Basic concept of chemistry/unit-1

Basic Concept of Chemistry:

Matter: – Matter is anything that occupies some space, posses mass, and offer resistance to any stress applied on it. Different kind matter is made up of different substance. Matter may exists in three physical state namely solid, liquid and gases.

1) Solid State: – In solid, the particles are closely packed and bound by strong interparticle attraction. This makes solids rigid and geometrical. A solid has definite shape and possess definite volume. E.g. wood, table, copper rod, common salt, etc.

2) Liquid State: – In liquid state, the particles are loosely packed and are bound to each other with forces weaker than those in solids. This makes liquid mobile and shapeless. A liquid has definite volume but not definite shape. A liquid takes the shape of the container in which it is placed. e.g. water, milk, oil, etc.

3) Gaseous State: – In gases, the particles are separated from each other by much greater distance, almost 10 to 100 times the size of the particles. Therefore, there is virtually no force of attraction between the particles. In fact, in gases the operating forces are very weak. As a result, the particles in gases are so loosely packed that they force to move in any direction independently. This makes gases shapeless and highly compressible i.e. a gas neither possesses a definite shape nor a definite volume. It occupies the whole of the volume of the vessel in which it is placed. e.g. oxygen, hydrogen, carbon dioxide, etc.

On the basis of chemical composition matter is classified into- element, compound and mixture.

Element:

An element is the simplest form of matter. “A pure substance which can neither be decomposed into nor built from simpler substances by ordinary physical or chemical methods”. In other wards, elements are made up of same kind of atoms.

The common examples of elements are hydrogen, oxygen, nitrogen, sulphur, iron, lead, gold, mercury, etc. There are about 118 elements known at present. Out of these, 92 have so far been found to occur naturally and the remaining has been prepared in the laboratory. All the elements do not occur in the crust of earth in equal proportions. About twenty elements make up 99% of the earth’s crust. The most abundant elements in the earth’s crust are oxygen, silicon, aluminium, iron, calcium, sodium, potassium, hydrogen, chlorine, carbon, etc.

Elements are further classified into following types:

1) Metals: – Metals are electropositive elements, generally solids and possess characteristics such as bright lustre, hardness and ability to conduct electricity and heat. Metals are generally malleable (can be beaten into thin sheets) and ductile (can be drawn into wires). Some common examples of metals are copper, iron, silver, gold, aluminium, etc. About 80% of the known elements are metals.

2) Non-metals:- Non-metals are electronegative elements, generally non lustrous, brittle and poor conductors of heat and electricity. The common examples of non-metals are carbon, hydrogen, oxygen, nitrogen, etc.

3) Metalloids: – These are elements which have characteristics common to both metals and non-metals. The common examples of metalloids are silicon, arsenic, bismuth, antimony, etc.

Compound:

A compound is a substance which is formed by the union of two or more elements in a definite proportion by mass and into which it may be decomposed by suitable chemical methods.

A compound always contains the same elements in fixed proportion by mass. For example, water always contains hydrogen and oxygen in the ratio of 1: 8 by mass. Similarly, carbon dioxide always contains carbon and oxygen in the ratio of 3: 8 by mass. The properties of the compounds are totally different from the elements from which these are formed. For example hydrogen is combustible while oxygen is supporter of combustion but water is normally used for extinguishing fire.

The compounds may be classified into two types-

1) Inorganic compounds: – These are the compounds which are obtained from non-living sources such as rocks, minerals, etc. For example, common salt, marble, washing soda, etc.

2) Organic compounds: – These are the compounds which are obtained from living sources such as plants and animals. All, these contain carbon. For example, carbohydrates, oils, fats, waxes, proteins, etc.

Mixture:

A mixture is a substance which is formed by the combination of two or more elements or compounds in any ratio so that the components do not lose their identity”. For example, air is a mixture of several gases. The components of the mixture can be separated by ordinary methods.

Mixtures are of two types- homogeneous and heterogeneous

1) Homogeneous mixtures: – These have the same composition throughout the sample. The components of a mixture cannot be seen even under a powerful microscope. Homogeneous mixtures are also called solutions. Some examples of homogeneous mixtures are air, gasoline, sea water, brass, etc. Air is a homogeneous mixture of a number of gases such as oxygen, nitrogen, carbon dioxide, water vapour, etc.

2) Heterogeneous mixtures: – These consist of two or more parts (called phases) which have different compositions. These have visible boundaries of separation between the different constituents and can be easily seen even with naked eye. For example, mixture of iron filings, common salt and sulphur gives a heterogeneous mixture in which the components can be seen lying side by side very easily.

Atom:

An atom is the smallest particle of an element which can take part in a chemical reaction and it may or may not have independent existence. E.g. hydrogen, nitrogen, oxygen, neon, argon, helium, etc.

Molecules:

Molecules are the smallest particle of matter (either element or compound) which is capable of independent existence. Thus a molecule contains two or more atoms. Molecules can be classified as-

1) Homoatomic Molecules: – These molecules are made up of the atoms of same element. e.g. hydrogen (H2), nitrogen (N2), oxygen (O2), etc.

2) Heteroatomic Molecules: – Heteroatomic molecules are made up of atoms of different elements. They are also classified as di, tri …etc. depending upon the number of atoms present. E.g. HCl, CO2, NH3, PCl5, etc.

Symbol:

A symbol is an abbreviation or shorthand notation for the full name of an element. For example, Boron (B), carbon (C), Iodine (I), Silver (Ag), etc.

Formula:

The representation of a molecule of a substance (element or compound) in terms of symbol and subscript number is known as formula or symbolic representation of compound. For examples, H2 is the formula of hydrogen, HCl is the molecular formula of hydrogen chloride, etc.

Valency:

According to the modern concept, valency is the number of electrons lost, gain or shared with one atom of an element in order to acquire the stable electronic configuration of nearest inert gas element. In other wards, it is a combining capacity of an element. For example valency of H =1, O =-2, SO4-2=-2 etc.

Ions:

An ion is an atom or groups of atom having either positive or negative charge and behave as a single unit in chemical change. Ions are two types-

  • Electropositive ions: – The ions having positive charge on it is known as an electropositive ions. It is also called cations or basic radical. e.g. K+, Ag+, Pb+2, NH4+, etc.
  • Electronegative Ions: – The ions having negative charge on it is known as an electronegative ions. It is also called anion or Acid radical. e.g. Cl, CO3-2, S-2, SO4-2, etc.

Chemical Equation:

A chemical equation is a symbolic representation of a chemical change. For example, when AgNO3 is reacts with NaCl to give AgCl and NaNO3 may be represented as-

AgNO3 + NaCl ——–> AgCl + NaNO3

    Reactants                      Products

The substances which react among themselves to bring about the chemical changes are known as reactants, whereas the substances which are produced as a result of chemical change are known as products.

A chemical equation must fulfill the following conditions-

  •  It must be consistant with experimental facts.
  • It must be balanced i.e. the total no. of atoms on both sides of the equation must be equal.
  • It must be molecular i.e. the elementary gases like hydrogen, oxygen, nitrogen etc must be represented in molecular form a H2, O2, N2 etc.

Limitations of chemical equation:

The chemical equations as such do not give us the following information:

  • The physical and chemical states of the reactant and products.
  • The concentration of reactants and products.
  • The reaction condition such as temperature, pressure or catalyst etc. which affects the reaction.
  • The heat changes accompanying the reaction.
  • The speed of the reaction.
  • The nature of the reaction i.e. wheather the chemical reaction is reversible or irreversible.
  • The formation of precipitated (ppt) or the evolution of gas in the reaction.

Laws Of Chemical Combination:

On the basis of various quantitative experiments on chemical changes, Lavosier and other scientists made certain generalization known as Law of chemical combination. These are-

  • Law of conservation of mass
  • Law of constant or definite proportion
  • Law of multiple proportion.
  • Law of reciprocal proportion
  • Law of gaseous volume

1) Law of conservation of mass:

It states that the total mass of matter in any chemical or physical change remain constant, though the matter may change its form.

Or

In other wards, it may also be state that in any chemical reaction the total mass of the reactants and products before and after the reaction is the same.

Or

Matter is neither created nor destroyed as a result of any physical or chemical change. Therefore,

A  +  B —–>  C + D

Let a gram of A and b gram of B react to give c gram of C and d gram of D, then

Total mass of reactants = Total mass of products

(a + b)  =  (c + d)

Landolt’s experiments:

This law can be verified with the help of Landolt’s experiment. Landolt took the solutions of sodium chloride (NaCI) and silver nitrate (AgNO3) separately in two limbs of a ‘H’ shaped tube (known as Landolt’s tube). The tube was sealed and weighed. After weighing, the two solutions were mixed thoroughly by shaking the tube. As a result, the reaction occurred between silver nitrate and sodium chloride and white ppt. of silver chloride is formed as:                                                                              AgNO3 + NaCl   →   AgCl + NaNO3                                                                                                                                   White ppt.

After the reaction, the tube was again weighed. It was observed that the weight remained practically unchanged. This verified the law of conservation of mass.

2) Law of constant or definite proportion:-

The law of constant composition deals with the composition of various elements present in a compound. It was stated by a French chemist, Louis Proust, in 1799. Law of constant proportion states that “a chemical compound always contains same elements combined together in the same definite proportion by weight irrespective of method of preparation or the source from where it is taken”. For example, carbon dioxide can be obtained by a number of methods, such as

  • By burning coal or candle,

C + O2 —–> CO2

  • By heating limestone,

CaCO3 ——-> CaO  +  CO2

  • By the action of dilute hydrochloric acid on marble pieces.

CaCO3 + HCl ——-> CaCl2  +  CO2  + H2O

It has been observed that each sample of carbon dioxide contains carbon and oxygen elements in the ratio of 3: 8 by weight.

Similarly, pure water can be obtained from many sources. Irrespective of the source, water always contains hydrogen and oxygen elements combined together in the ratio of 1: 8 by weight.

3) Law of multiple proportion:-

This law was proposed by Dalton in 1803. It states that –“when two elements combine to form two or more than two compounds, then the weights of one of the elements which combine with a fixed weight of the other, bear a simple multiple ratio”. For example, when nitrogen and oxygen combine to form five oxides namely nitrous oxide (N20), nitric oxide (NO), nitrogen trioxide (N2O3), nitrogen tetraoxide (N204) and nitrogen pentaoxide (N205). The different weights of oxygen which combine with the fixed weight of nitrogen in all these oxides are calculated.

Name of Compound           Mass ratio of nitrogen: Oxygen

  • Nitrous oxide (N20),                              2 x 14 : 1 x16
  • Nitric oxide (NO),                                  2 x 14 : 2 x 16
  • Nitrogen trioxide (N2O3),                      2 x 14 : 3 x 16
  • Nitrogen tetraoxide (N204)                   2 x 14 : 4 x 16
  • Nitrogen pentaoxide (N205)                 2 x 14 : 5 x 16

The ratio between the different weights of oxygen in different compounds which combine with the fixed weight of nitrogen (28) is 1: 2: 3: 4: 5. This is a simple multiple ratios and thus, verified the law.

4) Law of reciprocal proportion:-

This law was put forward by Richter in 1792. It states that-“When two different elements combine separately with the same weight of a third element, the ratio in which they do so will be the same or some simple multiple of the ratio in which they combine with each other”. In other words, the weight ratio of two elements A and B which combine with the fixed weight of C separately, is either the same or some simple whole number multiple of the weight ratio in which A and B combine together. This law may be illustrated by the following examples:

Consider three elements sulphur, oxygen and hydrogen. Both sulphur and oxygen separately combine to form hydrogen sulphide (H2S) and water H2O) respectively. They also combine with each other to form sulphur dioxide (SO2) as shown in below.

According to the law, the ratio of weights of S and O which combine with the same weight of H will either be same or a simple multiple of the ratio in which S and O combine with each other. This may be verified as:

In hydrogen sulphide (H2S), 2 parts by weight of hydrogen combine with 32 parts by weight of sulphur.

In water molecule (H2O), 2 parts by weight of hydrogen combine with 16 parts by weight of oxygen.
Therefore, the ratio of the weights of sulphur and oxygen which, combine separately with the fixed weight (2 parts) of hydrogen is, 32:16 or 2:1———— (i)

Now, let us calculate the ratio of sulphur and oxygen in SO2.

In sulphur dioxide (SO2),

Sulphur and oxygen combine in the ratio of 32:32 or 1:1 ——— (ii)

The two ratios (i) and (ii) are related to each other as-

2/1: 1/1 or 2:1

Which are simple multiple of each other.

5) Law of gaseous volume:-

According to this law, “under constant temperature and pressure the volumes of the gaseous reactants bear a simple ratio between themselves and also with the volumes of products, if gaseous.” For examples, in the reaction-

H2 (g)  +    Cl2 (g)  ———>  2HCl (g)

1-volume    1-volime               2-volume

Dalton’s Atomic Theory:

To provide theoretical justification to the laws of chemical combination, John Dalton postulated a simple theory of matter. The basic postulates of the theory are given below:

  • Matter is made up of extremely small indivisible and indestructible ultimate particles called atoms.
  • Atoms of the same element are identical in all respects i.e. in shape, size, mass and chemical properties.
  • Atoms of different elements are different in all respects and have different masses and chemical properties.
  • Atom is the smallest particle that takes part in chemical combinations.
  • Atoms of two or more elements combine in a simple whole number ratio to form compound atoms (now-a-days called molecules).
  • Atoms can neither be created nor destroyed during any physical or chemical change.
  • Chemical reactions involve only combination, separation or rearrangement of atoms.

 Limitations of Dalton’s Atomic Theory:

Dalton’s atomic theory explains the various laws of chemical combination successfully, but this theory could not explain the following facts:

  • It could explain the laws of chemical combination by mass but failed to explain the law of gaseous volumes.
  • It could not explain why atoms of different elements have different masses, sizes, valencies, etc.
  • It failed to explain the cause of chemical combination.
  • It gave no satisfactory explanation between the ultimate particle of an element and that of a compound.
  • It could not give any idea of isotopes and isobar.

Atomic Mass:

“The atomic mass of an element expresses as a number which shows how many times an atom of the element is heavier than one atom of hydrogen, taking the mass of an atom of hydrogen as unity”.

On oxygen scale, the atomic mass of an element is defined as follows:

The atomic mass of an element is the mass of an atom of the element as compared with the mass of one atom of oxygen taken as 16. Alternatively, the atomic mass of an element is a number which represents how many times an atom of the element is heavier than 1/16th part of the mass of an atom of oxygen.Therefore,

Atomic Mass = Mass of 1-atom of an element / 1/16th part of the mass of one atom of oxygen

Atomic Mass =(Mass of 1-atom of an element /Mass of one-atom of Oxygen) x 16

According to modern concept atomic mass may be defined as the relative atomic mass of one atom of an element with respect to the mass of one carbon atom taken as 12.

Atomic Mass = Mass of 1-atom of an element /1/12th part of the mass of one atom of carbon-12.

Gram Atomic Mass:

A quantity in grams which is numerically equal to the atomic mass of an element is called gram atomic mass of the particular element. In short, gram atomic mass of an element is its atomic mass expressed in grams.

Thus, 16 g of oxygen stands for 1-gram atomic of oxygen.

1-gram atom of hydrogen = 1.008 g of hydrogen

1-gram atom of carbon     = 12 g of carbon

Molecular Mass:

The molecule of an element is composed of atoms of the same element and the molecule of a compound is formed by the union of atoms of different elements. So from the practical point of view, the molecular mass may be regarded simply as the sum of atomic mass of all the atoms present in the molecules. As for example, hydrogen, nitrogen, oxygen, ammonia, sulphuric acid, cane sugar, sodium hydroxide, etc.

Molecular mass of hydrogen = 1.008 x 2 = 2.016.

Molecular mass of oxygen = 16 x 2 = 32

Molecular mass of nitrogen = 14 x 2 = 28

Molecular mass of ammonia = 14 + (1.004 x 3) = 17

Molecular mass of sulphuric acid = (1.008 x 2) + 32 + (16 x 4) = 98

Molecular mass of sodium hydroxide = 23 + 16 +1.008 = 40

Molecular mass of cane sugar = 12 x 12 + 1.008 x 22 + 16 x 11 = 342.176.

Or

The molecular mass of a substance (either an element or a compound) may be defined as the mass of a molecule of the substance compared with the mass of an atom of hydrogen taken as 1.008 or oxygen as 16 or carbon as 12. Therefore,

Molecular Mass = Mass of one molecule of a substance /Mass of one atom of hydrogen

Gram Molecular Mass:

The molecular mass of any substance express in grams is referred to as the gram molecular mass or frequently as gram-molecule of the substance. Thus,

1 Gram-molecule of CO2 = 44 gram of CO2

1 Gram-molecule of H2O = 18 gram of H2O

1 Gram-molecule of H2SO4 = 98 gram of H2SO4

Mole Concept:

Quite commonly, we use different units for counting such as dozen for 12 articles, score for 20 articles and gross for 144 articles, irrespective of their nature. For example, one dozen books means 12 books whereas, one dozen apples means 12 apples. In a similar way, chemists use the unit mole for counting atoms, molecules, ions, etc. A mole is a collection of 6.022 x 10 particles. Thus, a mole represents 6.022 x 1023 particles.

The number 6.022 x 1023 is called Avogadro number and is symbolised as N. In other words, a mole is an Avogadro number of particles.

Thus, ‘a mole is the amount of substance that contains the same number of entities (atoms, molecules, ions or other particles), as the number of atoms present in 12 g (or 0.012 kg) of the carbon-12 isotope’. For example,

1 mole of hydrogen atoms = 6.022 x 1023 hydrogen atoms.
1 mole of hydrogen molecules = 6.022 x 1023 hydrogen molecules.
1 mole of sodium ions = 6.022 x 1023 sodium ions.
1 mole of electrons = 6.022 x 1023 electrons.

Percentage Composition:

The percentage composition of an element in a compound is the number of parts by mass of it present in 100 parts by mass of the compound. Therefore,

Mass percentage of element = (Mass of element / Total mass of the compound) x 100

The percentage composition of a compound may be determined from the molecular formula of the compound as follows:

  • Write down the molecular formula of the compound correctly.
  • Calculate the mass of each element present in the molecule of the substance by multiplying the atomic mass of that element by the number of atoms in the molecule.
  • Calculate the molecular mass of the compound.
  • Calculate the mass P.C. of each element by using the relationship,

Mass percentage of element = (Mass of element / Total mass of the compound) x 100

Problem 1. Determine the p.c. of potassium, nitrogen and oxygen in potassium nitrate. Given K=39, N=14, O=16.

Empirical And Molecular Formula:-

The chemical formula may be of two types:

  •             Empirical formula
  •             Molecular formula

Empirical formula:

The formula which gives the simple whole number ratio of the atoms of various elements present in one molecule of the compound is called empirical formula. For example, empirical formula of hydrogen peroxide is HO. It represents that hydrogen and oxygen (i.e., H: O) are present in the ratio of 1: 1 in hydrogen peroxide. Similarly, empirical formula of benzene is CH which indicates that atomic ratio of C: H in benzene is 1: 1.

Molecular formula:

The formula which gives the actual number of atoms of various elements present in one molecule of the compound is called molecular formula. For example, the molecular formula of hydrogen peroxide is H202, because one molecule of hydrogen peroxide contains two atoms of hydrogen and two atoms of oxygen. Similarly, molecular formula of benzene is C6H6 which tells that one molecule of benzene contains 6 atoms of carbon and 6 atoms of hydrogen.

Relation between Empirical and Molecular Formulae: – Molecular formula and empirical formula are related as-

Molecular formula = n x (Empirical formula)

Where n is a simple whole number and may have values 1, 2, 3… It is equal to

n = (Molecular mass) /Empirical formula mass

For example, the molecular mass of benzene is 78. The empirical formula of benzene is CH and therefore, its empirical formula mass is 13. Thus,

n = (Molecular mass) /Empirical formula mass

n = 78 /13 = 6

Therefore, molecular formula of benzene = 6(CH) = C6H6

Note:- It may be noted that in many cases, the value of n comes out to be one and, therefore, empirical formula and molecular formula are same in these cases.

Determination of the Empirical Formula of a Compound:

The empirical formula of a compound can be determined from the percentage composition of different elements and atomic masses of the elements.

The various steps involved in determining the empirical formula are:

Step-1: Divide the percentage of each element by its atomic mass. This gives the moles of atoms of various elements in the molecule of the compound.

Step-2: Divide the result obtained in the above step by the smallest value among them to get the simplest ratio of various atoms.

Step-3: Make the values obtained above to the nearest whole number and multiply, if necessary, by a suitable integer to make the values whole numbers. This gives the simplest whole number ratio.

Step-4: Write the symbols of the various elements side by side and insert the numerical value at the right hand lower corner of each symbol. The formula thus obtained represents the empirical formula of the compound.

Determination of the Molecular Formula of a Compound:

Step-1: Determine the empirical formula as described above.

Step-2: Calculate the empirical formula mass by adding the atomic masses of the atoms in the empirical formula.

Step-3: Determine the molecular mass by a suitable method.

Step-4: Determine the value of n as

Molecular mass = n x Empirical formula mass

Change n to the nearest whole number.

Step-5: Multiply empirical formula by n to get the molecular formula.

Molecular formula = n x (Empirical formula).

Question.1: Determine the empirical formula of compound whose p.c. are given as- carbon 10%, hydrogen 0.80% and chlorine 89.2% respectively.                                                       Ans. CHCl3

Question 2: The molecular mass of an organic compound is 78 and its p.c. is carbon 92.4%, and hydrogen 7.6%. Determine the molecular formula of the compound.                        Ans. C6H6

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