 x 10-23g

Hence, 6.023 x 1023 atoms of element Z have mass

= 2.65 x 10-23 x 6.023 x 1023

= 15.69 g


27

Question 10

Back to Top

Question: What is gram molecular mass?

Answer: The amount of a substance whose mass in grams is numerically equal to its molecular mass is called gram molecular mass.

Question 11

Back to Top

Question: Calculate the molar mass of HNO3. [N = 14, O = 16, H = 1]

Answer: Molar mass of HNO3.

H = 1 x 1 = 01

N = 14 x 1 = 14

O = 16 x 3 = 48

Total mass = 63 grams

Molar mass of HNO3= 63 grams

Question 12

Back to Top

Question: Calculate the formula mass of CaCl2. [Ca = 40, Cl = 35.5]

Answer: 1(Ca) + 2(Cl) 40 + 2x(35.5) = 111 amu

The formula mass of CaCl2 is 111 amu.

Question 13

Back to Top

Question: A certain non-metal X forms two oxides I and II. The mass percentage of oxygen in oxide I (X4O6) is 43.7, which is same as that of X in oxide II. Find the formula of the second oxide.

Answer:

Now 43.7 parts of oxygen in I corresponds to = 6 oxygen atoms

Also 56.3 parts of X in I correspond to = 4 X atom


28

Now the atomic ratio X : O in the second

The formula of the second oxide is X2O5.

Question 14

Back to Top

Question: (i) Calculate the mass of 0.2 moles of water (O=16, H=1).

(ii) What is the volume of 7.1 g of chlorine (Cl=35.5) at S.T.P.

Answer: (i) Gram Molecular Weight of H2O = 2 x 1 + 16 = 18 g

1 mole of water weighs 18 g

(ii) Gram Molecular Weight of Cl2 (one mole)= 35.5 x 2 = 71 g.

71 g of Cl2 at S.T.P occupies 22.4 litres

Question 15

Back to Top

Question: The reaction between aluminium carbide and water takes place according to the following equation:

Calculate the volume of CH4 released from 14.4 g of Al4C3 by excess water at S.T.P. (C = 12, Al = 27)

Answer: Molecular weight of Al4C3 is (27 x 4) + (12 x 3) = 144

144 g of Al4C3 produces 3 x 22.4 litres of CH4 at S.T.P

Question 16


29

Back to Top

Question: A compound of sodium, sulphur and oxygen has the following percentage composition. Na=29.11%, S=40.51%, O=30.38%. Find its empirical formula (O=16, Na=23, S=32).

Answer:

Empirical formula is NaSO1.5 or to its nearest whole number i.e., the formula is Na2S2O3.

Question 17

Back to Top

Question: Solid ammonium dichromate with relative molecular mass of 252 g decomposes according to the equation.

(i) What volume of nitrogen at S.T.P will be evolved when 63 g of (NH4)2Cr2O7 is decomposed?

(ii) If 63 g of (NH4)2Cr2O7 is heated above 1000C, what will be the loss of mass? (H=1, N=14, O=16, Cr=52).

Answer: 252 g of (NH4)2Cr2O7 gives one mole or 22.4 litres of N2 at S.T.P as per the given equation.

(ii) At temperatures above 1000C water is in the form of steam.

Products as vapours are N2 and H2O.

The transformation of solids and liquids into gaseous substances results in loss of mass.

Total weight of gaseous products = {(2 x 14) + 4 (2 x 1) + 16}

= 28 + 72 = 100 g

Heating 252 g of (NH4)2Cr2O7 causes 100 g loss of mass.

The loss of mass is 25 g

Question 18


30

Back to Top

Question: How many litres of ammonia are present in 3.4 kg of it? (N = 14, H = 1)

Answer: Gram molecular weight of NH3 = 14 + (1 x 3) = 17 g.

17 g of NH3 = 22.4 litres

=

= 4480 litres.

Question 19

Back to Top

Question: About 640 mL of carbon monoxide is mixed with 800 mL of oxygen and ignited in an enclosed vessel. Calculate the total volume of gases after the burning is completed. All volumes are measured at S.T.P.

Answer: The chemical reaction actually taking place is:

Volume of O2 used = 320 mL.

Volume of O2 left = 800 - 320 = 480 mL.

Volume of CO2 formed = 640 mL.

Therefore the total volume of gases after burning is 480 + 640 = 1120 mL.

Question 20

Back to Top

Question: Calculate the number of moles of ammonium sulphate present in 15.84 kg of it. (H=1, N=14, O=16, S=32)

Answer: Molecular weight of (NH4)2SO4

= (2 x 14) + (2 x 4) + 32 + (16 x 4)

= 132 a.m.u.

132 g of (NH4)2SO4 = 1 mole


31

The number of moles of ammonium sulphate present in 15.84 kg is 120 moles.

Question 21

Back to Top

Question: What is the mass of 0.2 mole of lead nitrate? (N=14, O=16, Pb=207).

Answer: Gram molecular weight of Pb(NO3)2 = 207 + (2 x 14) + 2(16 x 3)

= 207 + 28 + 96

= 331

1 mole of Pb(NO3)2 is 331 g

Therefore 0.2 mole of Pb(NO3)2 is 331 x 0.2 = 66.2 g

Question 22

Back to Top

Question: Find the total percentage of oxygen in magnesium nitrate crystals i.e., Mg(NO3)2.6H2O (Atomic weight: H=1, N=14, O=16, Mg=24).

Answer: Molecular weight of Mg(NO3)2.6H2O

= 24 + 2(14 + 16 x 3) + 6(2 x 1 + 16)

= 24 + 124 + 108 = 256 a.m.u

Atomic mass of oxygen in Mg(NO3)2.6H2O is,

= 2 (16 x 3) + 6 (16) = 96 + 96 = 192

Question 23

Back to Top

Question: A compound has the following percentage composition H=2.04%, S=32.65%, O=65.31%. Relative molecular mass of the compound = 98. Calculate its molecular formula (H = 1, S = 32, O = 16).


32

Answer:

Empirical formula is H2SO4

Empirical formula mass = (2 x 1) + 32 + (16 x 4) = 98.

Relative molecular mass = 98

Question 24

Back to Top

Question: Calculate the amount of nitrogen supplied to soil by 1 quintal (100 kg) of ammonium nitrate (N=14, H=1, O=16).

Answer: Molecular weight of NH4NO3 = 14 + (4 x 1) + 14 + (16 x 3)

= 80 g

Molecular weight of N in the above formula = 14 x 2 = 28

80 units of NH4NO3 yield 28 units of Nitrogen.

Question 25

Back to Top

Question: Identify diatomic molecules from the following: (i) HCl (ii) P4 (iii) He (iv) O3 (v) H2S (vi) CO

Answer: HCl, CO are diatomic.

Question: In water, hydrogen and oxygen are present in the ratio of ________.

1. 1:8

2. 2:12

3. 2:3

4. 1:2

Answer: 1


33

Question 2

Back to Top

Question: An example of a triatomic molecule is ___________.

1. Ozone

2. Nitrogen

3. Carbon monoxide

4. Hydrogen

Answer: 1

Question 3

Back to Top

Question: The quantity of matter present in an object is called its _________.

1. Mass

2. Volume

3. Density

4. Vapour pressure

Answer: 1

Question 4

Back to Top

Question: Indivisibility of an atom was proposed by ___________.

1. Dalton

2. Rutherford

3. Thomson

4. Bohr

Answer: 1

Question 5

Back to Top

Question: The value of Avogadro constant is ___________.

1.

6.022 x 1024


34

2.

6.022 x 1022

3.

60.22 x 1023

4.

6.022 x 1023

Answer: 4

Question 6

Back to Top

Question: All samples of carbon dioxide contain carbon and oxygen in the mass ratio of 3:8. This is in agreement with the Law of ___________.

1.

Conservation of Mass

2.

Constant Proportion

3.

Multiple Proportion

4.

Reciprocal Proportion

Answer: 2

Question 7

Back to Top

Question: The atomic mass of sodium is 23. The number of moles in 46g of sodium is ________.

1.

2

2.

1

3.

4

4.

0

Answer: 1

Question 8

Back to Top

Question: The molecular formula of potassium nitrate is ________.


35

1. KNO

2. KNO3

3. KNO2

4. KON

Answer: 2

Question 9

Back to Top

Question: Kalium is the Latin name of ________.

1.

Potassium

2.

Krypton

3.

Calcium

4.

Phosphorus

Answer: 1

Question 10

Back to Top

Question: The smallest particle of a substance that is capable of independent existence is _________.

1.

Atom

2.

Molecule

3.

Electron

4.

Proton

Answer: 2

Question 11

Back to Top


36

Question:

1.

Atomic number

2.

Atomic mass

3.

Atomic mass scale

4.

Number of electrons

Answer: 2

Question 12

Back to Top

Question:

1. 1.25

2. 15

3. 14

4. 12

Answer: 2

Question 13

Back to Top

Question: The molecular mass of ammonia is _________.

1. 17 grams

2. 31 grams

3. 20 grams

4. 25 grams

Answer: 1

Question 14

Back to Top

Question: Atomic mass of calcium is 40. The mass of 2.5 gram atoms of calcium is _______.

1. 40 g


37

2. 2.5 g

3. 100 g

4. 80 g

Answer: 3

Question 15

Back to Top

Question: The number of atoms in a molecule of the elementary substance is called ________.

1.

Atomic number

2.

Avogadro number

3.

Atomic mass

4.

Atomicity

Answer: 4

Question 16

Back to Top

Question: Avogadro number represents the number of atoms in ________.

1.

12 grams of 12C

2.

320 grams of sulphur

3.

32 grams of oxygen

4.

1 gram of 12C

Answer: 1

Question 17

Back to Top

Question: The number of moles in 5 grams of calcium is _________.


38

1. 0.125 mole

2. 5 mole

3. 1.25 mole

4. 12.5 moles

Answer: 1

Question 18

Back to Top

Question: 2 molecules of nitrogen are represented by ________.

1. N

2. 2N2

3. N2

4. 2N

Answer: 2

Question 19

Back to Top

Question: One mole of H2O corresponds to

1.

22.4 litre at 1 atm and 250C

2.

6.02 x 1023 atoms of hydrogen and 6.02 x 1023 atoms of oxygen

3.

18 g

4.

1 g

Answer: 3

Question 20

Back to Top

Question: The number of molecules in 4.25 g of ammonia is approximately

1. 1.0 x 1023


39

2. 1.5 x 1023

3. 2.0 x 1023

4. 3.5 x 1023

Answer: 2

Question 21

Back to Top

Question: A sample contains 22 g of carbon dioxide. This is equal to

1. One molar volume of carbon dioxide

2. One mole of carbon dioxide

3. Half mole of carbon dioxide

4. Two moles of carbon dioxide

Answer: 3


matter in our surroundings and it.docx

Documents

pure water

simpler substances

different ratio

mixtures impure substances

different people

different places

different methods

different samples of