TYPING AND FORMATTING BY SANJAY KUMAR

INTRODUCTION

Chemistry is the science of substances, their properties, structures and their transformation. As all objects in the universe are made of matter. Chemistry is the branch of the science which deals with the study of material object. Study of chemistry is very interesting which covers various aspects of our culture and environment. All development in any science are based on scientific approach as in chemistry too. In order to achieve correct results, one has to rely upon the various skills connected with the measurements of quantities during a physical or chemical change. The degree of accuracy is closely linked with precision of the measuring instrument as well as on the skill of the person engaged in measurement. So we should be first familiar with some terminology used in chemistry.

Physical Property:

The property which can be measured without changing the chemical composition of the substance is known as physical property like mass, volume, density, refractive index etc.

Chemical Property:

The property which can be evaluated at the cost of matter itself is known as chemical property. For example combustible nature of hydrogen gas can be verified by burning of hydrogen. The sweet taste of sugar by consuming it.

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 (()

Kilogram m(3

Kg m(3

Energy (E)

Joule (J)

Kg m2s(2

Force (F)

Newton (N)

Kg ms(2

Frequency (()

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 ( 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 ( displacement = MLT-2 ( L = ML2T-2

1 joule = 1 kg (1 metre)2 ( (1see)-2

(

[

]

2

2

1000gm100cm

1Kg1metre1sec

1Kg1metre

-

éù

´´´´´

êú

ëû

( 100 gm ( (100)2( 1 em2( (1 sec)-2

( 1000(10000 ( 1 gm(1 cm2 ( 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,

33

3

10m101.325Pa

1Latm101.325Pam

1L1atm

-

æö

´´=

ç÷

èø

Knowing that

2

N

Pa

m

=

, we can write

33

2

N

101.325Pam101.325m= 101.325 N m =101.325

J

m

æö

=

ç÷

èø

Hence 1 L atm = 101.325 J

Illustration 1.What is the mass of 1 L of mercury in grams and in kilograms if the density of liquid mercury is

3

13.6gcm

-

?

Solution:

We know the relationship,

3

1L1000cm

=

and

Also,

mass

density

volume

=

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

Therefore, the mass of 1 L of mercury is equal to

3

3334

1000cm

(1L)(13.6gcm)(1000cm)(13.6gcm)1.3610g

1L

--

æö

==´

ç÷

èø

The mass in kilograms can be calculated as

44

1kg

1.3610g(1.3610g)13.6kg

1000g

æö

´=´=

ç÷

ç÷

èø

(Remember,

3

1000cm

1L

and

1kg

1000g

are conversion factors with which we have to multiply for getting our answer in appropriate units).

1kilogram1kilogram

1

2.205pond1000gm

==

Exercise 1.

Vanadium metal is added to steel to impart strength. The density of vanadium is 5.96 g/cm2. Express this in S.I. units (kg/m3).

Matter

Anything that exhibits inertia is known as matter. The quantity of matter is its mass. e.g. chalk table.

CLASSIFICATION OF MATTER

This classification of matter is based upon chemical composition of various substances. According to this matter can be further divided into two types, pure substance and mixture. Mixtures are also of two types, homogenous mixtures and heterogeneous mixtures.

Matter

Pure Substance

Mixture

Element

Compound

Homogenous

Heterogeous

Elements:

The primary stuff present in all the substance is known as element, whose smallest unit is known as atom. Total 112 elements are known till date of which 92 are naturally occurring elements rest are results of artificial transmutation. There are 88 metals, 18 non metals and 6 metaloids.

Compound:

A non elemental pure substance is called a compound in which more than one atom of elements are linked by chemical bonds formed due to chemical reaction. The resulting molecule is a electrically neutral particle of constant continuous composition.

Mixture:

Mixtures are the aggregate of more than one type of pure substance whose chemical identity remains maintained even in mixtures. Their constituent ratio may vary unlike compound.

For example – sugar + water = sugar syrup, Gun-powder 75 % KNO3 10% sulphur + 15% carbon

There are two types of mixture (a) homogeneous (b) heterogenous

(a)Homogeneous mixtures are those whose composition for each part remains constant. For example aqueous and gaseous solution.

(b)Heterogeneous mixtures are those whose composition may vary for each and every part. For example soil, concrete mixtures.

LAWS OF CHEMICAL COMBINATION

In order to understand the composition of the compounds, it is necessary to have a theory which accounts for both qualitative and quantitative observations during chemical changes. Observations of chemical reactions were most significant in the development of a satisfactory theory of the nature of matter. These observations of chemical reactions are summarized in certain statements known as laws of chemical combination.

Law of conservation of mass:

For any chemical change total mass of active reactants are always equal to the mass of the product formed. It is a derivation of Dalton’s atomic theory ‘atoms neither created nor destroyed’.

Total masses of reactants = Total masses of products + Masses of unreacted reactants

Illustration2. 5.2 g of CaCo

3

EMBED Equation.3 hen heated produced 1.99 g of Carbon dioxide and the residue (CaO) left behind weighs 3.2g. Show that these results illustrate the law of conservation of mass.

Solution:

Weight of CaCo

3

EMBED Equation.3 taken = 5.2 g

Total weight of the products (CaO +CO

2

)= 3.20+ 1.99 = 5.19 g

Difference between the wt. of the reactant and the total wt. of the products

= 5.20 – 5.19 =0.01 g.

This small difference may be due to experimental error.

Thus law of conservation of mass holds good within experimental errors.

Law of constant composition

A chemical compound always contains same elements in definite proportion by mass and it does not depend on the source of compound.

For example

22

COCO

+¾¾®

32

CaCOCaOCO

D

¾¾®+

2322

NaCO2HCl2NaClHOCO

+¾¾®++

4222

CH2OCO2HO

+¾¾®+

The composition of CO2 obtained by different means always having same C:O ratio =

12

0.375

32

=

.

Illustration 3. Ammonia contains 82.65 % N

2

and 17.65% H2. If the law of constant proportions is true, then the mass of zinc required to give 10 g Ammonia will be:

(A)8.265 g(B) 0.826 g

(C)82.65 g(D) 826.5 g

Solution:The mass of zinc required to give 10 g ammonia will be

4022.65

9.06gm

100

´

==

Hence (B) is correct

Illustration 4. Irrespective of the source, pure sample of water always yields 88.89% mass of oxygen and 11.11% mass of hydrogen. This is explained by the law of:

(A)conservation of mass(B)constant composition

(C)multiple proportion(D)constant volume

Solution:As in water

Hydrogen2

H:O1:8bymass

Oxygen16

=Þ=

So oxygen =

8

10088.89%

81

´=

+

Hydrogen =

1

10011.11%

81

´=

+

Both value always constant. Obey law of constant composition.

Hence (B) is correct.

Illustration 5. 6.488 g of lead combine directly with 1.002 g of oxygen to form lead peroxide

2

(PbO)

. Lead peroxide is also produced by heating lead nitrate and it was found that the percentage of oxygen present in lead peroxide is 13.38 percent. Use these data to illustrate the law of constant composition.

Solution:Step 1: To calculate the percentage of oxygen in first experiment.

Weight of peroxide formed

= 6.488 + 1. 002 = 7.490 g.

7. 490 g of lead peroxide contain 1.002 g of oxygen

\

100 g of lead peroxide will contain oxygen

1.002

10013.38g

7.490

´=

i.e. oxygen present = 13.38%

Step 2: To compare the percentage of oxygen in both the experiments.

Percentage of oxygen in

2

PbO

in the first experiment = 13.38

Percentage of oxygen in

2

PbO

in the second experiment = 13.38

Since the percentage composition of oxygen in both the samples of

2

PbO

is identical, the above data illustrate the law of constant composition.

Illustration 6.Copper oxide was prepared by the following methods:

(a) In one case, 1.75 g of the metal were dissolved in nitric acid and igniting the residual copper nitrate yielded 2.19 g of copper oxide.

(b)In the second case, 1.14 g of metal dissolved in nitric acid were precipitated as copper hydroxide by adding caustic alkali solution. The precipitated copper hydroxide after washing, drying and heating yielded 1.43 g of copper oxide.

(c)In the third case, 1.45 g of copper when strongly heated in a current of air yielded 1.83 g of copper oxide. Show that the given data illustrate the law of constant composition.

Solution:Step 1: In the first experiment.

2.19 g of copper oxide contained 1.75 g of Cu.

\

100 g of copper oxide contained

1.75

10079.91g

2.19

=´=

Step 2: In the second experiment.

1.43 g of copper oxide contained 1.14 g of copper.

\

100 g of copper oxide contained

1.14

10079.72g

1.43

=´=

.

Step 3: In the third experiment.

1.83 g of copper oxide contained 1.46 g of copper

\

100 g of copper oxide contained

1.46

10079.78g

1.83

=´=

Thus the percentage of copper in copper oxide derived from all the three experiments is nearly the same. Hence, the above data illustrate the law of constant composition.

Exercise 2.

5.06 g of pure cupric oxide (CuO), on complete reduction by heating in a current of hydrogen, gave 4.04 g of metallic copper. 1.3 g of pure metallic copper was completely dissolved in nitric acid and the resultant solution was carefully dried and ignited. 1.63 g CuO were produced in the process. Show that these results illustrate the law of constant proportions.

Exercise 3.

In an experiment, 2.4 g of iron oxide on reduction with hydrogen yield 1.68 g of iron. In another experiment 2.9 g of iron oxide give 2.03 g of iron on reduction with hydrogen. Show that the above data illustrate the law of constant proportions.

Exercise 4.

2.8 g of calcium oxide (CaO) prepared by heating limestone were found to contain 0.8 g of oxygen. When one gram of oxygen was treated with calcium, 3.5 g of calcium oxide were obtained. Show that the results illustrate the law of definite proportions.

Law of multiple proportion:

When two elements combine to form two or more than two different compounds then the different masses of one element B which combine with fixed mass of the other element bear a simple ratio to one another.

For example: Carbon forms two oxides in oxygen

Carbon monoxide

2

12

16

1

COCO

2

+=

Carbon oxide

22

12

32

COCO

+=

The ratio of masses of oxygen in CO and CO2 for fixed mass of carbon (12) is 16 : 32 = 1: 2.

Illustration 7. The law of multiple proportions is illustrated by the pair of compounds:

(A)sodium chloride and sodium bromide

(B)water and heavy water

(C)sulphur dioxide and sulphur trioxide

(D)magnesium hydroxide and magnesium oxide

Solution:

In SO2 32 gram of suphur react with 32 gram of oxygen. Similarly for SO3 fixed mass of sulphur (32 gram) react with 48 gram of oxygen. The ratio of oxygen’s mass = 32:48 = 2:3

Which support law of multiple proportions.

Hence (C) is correct answer.

Illustration 8.Carbon is found to form two oxides, which contain 42.9% and 27.3% of carbon respectively. Show that these figures illustrate the law of multiple proportions.

Solution:

Step 1: To calculate the percentage composition of carbon and oxygen in each of the two oxides

First oxideSecond oxide

Carbon42.9%

27.3%

(Given)

Oxygen57.1%

72.7%

(by difference)

Step 2:To calculate the weights of carbon which combine with a fixed weight i.e., one part by weight of oxygen in each of the two oxides.

In the first oxide, 57.1 parts by weight of oxygen combine with carbon = 42.9 parts.

\

1 part by weight of oxygen will combine with carbon

42.9

0.751

57.1

=

In the second oxide 72.7 parts by weight of oxygen combine with carbon = 27.3 parts.

\

1 part by weight of oxygen will combine with carbon

27.3

0.376

72.7

=

Step 3:To compare the weights of carbon which combine with the same weight of oxygen in both the oxides-

The ratio of the weights of carbon that combine with the same weight of oxygen (1 part) is 0.751: 0.376 or 2:1

Since this is a simple whole number ratio, so the above data illustrate the law of multiple proportions.

Exercise 5.

Metal M and chlorine combine in different proportions to form two compounds A and B. The mass ratio M : Cl is 0.895 : 1 in A and 1.791 : 1 in B. What law of chemical combination is illustrated?

Exercise 6.

By means of the following analytical results show that law of multiple proportions is true:

Mercurous chloride Mercuric chloride

Mercury = 84.92%

Mercury = 73.80%

Chlorine = 15.08%

Chlorine = 26.20%

Exercise 7.

1 g of a metal, having no variable valency, produces 1.67 g of its oxide when heated in air. Its carbonate contains 28.57% of the metal. How much oxide will be obtained by heating 1 g of carbonate?

Exercise 8.

Phosphorous and chlorine form two compounds. The first contains 22.54% by mass of phosphorous and the second 14.88% of phosphorous. Show that these data are consistent with law of multiple proportions.

Law of reciprocal proportion:

If two elements B and C react with the same mass of a third element (A), the ratio in which they do so will be the same or simple multiple if B and C reacts with each other.

The above law is the basis of law of equivalent masses, which will be described latter in details.

For example:

32

2

12

32

2

32

O

C

O

S

-

64

2

12

S

C

ratio of masses of carbon and sulphur which combine with fixed mass (32 parts) of oxygen is

12:32 or 3:8

…(1)

In

2

CS

ratio of masses of carbon and sulphur is

12:64 or 3:16

…(2)

The two ratios (1) and (2) are related to each other by

33

:

816

or 2:1

or

33

:2

816

æö

ç÷

èø

i.e. first ratio is integral multiple of second.

Illustration 9. One part of an element A combines with two parts of B (another element). Six parts of element C combine with four parts of element B. If A and C combines together, the ratio of their masses will be governed by:

(A)law of definite proportions(B)law of multiple proportions

(C)law of reciprocal proportions(D)law of conservation of mass

Solution:

B2B42

;

A1C63

===

from this when A combined with C the ratio is A/C = 1/3

Hence (C) is correct.

Illustration 10. Hydrogen combines with chlorine to form HCl. It also combines with sodium to form NaH. If sodium and chlorine also combine with each other, they will do so in the ratio of their masses as

(A)23 : 35.5(B)35.5 : 23

(C)1 : 1(D)23 : 1

Solution:

1atom of hydrogen combined with 1 atom of chlorine and 1 atom of sodium. So when sodium react with chlorine the ratio is 1 : 1 by atom. While 23 : 35.5 by mass.

Hence (A) is correct.

Illustration 11.Copper sulphide contains 66.6% Cu, copper oxide contains 79.9% copper and sulphur trioxide contains 40% Sulphur. Show that these data illustrates law of reciprocal proportions.

Solution:

In copper sulphides

Cu : S mass ratio is 66.6 : 33.4

In sulphur trioxide

Oxygen : Sulphur (O:S) mass ratio is 60:40

Now in CuS

33.4 parts of sulphur combines with Cu = 66.6 parts

3

66.6:33.5

40:60

CuSSO

-

1

.

20

:

9

.

79

CuO

40.0 parts of sulphur combines with cu

66.640

79.8parts

33.4

´

==

Now the ratio of masses of Cu and O which combines with same mass (40 parts) of sulphur separately is 79.8:60

…(1)

Cu : O ratio by mass in CuO is 79.9: 20.1 …(2)

Ratio 1: Ratio 2

79.820.1

1:3

6079.9

=´=

Which is simple whole number ratio, hence law of reciprocal proportion is proved.

Exercise 9.

Carbon combines with hydrogen to form three compounds A, B and C. The percentage of hydrogen in A, B and C are 25, 14.3 and 7.7 respectively. Which law of chemical combination is illustrated?

Exercise 10.

61.8g of A combines with 80 g of B. 30.9g of A combines with 106.5g of C, B and C combine to form compound

2

CB

. Atomic weight of C and B are respectively 35.5 and 6.6. Show that the law of reciprocal proportion is obeyed.

Exercise 11.

Carbon dioxide contains 27.27% carbon, carbon disulphide contains 15.97% carbon and sulphur dioxide contains 50% sulphur. Show that these figures illustrate the law of reciprocal proportions.

`

Exercise 12.

Aluminium oxide contains 52.9% aluminium and carbon dioxide contains 27.27% carbon. Assuming the validity of the law of reciprocal proportions, calculate the percentage of aluminium in aluminium carbide.

Gay Lussac’s law of Combining Volumes:

At given temperature and pressure the volumes of all gaseous reactants and products bear a simple whole number ratio to each other.

For example

[

]

2(g)2(g)(g)

1voume1volume2volumes

HCl2HCl1:1:2

+¾¾®

i.e. one volume of hydrogen reacts with one volume of chlorine to form two volumes of HCl gas. i.e. the ratio by volume which gases bears is 1:1:2 which is a simple whole number ratio.

Similarly other examples are:

[

]

223

2volume1volume

2volume

2SOO2SO2:1:2

+=

322

1volume3volume

2volume

2NHN3H[2:1:3]

+

+

ˆˆˆ†

‡ˆˆˆ

[

]

34

1volume

1volume

1volume

NHHClNHClVapour[1:1:1]

+

ˆˆˆ†

‡ˆˆˆ

Illustration 12. In the reaction,

223

N3H2NH,

+¾¾®

the ratio of volumes of nitrogen, hydrogen and ammonia is 1 : 3 : 2. These figures illustrate the law of:

(A)constant proportions(B)Gay-Lussac

(C)multiple proportions(D)reciprocal proportions

Solution:

The above ratio of 1 : 3 : 2 illustrates the Gay-Lussac law of combining volume.

Hence (B) is correct.

Illustration 13.How much volume of oxygen will be required for complete combustion of 40 ml of acetylene

22

(CH)

and how much volume of carbondioxide will be formed? All volumes are measured at NTP.

Solution:

22(g)2(g)2(g)2

2volume5volume4volume

2CH5O4CO2HO

+¾¾®+

40 ml

5

40ml

2

´

4

40ml

2

´

40 ml

100 ml

80 ml

So for complete combustion of 40 ml of acetylene, 100 ml of oxygen are required and 80 ml of

2

CO

is formed.

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.

actualyield

%yield100

theoreticalyield

=´

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

122211

CHO

, 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.

2

E = MC

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

Illustration 14. 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.

2223

H, O, N,He,HCl,CaCO

etc.

ATOMIC MASS AND MOLECULAR MASS

Atomic Mass: As atoms are very tiny particles, their absolute masses are difficult to measure. However it is possible to determine the relative masses of different atoms if small unit of mass is taken as standard (previously, this standard was mass of one atom of hydrogen and taken as unity. Later on it was

th

1/16

part of oxygen atom and now it is

th

1/12

part of C(12 atom).

The atomic mass of an element can be defined as the number which indicates how many times the mass of one atom of the element is heavier in comparison to the mass of one atom of hydrogen.

Atomic mass of an element

Massofoneatomoftheelement

1

MassofthpartofmassofC12atom

12

=

-

Massofoneatomoftheelement

1

MassofthpartofmassofOatom

16

=

Massofoneatomoftheelement

MassofoneatomofHydrogen

=

Atomic Mass Unit: The quantity

1

12

mass of an atom of carbon(12 is known as the atomic mass unit and is abbreviated as amu. The actual mass of one atom of carbon(12 is

-23

1.992410g

´

or

-26

1992410kg

´

Thus 1 amu

23

24

1.992410

1.6610g

12

-

-

´

==´

Atomic mass of an element

Massofoneatomoftheelement

1amu

=

Actual mass of an element

-24

= Atomic mass (amu)1.6610g

´´

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 ( 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 ( valency

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

Dulongs and Petit’s Law:

(

)

in cal/gm

Atomic massSpecific heat = 6.4

´

Illustration15. 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 =

6.46.4

0.2

atomicmass32

==

Hence (C) is correct.

Illustration 16.On dissolving 6 gm of metal in sulphuric acid, 13.53g of the metal sulphate was formed. The specific heat of metal is

-1

0.057 cal g

. What is equivalent mass of metal, valency and exact atomic mass?

Solution:

Equivalent mass of sulphate

ionicmass96

48

valency2

===

Eq. weight of metal

Massofmetal66

48484838.24g

Massofsulphate13.5367.53

=´=´=´=

-

Approximate atomic mass

6.46.4

112.5

Specificheat0.057

==´

Valency

Approximateatomicmass112.5

2.93

Equivalentmass38.24

===»

Exact atomic mass

= 338.24 = 114.72

´

Exercise 13.

Two oxides of a metal contain 63.2% and 69.62% of the metal. The specific heat of the metal is 0.117. What is the formula of the two oxides?

Exercise 14.

1 g of a metal which has specific heat of 0.06 combines with oxygen to form 1.08 g of oxide. What is the mass of M?

Exercise 15.

The chloride of a solid metallic element contains 57.89% by mass of the element. The specific heat of the element is 0.0324 cal deg(1g(1. Calculate the exact atomic mass of the element.

(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 ( 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.

Illustration 17.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

(iii)Law of Isomorphism

When two or more compound forms similar type of crystals or able to form mixed crystals, they are known as isomorphs. For examples: MgSO4.7H2O, ZnSO4.7H2O and FeSO4.7H2O are isomorphs of each other as their crystals posses same shape.

According to Mitscherlich [year 1819].

The valency of elements that are similarly placed to that of other elements in their isomorphs are always same.

In the above example Fe, Zn and Mg have same valency [2] and equal ratio of water molecule in each isomorphs.

If equivalent mass of one element is known then atomic mass can be calculated by knowing the valency of other isomorphs key element.

Illustration 18. 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.

Exercise 16.

White vitriol (hydrated sulphite) is isomorphous with MgSO4.7H2O. White vitriol contains 22.95% zinc and 43.9% of water of crystallization. Find the atomic mass of zinc.

Exercise 17.

Two oxides of metals A and B are isomorphous. The metal A whose atomic mass is 52, forms a chloride whose vapour density is 79. The oxide of the metal B contains 47.1% oxygen. Calculate the atomic mass of B.

(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

x

MCl

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

Atomic mass = xE

´

So

w

M = Ex + 35.5x = 2 V.D.

´

2VD

x

E35.5

=

+

(

Atomic weight = x E

´

Illustration 19.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

0.165

35.57.01g

0.835

=´=

Valency of metal

2V.D.285

4

E35.57.0135.5

´

===

++

Formula of chloride

4

= MCl

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

%abundanceatomicmass

100

´

=

å

Illustration 20.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 ( 0.071 + 37.96272 ( 0.163 + 39.96924 ( 0.766

= 39.352 g mol(1

Illustration 21.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

%abundanceatomicmass

100

´

=

å

i.e.

x12(100x)13

12.011

100

´+-

=

1201.1 = 12x + 1300 – 13x

x = 1300 – 1201.1 = 98.9%

Exercise 18.

Chlorine has isotopes

35

Cl

and 37Cl. There are three 35Cl isotopes for every one 37Cl isotopes in a sample of chlorine. Calculate the atomic mass of chlorine.

Exercise 19.

Natural hydrogen gas is a mixture of 1H and 2H in the ratio of 5000:1. Calculate the atomic mass of the hydrogen.

GRAM ATOMIC MASS OR GRAM ATOM

Atomic weight of an element in grams is called as Gram atomic mass of an element. Gram atomic mass is the weight of 1 mole atom

23

(6.02310 atoms)

´

of the element. It is also called as 1 gram atom. e.g. AW of C = 12, GAW of

23

C = 12g = 6.02310

´

atom of carbon = 1 mole atom of C = 1 gram atom of carbon.

Illustration 22.Calculate the weight of following samples

(a)100 atoms of carbon

(b) 0.5 mole atoms of sulphur

(c)8 gram atom of oxygen

(d) 25 amu of chlorine

(e)

20

1.5×10

atoms of N

Solution:

(a)1 mole atom of C = 12 g

1 atom of C

23

12

g

6.02310

=

´

100 atom of C

21

23

12100

g1.9910g

6.02310

-

´

==´

´

(b)AW of S = 32

GAW of S = 32g = 1 mole atom

0.5 mole atom

= 0.532g = 16gms

´

(c)GAW = 1 gm atom

GAW of oxygen = 16 g = 1 gm atom of oxygen

8 gm atom of oxygen

= 816 = 128g

´

(d)25 amu of chlorine

-24

1 amu = 1.6610g

´

-24-23

25 amu = 1.661025g = 4.1510g

´´´

(e)

20

1.510

´

atom of N

1 atom of N = 14 amu

20

2020

23

10

1.510atom141.510amu141.5g

6.02310

´=´´=´´

´

33

14

10gm3.510g

4

--

=´=´

Exercise 20.

Assuming the density of

2

HO

to be

3

1g/cm

. Calculate the volume of one molecule of

2

HO

.

Exercise 21.

Calculate the number of moles in each of the following:

(i) 392 gm of

24

HSO

; (ii) 44.8 litres of

2

CO

at STP.

Molecular Mass

Number of times a molecule of a compound is heavier than one atom of hydrogen or

1

th

12

part of C(12 is molecular mass of the compound. It is the sum of the atomic mass of atoms present in a molecule.

For exampleMolecular mass of

2

CO = 44

i.e 1 molecule of

2

CO

is 44 times heavier than one atom of hydrogen or

th

1/12

part of C(12.

Gram Molecular Mass

Molecular mass expressed in grams is called as gram molecular mass. It is the weight of one mole molecules of a compound. It is also called as one gram molecule.

For exampleGMW of

2

CO = 44g = 1 mole

molecule of

23

2

CO = 6.02310 molecule

´

= 1 gram molecule of

2

CO

Illustration 23.Calculate the actual mass of one molecule of

2

CO

Solution:

Molecular mass of

2

CO= 44

Weight of one molecule of

2

CO= 44 amu

Actual mass of

2

CO

molecule

-24-23

= 441.6610g = 7.30410g

´´´

Illustration 24.Calculate the mass and number of molecules of

24

HSO

in 1.5 gm molecules of sulphuric acid.

Solution:

Molecular mass of

24

HSO = 98

1.5 gm molecules of

24

HSO = 981.5g = 147.0g

´

No. of molecule of

23

24

HSO = 1.56.02310 molecules

´´

23

= 9.0310 molecules

´

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.

Illustration 25.A compound contains C = 71.23%, H = 12.95% and O = 15.81%. What is the empirical formula of the compound?

Solution:

Element’s name with their symbol

Atomic mass

Weight % in compound

Relative number of atom

Simplest atomic ratio

Empirical formula

Carbon (C)

12

71.93

71.93

5.936

12

=

5.936

6

0.988

=

613

CHO

Hydrogen (H)

1

12.95

12.95

12.95

1

=

12.95

13

0.988

=

Oxygen (O)

16

15.81

15.81

0.988

16

=

0.988

1

0.988

=

Illustration 26. The simplest formula of a compound containing 50% of element X (Atomic

mass = 10) and 50% of the element Y (Atomic mass = 20) is:

(A) XY(B)X2Y

(C) XY2(D)X2Y3

Solution:

Element’s name with their symbol

Atomic mass

Weight % in compound

Relative number of atom

Simplest atomic ratio

Empirical formula

X

10

50

50/10 = 5

5/2.5 = 2

X2Y

Y

20

50

50/20 = 2.5

2.5/2.5 = 1

Hence (B) is correct.

Illustration 27.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?

Solution:

Element

Element ratio

Atomic mass

Relative number of atoms

Simplest ratio

Carbon

9

12

9

0.75

12

=

0.75

3

0.25

=

Hydrogen

1

1

1

1

1

=

1

4

0.25

=

Nitrogen

3.5

14

3.5

0.25

14

=

0.25

1

0.25

=

Empirical formula = C3H4N

Empirical formula mass = (3 ( 12) + (4 ( 1) + 14

= 54

mol.mass

n

Emp.mass

=

108

2

54

==

Thus, molecular formula of the compound

= 2 ( empirical formula

= 2 ( C3H4N

= C6H8N2

Illustration 28.2.38 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).

Solution:

Step 1: To calculate the percentage of uranium and oxygen in the oxide.

2.806 g of the oxide contain uranium = 2.38 g

(

2.38

Percentageofuranium10084.82

2.806

=´=

Hence, the percentage of oxygen in the oxide

= 100.00 – 84.82 = 15.18

Step 2: To calculate the empirical formula

Element

Symbol

Percentage of elements

At. mass of elements

Relative no. of atoms

=

Percentage

At.mass

Simplest atomic ratio

Simplest whole no. atomic ratio

Uranium

U

84.82

238

84.82

0.3562

238

=

0.3562

1

0.3562

=

3

Oxygen

O

15.18

16

15.18

0.94875

16

=

0.94875

0.3562

2.666

=

8

Hence the empirical formula of the oxide is U3O8.

Illustration 29.Chemical analysis of a carbon compound gave the following percentage composition by weight of the elements present. Carbon 10.06%, hydrogen 0.84%, chlorine 89.10%. Calculate the empirical formula of the compound.

Solution:

Step 1: Percentage of the elements present

Carbon

Hydrogen

Chlorine

10.06

0.84

89.10

Step 2:Dividing the percentage compositions by the respective atomic weights of the elements

10.06

12

0.84

1

89.10

35.5

0.84

0.84

2.51

Step 3:Dividing each value in step 2 by the smallest number among them to get simple atomic ratio

0.84

0.84

0.84

0.84

2.51

0.84

Step 4:Ratio of the atoms present in the molecule C : H : Cl

1 : 1 : 3

\

The empirical formula of the compound

113

CHCl

or

3

CHCl

.

Illustration 30.A carbon compound on analysis gave the following percentage composition. Carbon 14.5%, hydrogen 1.8%, chlorine 64.46%, oxygen 19.24%. Calculate the empirical formula of the compound.

Solution:

Step1: Percentage composition of the elements present in the compound.

C:

H:Cl:O

14.5

1.8

64.46

19.24

Step 2:Dividing by the respective atomic weights

14.5

12

:

1.8

1

:

64.46

35.5

19.24

16

1.21

1.8 1.81

1.2

Step 3:Dividing the values in step 2 among them by the smallest number.

1.21

1.20

1.8

1.2

1.81

1.2

1.2

1.2

Step 4:Multiplication by a suitable integer to get whole number ratio.

(12)

´

(1.52)

´

(1.52)

´

(12)

´

2

3

3 2

\

The simplest ratio of the atoms of different elements in the compound.

C : H : Cl : O = 2 : 3 : 3 : 2

\

Empirical formula of the compound

2332

CHClO

.

Exercise 22.A crystalline hydrated salt, on being rendered anhydrous, loses 45.6% of its mass. The percentage composition of the anhydrous salt is: Al = 10.5%, K = 15.1%, S =24.8% and oxygen = 49.6%. Find the empirical formula of the anhydrous and crystalline hydrated salt.[K = 39; Al = 27; S = 32; O = 16; H = 1]

Exercise 23.A colourless crystalline compound has the following percentage composition: Sulphur 24.24%, nitrogen 21.21%, hydrogen 6.06% and the rest is oxygen. Determine the empirical formula of the compound. If the molecular mass is 132, what is the molecular formula of the compound? Name the compound if it is found to be sulphite.

Exercise 24.A gaseous hydrocarbon contains 85.7% carbon and 14.3% hydrogen. 1 litre of the hydrocarbon weighs 1.26 g at NTP. Determine the molecular formula of the hydrocarbon.

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.

n

Molecular formula = (Empirical formula)

Molecular weight = Empirical formula wei

ght n

´

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

2 Vapour density = Molecular weight

´

Illustration 31.The empirical formula of a compound is

2

CHO

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

Solution:

Empirical formula

2

= CHO

Empirical formula weight = (12 + 2 + 16) = 30

molecularweight

n

empiricalformulaweight

=

90

n3

30

\==

The molecular formula

23363

(CHO) = CHO

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.

Liquid -

(

)

l

Gas -(g)

Crystalline solid- (amorph)

Solution- (Sol)

Aqueous solution – (aq.)

Precipitate -

(

Light (gases) - (

Heating - (

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.

Illustration 32. 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.

Illustration 33.Write the chemical equation for the following reaction and balance the same by hit and trial method.

Solution:

The skeleton equation for the reaction is:

(

)

3223

2

CaNHOCaOHNH

+¾¾®+

The balancing is done in the following steps:

(i)Balancing the Ca atoms

(

)

3223

2

CaNHO3CaOHNH

+¾¾®+

(ii)Balance the N atoms:

(

)

3223

2

CaNHO3CaOH2NH

+¾¾®+

(iii)Balance the O atoms

(

)

3223

2

CaN6HO3CaOH2NH

+¾¾®+

Upon checking, the H atoms are already balanced.

( Final balanced chemical equation is:

(

)

3223

2

CaN6HO3CaOH2NH

+¾¾®+

Exercise 25.Balance the following chemical equations by hit and trial method

(i)

(

)

¾¾®

22

2

Ca+HOCaOH+H

(ii)

(

)

¾¾®

4324

3

AlC+HOAlOH+CH

(iii)

¾¾®

33343

HPOHPO+PH

(iv)

¾¾®

4222

KMnO+HClKCl+MnCl+HO+Cl

MOLE CONCEPT

Atoms and molecules are too small to count. To solve this problem their numbers are expressed in terms of Avogadro’s number (NA = 6.023 ( 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

12

C

. The number of atoms in 12 gm of

12

C

is called Avogadro’s number

A

(N)

.

23

A

N = 6.02310

´

One atomic mass unit (amu)

2427

23

A

11

gm1.6610gm1.6610kg

N

6.02310

--

===´=´

´

6 10

23

Particles

22.4 L of gas

at N.T.P or

Molar volume

1 gram

atom of

element

1 gram

mole of

substance

1 gram formula

mass of substance

1 MOLE

In terms of

particles

In terms of

volume

In terms of

mass

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

· Number of moles of molecules

Wt.ingm

Molecularmass

=

· Number of moles of atoms

Wt.ingm

Atomicmass

=

· Number of moles of gases

VolumeatSTP

Standardmolarvolume

=

(Standard molar volume at STP = 22.4 lit)

· Number of moles of particles e.g. atoms, molecules ions etc

Numberofparticles

Avogardronumber

=

·

Number of moles of solute = molarityvolu

me of solution in litres

´

EMBED Equation.DSMT4

molarityvolumeinml

1000

´

=

· for a compound

xy

AB

, 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)

E.g. for a mixture of substance A & B

AB

AB

ABAB

nn

X:X

nnnn

==

++

(n terms of denote number of moles)

ABAB

X + X = 1 & X= (1 - X)

Illustration 35. 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 H2O = 2 mole of H2O while other option has less mole.

Hence (C) is correct.

Illustration 36. The number of molecules in 89.6 litre of a gas at NTP are:

(A)6.02 × 1023(B)2 × 6.02 × 1023

(C)3 × 6.02 × 1023(D)4 × 6.02 × 1023

Solution:

89.6 litre of a gas at NTP = 89.6/22.4 = 3 mole = 3 ( NA

= 3 ( 6.02 ( 1023 molecules

Hence (D) is correct.

Illustration 37. The total number of protons in 10 g of calcium carbonate is:

(A)3.0115 × 1024(B)1.5057 × 1024

(C)2.0478 × 1024(D)4.0956 × 1024

Solution:The number of moles of CaCO3 present in 10 gram = 10/100 = 0.1 mole. So number of total proton = 0.1 ( 6.023 ( 1023 (20 + 6 + 8 ( 3) =

24

3.011510

´

Hence (A) is correct.

Illustration 38.Calculate the mass of (i) an atom of silver (ii) a molecule of carbon dioxide.

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

(

)

Atomicmassofsilver108u

=

Q

= 6.022 ( 1023 atoms

6.022 ( 1023 atoms of silver have mass = 108g

( Mass of one atom of silver

22

23

108

1.79310g

6.02210

-

==´

´

(ii)1 mole of CO2 = 44 g

(

)

2

MolecularmassofCO11221644u

=´+´=

Q

= 6.022 ( 1023 molecules

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

( 1 molecule of CO2 has mass

23

23

444410

g

6.022

6.02210

-

´

==

´

= 7.307 ( 10(23 g

Illustration 39.Calculate the number of molecules present

(i)in 34.20 grams of cane sugar (C12H22O11)

(ii)in one litre of water assuming that the density of water is 1 g/cm3

(iii)in one drop of water having mass 0.05 g.

Solution:(i)1 mole of C12H22O11 = 342 g

(

)

122211

MolecularmassofcanesugarCHO

éù

ëû

Q

= 12 ( 12 + 22 ( 1 + 11 ( 16 = 342 amu

= 6.022 ( 1023 molecules

Now 342 g of cane sugar contain 6.022 ( 1023 molecules.

( 34.2 g of cane sugar will contain

23

6.02210

34.2

342

´

=´

= 6.022 ( 1022 molecules

(ii)1 mole of water = 18 g = 6.022 ( 1023 molecules.

Mass of 1 litre of water = Volume ( density = 1000 ( 1 = 1000 g

Now 18 g of water contains = 6.022 ( 1023 molecules.

( 1000 g of water will contain =

23

6.022101000

18

´´

= 3.346 ( 1025 molecules

(iii)1 mole of H2O = 18 g = 6.022 ( 1023 molecules.

Mass of 1 drop of water = 0.05 g

Now 18 g of H2O contain = 6.022 ( 1023 molecules.

( 0.05 g of H2O will contain =

23

21

6.02210

0.051.67310molecules

18

´

´=´

Illustration 40.Calculate the number of moles in each of the following:

(i)392 grams of sulphuric acid (ii) 44.8 litres of carbon dioxide at STP

(iii) 6.022 ( 1023 molecules of oxygen (iv) 9.0 grams of aluminium (v) 1 metric ton of iron (1 metric ton = 103 kg ) (vi) 7.9 mg of Ca (vii) 65.5 (g of carbon.

Solution:(i)1 mole of H2SO4 = 98 g

(

)

24

MolecularmassofHSO213241698u

=´++´=

Q

Thus 98 g of H2SO4 = 1 mole of H2SO4

( 392 g of H2SO4 =

1

392

98

´

= 4 moles of H2SO4

(ii)1 mole of CO2 = 22.4 litres at STP

i.e. 22.4 litres of CO2 at STP = 1 mole

( 44.8 litres of CO2 at STP =

1

44.8

22.4

´

= 2 moles CO2

(iii)1 mole of O2 molecules = 6.022 ( 1023 molecules.

6.022 ( 1023 molecules = 1 mole of oxygen molecules.

(iv)1 mole of Al = 27 g of Al

(

)

Atomicmassofaluminium27g

=

Q

i.e. 27 g of aluminium = 1 mole of Al

( 9 g of aluminium =

1

9

27

´

= 0.33 mole of Al

(v)1 metric ton of Fe = 103 kg = 106 g

1 mole of Fe = 56 g of Fe

( 106 g of Fe =

6

10

molecules

56

= 1.786

´

104 moles

(vi)7.9 mg of Ca = 7.9

´

10(3 g of Ca

(

)

3

7.910

molatomicmassofCa40u

40

-

´

==

(vii)

6

65.5gofC65.510gofC

-

m=´

6

6

65.510

mol5.45810mol

12

-

-

´

==´

Exercise 26.

What is the mass of carbon present in 0.5 mole of K4[Fe(CN)6]?

Exercise 27.

Calculate how many methane molecules and how many hydrogen and carbon atoms are there in 25.0 g of methane?

Exercise 28.

One litre of gas at NTP weighs 1.97g. Find the molecular mass of gas.

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

32

2KClO2KCl3O

¾¾®+

. These is no limiting reactant.

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

3

(1 dm)

of the solution

1

w1000

M

mVinml

´

=

´

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

(

)

1

12

w1000

m

mwingrams

´

=

´

· Relation between molarity and molality

1

1000M

m

(1000dMm)

´

=

´-´

(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,

6

A

massofA

ppm10

massofsolution

=´

The pollution of the atmosphere is also reported in ppm but it is expressed in terms of volumes rather than masses, i.e. volume of the harmful gas (e.g. SO2) in cm3 present in 103 cm3 of the air.

(ii)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.

( Molality (m) =

2

M

1000

1000dMM

´

-

=

2

1000M

1000dMM

-

Thus m =

2

M

dMM/1000

-

or

22

MMmM

mdMorM1md

10001000

æöæö

-=+=

ç÷ç÷

èøèø

or

2

md

M

1mM/1000

=

+

(iii)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

(

)

2

1

m

Molefractionx

m1000/M

=

+

1

1

mM

1000mM

=

+

Thus

1

2

1

mM

x

1000mM

=

+

or

1

211

1000mM

11000

1

xmMmM

+

==+

or

21

1222

1xx

10001

1

mMxxx

-

=-==

or

2

11

1000x

m

xM

=

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

(

)

2

212

M

Molefraction,x

1000dMM/MM

=

-+

(

)

1

2

21

MM

orx

1000dMMMM

=

-+

=

(

)

1

12

MM

MMM1000d

-+

Thus x2 =

(

)

1

12

MM

MMM1000d

-+

Or rearrangement, we get

Or

2

1122

1000dx

M

xMxM

=

+

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.

Illustration 41.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:

0.5

0.5gofNaOHmoleofNaOH

40

=

(

)

Mol.massofNaOH40

=

Q

= 0.0125mole

Thus 500 cm3 of solution contain NaOH = 0.0125 mole

( 1000 cm3 of the solution contain NaOH

=

0.0125

1000

500

´

( 0.025 mole

Hence molarity of the solution = 0.025 M

Illustration 42.Calculate the molarity of a solution containing 9.8 gm of H2SO4 in 250 cm3 of the solution.

Solution:

Mass of H2SO4 dissolved = 9.8 g

Volume of the solution = 250 cm3

Calculation of molarity

Mol. mass of H2SO4 = 98

( No. of moles of H2SO4 =

Massing

Mol.mass

9.8

0.1

98

==

250 cm3 of the solution contain H2SO4

= 0.1 mole

( 1000 cm3 of the solution contain H2SO4

0.1

10000.4mole

250

=´=

Hence molarity of solution = 0.4 M

Illustration 43. 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 H2SO4 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 =

15

0.153moles

98

=

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

0.153

10001.8

85

=´=

Hence the molality of H2SO4 solution = 1.8 m

Calculation of molarity:

15 g of H2SO4 = 0.153 moles

Wt.ofsolution

Vol.ofsolution

Densityofsolution

=

3

100

98.04cm

1.02

==

Thus 98.04 cm3 of solution contain H2SO4 = 0.153 moles

Exercise 29.

Calculate the mole fraction of the solute in an aqueous solution containing 18.0 g of glucose (Mol. mass = 180) in 100 g of water.

Exercise 30.

Calculate the molality and mole fraction of the solute in an aqueous solution containing 23.4 g of NaCl in 90 g water. [Na = 23, Cl = 35.5].

Exercise 31.

Calculate the molarity and molality of 20 percent aqueous ethanol (C2H5OH) solution by volume (density of the solution = 0.960 g per cm3).

Exercise 32.

If 4 g NaOH are dissolved in 100 cm3 of the solution, what shall be the difference in its normality and molality?

Exercise 33.

A solution contain 2.80 mole of acetone and 8.20 moles f chloroform. Calculate the mole fraction of acetone.

Exercise 34.

A solution has 410.3 g of H2SO4 per litre of the solution at 20(C. If its density is 1.243 g cm(3, what will be its molality and molarity?

Exercise 35.

Determine the molality of a solution formed by dissolving 0.850 g of NH3 in 100 g of water.

Exercise 36.

A solution is prepared by adding 60 g of methyl alcohol to 120 g of water. Calculate the mole fraction of methyl alcohol in it.

ANSWERS TO EXERCISES

Exercise 1:

3

3

5.96g1kg100cm100cm100cm

5960kg/m

1000g1m1m1m

1cm

´´´´=

Exercise 2:

The ratio of copper and oxygen is 1:0.25. Hence, the law of constant proportions is illustrated.

Exercise 3.

Ratio of oxygen and iron in both the experiment is 1:2.33

Exercise 4:

0.36 g of Mg combine with chlorine to produce 1.425 g of magnesium chlorine. 9.50 g of another sample of anhydrous magnesium chloride gave, on electrolysis 2.24 litre of chlorine at NTP. Show that these data agree with the law of constant proportions.

Exercise 5:

Mass of metal which combine with 1 part of chlorine are in the ratio of 1:2, which is a simple ratio. Hence, law of multiple proportions is illustrated.

Exercise 6:

The masses of mercury which combine with 1 part of chlorine are in the ratio of 2:1 which is simple ratio. Hence, law of multiple proportions is illustrated.

Exercise 7:

0.477 g

Exercise 9:

Law multiple proportions.

Exercise 12:

75% aluminium

Exercise 13:

MO2 and M2O3

Exercise 14:

Molar mass of metal 100 g.

Exercise 15:

195.21

Exercise 16:

65.3 (atomic mass of zinc)

Exercise 17:

Atomic mass of B = 27 amu

Exercise 18:

335371

A35.5

4

´+´

==

Exercise 19:

1.000199

Exercise 22:

Empirical formula of anhydrous salt = KAS2O8 Hydrated salt composition; % anhydrous part = 54.4% and % H2O = 45.6%; Empirical formula of hydrated salt = KAIS2O8.12H2O.

Exercise 23:

Empirical formula SN2H8O4; Molecular formula = SN2H8O4; Name – Ammonium sulphite (NH4)2SO3.

Exercise 24:

C2H4

Exercise 25:

(i)

(

)

22

2

Ca2HOCaOHH

+¾¾®+

(ii)

(

)

4324

3

AlC12HO4AlOH3CH

+¾¾®+

(iii)

33343

4HPO3HPOPH

¾¾®+

(iv)

4222

2KMnO16HCl2KCl2MnCl8HO5Cl

+¾¾®+++

Exercise 26.36 gm

Exercise 27:

2323

4

9.4110CHmolecules9.4110

´´

carbon atoms and

23

37.6410hydrogenatoms

´

.

Exercise 28:

44.128

Exercise 29:

0.018

Exercise 30:

4.44 m, 0.074

Exercise 31:

3.48 M, 4.35 m

Exercise 32:

No difference

Exercise 33:

0.255

Exercise 34.

Molality = 5.028, Molarity = 4.18

Exercise 35.

0.5 m

Exercise 36.

0.22

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).

ANSWER TO MISCELLANEOUS EXERCISES

Exercise 1:

17.0 g

Exercise 3:

In NH3, 17.65 g of H combine with N = 82.35 g

(1 g of H combine with N

82.35

g4.6g

17.65

==

H

N

O

H

2

O

N

2

O

3

NH

3

Exercise 4:

Yes

Exercise 6:

Molecular formula of sulphur = S8

(Molecular mass of sulphur (S8) = 32 ( 8 = 256.0 (

1 mole of sulphur molecules = 256 g = 6.023 ( 1023 molecules of sulphur

Now, 256 g of sulphur contain 6.022 ( 1023 molecules

(64 g of sulphur will contains

23

23

6.0221064

1.50610molecules

256

´´

==´

1 molecule of sulphur (S8) contains 8 atoms of sulphur

(1.056 ( 1023 molecules of sulphur will contain sulphur atoms

= 8 ( 1.506 ( 1023

= 1.2048 ( 1024 atoms

Exercise 7:

(i)1 mole of KNO3 = 101 g ((Formula mass of KNO3 = 1 ( 39 + 1 ( 14 + 3 (

16 = 101 ()

(0.1 mole of KNO3 = 101 ( 0.1 = 10.1 g of KNO3

(ii)1 mole of CH4 = 16 g = 6.022 ( 1023 molecules

i.e., 6.022 ( 1023 molecules of methane have mass = 16 g

(1 ( 1023 molecules of methane have mass

23

23

16

102.657g

6.02210

=´=

´

(iii)1 mole of H2 = 2 g = 22400 cm3 at STP

i.e., 22400 cm3 of H2 at STP have mass = 2 g

(112 cm3 of H2 at STP will have mass

2

1120.01g

22400

=´=

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

SOLVED PROBLEMS

Subjective:

Board Type Questions

Prob 1.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?

Sol.1 g C will combine with S =

84.21

5.33g

15.79

=

1 g C will combine with O =

72.73

2.67g

27.27

=

S

O

C

CS

2

CO

2

SO

2

(

)

15.79%

(

)

27.27%

(

)

72.73%

(

)

84.21%

50%

50%

( Ratio of masses of S and O which combine with fixed mass of carbon (viz 1 g)

= 5.33 : 2.67

= 2 : 1

Ratio of masses of S and O which combine directly with each other = 50:50=1:1.

Thus the two ratio are simple multiple of each other.

Prob 2.Calculate the mass of iron which will be converted into its oxide (Fe3O4) by the action of 18 of steam on it.

Sol.The chemical equation representing the reaction is

2342

356

418

168g

72

3Fe4HOFeO4H

´

´

=

=

+¾¾®+

Now 72 g of steam react with 168 g of iron

( 18 g of steam will react with

168

1842gofiron

72

´=

Thus the mass of iron required = 42 g.

Prob 3.What mass of slaked lime would be required to decompose completely 4 grams of ammonium chloride and what would be the mass of each product?

Sol.The equation representing the decomposition of NH4Cl by slaked lime, i.e. Ca(OH)2 is,

(

)

(

)

(

)

(

)

(

)

4232

2

214435.540235.522116

21431

402116

111g

107g36g

34g

74g

CaOH2NHClCaCl2NH2HO

+++´´+

+´

++

=

==

=

=

+¾¾®++

(i)To calculate the mass of Ca(OH)2 required to decompose 4g of NH4Cl.

From the above equation,

107 g of NH4Cl are decomposed by 74 g of Ca(OH)2.

( 4 g of NH4Cl will be decomposed by

(

)

2

74

42.766gofCaOH

107

´=

Thus the mass of slaked lime required = 2.766 g

(ii)To calculate the mass of CaCl2 formed.

107 g of NH4Cl when reacted with Ca(OH)2 will produce

2

111

44.15gofCaCl

107

´=

Hence the mass of CaCl2

= 4.15 g

(iii)To calculate the mass of NH3 produced.

107 g of NH4Cl when reacted with Ca(OH)2 give 34 g of NH3.

( 4 g of NH4Cl when reacted with Ca(OH)2 will produce

3

34

41.271gofNH

107

´=

Hence, the mass of NH3 produced

= 1.271 g

(iv)To calculate the mass of H2O formed

107 g of NH4Cl when reacted with Ca(OH)2 yield 36 g of H2O.

( 4 g of NH4Cl when reacted with Ca(OH)2 will yield =

2

36

41.3458gofHO

107

´=

So the mass of H2O formed = 1.3458 g

Prob 4.1.5 g of an impure sample of sodium sulphate dissolved in water was treated with excess of barium chloride solution when 1.74 g of BaSO4 were obtained as dry precipitate. Calculate the percentage purity of the sample.

Sol.

2

treatedwith

244

BaClgave

Giventhat1.5gofimpureNaSO1.74gofBaSO

¾¾¾¾¾®

The chemical equation representing the reaction is

2424

2233241613732416

142g233g

NaSOBaClBaSO2NaCl

´++´++´

==

+¾¾®+

Step 1. To calculate the mass of Na2SO4 from 1.74 of BaSO4. From the chemical equation:

233 g of BaSO4 are produced from 42 of Na2SO4.

( 1.74 of it would be obtained form

142

1.741.06g

233

´=

The mass of pure Na2SO4 present in 1.5 g of impure sample = 1.06 g.

Step 2. To calculate the percentage purity of impure sample.

1.5 g of impure sample contains 1.06 g of pure Na2SO4.

( 100 g of the impure sample will contain

1.06

10070.67

1.5

=´=

g of pure Na2SO4.

Prob 5.A solution contains 25% water, 25% ethanol and 50% acetic acid by mass. Calculate the mole fraction of each component.

Sol.Let us suppose the total mass of the solution is 100 g. Then

Amount of water = 25 g

Amount of ethanol = 25 g

Amount of acetic acid = 50g

[

]

Percentagebymassaregiven

Q

25 g of water =

25

1.388moles

18

=

[

]

Mol.massofwater18

=

Q

25 g of ethanol =

25

0.543moles

46

=

(

)

25

Mol.massofethanolCHOH46

éù

=

ëû

Q

50 g of acetic acid =

50

0.833moles

60

=

(

)

3

Mol.massofaceticacidCHCOOH60

éù

=

ëû

Q

( Mole fraction of water

1.3881.388

0.503

1.3880.5430.8332.764

===

++

0.543

Molefractionofethanol0.196

2.764

==

and mole fraction of acetic acid =

0.833

0.301

2.764

=

IIT Level Questions

Prob 6.In an experiment, 2.4g of iron oxide on reduction with hydrogen yield 1.68g of iron. In another experiment 2.9g of iron oxide give 2.03g of iron in reduction with hydrogen. Show that these data, illustrate the law of constant composition.

Sol.In the first experiment

The mass of iron oxide = 2.4g

Mass of iron after reduction = 1.68

So mass of oxygen = Mass of oxide – Mass of iron

= 2.4 – 1.68 = 0.72

O: Fe = 0.72: 1.68 = 1:2.33

In second experiment mass of iron oxide = 2.9g

Mass of iron after reduction = 2.03g

Mass of oxygen = 2.9 – 2.03 = 0.87g

O : Fe = 0.87 : 2.03 = 1:2.33

This dated illustrate the law of constant proportion.

Prob 7.Certain non metal X forms two oxides I and II. The mass % of oxygen in 1st

46

(XO)

is 43.7, which is same as that of X in

nd

2

oxide. Find the formula of

nd

2

oxide.

Sol.

X

Oxygen

Ist oxide

42.7

56.3

2nd oxide

56.3

43.7

Now 43.7 parts of oxygen corresponds to = 6 oxygen atoms

\

56.3 part of oxygen corresponds to

6

56.37.73

43.7

=´=

oxygen atom

56.3 part of X in I correspond to = 4 X atoms

\

43.7 part of X in II will correspond

4

43.73.1Xatom

56.3

=´=

Now atomic ratio X: oxygen

3.17.73

:1:2.52:5

3.13.1

===

Formula of

nd

2

oxide

25

= XO

Prob 8. Two oxides of a metal contain 27.6% and 30.0% of oxygen respectively. If the formula of the first oxide is

34

MO

, find that of the second.

Sol.In the first oxide, oxygen = 27.6, metal = 100-27.6 = 72.4 parts by weight.

As the formula of the oxide is

34

MO

, this means 72.4 parts by wt. of metal = 3 atoms of metal and 4 atoms of oxygen = 27.6 parts by weight.

In the second oxide, oxygen = 30.0 parts by weight and metal = 100-30 =70 parts by weight.

But 72.4 parts by weight of metal = 3 atoms of metal

\

70 parts by weight of metal

3

70atomsofmetal

72.4

=´

= 2.90 atoms of metal

Also, 27.6 parts by weight of oxygen = 4 atoms of oxygen

\

30 parts by weight of oxygen

4

30atomsofoxygen

27.6

=´

= 4.35 atoms of oxygen

Hence, ratio of M: O in the second oxide = 2.90: 4.35 = 1:1.5 = 2:3

\

Formula of the metal oxide is

23

MO

.

Prob 9.A crystalline salt on being rendered anhydrous loses 45.6% of its weight. The percentage composition of the anhydrous salt is

Aluminium = 10.50%; Potassium = 15.1%; Sulphur = 24.96%; Oxygen = 49.92%

Find the simplest formula of the anhydrous salt.

Sol.Step 1: To calculate the empirical formula of the anhydrous salt.

Element

Symbol

Percentage of elements

At. mass of elements

Relative no. of atoms

=

Percentage

At.mass

Simplest atomic ratio

Simplest whole no. atomic ratio

Potassium

K

15.1

39

15.10

0.39

39

=

0.39

1

0.39

=

1

Aluminium

Al

10.50

27

10.50

0.39

27

=

0.39

1

0.39

=

1

Sulphur

S

24.96

32

24.96

0.78

32

=

0.78

2

6.39

=

2

Oxygen

O

49.92

16

49.92

3.12

16

=

3.12

8

0.39

=

8

Thus the empirical formula for the anhydrous salt is K AlS2O8.

Step 2: to calculate the empirical formula mass of the anhydrous salt.

Empirical formula mass of the anhydrous salt (KAIS2O8).

= 1 ( 39.0 + 1 ( 27.0 + 2 ( 32.0 + 8 ( 16.0

= 258.0 u

Step 3: To calculate the empirical formula mass of the hydrated salt.

Let of weight due to dehydration = 45%

( Empirical formula mass of the anhydrous salt = 100 – 45.6 = 54.4 u

Now, if the empirical formula mass of the anhydrous salt is 54.4 that of hydrated is

= 100

( If the empirical formula mass of the anhydrous salt is 258, that of hydrated is

=

100

258474.3u

54.4

´=

Step 4: To calculate the number of molecules of water in the hydrated salt.

Total loss in mass due to dehydration

= 474.3 – 258.0 = 216.3 u

Loss in mass due to one molecule of water = 18.0 u

( No. of molecule of water in the hydrated sample =

216.3

12

18

=

Step 5: To calculate the empirical formula of the hydrated salt.

Empirical formula of the anhydrous salt = KAS2O8.

No. of molecules of water of crystallization = 12

( Empirical formula of the hydrated salt = KAIS2O8.12H2O.

Prob 10.A carbon compound contains 12.8% carbon, 2.1% hydrogen, 85.1% bromine. The molecular weight of the compound is 187.9. Calculate the molecular formula.

Sol. Step 1: Percentage composition of the elements present in the compound.

C

H

Br

12.8

2.1

85.1

Step 2: Dividing with the respective atomic weights of the elements.

12.8

12

2.1

1

85.1

80

1.067

2.1

1.067

Step 3: Dividing by the smallest number to get simple atomic ratio.

1.067

1.067

2.1

1.067

1.067

1.067

1

2 1

\

The empirical formula is

2

CHBr

.

Empirical formula weight

12 + (21) + 80 = 94

´

The molecular weight = 187.9 (given)

187.9

n2

94

\==

2

The molecular formula = (empirical formu

la)

\

22242

(CHBr) = CHBr

Prob 11.Calculate the number of molecules present in 350 cm3 of NH3 gas at 273 K and 2 atomsphere pressure.

Sol.First of all, we have to determine the volume of the gas at STP.

Given conditions

At STP

V1 = 350 cm3

V2 = ?

T1 = 273 K

T2 = 273 K

P1 = 2 atomsphere

P2 = 1 atm

1122

12

PVPV

Applyinggasequation:

TT

=

We get

2

1V

3502

273273

´

´

=

or

3

2

3502273

V700cm

2731

´

=´=

By mole concept,

1 mole of NH3 = 6.022 ( 1023 molecules = 22400 cm3 at STP.

Thus, 22400 cm3 of NH3 at STP contain 6.022 ( 1023 molecules.

( 700 cm3 of NH3 at STP will contain

23

6.02210

700

22400

´

=´

= 1.882 ( 1022 molecules

Prob 12.In order to find the strength of a sample of sulphuric acid, 10 g were diluted with water and a piece of marble weighing 7g placed in it. When all action had ceased, the marble was removed, washed, dried and was found to weigh 2.2 g. What was the percentage strength of sulphuric acid?

Sol.Mass of marble taken = 7.0 g

Mass of marble left unused = 2.2 g

( Mass of marble reacted = 7.0 – 2.2 = 4.8 g

The chemical equation involved in the above problem is

324422

23241698g

4012163100g

CaCOHSOCaSOHOCO

++´=

++´=

+¾¾®++

Step 1. To calculate the mass of pure H2SO4 required to react 4.8 g of marble.

100 g of marble react with H2SO4 = 98 g

( 4.8 g of marble will react with H2SO4

98

4.84.704g

100

=´=