CBSE Board Questions Ch- 5 Periodic Classification of Elements

Question.  Which physical and chemical properties of the elements were used by Mendeleev in creating his periodic table? List two observations which posed a challenge of Mendeleev’s periodic law. (C.B.S.E 2009)

Ans. The creation of Mendeleev’s periodic table was based upon certain physical and chemical properties.

Physical properties:
The atomic mass of the elements was taken into account and the elements were
arranged in order of increasing atomic masses. The influences of their physical properties such as melting
points, boiling points, density etc.

Chemical properties: 
The distribution of the elements into different groups was linked with formation of hydrides by combining with hydrogen and formation of oxides by combining with oxygen. This is linked with
the valency of the elements.

The two main observations which posed challenge to Mendeleev’s periodic table are as follows.

(i) Position of isotopes: Since the isotopes of an element differ in their atomic masses, they must be assigned
separate slots or positions in the periodic table.
(ii) Anomalous positions of some elements: In the Mendeleev’s periodic table, certain elements with higher
atomic masses precede or placed before the elements with lower atomic masses. For example, the element Ar (Atomic mass = 39.9) is placed before the element K (Atomic mass = 39.1)

Question.  Using the part of the periodic table given below, answer the questions that follow.

Group   Period	I	II	III	IV	V	VI	VII	Zero
1	H							He
2	Li	Be	B	C	N	O	F	Ne
3	Na	Mg	Al	Si	P	S	Cl	Ar
4	K	Ca

(i) Na has physical properties similar to which elements and why?

(ii) Write the electronic configuration of N and P

(iii) State one property common to fluorine and chlorine. (C.B.S.E. All India 2008)

Ans. (i) Na has physical properties similar to Li and K. All the three elements have one electron each in the valence of their atoms. These are known as alkali metals.

(ii) Electronic configuration of N(z = 7) = 2,5

Electronic configuration of P (z = 15) = 2,8,5

(iii) Both the elements have seven electrons in the valence shells as their atoms

Fluorine (z = 19) = 2, 7

Chlorine (z = 17) = 2,8,7

Question. .How and why does the atomic size vary as you go :

(i) from left to right across a period? (ii) down a group? [2009, 2011 (T-II)]

Ans. (i) Atomic size decreases on moving from left to right across a period. This is due to the increase in nuclear charge which tends to pull the electrons closer to the nucleus and reduces the size of the atom.

(ii) Atomic size increases on moving down a group. This is due to addition of new shells which increases the distance between the outermost electrons and the nucleus even though nuclear charge increases

Question. How will the tendency to gain electrons change as we go from left to right across a period?

Why? [2009, 2011 (T-II)]

Ans. On moving from left to right across a period, metallic character decreases and non-metallic

character increases.Since metals tend to lose electrons and non-metals tend to gain electrons, the tendency to gain electrons increases as we move from left to right across a period.

Question. How does electronic configuration of atoms change in a period with increase in atomic

number? [2009]

Ans. On moving across a period from left to right, the atomic number of the elements increases, therefore, the number of electrons in the valence shell increases from 1 to 8, i.e, the first element in the given period will have one electron in its valence shell and the last element in the same period will have eight electrons.

Question.  Lithium, sodium and potassium form a Dobereiner’s triad. The atomic masses of lithium and potassium are 7 and 39 respectively. Predict the atomic mass of sodium. [2009]

Ans. Atomic mass of sodium = (At. mass of lithium + At. mass of potassium)/2

                                            =(7+39)/2 =23

State the first limitation of Mendeleev’s Periodic Table. [2009]

Ans. Hydrogen was not given a fixed position in Mendeleev’s Periodic Table.

Question.  Why was the system of classification of elements into triads not found suitable? [2009]

Ans. (i) Only a few similar elements could be grouped into triads and quite a large number of elements were left out.

(ii) It was also possible to group quite dissimilar elements into triads.

Question. . Why could no fixed position be given to hydrogen in Mendeleev’s Periodic Table? [2009]

Ans. Hydrogen resembled both the alkali metals and the halogens. So, it was placed above both the groups and could not be given a fixed position in Mendeleev’s Periodic Table. 

Question. What are ‘groups’ and ‘periods’ in the periodic table? [2009]

Ans. The vertical columns in the periodic table are known as groups while the horizontal rows are

known as periods.  

Question. Three elements A,B and C with similar properties have atomic masses X,Y and Z respectively.  The mass of Y is approximately equal to the average mass of X and Z. What is such an arrangement of elements called? Give one example of such a set of elements. [HOTS]

Ans. The arrangement of these elements is known as Dobereiner’s triad. Example, lithium, sodium

and potassium.

Question.  Elements have been arranged in the following sequence on the basis of their increasing atomic masses. [HOTS] F, Na, Mg, Al, Si, P, S, Cl, Ar, K

(a) Pick two sets of elements which have similar properties. 

(b) The given sequence represents which law of classification of elements?

Ans. (a) (i) F and Cl (ii) Na and K.

(b) Newland’s Law of Octaves.

Question. Two elements M and N belong to groups I and II respectively and are in the same period of

the periodic table. How do the following properties of M and N vary? [2009, 2011 (T-II)]

(i) Sizes of their atoms                         (ii) Their metallic characters

(iii) Their valencies in forming oxides  (iv) Molecular formulae of their chlorides

Ans. (i) The atomic radii of M is greater than N.

(ii) M is more metallic than N.

(iii) M has a valency of 1 and N has a valency of 2.

(iv) MCl, MCl2

Question. An element X belongs to group 17 and third period of the periodic table.

(a) Write electronic configuration of the element. What is its valency?

(b) Predict its nature, whether it is a metal or non-metal.

(c) Give the formula of the compound formed when it combines with an element Y having valency three. [2011 (T-II)]

Ans. (a) Electronic configuration — 2, 8, 7. Its valency is one.

(b) It is a non-metal (c) YX3

Question. How do atom exists ?

Answer: Atoms of most of the elements are chemically very reactive and do not exist in the free state (as single atoms ). Atoms usually exist in two ways :
(a) In the form of molecules
(b)In the form of ions

Periodic Properties of Elements in modern periodic tables Class X (10^th)

10th Periodic Classification Notes                                 Download File

Periodic Classifications Solved Questions Paper             Download File

Periodic classification of elements question bank            Download File

Dalton’s atomic theory,Atomic mass and Relative Atomic mass (9th CBSE Ch.Atoms and Molecules)

In 1803, A British school teacher John Dalton provided the basic theory about the nature of matter which provides explanation for the law of conservation of mass and the law of definite proportions.

According to Dalton’s atomic theory, all matter, whether an element, a compound or a mixture is composed of small particles called atoms

The postulates of this theory may be stated as follows:

 (i) All matter is made of very tiny particles called atoms.

(ii) Atoms are indivisible particles, which cannot be created or destroyed in a chemical reaction.

(iii) Atoms of a given element are identical in mass and chemical properties.

(iv)  Atoms of different elements have different masses and chemical properties.

(v)  Atoms combine in the ratio of small whole numbers to form compounds.

(vi) The relative number and kinds of atoms are constant in a given compound.

Drawbacks of Dalton’s Atomic Theory:

ð  The atom is further subdivided into protons, neutrons and electrons.

ð  The atoms of same elements are not similar in all respect. They may vary in mass and density. These are known as isotopes. For example: chlorine has two isotopes having mass numbers 35 a.m.u and 37 a.m.u.

ð  Atoms of different elements are not different in all respects. Atoms of different elements that have the same atomic mass are called isobar.

ð  According to Dalton atoms of different elements combine in simple whole number ratio to form compounds. This is not seen in complex organic compounds like sugar C₁₂H₂₂O₁₁.

ð  The theory fails to explain the existence of allotropes like Diamond and Graphite which having different properties even these are made up of same kind ot atom namely Carbon.

The introduction of matter wave concept by de Broglie, the principle of uncertainty by Heisenberg etc., paved the way for modern atomic theory [Modification in Dalton’s atomic theory]

Modifications in Dalton’s atomic theory - Modern atomic theory are as follows.

ð  Atom is considered to be a divisible particle.

ð  Atoms of the same element may not be similar in all respects. eg: Isotopes (¹⁷Cl₃₅,¹⁷Cl₃₇ )

ð   Atoms of different elements may be similar in some respects eg. Isobars (¹⁸Ar ₄₀ , ²⁰Ca 40 )

ð   Atom is the smallest particle which takes part in chemical reactions.

ð  The ratio of atoms in a molecule may be fi xed and integral but may not be simple e.g., C₁₂H₂₂O₁₁ is not a simple ratio (Sucrose)

ð  Atoms of one element can be changed into atoms of other element by transmutation.

ð  The mass of an atom can be converted into energy. This is in accordance with Einstein’s equation E = mc²

Atom: It is the smallest particle of an element which may or may not have independent existence. The atoms of certain elements such as hydrogen, oxygen, nitrogen, etc .do not have independent existence whereas atoms of helium, neon, argon, etc. do have independent existence. Thus we can say that all elements are composed of atoms.

Q. How do we know the presence of atoms if they do not exist independently for most of the elements?

Answer: Atom join in different way to form matter (neutral molecules or ion) that we are able to touch, feel and see.

How big are atoms?

Atoms are extremely small. They are so small, that you cannot see them with most microscopes. Now, Scan tunneling Microscope (STM) is the modern instrument that made it possible to take photograph of atom. The size of an isolated atom can 't be measured because we can 't determine the location of the electrons that surround the nucleus. We can estimate the size of an atom, however, by assuming that the radius of an atom is half the distance between adjacent atoms in a solid. This technique is best suited to elements that are metals, which form solids composed of extended planes of atoms of that element. The results of these measurements are therefore often known as metallic radii.

Q. What is the unit of measurement of atomic radius?

Ans: Picometers (pm) or Angstroms (Å)

Q.The size of sodium atom is bigger than that of hydrogen atom. Why?

Answer: Size of atom is the distance between the nucleus and outermost shell ( valence shell ) of an atom. The atomic number of sodium is greater than that of hydrogen. So, it needs more number of shells to fill electrons and hence will have more number of shells than hydrogen. Hence, atomic size of sodium is bigger than that of sodium.

Naming of an element

Dalton was the first scientist to use the symbols for elements in a very specific sense.

Q. Why are Dalton’s symbols not used in chemistry?

Answer: Dalton was the first scientist to use the symbol for the name of the elements a specific sense but it was difficult to memorize and in uses so Dalton's symbol are not used in chemistry

Berzelius suggested that the symbols of elements be made from one or two letters of the name of the element.

IUPAC (International Union of Pure and Applied Chemistry) approves names of elements. Many of the symbols are the first one or two letters of the element’s name in English. The first letter of a symbol is always written as a capital letter (uppercase) and the second letter as a small letter (lowercase)

For example:  (i) hydrogen, H (ii) aluminum, Al and not AL (iii) cobalt, Co and not CO.

Symbols of some elements are formed from the first letter of the name and a letter, appearing later in the name. Examples are: (i) chlorine, Cl, (ii) zinc, Zn etc.

Other symbols have been taken from the names of elements in Latin, German or Greek. For example, the symbol of iron is Fe from its Latin name ferrum, sodium is Na from natrium, and potassium is K from kalium. Therefore, each element has a name and a unique chemical symbol.

Molecule: A molecule is the smallest or the simplest structural unit of an element (or) a compound which contains one (or) more atoms. It retains the characteristics of an element. A molecule can exist freely and it is a combined form of bonded units whereas an atom is a singular smallest form of non bonded unit.

Molecules are of two types, namely homo atomic molecules and hetero atomic molecules.

Homo atomic molecules: These are the molecules which are made up of atoms of the same element. For example hydrogen gas consists of two atoms of hydrogen (H₂).Similarly oxygen gas consists of two atoms of oxygen (O₂).

HETERO ATOMIC MOLECULES : The hetero atomic molecules are made up of atoms of different elements. They are also classified as diatomic, triatomic, or polyatomic molecules depending upon the number of atoms present. H₂O, NH₃, CH₄, etc., are the examples for hetero atomic molecules.

Atomicity:The number of atoms present in one molecule of an element is called the atomicity of an element. Depending upon the number of atoms in one molecule of an element, molecules are classified into monoatomic, diatomic, triatomic or poly atomic molecules containing one, two, three, or more than three atoms respectively.

Mono atomic molecules: Helium (He) Neon (Ne) Metals ;  Di atomic molecules:   Hydrogen H₂ Chlorine Cl₂

Tri atomic molecules: Ozone (O₃) ; Poly atomic molecules: phosphorous P₄ Sulphur S₈

Atomicity = Molecular Mass/Atomic mass

MORE TO KNOW:

Isotopes ⇒ These are the atoms of same element with same atomic number (Z) but different mass number (A). Example (¹⁷Cl₃₅,¹⁷Cl₃₇ )

Isobars ⇒ These are the Atoms of the different element with same mass number but different atomic number. Example (¹⁸Ar₄₀, ²⁰Ca ₄₀ )

 Isotones ⇒ These are the atoms of different elements with same number of neutrons Example ( ⁶C₁₃, ⁷N₁₄ )

AVOGADRO’S HYPOTHESIS: Amedeo Avogadro put forward hypothesis and is based on the relation between number of molecules and volume of gases that is “volume of a gas at a given temperature and pressure is proportional to the number of particles”.

Avogadro’s Law: Equal volumes of all gases under the same conditions of temperature and pressure. contain the equal number of molecules.

TEST YOUR UNDERSTANDING SKILL

 (a) Find the atomicity of chlorine if its atomic mass is 35.5 and its molecular mass is 71 

(b) Find the atomicity of ozone if its atomic mass is 16 and its molecular mass is 48

WHAT IS AN ION? 

An ion is a charged particle and can be negatively or positively charged.

A negatively charged ion is called an ‘anion’ and the positively charged ion, a ‘cation’. For example, sodium chloride (NaCl). Its constituent particles are positively charged sodium ions (Na+) and negatively charged chloride ions (Cl–).

Ions may consist of a single charged atom or a group of atoms that have a net charge on them. A group of atoms carrying a charge is known as a polyatomic ion e.g. Calcium oxide (Ca⁺² O^-2)

Atomic mass and Relative Atomic mass (RAM):

Q. Each element had a characteristic atomic mass even then we are using Relative Atomic mass. Give reason?

Answer: Since determining the mass of an individual atom was a relatively difficult task due to extremely smaller size, relative atomic masses were determined using the laws of chemical combinations and the compounds formed.

Relative Atomic mass (RAM): In 1961 IUPAC selected an isotope of carbon (¹²C) as a slandered for comparing atomic and molecular mass of element and compound.

Relative atomic mass of an element is the ratio of mass of one atom of element to the 1/12^th part of mass of one atom of carbon. Relative atomic mass is a pure ratio and has no unit. If the atomic mass of an element is expressed in grams, it is known as gram atomic mass. e.g. 

Gram atomic mass of hydrogen = 1g where as gram atomic mass of carbon = 12g

Atomic mass is expressed in atomic mass unit (amu). One atomic mass unit is defined as 1/12th part of the mass of one atom of carbon.

Q. The atomic mass of an element is in fraction .What does it mean?

Ans If the atomic mass of an element is in fraction, this mean that it exists in the form of isotopes. The atomic mass is the average atomic mass and is generally fractional.

Chemical Formulae: The chemical formula is a symbolic representation of a compound of its composition.

For writing Chemical Formulae the name or symbol of the metal is written first then non-metals with their valencies. Then we must crossover the valencies of the combining atoms. For example:

 (a) Formula for aluminium oxide: Al³⁺ O^-2 Þ Al₂O₃  (b) calcium hydroxide :  Ca⁺² OH^-1 Þ Ca(OH)₂

Valency: The combining power (or capacity) of an element is known as its valency. Valency can be used to find out how the atoms of an element will combine with the atom(s) of another element to form a chemical compound.

RELATIVE MOLECULAR MASS (RMM) : The relative molecular mass of an element or a compound is the ratio of mass of one molecule of the element or a compound to the mass of 1/12 th part of mass of one atom of carbon. Relative Molecular mass is a pure ratio and has no unit. If the molecular mass of a given substance is expressed in gram, it is known as gram molecular mass of that substance.

Molecular mass is the sum of the masses of all the atoms present in one molecule of the compound or an element.

Test your numerical skill:

Problem:    Find the gram molecular mass of water (H2O)

Solution:  Þ 2(H) = 2 x 1 = 2  and   1(O) = 1 x 16 = 16    ;  Gram molecular mass of H2O = 2 + 16 = 18g

Problem:  Find the gram molecular mass of carbon dioxide

Solution:  Þ  (CO2) 1(C) = 1 x 12 = 12 and  2(O) = 2 x 16 = 32;  Gram molecular mass of CO2 = 12 + 32 = 44 g

Calculate the percent by mass (weight) of sodium (Na) and chlorine (Cl) in sodium chloride (NaCl)

Relative molecular mass of NaCl  = 22.99 + 35.45 = 58.44

The total mass of Na present:

1 Na is present in the formula, mass = 22.99

Tthe percent by mass (weight) of Na in NaCl: 

%Na = (22.99 ÷ 58.44) x 100 = 39.34%

The total mass of Cl present:

Cl is present in the formula, mass = 35.45

The percent by mass (weight) of Cl in NaCl:

%Cl = (35.45 ÷ 58.44) x 100 = 60.66%

 Here,   39.34 + 60.66 = 100.

The answers above are probably correct if %Na + %Cl = 100, that is, 

Empirical formula and molecular formula: 

Empirical formula: The empirical formula is the simplest formula for a compound in which atoms of different elements are present in simple ratio. It shows the relative number of atoms of each element. For example CH₂O is the empirical formula of Glucose C₆H₁₂O₆

Molecular formula: It is the formula in which the actual number of atoms of  different element are present. For example, if the empirical formula of benzene is CH where as molecular formula is C₆H₆, etc.

Further reading Empirical formula and molecular formula 

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9th(CBSE) Ch. Atoms and Molecules - Mole Concept and Related Problems

MOLE CONCEPT [CBSE Class IX Chemistry]

While performing a reaction, to know the number. of atoms (or) molecules involved, the concept of mole was introduced. The quantity of a substance is expressed in terms of mole.

Definition of mole:  Mole is defined as the amount of substance that contains as many specified elementary particles as the number of atoms in 12g of carbon-12 isotope.

One mole is also defined as the amount of substance which contains Avogadro number (6.023 x 10²³) of particles.

MORE TO KNOW

Avogadro number: Number of atoms or molecules or ions present in one mole of a substance is called Avogadro number. Its value is 6.023x10²³.

Therefore, one mole of any substance = 6.023 x 10²³ particles may be atoms, molecules, ions

For example: One mole of oxygen atoms represents 6.023 x 10²³ atoms of oxygen and 5 moles of oxygen atoms contain 5 x 6.023x10²³ atoms of oxygen.

Questions based on mole concept:

1. When the mass of the substance is given:  Use this formula: Number of moles = given mass/ atomic mass

(a). Calculate the number of moles in  (i) 81g of aluminium ii) 4.6g sodium  (iii) 5.1g of Ammonia  (iv) 90g of water  (v) 2g of NaOH 

Solution: (i) Number of moles of aluminum = given mass of aluminium / atomic mass of aluminium = 81/27 = 3 moles of aluminum [Rest Question do yourself]

(b) Calculate the mass of 0.5 mole of iron

Solution: mass = atomic mass x number of moles = 55.9 x 0.5 = 27.95 g

Do yourself:  Find the mass of 2.5 mole of oxygen atoms [ Mass = molecular mass x number of moles]

2. Calculation of number of particles when the mass of the substance is given:

Number of particles =( Avogadro number x given mass)/gram molecular mass

Problem: Calculate the number. of molecules in 11g of CO₂

Solution: gram molecular mass of CO₂ = 44g

Number of molecules = (6.023 x 10²³ x 11) / 44 = 1.51 x 10²³ molecules

Do yourself: Calculate the number of molecules in 360g of glucose

3. Calculation of mass when number of particles of a substance is given:

Mass of a substance = (gram molecular mass x number of particles)/6.023 x 10²³

Problem: Calculate the mass of 18.069 x 10²³ molecules of SO₂

Solution: Gram molecular mass SO₂ = 64gm

The mass of 18.069 x 10²³ molecules of SO₂ = (64x18.069 x 10²³)/ (6.023 x 10²³) = 192 g

Do yourself: (a) Calculate the mass of glucose in 2 x 10²⁴ molecules (b) Calculate the mass of 12.046 x 1023 molecules in CaO

4. Calculation of number of moles when you are given number of molecules:

Problem:  Calculate the number moles for a substance containing 3.0115 x 10²³ molecules in it.

Solution: Number of moles = [Number of molecules/(6.023 x 10²³)]  = ( 3.0115 x 10²³)/( 3.0115 x 10²³) =0.5 moles

Do yourself: (a) Calculate number of moles in 12.046x 1022 atoms of copper (b) Calculate the number of moles in 24.092 x 1022 molecules of water.

Problem: Calculate the number of aluminum ions present in 0.051 g of aluminum oxide. (Hint: The mass of an ion is the same as that of an atom of the same element. Atomic mass of Al=27 u)

Solution: Mass of the 1 mole of Al₂ O₃  = 2x27 + 3x16 = 102gm

The number of ions present in 102 gm of aluminum oxide  = 6.023 x 10²³ ion

The number of ions present in 0.051g of aluminum oxide= (6.023 x 10²³ ion x 0.051g)/ 102 gm 

=  6.023 x 10²³ ion x0.0005 = 3.0115 x 10²⁰ ions

In Al₂ O₃, Aluminium and oxygen are in ratio 2:3

So, The number of aluminum ions present(Al³⁺) in 0.051g of aluminum oxide = 2 x 3.0115 x 10²⁰ ions =6.023 x 10²⁰ ion

MORE TO KNOW

Volume occupied by one mole of any gas at STP is called molar volume. Its value is 22.4 litres 22.4 litres of any gas contains 6.023 x 1023 molecules.

Problem: Calculate the volume occupied at STP by :- (a) 64 gram of oxygen gas (b) 6.02 x10²² molecules of CH₄ (c) 5 moles of nitrogen gas

Solution: (a) One mole of a gas occupies 22.4 L volume at STP

Now, number of moles in 64 g oxygen gas = 64/32 = 2

Therefore, volume occupied by 2 moles(64 g) of oxygen gas = 2 x 22.4 L = 44.8 L

(b) 1 mole = 6.02 x 10²³ molecules

Therefore, 1 mole (6.02 x 10²³ molecules) of CH₄ gas occupies 22.4 L volume at STP.

(c) One mole of a gas occupies 22.4 L volume at STP

Therefore, volume occupied by 5 moles of nitrogen gas = 5 x 22.4 L = 112 L

Problem: Calculate the volume at STP occupied by 10²¹ molecules of Oxygen?

Solution: The molar volume that is the volume occupied by one mole of gas is 22.4 L. We know there are 6.022 X 10²³ particles of a substance in one mole of that substance. Thus

volume occupied by 6.022 X 10²³ molecules of oxygen = 22.4 L

volume occupied by 10²¹ molecule of oxygen = [(22.4 X 10²¹) / (6.022 X 10²³)] 

                                                                              = 3.72 X 10^-2 L = 37.2 ml

Q. Calculate the mass and volume of oxygen at STP, which will be evolved on electrolysis of 1 mole of(18 g) of water.

Answer: Electrolysis of water will break it down in its component as Hydrogen and Oxygen

Balanced chemical reaction:

2H₂O → 2H₂ + O₂

From above equation, 2 mole of water will evolve 1 mole of oxygen gas upon electrolysis. Therefore 1 mole of water will produce 1/2 mole of Oxygen.

Mass of Oxygen evolved = number of moles of Oxygen evolved × Molecular wt. of Oxygen =1/2 × 32 = 16 g

At STP 1 mole of any gas will occupy 22.4l volume.

Volume of Oxygen evolved = No. of moles × 22.4l = 1/2 × 22.4l = 11.2l.

Q. How many molecules are present in 1 ml of water?

Answer: we know that density of water is 1gm/ml.

Hence, 1 gm water will = 1 ml water.

Now, we have molecular mass of water H₂O = 1x2 + 16 = 18 gm

18 gm of water contain 6.022 x 10²³ molecules

1 gm of water will contain = (6.022 x 10²³)/18 molecules = 0.33 x 10²³ molecules

So, the no. of molecules of water in 1ml of water = 3.3 x 10²²

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Laws of chemical combination and Landolt Experiment ( IX Atoms and Molecules)

How and why elements combine and what happens when they combine. Antoine L. Lavoisier laid the foundation of chemical sciences by establishing two important laws of chemical combination.

(a) The Law of conservation of mass was stated by Antoine L. Lavoisier in 1785 as” Mass can neither be created nor destroyed in a chemical reaction” [The Law of conservation of mass is the 2^nd  postulate of Dalton's atomic theory. It states that Atoms are indivisible particles, which cannot be created or destroyed in a chemical reaction.]

Example:
ð  Water forms by the union of hydrogen and oxygen. If we weigh the reactants (hydrogen and oxygen) before the chemical reaction, we find the weight of the product (water) equal to the combined weight of reactants.
ð  The weight of iron increases on rusting. The increase in weight is equal to the weight of oxygen added to iron.

ð  Carbon combines with Sulphur to form Carbon disulphide. The mass of reactants i.e. carbon and sulphur is same mass of products (carbon disulphide).

 Carbon + Sulphur   -----➜   Carbon DiSulphide

 C          +   S            -----➜     C₂S  

1g         +  5. 34g            =       6.34 g

LHS                                =     RHS

The verification of the Law of conservation of mass by Landolt Experiment

H. Landolt was German Chemist. He proved the law of conservation of mass by using an H-shaped glass tube. He filled silver nitrate in limb A and hydrochloric acid in limb B. The tube was sealed and weighed before the chemical reaction. The reactants were mixed by inverting and shaking the tube. A white precipitate of silver chloride was formed along with Sodium nitrate . The tube was weighed again. He found that there was no change in weight during the following chemical reaction. 

AgNO₃ + NaCl      ® AgCl₂ (white precipitate) + NaNO₃

Limitations [Is there any exception to law of conservation of mass?]

In all the chemical reactions, energy is evolved or absorbed which would be at the expense of change in mass. In ordinary chemical reactions, this change in mass is so small that it cannot be registered on the most sensitive balance. This suggests that some matter of the reaction mixture gets converted into energy such as light, heat etc. Thus mass and energy are interconvertible. The mass is converted to energy by Einstein’s relation E = mc².

(b) The law of constant proportions which is also known as the law of definite proportions was stated by Proust in 1799 as “In a chemical substance the elements are always present in definite proportions by mass”.

[The Law of constant proportions is the 6^th postulate of Dalton's atomic theory. The relative number and kinds of atoms are constant in a given compound.]

E.g. In a compound such as water, the ratio of the mass of hydrogen to the mass of oxygen is always 1:8, whatever the source of water. Thus, if 9 g of water is decomposed, 1 g of hydrogen and 8 g of oxygen are always obtained.

if the element ‘A’ and ‘B’ combine chemically to form the compound AB, then in whatever manner AB is formed, it is always composed of same two elements ‘A’ and ‘B’ combined together in the same fixed ratio or proportion by weight.
For example:  Sulphur dioxide can be obtained b following sources:

         (i).     Sulphur is burnt in air, 
                   S               +    O₂                    ---------®   SO₂
         (ii).    Copper is heated with conc. sulphuric acid
                   Cu            +   2H₂SO₄       ---------®  CuSO₄   +   2H₂O   +   SO₂
         (iii).   Dilute hydrochloric acid is added to sodium bisulphate
                   NaHSO₃   +   HCl            ----------®   NaCl   +   H₂O   +   SO₂
In each case, The ratio of sulphur and oxygen in the sulphur dioxide obtained is of 32 : 32 or 1 : 1 by mass.

The Law of Multiple Proportions
when two elements A and B combine to form more than one compounds, then the weight of one is constant and the other has a simple ratio.  [The Law of Multiple Proportions is the third postulate of Dalton's atomic theory. It states that the masses of one element which combine with a fixed mass of the second element are in a ratio of whole numbers.]
E.g. Two different compounds are formed by the elements carbon and oxygen.

C  (12gm)        +         ½ O₂   (16gm)                         ---------®         CO (28gm)

C  (12gm)        +         O₂       (32gm)            ---------®         CO₂ (44gm)

Here, 12 gm of carbon combine with 16g and 32gm of Oxygen to form Carbon monoxide and Carbon dioxide respectively. The ratio of oxygen combining with 12 gm of Carbon is 16: 32 or, 1:2 which is in a simple ratio.

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