Chemistry is defined as that branch of science which deals with the study of composition, structure and properties of matter. It also deals with the relation between changes in composition and changes in energy. Study of chemistry is very interesting which covers various aspects of our culture and environment. In ancient India, chemistry was called Rasayan Shastra, Rastantra, Ras Kriya or Rasvidya. It included metallurgy, medicine, manufacture of cosmetics, glass, dyes, etc. Due to the immense vastness of the subject, chemistry has been divided into a number of branches. Important branches of chemistry are:

1.      Inorganic Chemistry : Under this title we study all the elements and their compounds. Compounds of carbon are not dealt with in this branch of chemistry.

2.      Organic Chemistry : The study of carbon compounds is done in this branch of chemistry. Carbon oxides, carbonates, etc., are studied in inorganic chemistry because they have mineral origin.

3.      Physical Chemistry : The laws and principles of chemical reactions are studied in this branch of chemistry.

4.      Analytical Chemistry : This branch emphasizes the techniques that are used to find out the composition of matter.

Matter: Matter is defined as any thing that occupies space, possesses mass. Matter exists mainly in solid, liquid and gaseous state.

a) Solid: Particles are held very closely together in solids in an orderly fashion and there is not much freedom of movement. They have definite volume and shape.

b) Liquid: In liquids particles are close to one another but they can move around. Liquids have definite volume but do not have definite shape.

c) Gas: In gases, the particles are far apart as compared to those present in solid or liquid states and their movement is easy and fast. Gases have neither definite volume nor definite shape.

Classification of Substances :  A substance is a form of matter which has mass and possesses volume. It resists a change in itself. Substances have been classified as:

(i)   Heterogeneous Substances : Substances having two or more parts of different kinds are called heterogeneous substances. For instance, a mixture of ice and water, oil and water, milk, gun power, soil, smoke, etc.

(ii)  Homogeneous Substances : The substances which are uniform throughout, i.e., whose every part is of the same type, are called homogeneous substances. For example, mixture of gases, true solutions, all elements, etc., are homogenous substances. Homogeneous substances are again of two types:

(a)   Pure substances : Substances which have definite and constant chemical composition are known as pure substances. For example, all elements and compounds are pure substances.(b)   Solutions : A homogenous mixture of two or more pure substances is known as a solution. For example, air, a mixture of NaCl and water, alcohol and water, etc. A solution does not have a definite composition.

Element : An element is a pure substance made up of same kind of atoms. Examples are Na, K, Au, Cu, Hg, H, O, N, S, etc. There are about 109 elements known till today.

Compounds : They are formed by the chemical combination of atoms of different elements in a definite ratio. Elements lose their properties in the formation of during compound formation from their elements. An exothermic compound is one in whose formation energy is released in the form of heat; when energy is absorbed, the compound is called endothermic compound. Components of a compound can not be separated by simple methods. For example, NH₃, H₂O, CH₄, HaI, KCl, C₂H₅OH, etc. are compounds.           

Mixture : It is a material composed by mixing two or more pure substances (elements or compounds) in any proportion. It may be homogeneous or heterogeneous. The components of mixture may undergo temporary change in their physical properties such as physical state, colour etc. There may or may not be enthalpy change in the formation of a mixture. A mixture gives the properties of all of its components.

Units for Measurement

All physical quantities have to be measured. The value of a physical quantity is expressed as the product of the numerical value and the unit in which it is expressed.

Fundamental Units:

Fundamental units are those units which can neither be derived from one another nor they can be further resolved into any other units. 

The seven fundamental units of measurement in S.I. system.

	Name of unit	Abbreviation
Mass	Kilogram	Kg
Length	Meter	m
Temperature	Kelvin	K
Amount of substance	Mole	Mol
Time	Second	S
Electric current	Ampere	A
Luminous intensity	Candela	Cd

  Derived unit: Some quantities are expressed as a function of more than one fundamental units known as derived units. For example velocity, acceleration, work, energy etc.

Quantity with Symbol	Unit (S.I.)	Symbol
Velocity (v)	Metre per sec	ms–1
Area (A)	Square metre	m2­
Volume (V)	Cubic metre	m3
Density (r)	Kilogram m–3	Kg m–3
Energy (E)	Joule (J)	Kg m2s–2
Force (F)	Newton (N)	Kg ms–2
Frequency (n)	Hertz	Cycle per sec
Pressure (P)	Pascal (Pa)	Nm–2
Electrical charge	Coulomb (C)	A-s (ampere – second)

 

Units and Dimensional Analysis: Conversion of Units The simplest way to carry out calculations that involve different units is to use dimensional analysis. In this method a quantity expressed in one unit is converted into an equivalent quantity with a different unit by using conversion factor which express the relationship between units:

Original quantity x conversion factor = equivalent quantity

(in former unit)                                               (in other unit)

This is based on the fact that ratio of each fundamental quantity in one unit with their equivalent quantity in other unit is equal to one. For example in case of mass   So 1 kg = 2.205 pond = 1000 gm In this way any derived unit first expressed in dimension and each fundamental quantities like mass length time are converted in other system of desired unit to work out the conversion factor. For example: How unit of work / energy i.e. joule, in S.I. system is related with unit erg in C.G.S system

Dimension of work = force x displacement = MLT^-2 x L = ML²T^-2

                                                                              1 joule = 1 kg (1 metre)² x (1sec)^-2

\(\displaystyle 1\times Kg\times \frac{{1000\,\,gm}}{{1\,Kg}}\times {{\left[ {1\,\,metre\times \frac{{100cm}}{{1\,metre}}} \right]}^{2}}\times {{\left[ {1\,\,\sec } \right]}^{{-2}}}\)

            ⇒100 gm x (100)²x 1 em²x (1 sec)^-2

            ⇒ 1000×10000 x 1 gmx1 cm² x 1 sec^-1

            ⇒ 1 joule = 10⁷ erg

Similarly we can deduce other conversion factor for other quantity in different unit by the dimensional analysis method

Another interesting example is the conversion of litre – atmosphere to joule (the SI unit of energy) by multiplying with two successive unit factors. Thus,

\(\displaystyle 1\,\,L\,\,atm\,\,\times \left( {\frac{{{{{10}}^{{-3}}}\,\,{{m}^{3}}}}{{1\,\,L}}} \right)\times \frac{{101.325\,\,Pa}}{{1\,\,atm}}=101.325\,\,Pa\,\,{{m}^{3}}\)

Knowing that Pa = N/m², we can write

101.325 pa m³ = 101.325 (N/m²)m³ =101.325 Nm = 101.325 J

hence 1L atm = 101.325 J

 

Question.         What is the mass of 1 L of mercury in grams and in kilograms if the density of liquid mercury is 13.6 g/cm³?

Solution:            We know the relationship, 1L = 1000cm³ and also

                                 density = mass/volume

                                 We can write, mass = (volume)x (density)

                                 Therefore, the mass of 1 L of mercury is equal to \(\displaystyle (1\,\,L)\,\,\left( {\frac{{1000\,\,c{{m}^{3}}}}{{1\,\,L}}} \right)\,\,(13.6\,\,g\,\,c{{m}^{{-3}}})=(1000\,\,c{{m}^{3}})\,\,\,(13.6\,\,g\,\,c{{m}^{{-3}}})=1.36\times {{10}^{4}}\,\,g\)

                               The mass in kilograms can be calculated as \(\displaystyle 1.36\times {{10}^{4}}\,\,g=(1.36\times {{10}^{4}}\,\cancel{g})\,\left( {\frac{{1\,\,kg}}{{1000\,\,\cancel{g}}}} \right)=13.6\,\,kg\)

                              (Remember, 1000 cm³/1 L and 1Kg/1000 gm are conversion factors with which we have to multiply for getting our                                                  answer in appropriate units).

DALTON’S ATOMIC THEORY

By observing the laws of chemical combination discussed above, John Dalton (1808) proposed atomic theory of matter. The main points of Dalton’s atomic theory are as follows:

• Matter is made up of extremely small, indivisible particles called atoms.
• Atom of same substance are identical in all respect i.e. they posses same size, shape, mass, chemical properties etc.
• Atom of different substances are different in all respect i.e. they posses different shape, size, mass and chemical properties etc.
• Atom is the smallest particle that takes part in chemical reactions.
• Atom of different elements may combine with each other in a fixed, simple, whole number ratio to form compound atoms.
• Atom can neither be created nor destroyed i.e. atoms are indestructible.

 

Limitations

The main failures of Dalton’s atomic theory are:

• Atom was no more indivisible. It is made up of various sub-atomic particles like electrons, proton and neutron etc.
• It failed to explain how atoms of different elements differ from each other.
• It failed to explain how and why atoms of elements combine with each other to form compound atoms or molecules.
• It failed to explain the nature of forces that bind together different atoms in molecules.
• It failed to explain Gay Lussac’s law of combining volumes.
• It did not make any distinction between ultimate particle of an element that takes part in reaction (atoms) and the ultimate particle that has independent existence (molecules).

Modern Atomic theory or Modified Atomic Theory

• Atom is no longer supposed to be indivisible. Atom has a complex structure and is composed of sub-atomic particles such as electrons protons and neutrons.
• Atom of the same element may not be similar in all respects e.g. isotopes.
• Atom of different elements may be similar in one or more respects e.g. isobars.
• Atom is the smallest unit which takes part in chemical reactions.
• The ratio in which atoms unite may be fixed and integral but may not be simple. e.g. In sugar molecules, the ratio of C, H and O atoms is 12:22:11 which is not simple.
• Atom of one element can be changed into atoms of other element for e.g. transmutation.
• Mass of atom can be changed in energy.

            according to Eienstein mass energy relationship, mass and energy are inter-convertible. Thus atom is no longer indestructible.

Question.    An important postulate of Dalton’s atomic theory is:

                        (A) an atom contains electrons, protons and neutrons

                        (B) atom can neither be created nor destroyed nor divisible

                        (C) all the atoms of an element are not identical

                        (D) all the elements are available in nature in the form of atoms

Solution:         The statement written in (B) is about law of mass conservation which is true for all chemical reaction.

                        Hence (B) is correct.

Atom: Atom is the ultimate electrically neutral, made up of fundamental particle (Electron, neutron, Proton) which shows the characteristic properties of the element and exist freely in a chemical reaction.

Molecules: A molecule is the smallest particle made up of one or more than one atom in a definite ratio having stable and independent existence e.g.  etc. 

LAWS OF CHEMICAL COMBINATION

1.      Law of Conservation of Mass: This law was given by Russian Scientist Lomonosov in 1756. According to this law, “The total mass in chemical reactions remains unchanged.”

         Thus,

Total mass of reactants = Total mass of products

(Before reaction)              (After the chemical reaction)

For example, 12 gram carbon combines with 32 gram oxygen to give 44 gram carbon dioxide. Chemical equations are based on law of conservation of mass.

*      Chemical combination is a must for the validity of this law.

*      This law was experimentally verified by Lavoisier and Landolt.

2.      Law of Constant Proportion: This law was proposed by French Scientist J.L. Proust in 1797. This law states, “In a compound, elements of definite kind are combined in a definite ratio of their weights.”

                  It is clear from this law that the chemical composition of every compound is definite and it does not depend upon the source or method of its preparation. For instance, in pure water, hydrogen and oxygen are always combined together in the ratio 1 : 8 by weight, whether it is obtained from a river, well, spring or sea.

3.      Law of Multiple Proportions: This law was given by John Dalton  in 1803. According to this law, “when 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, are in a simple whole number ratio.”

                  For instance, carbon and oxygen react with each other to form carbon monoxide and carbon dioxide.

In carbon monoxide (CO): 12 parts by weight of carbon react with 16 parts by weight of oxygen.

In carbon dioxide (CO₂): 12 parts by weight of carbon combine with 32 parts by weight of oxygen. Ratio of the weights of oxygen which react with a fixed weight of carbon (12 parts) in these oxides is 16 : 32 which is a simple whole number (1 : 2) ratio. Other examples of law of multiple proportions: (i) H₂O and H₂O₂, (ii) CuO and Cu₂O, (iii) N₂O, NO, N₂O₃, N₂O₄  and N₂O₅ verify the law of multiple proportions.

4.      Law of Reciprocal Proportions or Equivalent Proportions: This law was put forward by German Scientist J.B. Richter in 1792. According to this law, “when two elements combine separately with a fixed weight of a third element, the ratio of their weight in which they do so is either the same or a whole number multiple of the ratio in which they react together.” Let us consider an example.

                  Carbon and sulphur react separately with oxygen (third element) to give carbon dioxide (CO₂) and sulphur dioxide (SO₂). They also react together to form carbon disulphide (CS₂).

Now, in carbon dioxide (CO₂) : 12 parts by weight of carbon are combined with 32 parts by weight of oxygen.

In sulphur dioxide (SO₂): 32 parts by weight of sulphur are combined with 32 parts by weight of oxygen.

Ratio of the weights of carbon and sulphur which combine with fixed weight (32 parts) of oxygen is 12 : 32.

i.e.,            3 : 8

                  In carbon disulphide (CS₂) : 12 parts by weight of carbon are combined with 64 parts by weight of sulphur. So, ratio of weight of carbon and sulphur in which they react together is 12 : 64.

or               3 : 16

Now, the ratios (i) and (ii) are related to each other as 3 : 8 and 3 : 16 or 3 : 3 and 8 : 16 or 1 : 2.

It indicates that ratio (i) and (ii) are whole number multiples of each other. Thus, the ratio of weights of ‘C’ and ‘S’ (two elements) which reacts with a fixed amount of oxygen (third element) is a whole number multiple of the ratio in which carbon and sulphur combine with each other. It verifies the law of reciprocal proportions.

5.      Gay-Lussac’s law of combining volumes: The law was proposed by J.F. Gay-Lussac in 1808. This law states, “whenever gases react together, the volumes of the reacting gases as well as the products, if they are gases, bear a simple whole number ratio provided all the volumes are measured under similar conditions of temperature and pressure.”

 Let us consider for instance, the combination of hydrogen and chlorine gases. It has been experimentally observed that:

One volume of hydrogen reacts with one volume of chlorine to give two volume of chlorine to give two volumes of hydrogen chloride (gas).

Hence, the volume ratio of H₂ : Cl₂ : HCl is 1 : 1 : 2. This ratio is in agreement with their molar ratios when they are involved in the reaction.

Thus, Gay-Lussac’s law stands verified.

ATOM AND ATOMIC WEIGHT : The smallest particle of an element which can not exist in free from is called an atom. Atoms have a tendency to exist as clusters (group of atoms or molecules) of same or different atoms. They take part in a chemical reaction without undergoing any decomposition. An atom is very small (atomic size of the order 10^–6 cm) and its real mass is very low. The smallest and lightest atom is of hydrogen. The mass of the hydrogen atom is 1.008 a.m.u. Weights of atoms of other elements are in the range of 2 to 260 a.m.u.

Atomic weight : The atomic weight of an element is a number which indicates how many times heavier is an average atom of that element as compared with 1/12 of the mass of an atom of carbon –12 (C¹²).

Thus, the atomic weights of elements are average relative weights.

Most often the atomic weight of elements are fractional. It is due to the existence of isotopes of the elements which are atoms with different masses. Since an element is a mixture of all of its isotopes, the atomic weight of an element is the average of weights of all its isotopes. So, atomic weight have non –integral (fractional) values.

Unit of atomic weight: The modern unit of atomic weight is 1/12 of the mass of one atom of C–12. The representation of unit of atomic weight is a.m.u.

1 a.m.u. =(1/12)x mass of C — 12 atom.

The real mass of one atom of C–12 has been determined as 1.9924 × 10^–23 gram. Gram-atomic weight : When atomic weights of elements are expressed in grams, they are called gram atomic weights. Thus, the weight of an element in grams which is equal to its atomic weight is known as the gram atomic weight (GAW) of that element. For example, atomic weight of sodium is 23 and its gram atomic weight is 23 grams. Similarly, the gram atomic weights of hydrogen, sulphur, nitrogen are 1.008 gram, 32 gram and 14 gram respectively.

                  Thus,         1 GAW P = 31.0 gram phosphorus,

                                       1 GAW K = 39.0 gram potassium,

                                       1 GAW Ca = 40.0 gram calcium.

Gram atom or mole atom: The atomic weight of an element taken in grams is one gram-atom or mole atom of that element. For instance, one gram atom of oxygen represents 16 gram oxygen. One gram atom of sulphur represents 32 gram sulphur.

Number of gram atoms of an element = (Wt. of the element in grams)/( gram atomic et. of the element)

One gram atomic wt. of every element contains 6.23 x 10²³ atoms of that element.

 

Examples:

Question        Approximately how many atoms are present in 10.8 gram of silver (At. Wt. 108).

Solution          Gram atomic weight of silver is 108 gm.                        

                            ⇒ 108 gram of silver contains 6.023 x 10²³ atoms of silver

                              ⇒ 1 gram of silver contains = (6.023 x 10²³)/108 atoms

                              ⇒ 10.8 gram of silver contains = [(6.023 x 10²³)/108] x 10.8 = 6.023 × 10²² atoms.

Question        If the density of silver is 10.8 gm cm^–3, what is the volume of one atom of silver ?

Solution          As calculated in the above example, number of atoms in 10.88 gm of silver = 6.023×10²².                        

                            Since volume of 10.8 gm of silver is 1 cm³,

                             ⇒Volume of 6.023 x 10²² atoms = 1 cm³

                            ⇒Volume of 1 atom = (1/6.023 x 10²²) cm³ = 1.66 × 10^–23 cm³.

Question        Find the number atoms present in (a) 4gm He                                          (b) 4 a.m.u He                                                                                                                                                                        (c) 8 gm 0-atom                                 (d) 16 a.m.u oxygen atom.

Solution          a) G.A.W of He = 4 gm = 1 mole = 6.022 x 10²³ He- atom.                          

                           b) Atomic mass of He = 4 a.m.u = 1 – He-atom

                           c) Gram Atomic mass of 0-atom = 16 gm = 1 mole = 6.022 x 10²³ atom = 8 gm =  mol = 3.011 × 10²³ atom

                          d) Atomic mass of oxygen         = 16 a.m.u = 1 – oxygen atom.

Question        If mass of one –atom of x is 3.24 x 10^–22 gm  then find atomic mass of atom x.

Solution          G.A.W of x means mass of 6.022 x 10²³ atoms

                        Mass of 1 – atom of x = 3.24 x 10^–22 gm

                        Mass of  6.022 x 10²³ atoms of x = 3.24 x 10^–22 x 6.022 x 10²³ =  195.1

Practice Questions

1. Find the number of atoms in followings :

(a) 2.3 gm Na                   (b) 20 gm Ca                      (c)   28 gm of –N         (d) 39 a.m.u of K

2. Find the number of mole in followings :

(a) 108 gm Ag                 (b)   23 gm Na                    (c)   4 a.m.u He            (d)  71 gm Cl-atom

3. Mass of 100 atoms of y is 3.11 x 10^–21 gm then find atomic mass of y.
   
   
4. The dot at the end of this sentence has a mass of about one microgram. Assuming that the black stuff is carbon, calculate the approximate number of atoms of carbon, calculate the approximate number of atoms of carbon needed to make such a dot. (1 microgram = 1 x 10^–6 g) (5x 10¹⁶ atoms).
   
   
5. How many atoms are in 100 amu of He?

PERCENTAGE YIELD Reactants, often yield quantities of products that are less than those calculated from the balanced chemical equation.

Reason behind such discrepancy may be:

(i)   Some of the reactants fail to undergo reaction due to lower amount at the end.

(ii)  Some of the reactant follows other route of reaction resulting unexpected, undesired side product.

(iii) Some of the expected product recombines to form other undesired product or reverting toward reactant also known as backward reaction.

(iv) The total recovery, isolation of products is not possible due to some unavoidable reason. So percentage yield is the ratio of actual yield (recovered) to theoretical yield multiplied by 100.

\(\displaystyle \%\,\,yield=\frac{{actual\,\,yield}}{{theoretical\,\,yield}}\times 100\)    

Determination of atomic mass

(i)   Applying Dulong and Petit’s law.

(ii)  Cannizzaro’s methods

(iii) By mitscherlich’s law of isomorphism.

(iv) By measurement of V.D. of volatile chloride or bromide.

(i)   Dulong & Petits Law: The product of specific heat of pure element and atomic mass of the element is equal to 6.4.

      i.e. Atomic mass x specific heat = 6.4 (approx)

      But this law is not applicable to lighter element like boron, carbon, silicon. To obtain correct atomic mass of element first of all equivalent mass of the element is known by any other method and their atomic mass = eq. weight x valency

      In which valency has whole number value which can be deduced by dividing approximate by equivalent mass.

            Dulongs and Petit’s Law: Atomic Mass x Specific Heat =6.4

Question.     The specific heat of a metal of atomic mass 32 is likely to be:

                        (A) 0.25                                                (B) 0.24

                        (C) 0.20                                                (D) 0.15

Solution:                     Specific heat  = (6.4/atomic mass) = 6.4/32 = 0.2

Question.         On dissolving 6 gm of metal in sulphuric acid, 13.53g of the metal sulphate was formed. The specific heat of metal is. What is equivalent mass of metal, valency and exact atomic mass?

Solution:                     Equivalent mass of sulphate = ionic mass/valency = 96/2 = 48

                                         Eq. weight of metal

                                        \(\displaystyle =\frac{{Mass\,\,of\,\,metal}}{{Mass\,\,of\,\,sulphate}}\times 48=\frac{6}{{13.53-6}}\times 48=\frac{6}{{7.53}}\times 48=38.24\,g\)

                                        Approximate atomic mass = \(\displaystyle =\frac{{6.4}}{{Specific\stackrel{\scriptscriptstyle\to}{\leftarrow}\stackrel{\scriptscriptstyle\to}{\leftarrow}heat}}=\frac{{6.4}}{{0.057}}\times 112.5\)

                                      Valency \(\displaystyle =\frac{{Approximate\stackrel{\scriptscriptstyle\to}{\leftarrow}\stackrel{\scriptscriptstyle\to}{\leftarrow}atomic\stackrel{\scriptscriptstyle\to}{\leftarrow}\stackrel{\scriptscriptstyle\to}{\leftarrow}mass}}{{Equivalent\stackrel{\scriptscriptstyle\to}{\leftarrow}\stackrel{\scriptscriptstyle\to}{\leftarrow}mass}}=\frac{{112.5}}{{38.24}}=2.9\approx 3\)

                                        Exact atomic mass = 114.72

(ii)  Cannizzaro’s methods

If an element has several compound with other same or different elements of known atomic mass then the compound that has minimum presence of former element indicate the atomic mass of former element.

Procedure

(i)   First of all the molecular mass of all compounds known by applying

      V.D x 2 = mol. weight

(ii)  By analysis the presence of the desired element in each compound is known.

(iii) The mass that is lowest among all the compound indicate the atomic mass of that element.

Question.         Estimate the atomic mass of nitrogen given that vapour density of NH₃ = 8.5, Nitrous oxide = 22, Nitric oxide = 15,                                              Nitrogen peroxide = 23, Nitrogen trioxide = 38.

 

Solution:	Compound	V.D.	Molecular weight of compound	Nitrogen mass in compound	Least presence of nitrogen
	Ammonia	8.5	17	14	So atomic mass of nitrogen = 14
	Nitrous oxide	22	44	28
	Nitric oxide	15	30	14
	Nitrogen peroxide	23	46	14
	Nitrogen trioxide	38	76	28

 

Question.         Which pair of the following substances is said to be isomorphous?

                        (A) White vitriol and blue vitriol              (B) Epsom salt and Glauber’s salt

                        (C) Blue vitriol and Glauber’s salt           (D) White vitriol and Epsom salt

Solution:         Epsom salt (MgSO₄.7H₂O) and White vitriol (ZnSO₄.7H₂O) contains divalent cation Mg²⁺ and Zn²⁺ and same number of water molecules as water of crystallization which hold criteria for isomorphism.

                                                Hence (D) is correct.

(iv) Atomic mass from vapour density of a chloride:

The following steps are involved in this method

• Vapour density of chloride of the element is determined
• Equivalent mass of the element is determined

            Let the valency of the element be x. The formula of its chloride will be

                        Molecular weight of chloride = Atomic mass of M + 35.5 x

                        Atomic Mass = x × E

                        So Mw = x × E + 35.5x = 2V.D

                      \(\displaystyle x=\frac{{2VD}}{{E+35.5}}\)

                    ⇒ Atomic Wt = x × E

Question.   One gram of chloride was found to contain 0.835g of chlorine. Its vapour density is 85. Calculate its molecular formula.

Solution:         Mass of metal chloride = 1g

                        Mass of chlorine = 0.835

                        Mass of metal = 1 – 0.835 = 0.165g

                        E mass of metal\(\displaystyle =\frac{{0.165}}{{0.835}}\times 35.5=7.01\,g\)

                        Valency of metal\(\displaystyle =\frac{{2V.D.}}{{E+35.5}}=\frac{{2\times 85}}{{7.01+35.5}}=4\)

                        Formula of chloride \(\displaystyle \text{= MC}{{\text{l}}_{\text{4}}}\)

AVERAGE ATOMIC MASS

Elements are found in different isotopic forms (atoms of same elements having different atomic mass), so the atomic mass of any element is the average of all the isotopic mass within a given sample.

Average atomic mass = (∑% abundance x atomic mass)/100

Question.         Use the date given in the following table to calculate the molar mass of naturally occurring argon.

Isotope	Isotopic molar mass	Abundance
36Ar	35.96755 g mol–1	7.1%
38Ar	37.96272 g mol–1	16.3%
40Ar	39.9624 g mol–1	76.6%

Solution:         Molar mass of Ar

                        = 35.96755 x 0.071 + 37.96272 x 0.163 + 39.96924 x 0.766

                        = 39.352 g mol^–1

Question.     Carbon occurs in nature as a mixture of carbon 12 and carbon 13. The average atomic mass of carbon is 12.011. What is the percentage abundance of carbon
12 in nature?

Solution:                     Average atomic mass = = (∑% abundance x atomic mass)/100

                                        ⇒ \(\displaystyle 12.011=\frac{{x\times 12+(100-x)13}}{{100}}\)

                                        ⇒1201.1 = 12x + 1300 – 13x

                                          ⇒ x = 1300 – 1201.1 = 98.9%

EMPIRICAL FORMULA

It is the formula which expresses the smallest whole number ratio of the constituent atom within the molecule. Empirical formula of different compound may be same. So it may or may not represent the actual formula of the molecule. It can be deduced by knowing the weight % of all the constituent element with their atomic masses for the given compound.

For example: C₆H₁₂O₆, CH₃COOH, HCHO

All have same empirical formula CH₂O, but they are different.

The empirical formula of a compound can be determined by the following steps:

• Write the name of detected elements in column-1 present in the compound.
• Write the corresponding atomic mass in column-2.
• Write the experimentally determined percentage composition by weight of each element present in the compound in column-3.
• Divide the percentage of each element by its atomic weight to get the relative number of atoms of each element in column-4.
• Divide each number obtained for the respective elements in step (3) by the smallest number among those numbers so as to get the simplest ratio in column-5.
• If any number obtained in step (4) is not a whole number then multiply all the numbers by a suitable integer to get whole number ratio. This ratio will be the simplest ratio of the atoms of different elements present in the compound. Empirical formula of the compound can be written with the help of this ratio in column-6.

Question. A compound contains C = 71.23%, H = 12.95% and O = 15.81%. What is the empirical formula of the compound?

Question . A compound of carbon, hydrogen and nitrogen contains these elements in the ratio 9:1:3.5. Calculate the empirical formula. If its molecular mass is 108, what is the molecular formula?

Question.      2.38 gm of uranium was heated strongly in a current of air. The resulting oxide weighed 2.806 g. Determine the empirical formula of the oxide. (At. mass U = 238; O = 16).

MOLECULAR FORMULA

The formula which represents the actual number of each individual atom in any molecule is known as molecular formula.

 For certain compounds the molecular formula and the empirical formula may be same.

If the vapour density of the substance is known, its molecular weight can be calculated by using the equation.

Question .The empirical formula of a compound is. Its molecular weight is 90. Calculate the molecular formula of the compound. (Atomic weights C = 12, H = 1, O = 16)

CHEMICAL EQUATION AND STOICHIOMETRY

Chemical equation tell about what substances react and what substances are produced along with the form of aggregation of the substance formed and reacted by the use of following symbols.

According to general assumptions reactants on left hand side and products (although they are informed as per experimental facts) on right hand side in corporating on arrow in the middle to indicate the progress of reaction [is forward] and = in the middle if equation is balanced, are written.

If any physical parameters to procure the desired reaction is necessary they are indicated in the middle along the arrow or equal sign like temperature, pressure, catalyst medium etc.

As atoms are neither created nor destroyed in a chemical reaction according to law of mass conservation, balanced chemical equation must contain equal no of atoms for each elements across the sign of   equality maintaining right chemical formula for each component of reaction system.

A balanced equation is a statement of the both qualitative and quantitative relation between reactant and products involved in any chemical change.

We balance the equation by making the number of atoms of each element participating in the reaction the same as that appearing in the products. This can be accomplished by placing the required number stoichio-coefficient before each formula by taking consideration that subscripts in the formulas must not be altered.

Hit and Trial Method

This method as the name suggests, is not based on any definite procedure. However, the following guidelines may be helpful.

(i)   Write the skeleton equation by mentioning the symbols and formulae of the reactants and products.

(ii)  If an elementary gas like H₂, O₂ or N₂ etc. appears on either side of the equation, write the same in the atomic form.

(iii) Select the formula which contains the maximum number of atoms and start balancing.

(iv) In case the above method is not convenient, then start balancing the atoms which appear the  least number of times.

(v)  Balance the atoms of the elementary gas in the last.

(vi) When the balancing is complete, convert the equation into molecular form.

The balancing of a chemical equation by the above method is illustrated by the following examples.

Question . A chemical equation is balanced according to the law of:

                        (A) multiple proportions                         (B) constant proportions

                        (C) reciprocal proportions                      (D) conservation of mass

Solution:         A balanced chemical equation is another statement of law of mass conservation.

                        Hence (D) is correct.

Question . Write the chemical equation for the following reaction and balance the same by hit and trial method.

MOLE CONCEPT Atoms and molecules are too small to count. To solve this problem their numbers are expressed in terms of Avogadro’s number (N_A = 6.023 x 10²³ ).

Mole is the number equal to Avogadro’s number just like a dozen is equal to 12, a century means 100, a score means = 20. A mole (symbol mol) is defined as the amount of substance that contains as many atoms, molecules, ions, electrons or any other elementary entities as there are carbon atoms in exactly 12 gm of ¹²C. The number of atoms in 12 gm of ¹²C is called Avogadro’s number.

N_A = 6.023 x 10²³

The number of moles of a substance can be calculated by various means depending on data available, as follows.

• Number of moles of molecules \(\displaystyle =\frac{{Wt.\stackrel{\scriptscriptstyle\to}{\leftarrow}\stackrel{\scriptscriptstyle\to}{\leftarrow}in\stackrel{\scriptscriptstyle\to}{\leftarrow}\stackrel{\scriptscriptstyle\to}{\leftarrow}gm}}{{Molecular\stackrel{\scriptscriptstyle\to}{\leftarrow}\stackrel{\scriptscriptstyle\to}{\leftarrow}mass}}\)
• Number of moles of atoms \(\displaystyle =\frac{{Wt.\stackrel{\scriptscriptstyle\to}{\leftarrow}\stackrel{\scriptscriptstyle\to}{\leftarrow}in\stackrel{\scriptscriptstyle\to}{\leftarrow}\stackrel{\scriptscriptstyle\to}{\leftarrow}gm}}{{Atomic\stackrel{\scriptscriptstyle\to}{\leftarrow}\stackrel{\scriptscriptstyle\to}{\leftarrow}\stackrel{\scriptscriptstyle\to}{\leftarrow}mass}}\)
• Number of moles of gases \(\displaystyle =\frac{{Volume\stackrel{\scriptscriptstyle\to}{\leftarrow}\stackrel{\scriptscriptstyle\to}{\leftarrow}at\stackrel{\scriptscriptstyle\to}{\leftarrow}\stackrel{\scriptscriptstyle\to}{\leftarrow}STP}}{{S\tan dard\stackrel{\scriptscriptstyle\to}{\leftarrow}\stackrel{\scriptscriptstyle\to}{\leftarrow}molar\stackrel{\scriptscriptstyle\to}{\leftarrow}\stackrel{\scriptscriptstyle\to}{\leftarrow}volume}}\) 
• (Standard molar volume at STP = 22.4 lit)
• Number of moles of particles e.g. atoms, molecules ions etc \(\displaystyle =\frac{{Number\stackrel{\scriptscriptstyle\to}{\leftarrow}\stackrel{\scriptscriptstyle\to}{\leftarrow}of\stackrel{\scriptscriptstyle\to}{\leftarrow}\stackrel{\scriptscriptstyle\to}{\leftarrow}particles}}{{Avogardro\stackrel{\scriptscriptstyle\to}{\leftarrow}\stackrel{\scriptscriptstyle\to}{\leftarrow}number}}\)
• \(\displaystyle \text{Number of moles of solute = molarity}\times \text{volume of solution in litres }\)
  \(\displaystyle =\frac{{molarity\times volume\stackrel{\scriptscriptstyle\to}{\leftarrow}\stackrel{\scriptscriptstyle\to}{\leftarrow}in\stackrel{\scriptscriptstyle\to}{\leftarrow}\stackrel{\scriptscriptstyle\to}{\leftarrow}ml}}{{1000}}\)
• for a compound AxBy, y moles of A = x moles of B
• Mole fraction = fraction of the substance in the mixture expressed in terms of mol is called its mol fraction (X)
  \(\displaystyle {{X}_{A}}=\frac{{{{n}_{A}}}}{{{{n}_{A}}+{{n}_{B}}}}\,\,\,:\,\,{{X}_{B}}=\frac{{{{n}_{B}}}}{{{{n}_{A}}+{{n}_{B}}}}\)
  \(\displaystyle {{\text{X}}_{\text{A}}}\text{ + }{{\text{X}}_{\text{B}}}\text{ = 1 }\!\!\And\!\!\text{ }{{\text{X}}_{\text{A}}}\text{= (1 – }{{\text{X}}_{\text{B}}}\text{)}\) 
  
  
  

  Question.   The largest number of molecules is in:

                          (A) 28 g of CO                                      (B) 46 g of C₂H₅OH

                          (C) 36 g of H₂O                                     (D) 54 g of N­₂O₅

  Solution:         36 g H_2­O = 2 mole of H₂O while other option has less mole.

                          Hence (C) is correct.                            

  Question. The number of molecules in 89.6 litre of a gas at NTP are:

                          (A) 6.02 x 10²³                                      (B) 2 x 6.02 x 10²³

                          (C) 3 x 6.02 x 10²³                                 (D) 4 x 6.02 x 10²³

  Solution:         89.6 litre of a gas at NTP = 89.6/22.4 = 3 mole = 3  xN_A

                          = 3 x 6.02 x 10²³ molecules

                          Hence (D) is correct.

  Question.   The total number of protons in 10 g of calcium carbonate is:

                          (A) 3.0115 x 10²⁴                                  (B) 1.5057 x 10²⁴

                          (C) 2.0478 x 10²⁴                                  (D) 4.0956 x 10²⁴

  Solution:         The number of moles of CaCO₃ present in 10 gram = 10/100 = 0.1 mole. So number of total proton = 0.1 x 6.023 x 10²³ (20 + 6 + 8 x 3) = \(\displaystyle 3.0115\times {{10}^{{24}}}\)
  

  Hence (A) is correct.

 

            Question.         Calculate the mass of (i) an atom of silver (ii) a molecule of carbon dioxide.

            Solution:         (i)   1 mole of Ag atoms = 108 g

                            (Atomic mass of Silver is 108 U)

                                           = 6.022 x 10²³ atoms

                                               6.022 x 10²³ atoms of silver have mass = 108g

                                              ∴Mass of one atom of silver \(\displaystyle =\frac{{108}}{{6.022\times {{{10}}^{{23}}}}}=1.793\times {{10}^{{-22}}}g\)

                                                (ii)        1 mole of CO₂ = 44 g 

                                              (Molecular Mass CO₂ = 44 u)

 

                                                 = 6.022 x 10²³ molecules

                                                  Thus, 6.022 x 10²³ molecules of CO₂ has mass = 44 g

                                                    ∴ 1 molecule of CO₂ has mass

                                                  \(\displaystyle =\frac{{44}}{{6.022\times {{{10}}^{{23}}}}}=\frac{{44\times {{{10}}^{{-23}}}}}{{6.022}}g\)

                                                  = 7.307 x 10^-23 g 

CONCEPT OF LIMITING REAGENT

Limiting Reactant

The reactant which is totally consumed during the course of reaction and when it is consumed reaction stops. The concept of limiting reactant is applicable to reaction other than monomolecular i.e., when more than one type reactant involved. For example

These is no limiting reactant in the above example.

To determine the limiting reagent amount of all reactants and mole ratio of reactants must be known. If the ratio of moles of reactant A with respect to reactant B is greater than the ratio of the moles of A to moles of B for a balanced chemical equation then B is the limiting reactant. All other terms like left (unused) mass of other reactant, amount of formed product can be known stoichiometrically by knowing the amount of limiting reactant

Method of Expressing Concentration of Solution

Molarity (M): The molarity of a solution is the number of moles of solute present in one litre (1dm³) of solution.

 

The molality (m): The molality is the number of moles of solute present in one Kg of solvent.

Relation between molarity and molality

where d=density of solution

Parts per million parts (ppm): For every dilute solution, i.e., when a very small quantity of a solute is present in large quantity of a solution, the concentration of the solute is expressed in terms of ppm. It is defined as the mass of the solute present in one million (10⁶) parts by mass of the solution. Thus for a solute A,

Relationship between molarity (M) and molality (m): Molarity M means M moles of solute are present in 100 cc. of the solution. If density of the solution is d g/cc, mass of solution = 1000 f grams. Mass of solute = MM₂g (M₂ is mol mass of solute).

Hence mass of solvent = 1000d – MM₂g.

Relationship between molality (m) and mole fraction (x₂): Molality (m) means m moles of the solute in 1000 g of the solvent = 1000 / M₁ moles (M₁ = mol of mass of the solvent). Hence

Relationship between molarity (M) and mole fraction (x₂): Referring to calculations above,

\(\displaystyle Mole\,\,fraction,\,\,{{x}_{2}}=\frac{M}{{\left( {1000d-M{{M}_{2}}} \right)/{{M}_{1}}+{{M}_{2}}}}\)

\(\displaystyle or\,\,\,{{x}_{2}}=\frac{{M{{M}_{1}}}}{{\left( {1000d-M{{M}_{2}}} \right)+M{{M}_{1}}}}\)

= \(\displaystyle \frac{{M{{M}_{1}}}}{{M\left( {{{M}_{1}}-{{M}_{2}}} \right)+1000d}}\)

Thus x₂ = \(\displaystyle \frac{{M{{M}_{1}}}}{{M\left( {{{M}_{1}}-{{M}_{2}}} \right)+1000d}}\)

Or rearrangement, we get

            Or        \(\displaystyle M=\frac{{1000d{{x}_{2}}}}{{{{x}_{1}}{{M}_{1}}+{{x}_{2}}{{M}_{2}}}}\)

Note:

      If molarity (M) is in moles / litre and density d is kg/litre and molality m is in moles / kg of the solvent, 1000 will be replaced by 1 in the above formulae..

Question.   Calculate the molarity of a solution containing 0.5 g of NaOH dissolved in
      500 cm³.

Solution:         Weight of NaOH dissolved = 0.5 g

                        Volume of the solution = 500 cm³

                        Calculation of molarity:

                        \(\displaystyle 0.5\,\,g\,\,of\,\,NaOH=\frac{{0.5}}{{40}}mole\,\,of\,\,NaOH\)

                           (Molecular mass of NaOH = 40)

                             = 0.0125mole

                        Thus 500 cm³ of solution contain NaOH = 0.0125 mole

                        ∴ 1000 cm³ of the solution contain NaOH

                         =\(\displaystyle \frac{{0.0125}}{{500}}\times 1000\) = 0.025 Mole

                              Hence molarity of the solution = 0.025 M 

Question..  Calculate the molarity of a solution containing 9.8 gm of H₂­SO₄ in 250 cm³ of the       solution.

Solution:         Mass of H₂SO_­4 dissolved = 9.8 g

                        Volume of the solution = 250 cm³

                        Calculation of molarity

                        Mol. mass of H_­2SO₄ = 98

                                                ∴ No. of moles of H_­2SO₄ = \(\displaystyle \frac{{Mass\,\,in\,\,g}}{{Mol.\,\,mass}}\)

\(\displaystyle =\frac{{9.8}}{{98}}=0.1\)

250 cm³ of the solution contain H₂SO₄

                        = 0.1 mole

                        ∴1000 cm³ of the solution contain H₂SO₄

\(\displaystyle =\frac{{0.1}}{{250}}\times 1000=0.4\,mole\)

Hence molarity of solution = 0.4 M

Question.. Find the molarity and molality of a 15% solution of H₂SO₄ (density of H₂SO₄ = 1.020 g cm^–3) (Atomic mass: H = 1, O = 16, S = 32 amu).

Solution:         15% of solution of H₂SO₄ means 15 g of H₂SO_­4 are present in 100 g of the solution i.e.

                        Mass of H₂SO₄ dissolved = 15 g

                        Mass of the solution = 100 g

                        Density of the solution = 1.02 g/cm³ (given)

                        Calculation of molality:

                        Mass of solution = 100 g

                        Mass of H₂SO₄ = 15g

                        Mass of water (solvent) = 100 – 15 = 85 g

                        Mol. mass of H₂SO₄ = 98

                                                ∴ 15 g H₂SO₄ = \(\displaystyle \frac{{15}}{{98}}=0.153\,\,moles\)

                          Thus 85 g of the solvent contain 0.153 moles 100 g of the solvent contain

                            = (0.153/85) x 1000 = 1.8= Molality

                        Calculation of molarity : 15 g of  H₂SO₄   = 0.153 moles

                         Vol of solution = Wt.of solution/Density of solution =100/1.02 = 98.04 cm³

                          Thus 98.04 cm³of solution contain H2SO4 = 0.153 moles

 

MISCELLANEOUS EXERCISES

Exercise 1:       What mass of silver nitrate will react with 5.85 g of sodium chloride to produce
14.35 g of silver chloride and 8.5 g of sodium nitrate, if the law of conservation of mass is true?

Exercise 2:      Elements X and Y form two different compounds. In the first, 0.324 g of X is combined with 0.471 g of Y. In the second, 0.117 g of X is combined with 0.509 g of Y. Show that these data illustrate the law of multiple proportions.

Exercise 3:      Ammonia contains 82.35% of nitrogen and 17.65% of hydrogen. Water contains 88.90% of oxygen and 11.10% of hydrogen. Nitrogen trioxide contains 63.15% of oxygen and 36.85% of nitrogen. Show that these data illustrate the law of reciprocal proportions.

Exercise 4:      Carbon dioxide contains 27.27% of carbon, carbon disulphide contains 15.79% of carbon and sulphur dioxide contains 50% of sulphur. Are these figures in agreement with the law of reciprocal proportions?

Exercise 5:      The phosphorous trichloride contains 22.57% of phosphorous, phosphine (PH₃) contains 91.18% of phosphorous while hydrogen chloride gas contains 97.23% of chlorine. Prove by calculations, which law is illustrated by these data.

Exercise 6:      How many atoms and molecules of sulphur are present in 64.0 g of sulphur (S­₈)?

Exercise 7:      Calculate the mass of (i) 0.1 mole of KNO₃ (ii) 1 ´ 10²³ molecules of methane and (iii) 112 cm³ of hydrogen at SP.

Exercise 8:      0.5 mole of calcium carbonate is decomposed by an aqueous solution containing 25% HCl by mass. Calculate the mass of the solution consumed.

Exercise 9:      How much marble of 96.5 % purity would be required to prepare 10 litres of carbon dioxide at STP when the marble is acted upon by dilute hydrochloric acid?

Exercise 10:     Calculate the volume of air containing 21% oxygen by volume at STP, required in order to convert 294 cm³ of sulphur dioxide to sulphur trioxide under the same conditions.

Exercise 11:     What is the molarity of a solution of sodium chloride (At. wts. Na = 23, Cl = 35.5) which contains 60 g of sodium chloride in 2000 cm³ of a solution?

Exercise 12:     Calculate the mole fraction of ethyl alcohol and water in a solution in which 46 g of ethyl alcohol and 180 g of water are mixed together.

Exercise 13:     A 100 cm³ solution of sodium carbonate is prepared by dissolving 8.653 g of he salt in water. The density of solution is 1.0816 g per millitre. What are the molarity and molality of the solution? (Atomic mass of Na is 23, of Cl is 12 and of O is 16).

Exercise 14:     4.0 g of NaOH contained in one deciliter of a solution. Calculate the following in this solution.

                        (i)   Mole fraction of NaOH (ii) Molality of NaOH (iii) Molarity of NaOH.

                        (At. wt. of Na = 23, O = 16; Density of NaOH solution is 1.038 g/cm³)

Exercise 15:     The percentage composition (by weight) of a solution is 45% X, 15% Y, 40% Z. Calculate the mole fraction of each component of the solution. (Molecular mass of X = 18, Y = 60, Z = 60).

Exercise 8:      146

Exercise 9:      46.26 gm

Exercise 10:     700 cc

Exercise 11:     0.513

Exercise 12:     0.09, 0.91

Exercise 13:     0.816M, and 0.820 m

Exercise 14:     (i) 0.0177   (ii) 1 M  (iii) 1.002 ml

Exercise 15:     X = 0.732, Y = 0.073, Z = 0.19