5.stoichiometry

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5. STOICHIOMETRY

SynopsisLAWS OF CHEMICAL COMBINATION

• The mass relation between the reactants and products in a chemical reaction is called stoichiometry.

• There are four important laws of chemical combinations

• THE LAW OF CONSERVATION OF MASS:

• This law was proposed by Lavoisier in 1789 by carrying several experiments.

• The law states that matter can neither be created nor destroyed during a chemical change.

• The law may also be stated as the total mass of the products formed during a chemical change is exactlyequal to the total mass of the reactants.

• Weighed amounts of solid and solid KI are dissolved in water separately and their solutions are mixed. Thefollowing reaction takes place

• Total mass of is equal to the total mass of .

• LAW OF DEFINITE PROPORTIONS:

• Proposed by Proust. Verified by Stress and Richards.

• It is also known as Law of constant proportions.

• A given compound always contains the sameelements combined in a fixed proportions by weight.

• What ever the method a compound is prepared, it contains the same elements combined in a fixed ratio byweight

• Eg: CO2 can be prepared by many ways i.e., by combining of carbon with oxygen or by

heating lime stone etc., but what ever the method CO2 is prepared; The ratio of carbon and oxygen by mass is

12 : 32 = 3 : 8

• LAW OF MULTIPLE PROPORTIONS:

• Proposed by Dalton. Verified by Berzelius.• If two elements chemically combine to give two or more compounds, then the weight of one

element which combines with the fixed weight of the other element in those compound bear a simple multipleratio to one another.

• Eg: Nitrogen forms the oxides; N2O, NO, N

2O

3, NO

2, N

2O

5

• In these compounds 28 gm of Nitrogencombines with 16, 32, 48, 64, 80 gm of oxygen respectively. The weight of oxygen in thesecompounds are in the ratio 16:32:48:64:80 or 1:2 : 3 : 4 : 5 a simple multiple ratio.

LAW OF RECIPROCAL PROPORTIONS:

• This law was proposed by Richter (1792) which states as “when two elements combine separately

with a fixed mass of a third element, then the ratio of their masses in which they do so is either same

or some whole number multiple of the ratio in which they combine with each other.

• GAY-LUSSAC’S LAW OF COMBINING VOLUMES

• According to this law gases combine in the simple whole number ratio of their volumes under similarconditions of temperature and pressure. If products are also gases, the simple whole number ratio alsoextends to the products.

• Eg:

• Under similar conditions, 2 lts of Hydrogen combines with 1lt of oxygen to give 2 lts of water vapour.

• It is applicable only to gaseous reaction.

• Law of combining volumes can be derived from Law of defininte proportions when expressed in terms ofvolumes.

• AVOGADRO’S LAW:



Stoichiometry

2

• At the same T, P equal volumes of all gasescontain equal number of moles or molecules.

• No. of molecules = no. of moles × N • The study of mass relation or quantity relation between reactants and products is called

Stoichiometry.

• According to law of conservation of mass, proposed by Lavoisier, the mass of reactants should be equal

to the mass of products.

• The balanced chemical equation which gives correct relation between reactants and products is called

stoichiometric equation.

• The definite quantities which are involved in chemi-cal reaction and present in balanced stoichiometric

equation are called stoichiometric quantities.

1) H2 + O2 → H2O

2g + 32g → 18g (x)

2) 2H2 + O2 → 2 H2O

4g + 32g → 36 gEquation 1) is not stoichiometric equation and will not obey law of conservation of mass, because

masses are not stoichiometric quantities

Equation 2) is stoichiometric equation and will obey law of conservation of mass, because masses are

stoichiometric quantities.

Isotopic abundance:-

It is the percentage availability of an Isotope in the nature.

• The atomic wt of Cl is fractional i.e., 35.5 because it has two isotopes and both have significant

abundance 35Cl is available by 75% and 37Cl is available by 25%

∴ Atomic weight of Cl (average) =

5.35100

37253575 =×+×

Though many elements have isotopes, their atomic weights are whole numbers, because of negligible

abundance of their isotopes.

Eg.: HHH 31

21

11

Protium deuterium tritium

Atomic weight = 1

Here, abundances of HH 31

21

are negligible.

∴ Its atomic weight is 1.

Atomic weights:-

The relative atomic weights of elements are expressed in a.m.u.

1 amu = 1.66 × 10 –24 g.

weight of one H-atom = 1 amu

weight of 1/16 of O-atom = 1 amu

weight of 1/12 of C-atom = 1 amu

The atomic weights are measured by C-12 scale.

The atomic weight of an element is the number of times heavier when compared to 1/12 th of C atom.

Atomic weight =weightatomiccarbonof 1/12

elementof atomof weight

Since, the atomic weight of an element is ratio, it has no units.



Stoichiometry

3

Molecular weights:- Even the molecular weights can be measured with the help of C-12 scale. The

molecular weight of a substance is defined as the number of times when compared to 1/12 th of carbon atom.

molecular weight =carbonof 1/12thof weight

moleculeoneof weight

Thus, relative atomic weight and molecular weights are expressed in amu, but they don't have units.Eg.:

1. Atomic weight of Hydrogen = 1 amu.

Actual weight of Hydrogen = 1.66 × 10 –24 g

Actual weight of 10 H-atoms= 10×1.66×10 – 24 g

2. Atomic weight of oxygen = 16 amu

Actual weight of oxygen = 16×1.66×10 –24g

Actual weight of 100 oxygen atoms=

100×16×1.66×10 –24g

Gram atom (Gram atomic weight):

If atomic weight is expressed in grams it is called gram atomic weight or gram atom.

1 gram atom of Hydrogen = 1 g

2 gram atom of Hydrogen = 2 g

1 gram atom of Oxygen = 16 g



Hydrogen : Atomic weight = 1

Gram Atomic weight = 1 g

Oxygen : Atomic weight = 16

Gram Atomic weight = 16 g

Gram molecule

(Gram molecular weight or Gram mole):

If molecular weight is expressed in grams, it is called gram molecule or gram molecular weight.

Ex.: Hydrogen : Molecular weight = 2

Gram Molecular weight = 2 g

Oxygen : Molecular weight = 32

Gram Molecular weight = 32 g

CO2 : Molecular weight = 44

Gram Molecular weight = 44g

64g of Oxygen = 2 gram molecule of O2

22g of CO2 = 1/2 gram molecule of CO2

360g of H2O = 20 gram molecule of H2O

Mole Concept:-

The amount of substance which contains Avagadro's number of particles is called mole. (or)

Mole is the amount of substance containing as many particles as the number of atoms in 12g of carbon.

1 mole of Hydrogen = 6.023 × 1023 molecules

1 gm molecule weight = 2 g.

1 mole of Carbon = 6.023 × 1023 atoms

1 gm atom weight = 12 g

1 mole of Carbondioxide = 6.023 ×1023 molecules



Stoichiometry

4


1 mole of Sodium = 6.023 × 1023 atoms

1 gm atom weight = 23 g.

1 mole of Sulphur dioxide = 6.023×1023 molecules

1 gm molecule weight = 64 g.1 mole of Oxygen = 6.023 × 1023 molecules


1 mole of H2SO4 = 6.023 × 1023 molecules


Gram Molar volume (or)

Gram Molecular Volume (GMV):

It is the volume occupied by one mole of gas at STP

i.e. 22.4 lit.

44 g of CO2 occupies 22.4 lit at STP

11 g of CO2 occupies 5.6 lit at STP

8 g of CH4 occupies 11.2 lit at STP

4 g of He occupies 22.4 lit at STP

142 g of Cl2 occupies 2 × 22.4 lit at STP

Weight of 5.6 lit of Ethane at STP is 7.5 g

Weight of 44.8 lit of SO2 at STP is 128 g.

22.4 lit of any gas at STP contains 6×1023 molecules

11.2 lit of Chlorine contains 3 × 1023 molecules

5.6 lit of Hydrogen contains 1.5 × 1023 molecules

3 × 1023 molecules of CO2 will occupy 11.2 lit

6 × 1023 molecules of O2 will occupy 22.4 lit

6 × 1024 atoms of O2 will occupy 224 lit

3 × 1024 atoms of O2 occupies a volume of 112 lit.

Empirical and Molecular formula:

Empirical formula:

It is the formula which gives the simplest ratio of atoms of various elements.

Molecular formula:

It is the formula which gives the actual number of atoms of various elements present in the molecule.

NameMolecular

formula

Empirical

formulaAcetylene C2H2 CH

Benzene C6H6 CH

Methane CH4 CH4

Glucose C6H12O6 CH2O

Borazol B3 N3H6 BNH2

Sulphuric

acid

H2SO4 H2SO4

Methyl

alcohol

CH3OH CH3OH



Stoichiometry

5

Ethylene C2H4 CH2

Ethane C2H6 CH3

Sugar C12H22O11 C12H22O11

Acetic acid CH3COO

H

CH2O

Ethyl alcohol C2H5 –

OH

C2H5 –

OH

Formic acid HCOOH HCOOH

Calculating Empirical formula:-

1) Divide the percentage weight of each element with its atomic weight.

2) Divide the above values with least of them.

3) Multiply with a suitable co-efficient to get whole number ratio. (If necessary)

Determining the molecular formula of gaseous hydrocarbon:

CxHy + ⎟⎟ ⎠

⎞⎜⎜⎝

⎛ + 4

yx O2 → xCO2 + 2

y H2O

1 ml + ⎟⎟ ⎠

⎞⎜⎜⎝

⎛ +

4

yx ml → x ml

10 ml+ 10 ⎟⎟ ⎠

⎞⎜⎜⎝

⎛ +

4

yx ml → 10x ml

1. 10 ml of gaseous hydrocarbon is completely burnt in 80 ml of O2 at STP. The gaseous residue occupied a

volume of 70 ml. This when passed through caustic potash has become 50 ml. Calculate the empirical

formula of the hydrocarbon.

CxHy + ⎟⎟ ⎠

⎞⎜⎜⎝

⎛ +

4

yx O2 → xCO2 +

2

yH2O

10 ml+10 ⎟⎟ ⎠

⎞⎜⎜⎝

⎛ +

4

yx ml → 10x ml

Volume of CO2 = contraction in volume by KOH = 20 ml.

∴ 10x = 20 ⇒ x = 2.

At the end of the reaction, the residual gases will consist of volume of CO2 produced and the volume of

unreacted O2 volume of CO2 + volume of unreacted O2 = 70 ml.

Volume of unreacted oxygen= 70–volume of CO2

Volume of unreacted oxygen = 70 –20 = 50 ml

Volume of O2 reacted = 80 – 50 = 30 ml.

∴10 ml ⎟

⎟ ⎠

⎞

⎜⎜⎝

⎛ +

4

yx

= 30 ml.

4

yx + = 3;

4

y2 + = 3; y = 4

∴Molecular formula is C2H4; Empirical formula is CH2

Victor Meyer’s method to determine Mol. Wt.:

1. 0.2 g of a volatile organic compound on vapourisation gave 200 cc at STP it’s mol.wt is

2g → 2000 cc

2 x Vapour Densi ty = Molecular



Stoichiometry

6

? → 22,400 cc

2000

224002× = 22.4 g

∴ mol. Wt is 22.4 g

Oxidation states (or) oxidation number:The charge present on an atom is called oxidation state (or)

The charge that appears to have been present on the atom when electrons are counted according to some

arbitary rules is called oxidation state.

• Oxidation state could be positive, negative, zero, whole number or fraction.

• More electronegative atom is assigned negative Oxidation State and less electronegative atom is

assigned positive Oxidation State.

• The sum of Oxidation State’s of all atoms in a molecule is equal to zero.

• The sum of oxidation state’s of all atoms in an ion is equal to the charge present on the ion.

• IA group elements always exhibit +1 oxidation state

• IIA group elements always exhibit +2 oxidation state• F always exhibits –1 oxidation state

• Common oxidation state of hydrogen is +1

• In metal hydride, hydrogen exhibits –1 oxidation state

• Common oxidation state of oxygen is –2.

• In perioxides, oxygen exhibits –1 oxidation state.

• In superoxides, oxygen exhibits –1/2 oxidation state.

• In O2F2 and OF2 oxygen oxidation states are +1 and +2 respectively.

• Transition elements exhibit more than one oxidation state.

• Osmium and Ruthenium show the highest oxidation state i.e. +8.• The oxidation state of any atom in its elementary state is zero.

• Nitrogen exhibits large number of oxidation state such as –3, –2, –1, 0, +1, +2, +3, +4, +5 and even –

1/3.

• Oxidation state of an element will not exceed it’s group number. Thus, maximum oxidation state of an

element may be equal to its group number.

• Maximum oxidation state of sulphur is +6

(VIA group)

• Maximum oxidation state of chlorine is +7

(VIIA group)

• Maximum oxidation state of nitrogen is +5

(VA group)

(1) H2SO4 (2) PCl3

2 + x + 4(–2) = 0 x+3(–1) = 0

2 + x – 8 = 0 x = 3 x – 6 = 0

x = 6

(3) SO2 (4) N2O5

x+2(–2) = 0 2(x)+5(–2) = 0

x = 4 x = +5



Stoichiometry

7

(5) N2H4 (6) HClO4

2(x)+4(1) = 0 1+x+4(–2) = 0

2x = 4 x = +7

x = +2

(7) Co(NH3)6Cl3 (8) [Fe(H2O)6]3+

x + 6(0) + 3(–1) = 0 x + 0 = 3

x = 3. x = 3

(9) SO32–

(10) H3N

x + (– 6) = –2 3 + x = 0

x = +4 x = –3

(11) SO42–

(12) OCl–

x + (–8) = –2 –2 + x = –1

x = +6 x = +1

(13) N3H (14) ClO2–

3x = –1 x + 2(–2) = –1x = –1/3 x – 4 = –1; x = 3

(15) Cl2O3–

(15) ClO4–

2x + (–6) = –1 x + (–8) = –1

2x = 5 x = 8 – 1

x = 5/2 x = 7

(17) NO2– (18) NO3

–

x + 2(–2) = –1 x + 3(–2) = –1

x – 4 = –1 x – 6 = –1

x = +3 x = 6 – 1

x = +5Sum of oxidation states of all C-atoms in glucose molecule is

C6H12O6

6x + 12 + (–12) = 0

6x = 0

The oxidation state of N-atom in hydrazoic acid is

N3H

3x + 1 = 0

3x = –1

x = –1/3

The sum of oxidation states of all C-atoms in benzaldehyde is

C6H5CHO

7x + 6 – 2 = 0

7x = –4

x = – 4/7

The sum of oxidation states of all carbon atoms in

nitrobenzene is

C6H5 NO2

6x + 5 +2(–2) + 3 = 0

6x + 8 – 4 = 0



Stoichiometry

8

6x = –4 ⇒ x = –4/6

Sum of oxidation states of all carbon atoms in aniline

(aminobenzene) is

C6H5 NH2

6x + 7 – 3 = 06x = –4 ⇒ x = –4/6

Calculate the oxidation states of underlined atoms in the following?

1) H2S2O3 2) CuSO4 . 5H2O

3) K3[Fe(CN)6] 4) K4[Fe(CN)6]

5) Na2S4O6 6) [Cu(NH3)4]SO4

7) (NH4)2CO3 8) Na2S2O3

9) H2S2O7 10) PCl5

Ans:

1) +2 2) +6

3) +3 4) +2

5) +2.5 6) +2

7) +4 8) +2

9) +6 10) +5

Redox reactions:

Based on the involvement of electrons, chemical reactions are classified into two types.

(i) Reactions which donot involve electron transfer:

Precipitation ionic reactions and neutralisation reactions do not involve electron transfer

a) AgNO3 + NaCl → AgCl↓ + NaNO3(aq)

b) BaCl2 + Na2SO4 → 2NaCl + BaSO4↓

Spectator ions:

These ions do not undergo any change in a chemical reaction. These ions are present commonly on both

sides of the equation.

Spectator ions:

a) Na+, NO3 ⎯

b) Na+, Cl ⎯

NaOH + HCl → NaCl + H2O

Spectator ions: Na+, Cl ⎯

2KOH + H2SO4 → K 2SO4 + 2H2O

Spectator ions: K +, SO42–

Reactions which involve electron transfer:

The reactions which involve e – transfer are commonly known as Redox reactons.

Na +2

1Cl2 → NaCl



Stoichiometry

9

Zn + CuSO4 →Cu + ZnSO4

H2 + Cl2 → 2HCl

Oxidation: It is the process in which

1) electrons are removed

2) de-electronation occurs.

3) addition of oxygen.

4) removal of hydrogen

5) increase in the oxidation state

Ex: Na → Na+ + e ⎯

Zn → Zn2+ + 2e –

Ca +1

2O2 → CaO

4HCl + O2 → 2H2O + 2Cl2

Reduction: It is process in which1) electrons are added

2) electronation occurs

3) oxygen is removed

4) hydrogen is added

5) decrease in the oxidation state

Cl2 + 2e ⎯ → 2Cl –

F2 + H2O → 2HF +1

2O2

CO2 + C → 2CO

CuO + H2 → Cu + H2ODisproportionantion Reactions:

If same substance undergoes both oxidation and reduction, it is called disproportionation. Thus,

disproportionation reactions are also redox reactions.

Cl2 + NaOH → NaCl + NaOCl + H2O

P + NaOH → PH3 + NaH2PO2

Oxidising agent (Oxidant): It is the substance

1) Which gains electrons

2) Which undergoes reduction

3) Which oxidises other substances.

Reducing agent: It is the substance

1) Which loses electrons.

2) Which undergoes oxidation

3) Which reduces other substances.



Stoichiometry

10

ReactionHalf-oxidation

reaction

Half-reduction

reaction

Oxidising

agent

Reducing

agent

1) KMnO4+H2C2O4+H2SO4 →

K 2SO4+MnSO4+H2O+CO2

H2C2O4 → CO2 KMnO4 →

MnSO4

KMnO4 H2C2O4

2) K 2Cr 2O7+FeSO4+H2SO4 →

K 2SO4+ Cr 2(SO4)3 + Fe2(SO4)3 +

H2O

FeSO4 →

Fe2(SO4)3

(or) Fe2+ → Fe3+

K 2Cr 2O7 →

Cr 2[SO4]3

2Cr +6 → 2Cr 3+

K 2Cr 2O7 FeSO4

3) KMnO4+FeSO4+H2SO4 →

K 2SO4+ Fe2(SO4)3 + MnSO4 + H2O

KMnO4 →

MnSO4

Mn+7 → Mn2+

FeSO4 →

Fe2(SO4)3

Fe2+ → Fe3+

KMnO4 FeSO4

4) K 2Cr 2O7 + NaNO2 + H2SO4 →

K 2SO4 + Cr 2(SO4)3 + NaNO3 + H2O

NO2 – → NO3

– Cr 2O72– → Cr 3+ K 2Cr 2O7 NaNO2

5) Cu + HNO3 →

Cu(NO3)2 + NO2 +H2O Cu→

Cu(NO3)2 Cu → Cu2+ HNO3 →

NO2 NO3 – → NO2

HNO3 Cu

6) C + HNO3 → CO2 + NO2 + H2O C → CO2 HNO3 → NO2

NO3 – → NO2

HNO3 C

7) P + HNO3 → H3PO4 + NO2 +

H2O

P0 → P+5 HNO3 → NO2 HNO3 P

8) F2Cl2 → 2ClF F2 → ClF

F0 →F –

Cl2 → ClF

Cl2 → 2Cl+

F2 Cl2

9) 2Na + Cl2 → 2NaCl Na → NaCl

Na → Na+

Cl2 → NaCl

Cl2 → 2Cl –Cl2 Na

10) SO2 + H2S → S + H2O H2S → S SO2 → S SO2 H2S

11) Na2S2O3 + Cl2 + H2O →

Na2SO4 + 2HCl + S

Na2S2O3 →

Na2SO4

S+2 → S+6

Cl2 →2HCl

Cl20 → 2Cl –

Cl2 Na2S2O3

12) Na2S2O3 + I2 → Na2S4O6 + 2NaI Na2S2O3→ Na2S4

O6

S+2 → S+5/2

I2 →2NaI

I0 → I –

I2 Na2S2O3

Balancing of redox reactions:

There are two methods of balancing an equation

(i) Oxidation number method:

1) Write the skeleton equation

Skeleton equation:

KMnO4 + H2C2O4 + H2SO4 → K 2SO4 + MnSO4 + CO2 + H2O

2) Identify the elements for which oxidation states are changing

KMn + 7O4 + H2C2 + 3O4 + H2SO4 →K 2SO4 + Mn + 2SO4 + C + 4O2 + H2O

3) Locate the change in oxidation states

(1)

(–5)

KMnO +H C O +H SO →K SO +MnSO +CO



Stoichiometry

11

4) Interchange the coefficient so that total decrease in the oxidation state becomes equal to total increase in

the oxidation state.

2KMnO4 + 5H2C2O4 + H2SO4 →

K 2SO4 + MnSO4 + CO2 + H2O

5) Balance elements other than oxygen and hydrogen by inspection (hit and trail) method.

2KMnO4 + 5H2C2O4 + H2SO4 →

K 2SO4 + 2MnSO4 + 10CO2 + H2O

6) Balance oxygen and hydrogen

2KMnO4 + 5H2C2O4 + 3H2SO4 →

K 2SO4 + 2MnSO4 + 10CO2 + 8H2O

Half-reaction method: (Ion electron method)

This method is for balancing of ionic form of equations.

Molecular form or ionic form of equations can be balanced by this method.

Balancing of Redox reaction equations by the half reaction method or ion-electron method is as follows:

Represent first the ionic equation.

a) Indicate the oxidation half-reaction and reduction half-reaction separately.

b) Balance the half reactions separately. While balancing the half reactions, balance the atoms other than

oxygen and hydrogen first.

c) In acid medium include enough number of water molecules where there is a deficiency of oxygen and

include enough number of H + ions on the side where there is a deficiency of Hydrogen.

d) In alkaline medium include enough number of OH ⎯ ions on the side where there is a deficiency of

oxygen and enough number of water molecules where there is a deficiency of hydrogen. After balancing

the atoms, balance the charge.

e) To balance the charge include enough number of electrons wherever necessary.

f) Multiply the reduction half reaction with the number of electrons lost in oxidation and multiply the

oxidation half reaction with the number of electrons gained in reduction.

g) Add the two half reactions after canceling the electrons.

In the final balanced equation not only the atoms but the net charge also should be balanced.

How many electrons and on which side are to be added in the following half-reaction

i) CH4 → CH2O; CH4 → CH2O + 4e –

ii) Na4S2 + 2

O3 → Na2 S4 + 2.5

O6 Na4S2O3→ Na2S4O6 + 1e –

iii) CH3OH → HCOOH ; CH3OH → HCOOH + 4e –

iv) MnO4 – → Mn2+ ; MnO4

– + 5e – → Mn2+

v) S2O32– → S0; S2O3

2– + 4e – → S

• If an element is in higher oxidation state, it acts as oxidising agent.

• If an element is in its lowest oxidation state, it acts as reducing agent.

• If the element is in intermediate oxidation state, it acts as both O.A and R.A.

Eg:

i) Na2S (R.A.); Na2SO3 (Both);



Stoichiometry

12

Na2S2O3(both); Na2SO4(O.A.)

ii) SO2 (R.A.) SO3(O.A.)

iii) KMnO4(O.A.); K 2MnO4 (both); MnO2 MnSO4 (R.A)

iv) HNO2; HNO3(O.A.); N3H; NH3(R.A); N2H4 (O.A)

v) N2O; NO; N2O3; NO2; N2O4; N2O5 (O.A.)vi) K 2Cr 2O7(OA); H2CrO4(O.A)

vii) O2(OA); H2O(RA); H2O2(both)

Equivalent weight of an element:

• It may be defined as the number of parts by weight of that element which combines with (or) displaces

directly (or) indirectly 1.008 parts by weight of hydrogen (or) 8 parts by weight of oxygen (or) 35.45

parts by weight of chlorine.

Eq. wt of an element = At.wt. of an element

Valency

• The valency of an element may vary though the atomic weight of the element is constant.

• One element may possess more than one equivalent weight.

Eq. wt. of Fe+2 (Ferrous ion) =55.84

2= 27.92

Eq. wt of Fe+3 (Ferric ion) =55.84

3 = 18.613

• Radicals (ions) also possess equivalent weights.

• A radical behaves like an d element.

Eq. wt. of radical =Molecular weight of radical

Valency

Eq. wt. of 24

96SO 482

− = =

• If the equivalent weight of an element (or) radical expressed in grams is called gram equivalent weight.

• Ex. Gram equivalent weight of oxygen = 8 g.

Gram equivalent weight of carbonate ion =60

2=30 g

• Equivalent weight of substance depends on whether one needs to calculate the equivalent weight of an

element (or) acid (or) base (or) oxidant (or) a reductant.

• Equivalent weight of substances may have different values.

• Eq. wt. of KMnO4 acidic medium = 31.6

• Eq. wt. of KMnO4 in Alkaline medium = 52.6

• Eq. wt. of KMnO4 in neutral medium = 158

Equivalent weights of Acids :

• Equivalent weight of an acid is the number of parts by weight of it that contains 1.008 parts by weight

of replaceable hydrogen. Alternately it is the quantity by weight that supplies one mole of H+.

• The number of hydrogen atoms that get displaced from a molecule of acid is equal to the number of

equivalent per molecular weight of the acid.

• Eq.wt. of an acid =Mol.wt (or) Formula wt.of acid

Basicity of the acid

• Basicity of an acid is the number of replaceable hydrogen atoms present in one molecule of the acid.



Stoichiometry

13

• Eq.wt of HCl =M 36.5

36.51 1

= =

• Eq.wt of HNO3 =M 63

631 1

= =

• Eq.wt of H2SO4 = M 98 492 2= =

• Eq.wt of H3PO4 =M 98

32.63 3

= =

• Eq.wt of H3PO2 =M 82

412 2

= =

• Eq.wt of H2C2O4.2H2O =M 126

632 2

= =

• Eq.wt of H2C2O4 =M 90

452 2

= =

Equivalent weight of Bases :

• Equivalent weight of a base is the number of parts by weight of it that completely neutralises one gram

equivalent of an acid. Alternately it is the quantity of weight that reacts with one mole of H +.

• Eq. wt. of base

=Mol.wt.(or)Formula wt. of the base

Acidity

• The number of hydroxyl groups present in a molecule of the base is known as its acidity.

• Eq. wt. of NaOH =M 40

401 1

= =

• Eq. wt of KOH =M 56

56

1 1

= =

• Eq. wt. of Ca(OH)2 =M 74

372 2

= =

• Eq. wt. of Ba(OH)2 =M 171

85.52 2

= =

• Eq. wt. of Fe(OH)3 =M 107

35.73 3

= =

Equivalent weight of oxidising and reducing agents:

• Equivalent weight of an oxidising agent is that weight of the compound which can supply 8 parts by

weight of oxygen (or) can react with 1.008 g parts by weight of hydrogen similarly.

• Equivalent weight of a reducing agent is that weight of the compound which can supply 1.008 parts byweight of hydrogen (or) can be oxidised by 8 parts by weight of oxygen.

• In redox reaction electrons are transfered from the reducing agent to the oxidising agent.• In a stoichiometric equation if the number of electrons (n) transferred from one mole of reducing agent (reductant)

to one mole of oxidising agent (oxidant).

• Eq. wt. of oxidant =Formula wt. of oxidant

No.of electrons gained by oxidant

• Eq. wt. of reductant =Formula wt. of reductant

No.of electrons lost by reductant

Example 1:

• FeSO4 is reducing agent. It is oxidised to e2(SO4)3. The reaction can be written as Fe

2+

→

Fe

3+

+ 1e

–

.



Stoichiometry

14

• One mole of Ferrous ion loses one mole of electrons.

• Eq. wt. of Ferrous ion =wt. of Ferrous ion 55.84

55.841No.of e involved in the reaction−

= =

Example 2:

• Oxalate ion (C2O4

–2

) is another reducing agent. When it is oxidised the reaction written asC2O4

–2 → 2CO2 + 2e –

• One mole of oxalate ion loses two moles of electrons.

• Eq. wt. of C2O4 –2 ion =

22 4Formula wt. of C O 88

442No.of e involved in the reaction

−

− = =

• Eq.wt of Na2C2O4 . 2H2O = 2 2 4 2Molecular wt. of Na C O .2H O

2=

17085

2=

Example 3:

• In acid medium KMnO4 is a strong oxidising agent.

KMnO4 + 8 H+ + 5e – → K + + Mn+2 + 4H2O

One mole of KMnO4 takes five moles of electrons.• Eq. wt. of KMnO4 =

4

4

Molecular weight of KMnO

No.of e involved in the reaction per mole of KMnO−

= 4Molecular weight of KMnO 15831.6

5 5= =

(or) Eq. wt. of 44

wt.of MnOMnO

5

−− = =

11923.8

5= .

Example 4 :

• KMnO4 in function as oxidising agent in neutral as well as basic medium.

• In Alkaline medium the reaction isKMnO4 + 2H2O + 3e – → K + + 4OH – + MnO2

Eq. wt. of KMnO4 = = 4

4

Molecular weight of KMnO

No.of.e involved in the reaction per mole of KMnO−

=158

52.63

=

• In neutral medium the reaction is 24 4KMnO e K MnO− + −+ → +

• Eq. wt of 44

Mol.wt.of KMnOKMnO 158

1= =

• Equivalent weight of the substances changes with chemical reaction.

Example 5 :• K 2Cr 2O7 functions as oxidising agent in acidic medium Cr 2O7

–2 + 14H+ + 6e – → 2Cr +3 + 7H2O.

• Change in oxidation number for one Cr atom = 3.

• Change in oxidation number for two Cr atoms = 6.

• No. of electrons gained = 6.

• Eq. wt. of K 2Cr 2O7 =2 2 7Molecular weight of K Cr O 294

496 6

= =

Equivalent weight of salts :

• Salts consist of cations and anions.

• Salts are neutral substances.



Stoichiometry

15

• For a salt total positive charge is equal to total negative charge.

• Eq. wt. of salt =Formula weight of salt

Total charge of the cation (or) anion of the salt

• Eq. wt. of NaCl =F 58.5

58.5

1 1

= =

• Eq. wt. of Na2CO3 =F 106

532 2

= =

• Eq. wt. of CaCO3 =F 100

502 2

= = .

TYPES OF REDOX REACTIONS

Chemical combination reactions: The redox reactions in which the compounds are formed by combining two or

more elements with eath other or with a compound →A + B AB .

1)( ) ( ) ( )

0 0 2

2 2O

−+ →g g

+4

sC CO

2)( ) ( ) ( )

0 0 2

2 2−+ →

g g

+4

sS O SO

3)( ) ( ) ( ) ( )

0 2 24 22 2 l

− −+ → +g

g g

-4 +1 +4 +1

CH O CO H O

Decomposit ion reactions: The redox reactions in which the chemical compounds chemically split into two or more

simple substances. AB A B→ +

Eg: 1)2 0

3 22 2K K − →

+1 +5 +1 -1

C O C +3Ol l

2)1 2 0 0

2 2 2

+ − → +H O H O

Displacement reactions: The redox reactions in which the place of one species in it’scompound is taken up by other species

Eg: →A+BC AC+B

( ) ( ) ( ) ( )aq aq

0 2 2 2 2 0

4 4s sZn Cu SO Zn SO Cu+ − + −+ → +

l In non metal displacement redox reactions,

generally hydrogen gets displaced and rarelyoxygen.

Eg: ( ) ( ) ( ) ( )g

0 1 2 1 2 1 0

2 2s aq Na 2H O 2Na O H H+ − + − ++ → +

l

( ) ( ) ( ) ( )aq g

20 1 1 1 0

2 2s aq Zn 2H C ZnC H

++ − −+ → +l l

( ) ( ) ( ) ( )g

50 1

2 3 2aq aq aq

3C 6OH C O 5C 3H O

−+− −⎡ ⎤

+ → + +⎢ ⎥⎣ ⎦ l

l l l

( ) ( ) ( )

−− + −⎡ ⎤+ → + +⎣ ⎦g

1

2 2aq aq aq C 2OH C O C H Ol l l

Comproportionation reactions: In these reactions, two species with the same element in two different oxidation states form a single product inwhich the element is in an intermediateoxidation state. It is reverse of disproportionation.

Eg:



Stoichiometry

16

( ) ( ) ( )2

aq s aq Ag Ag 2Ag+ ++ →

Reduction

Oxidation

APPLICATIONS OF REDOX REACTIONS IN QUANTITATIVE ANALYSIS AND ELECTRODE PROCESS

In titrimetric quantitative analysis:

• The solutions of accurately known concentration is called standard solution.

• In titrimetric analysis, the substance of known concentration is called titrant.

• Substance being titrated is called titrand.

• The process of adding the standard solution until the reaction gets completed is called titration.

• The point at which the titrand just completelyreacts is called equivalence point or theoretical point or Stoichiometric end point.

• Completion of titration is detected by

o Observing physical change

o By using an Indicator.

• In2

2 7Cr O −

(dichromate) titrations, diphenyl amine is used as a reagent. End point is intense blue colour due

to oxidation of2

2 7Cr O −

.

• In titration of Cu2+ and I – to I2 it gives deep blue Colour with starch solution.

• ( ) ( ) ( ) ( )2

2 2aq aq 2 aq Cu 4I Cu I s I+ −+ → +

• In the titration of I

2 with thiosulphate, blue colour disappears due to the conversion of I

2 to I – .

• ( ) ( ) ( ) ( )aq aq aq

2 2

2 2 3 4 6aq I 2S O 2I S O− − −+ → +

• Br

2 and I

2 dissolve in CCl

4 solvent to give reddish brown and purple colour respectively. They are detected by

extracting into CCl4 layer and the test is called ‘Layer test’.

• Electrode process: • Galvanic, Daniel cells were constructed based on redox reactions. They convert chemical

energy to electrical energy.

• In Galvanic cell Cu acts as cathode, Zn acts as anode.

• It is represented as: ( ) ( )

+ +

2 2

aq aq Zn Zn Cu Cu

.

• Daniel cell is represented as ( ) ( )+ +

2 2

aq aq Zn, Zn Cu , Cu

.

• Salt Bridge is an inverted U shaped rubber tube containing an inert electrolyte solutions like KCl or KNO3 or

NH4NO

3 made in the form of semi solid in agar agar. It is denoted by in galvanic and in Danielcell.

• When the concentration of the electrolyte is 1M at 298 K, the potential obtained for the electrode is calledstandard electrode potential

• Standard electrode potential for hydrogen electrode is assumed as zero.

• The series in which the electrods are arranged in ascending or descending order of reductionpotential values is known as electro chemical series.

5.stoichiometry

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