Chemistry For Class IX - Chapter No.2 - Questions And Answers

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Chapter No.2: Laws Of Chemical Combinations
Questions And Answers

CHEMICAL REACTIONS:

 "When two or more substances combined together or a single substance changes up to produce one or more substances with entirely different properties, such a change is called chemical reaction."

• 4Fe + 3O₂ →2Fe₂O₃
• 2MG + O₂ →2MgO

 LAWS OF CHEMICAL COMBINATION:

 Chemistry deals with chemical reactions. Chemist had found that these changes are governed by some empirical law known as Laws of Chemical Combinations.
These laws are:

• Law of conservation of mass.
• Law of constant composition OR Law of definite proportions.
• Law of multiple proportions.
• Law of reciprocal proportions.

1. LAWS OF CONSERVATION OF MASS:

 Lavoisior in 1785 gave the law of conversation of Mass. This law states that:

 “Matter can neither be created nor be destroyed by chemical change.”

 OR

 “in any chemical reaction the initial mass of reacting substance is equal to the final mass of products.”

 After latest research law can be stated as:

 “There is no detectable gain or loss of mass in an ordinary chemical reaction.”

PRACTICAL VERIFICATION: (Landolt Experiment)

This law was verified by many experiments performed by H. Landolt. He was a German chemist. His most popular and the simplest experiment is as follows:
Experiment: He took an H- shaped tube having two limbs ‘A’ and ‘B’ as shown in figure. One limb ‘A’ was filled with AgNO3solution and the other limb was filled with HCl solution. The upper portion of the limbs were sealed to avoid the escaping of any material. Both solutions are colourless. The H-shaped tube was weighed in vertical position to avoid mixing. Then the tube was inverted and shaken to mix the two solutions. Following reaction took place.

AgNO₃ + HCl →AgCl + HNO₃ (white ppt)

       "Due to formation of AgCl precipitates of white color, the entire tube became" white. The reaction was completed by shaking and inverting the tube.
        H-shaped tube was weighed again. It was observed that the tube mass of the substances before the reaction was equal to the total mass of the substances after reaction.

CONVERSATION OF MASS TO ENERGY: 

   Certain radioactive substances like uranium undergo changes of such nature that very small quantity of mass is conventional to energy by thy following equation given by Albert Einstein in 1906.

E = mc²

 Where “m” is mass of the substance in gm and
 “c” is the velocity of light in cm/sec (3x10² cm/sec).
By putting these values in equation (A) we can calculate the amount of Energy obtained by the conversation of the mass. This change of mass in ordinary experiments is so small that it cannot be detected by ordinary weighing techniques.

FINALLY THE LAW OF CONSERVATION OF MASS

    Law of conservation of mass can be stated as:

 “There is no detectable gain or loss of mass in an ordinary chemical reaction.” 

 2. LAW OF CONSTANT COMPOSITION OR LAW OF DEFINITE PROPORTIONS: 

 Statement:

 “Different samples of the same compound always contain the same 

element combined together in the same proportion by mass” 

Example:
 "Water obtained from any source( prepared in laboratory, or obtained from rain, river or water pump), but if it is pure water always contain Hydrogen and Oxygen in the ratio of 1:8 by mass."
 H₂O
 2:16 (Atomic mass of H is 1 and O is 16)
1:8 (parts by mass)

 Experimental verification: 

 Swedish chemist J.J Berzelius performed an experiment to prove this law.
10gm of lead (Pb) was heated with excess of sulphur but only 1.56gm of sulphur combined to give 11.56 gm of PbS.
    "Again this experiment was repeated by heating 18gm of Pb and 1.56gm of ”S”, it was observed that 11.56gm of PbS was prepared and 8g of Pb remained unused."
   This indicates that Pb and S always combine in the fixed ratio by mass. This is according to the law of constant composition.

3. LAW OF MULTIPLE PROPORTIONS: 

Statement:

“If two element combined to form more than one compound, the masses of one 

element that combines with a a fixed mass of the other element are" in the ratio 

of small whole numbers or some multiple of it”

 Example: Carbon and oxygen combine to form two stable compounds CO and CO₂

COMPOUND	MASS OF CARBON (C)	MASS OF OXYGEN (O)	RATIO OF (O)
Carbon monoxide CO	12	16	1
Carbon dioxide CO2	12	32	2

  From above chart it is very clear that 12g ‘C’ combines with 16g of ’O’ in CO and 32gm of ‘O’ in CO2 . Hence the ratio of ‘O’ is 16:32 or 1:2 which is the simple multiple ratio.

4. LAW OF RECIPROCAL PROPORTIONS: 

Statement:

 “when two different elements separately combine with the fixed mass of

 third element the proportions in which they combine with one another

 shall be either in the simple ratio or some multiple of it.” 

 Example: "‘C’ , ’H’ and ‘O’ combine separately to form CH₄, H₂O and CO₂"

METHANE CH4	MASS OF C-12	MASS OF H-4	RATIO OF H:C 4:12 OR 1:3
Water H2O	Mass of H ‘2’	Mass of O-16	RATIO OF H:O 2:16 OR 1:8
Carbon dioxide CO2	Mass of C 12	Mass of O-32	RATIO OF C:O 12:32 OR 3:8

 From the above chart It is clear that in CH₄ 1gm H combines with 3 gm of C and in H₂O 1gm H combines with 8 gm O. Also in CO₂ the ratio of C and O is 3:8 which is according to the law of Reciprocal proportion.

 ATOMIC MASS

 “ The atomic mass of an element is the average mass of natural mixture of isotopes 

which is compared to the mass of 1/12^th part of an atom C-12.”

Its unit is a.m.u.
The mass of an atoms depends upon Neutrons and protons present in nucleus The mas of an atom is so small that it cant be measured by any ordinary weighing instrument.
The mass of Hydrogen atom is 1.6 x 10^-24 g (a very small mass)
C-12 has 6 protons and 6 neutrons, so the mass of C is 12 atomic mass unit (a.m.u).
The mass of C-12 is taken as a standard. Hence (a.m.u is 1/12^th of the mass of C-12 atom. For example the mass of H is 1/12^th mass of C-12, so its atomic mass is 1 a.m.u. Most of the elements consists of its isotopes. The average of the mass of isotopes give the atomic mass of the atom.
The atomic mass of O=16, S=32, H=1, Ca=40, Mg=24.

EMPIRICAL FORMULA (E.F) OR SIMPLEST FORMULA

“The formula that shows the simplest ratio between the atoms of different elements of a 

compound is called empirical formula.”

COMPOUND	MOLECULAR FORMULA	EMPIRICAL FORMULA
Benzene	C6H6	CH
Glucose	C6H1206	CH20
water	H20	H20

The empirical formula of Benzene, Glucose and Hydrogen Peroxide are different from their Molecular Formula.
Hence for a compound, Empirical and Molecular Formula may be similar or different.

MOLECULAR FORMULA

Molecular Formula is the formula which represents a molecule of an element or a compound 

with exact number of atoms.

Relationship between molecular and empirical formula:
Molecular formula = n x Empirical Formula
OR  n = (Molecular Formula)/(Empirical Formula)
Similarly n = (Molecular Formula mass)/(Empirical Formula mass)

MOLECULAR FORMULA MASS OR MOLECULAR MASS

“Molecular Formula Mass of a substance is the sum of the atomic mass of all atoms present in the molecular formula of a substance or molecule.”

COMPOUND	MOLECULAR FORMULA	MOLECULAR MASS
Carbon dioxide	CO2	12+2(16)=12+32=44
Glucose	C6H1206	6(12)+12(1)+6(16)=180
water	H20	(1)+16=2+16=18

FORMULA MASS:

“Formula mass of a substance is the sum of the atomic masses of all atoms in a

 formula unit of the substance”.

Explanation:
Some compounds are not available in molecular form. For example NaCl is available in ionic form NaCl, so we can consider its formula mass and not molecular mass.

MOLAR MASS

“Molar mass of the substance is its relative Atomic mass, Molecular mas or 

Formula mass expressed in grams.”

MOLE

“The atomic mass, Molecular mass or Formula mass of a substance expressed in grams

 is known as Mole.”

AVOGADRO’S NUMBER (NA)

“One mole of any substance contains 6.02x10?? particles (atoms, molecules, ions or formula units). This constant number is called Avogadro’s number.”

INTER-CONVERSION OF MASS AND MOLE

Number of moles = (Mass of substance in grams)/(Gram Atomic mass OR formula mass)
Mass of substance = Number of moles x Gram Atomic mass or Formula mass

PROBLEM:  Calculate the number of moles in 50g of each: (a) Na (b) H2O
Solution:
Method # 1
Given:
Given mass of Na = 50g
Atomic mass of Na = 23 a.m.u
Required:
Number of moles of Na = ?
Formula:
Number of moles of Na = (Mass of Na in grams)/(Gram Atomic mass of Na) = 50/23
Number of moles of Na = 2.173 moles of Na Ans.

Solution:
Method # 2
Given:
Given mass of H₂0 = 50g
Atomic mass of H₂0 = 18 a.m.u
Required:
Number of moles of H₂O = ?
Formula:
Number of moles of H₂0 = (Mass of H₂0 in grams)/(Gram Atomic mass of H₂0)
= 50/18
Number of moles of H₂0 = 2.777 moles of H₂0 Ans.

USE OF AVOGADRO’S NUMBER:

PROBLEM: Calculate the number of Atoms in 10g of Al.
Solution:
Method # 1
According to Avogadro’s number.
1 mole of Al = 27g = 6.02x10²³ atoms.
This shows that.
27g of Al contain 6.02x10²³ atoms of Al 1g of Al contain 6.02x10²³/ 27g
10g of Al contain (6.02x10²³ x 10) / 27g = 2.23x10²³ atoms of Al Answer.

Solution:
Method # 2
Number of atoms = (NA x Mass of substance in grams) / (Gram Atomic mass)
Number of atoms of Al = (6.02x10²³ x 10g) / 27g
Number of atoms of Al = 2.23x10²³ atoms of Al Answer

CHEMICAL REACTION OR CHEMICAL CHANGE:

“A chemical reaction is that change in which the chemical composition of a substance is altered.”
During a chemical reaction, the original substances produce one or more new substances.
Examples:
i) Rusting of Iron 
4Fe + 3O₂→2Fe₂O₃
Iron + Oxygen → Iron oxide (rust)

ii) Burning of Coal
C + O₂ → CO₂
Carbon + Oxygen → Carbon dioxide

TYPES OF CHEMICAL REACTION:

Chemical reaction is divided into five different types.
i) Decomposition reactions.
ii) Addition reaction (Combination reaction)
iii) Single displacement reaction
iv) Double displacement reaction
v) Combustion reaction

1. DECOMPOSITION REACTIONS:
“Decomposition reactions is that chemical reaction in which a substance is divided into two or more simpler substances.”
Example:

CaCO₃(s) → CaO(s) + CO₂(g)

In this reaction Calcium Carbonate on heating is divided into Calcium oxide and Carbon dioxide gas.

2. ADDITION REACTION (Combination Reaction):
“Addition reaction is that reaction in which two or more substances combine to form a single substance.”
Addition reaction is the reverse of decomposition reaction.
Example:

CaO(s) + CO₂(g) → CaCO₃(s)

In this reaction Calcium Oxide and Carbon dioxide are added to give Calcium carbonate.

3. SINGLE DISPLACEMENT REACTION:
“Displacement reaction is that reaction in which one atom or group of atoms of a compound is replaced by another atom or group of atoms.”
Example:

Zn + 2HCl → ZnCl₂ + H₂(g)

In this reaction H of HCl is replaced by Zn

4. DOUBLE DISPLACEMENT REACTION:
“In double displacement reaction the two compound exchange their partners so that new compounds are formed.”
Example:

NaCl + AgNO₃ →NaNO₃ + AgCl

In the above reaction the exchange of partners takes place.

5. COMBUSTION REACTION:
“In combustion reaction substances react with oxygen (free oxygen or oxygen of air) to produce eat energy and flame.”
Example:

C + O₂→CO₂ + ∆H

In this reaction carbon is burnt with oxygen to form carbon dioxide along with evolution of heat and flame.

CHEMICAL EQUATION

“Chemical equation is a method of expressing the chemical reaction in terms of symbols and formula of the substances involved in the chemical equation.”
Example:

C + O₂ → CO₂

POINTS TO REMEMBER:
1. Reactants are on the left hand side of the reaction and product are on the right hand side of the reaction.
2. Balancing of Equation is done by coefficients.
3. Delta over an arrow indicates that reactant is heated to give product.
4. Catalyst is represented by its symbol written over arrow.

BALANCE CHEMICAL EQUATION:
Balanced chemical equation gives the following information.
i) The nature of reactants and products.
ii) The relative number of each reactants and products.

RULES OF BALANCE CHEMICAL EQUATION:
i) Write the given formula for all reactants on the left hand side and the formula of products on right hand side of an equation.
ii) Mention the number of atoms on both sides of chemical equation.
iii) If the number of atoms appear more on one side than the other, balance the equation by inspection method. For this purpose multiply the formula by coefficient so as to make the number of atoms, same on both sides of an equation.
iv) The covalent molecule of hydrogen, oxygen, nitrogen and chlorine exist as diatomic molecule e.g H₂, N₂, O₂ and Cl₂, rather than isolated atoms hence we must write them as such in chemical equation.
v) Finally, check the balanced equation, to be sure that the number and kind of atoms are the same on both sides of the equation.

BALANCE THE CHEMICAL EQUATION:

Example:
1. KClO₃→KCl + O₂
Write down the number of atoms on each side.

Reactants → Products

K (1)	K (1)
Cl (1)	Cl (1)
O (3)	O (2)

It is clear that K and Cl elements have same number of atoms on both sides of equation, but the Oxygen atoms are not balanced so we place 2 on left hand side and 3 on right hand side (cross multiply) to balance the Oxygen atoms.

2KClO₃→KCl + 3O₂

Reactants → Products

K (2)	K (1)
Cl (2)	Cl (1)
O (6)	O (6)

Now we simply balance K by placing 2 in front of KCl.

2KClO₃ → 2KCl + 3O₂

Reactants → Products

K (2)	K (2)
Cl (2)	Cl (2)
O (6)	O (6)

The Equation is now balanced.

2KClO₃ → 2KCl + 3O₂

CONCEPT OF MOLE RATIO TO CALCULATE THE AMOUNT OF REACTANTS:

Q. Consider the following reaction.

2H₂ + O₂ → 2H₂O

i) How many moles of Oxygen are needed to react with 4.5 moles of Hydrogen.
ii) How many grams of Hydrogen will completely react with 100g of Oxygen to form water?

ATOMIC MASSES : H=1 and O= 16

Solution
(i)

2H₂+ O₂ → 2H₂O

2 moles + 1 mole → 2 moles

Given:
4.5 moles
Required:
Moles of Oxygen

2 moles of H₂ react with 1 mole of O₂
1 mole H₂ react with ½ mole of O₂
4.5 moles H₂ react with ½ x 4.25 moles of O₂
4.5 moles H₂ reacts with 2.25 moles of O₂ ANS.

Solution
(ii)

2H₂ + O₂ → 2H₂O

2 moles + 1 mole → 2 moles

2x2g  +  32g  →  2x18g

4g  + 32g  →  36g

32g of O₂ react with 4g H₂
1g of O₂ react with 4/32 g H₂
100g of O₂ react with 4/32 x 100g H₂
100g of O₂ react with 12.5g H₂ ANS

RESULT:
i) Number of moles of Oxygen = 2.25 moles
ii) 100 g of Oxygen require 12.5g of H₂