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04/11/2304/11/23 MATTERMATTER 11
1.2 MOLE CONCEPT1.2 MOLE CONCEPT
04/11/2304/11/23 MATTERMATTER 22
Learning OutcomeLearning Outcome
At the end of this topic, students should be
able :
(a) Define mole in terms of mass of
carbon-12 and Avogadro constant, NA.
(b) Interconvert between moles, mass, number of particles, molar volume of gas at s.t.p. and room temperature.
04/11/2304/11/23 MATTERMATTER 33
(c)(c) Determine empirical and molecular
formulae from mass composition or from mass composition or
combustion data.combustion data.
04/11/2304/11/23 MATTERMATTER 44
(d) (d) Define and perform calculation for each for each
of the following concentration of the following concentration
measurements :measurements :
i) molarity (M)
ii) molality (m)
iii) mole fraction, X
iv) percentage by mass, % w/w
v) percentage by volume, %V/V
04/11/2304/11/23 MATTERMATTER 55
(e) (e) Determine the oxidation number of an of an
element in a chemical formula.element in a chemical formula.
(f) (f) Write and balance : :
i) chemical equation by i) chemical equation by inspection inspection
methodmethod
ii) redox equation by ii) redox equation by ion-electron ion-electron
methodmethod
04/11/2304/11/23 MATTERMATTER 66
(g) (g) Define limiting reactant and and percentage
yield.
(h) (h) Perform stoichiometric calculations
using mole concept including reactant using mole concept including reactant
and percentage yield.and percentage yield.
04/11/2304/11/23 MATTERMATTER 77
1.2 Mole Concept1.2 Mole Concept
A mole is defined as the amount of substance which contains equal number of particles (atoms / molecules / ions) as there are atoms in exactly 12.000g of carbon-12.
04/11/2304/11/23 MATTERMATTER 88
One mole of carbon-12 atom has a mass of exactly 12.000 grams and contains 6.02 x 1023 atoms.
The value 6.02 x 1023 is known as Avogadro Constant.
NA = 6.02 x 1023 mol-1
04/11/2304/11/23 MATTERMATTER 99
ExampleExample
1.0 mole of chlorine atom = 6.02 x 1023 chlorine atoms
= 35.5 g Cl
1.0 mole of chlorine molecules
= 6.02 x 1023 chlorine molecules
= 71.1 g Cl2= 6.022 x 1023 x 2 chlorine atoms
1.0 mole of NH3 = 6.02x 1023 molecules
= 6.02 x 1023 x 4 atoms
= 6.02 x 1023 N atom
= 6.02 x 1023 X 3 H atoms
04/11/2304/11/23 MATTERMATTER 1010
Molar MassMolar Mass
The mass of one mole of an element or one mole of compound is referred as molar mass.
Unit : g mol-1
Example:- molar mass of Mg = 24 g mol-1
- molar mass of CH4 = (12 + 4) gmol-1
= 16 g mol-1
04/11/2304/11/23 MATTERMATTER 1111
Number of MoleNumber of Mole
)mol (g Mass
(g) mole of
1-Molar
MassNumber
)mol (g Mass
(g) mole of
1-Molar
MassNumber
04/11/2304/11/23 MATTERMATTER 1212
Example 1Example 1
atoms ofNumber iii.
molecule ofNumber ii.
mol g 28massmolar if molecule moles ofNumber i.
calculate; , N of g 14In 1-
2
atoms ofNumber iii.
molecule ofNumber ii.
mol g 28massmolar if molecule moles ofNumber i.
calculate; , N of g 14In 1-
2
04/11/2304/11/23 MATTERMATTER 1313
mol 0.5
mol g 28
g 14
)mol (g Mass
(g) N molecules mole of
gmol 28massmolar if molecule moles ofNumber i.
:Solution
1-
1-2
1-
Molar
MassNumber
mol 0.5
mol g 28
g 14
)mol (g Mass
(g) N molecules mole of
gmol 28massmolar if molecule moles ofNumber i.
:Solution
1-
1-2
1-
Molar
MassNumber
04/11/2304/11/23 MATTERMATTER 1414
Example 1 (cont…)Example 1 (cont…)
atoms 10 x 6.022
N of atoms 10 x 3.011 x 2 contains N of molecule 10 x 3.011
N of atoms 2 contains N of molecule 1
atoms ofNumber iii.
molecules 10 x 3.011
10 x 6.022 x 0.5
N x mole of NumberN molecules of Number
molecule ofNumber ii.
23
232
23
2
23
23
A2
atoms 10 x 6.022
N of atoms 10 x 3.011 x 2 contains N of molecule 10 x 3.011
N of atoms 2 contains N of molecule 1
atoms ofNumber iii.
molecules 10 x 3.011
10 x 6.022 x 0.5
N x mole of NumberN molecules of Number
molecule ofNumber ii.
23
232
23
2
23
23
A2
04/11/2304/11/23 MATTERMATTER 1515
Example 2Example 2
atoms10 x 1.807
10 x 6.022 x 3atoms ofNumber
Natoms ofNumber
moles ofNumber
atoms H of moles 3 contains NH of mole 1
NH of mole 1 in atom Hnumber theCalculate
24
23
A
3
3
atoms10 x 1.807
10 x 6.022 x 3atoms ofNumber
Natoms ofNumber
moles ofNumber
atoms H of moles 3 contains NH of mole 1
NH of mole 1 in atom Hnumber theCalculate
24
23
A
3
3
04/11/2304/11/23 MATTERMATTER 1616
Example 3Example 3
10 x 1.2046
10 x 6.022 x 4ions bromide ofNumber
so,
ions bromide of moles 4 CaBr of mole 2
ions bromide of moles 2 CaBr of mole 1
CaBr of moles 2in ions bromide ofnumber theCalculate
24
23
2
2
2
ions
contains
contains
10 x 1.2046
10 x 6.022 x 4ions bromide ofNumber
so,
ions bromide of moles 4 CaBr of mole 2
ions bromide of moles 2 CaBr of mole 1
CaBr of moles 2in ions bromide ofnumber theCalculate
24
23
2
2
2
ions
contains
contains
04/11/2304/11/23 MATTERMATTER 1717
1.2.1 Mole Concept of Gases1.2.1 Mole Concept of Gases
Molar volume of any gas at STP = 22.4 dm3 mol-1
s.t.p. = Standard Temperature and Pressure
Where,
T = 273.15 K
P = 1 atm
04/11/2304/11/23 MATTERMATTER 1818
1 mole of gas has a volume of 22.4 dm3 at s.t.p
At s.t.p,volume of gas (dm3) = number of mole X 22.4 dm3 mol-1
1 mole of gas has a volume of 24.0 dm3 at room temperatureAt room temperature,volume of gas (dm3) = number of mole X 24.0 dm3 mol-1
04/11/2304/11/23 MATTERMATTER 1919
Example 1Example 1
mol 1.0mole ofNumber
so,
gashydrogen of mol 2.24x 22.4
1 consists dm 2.24
gashydrogen of mol 1 consists dm 22.4
1, Solution
gas.hydrogen of (mole)amount the
calculate ,dm 2.24 isballon theof volume theIf
s.t.p.at gashydrogen with filled isballoon A
3
3
3
mol 1.0mole ofNumber
so,
gashydrogen of mol 2.24x 22.4
1 consists dm 2.24
gashydrogen of mol 1 consists dm 22.4
1, Solution
gas.hydrogen of (mole)amount the
calculate ,dm 2.24 isballon theof volume theIf
s.t.p.at gashydrogen with filled isballoon A
3
3
3
04/11/2304/11/23 MATTERMATTER 2020
Cont… from example 1Cont… from example 1
mol 0.1 4.22
2.24dm
4.22
)(dm gas of mole ofNumber
2, Solution
13
3
13
3
moldm
moldm
volume
04/11/2304/11/23 MATTERMATTER 2121
ExerciseExercise
A sample of CO2 has a volume of 56 cm3 at STP. Calculate:
a. The number of moles of gas molecules0.0025 mol
b. The number of molecular1.506 x 1021 molecules
c. The number of oxygen atoms in the sample3.011x1021atoms
Note:1 dm3 = 1000 cm3
1 dm3 = 1 L
04/11/2304/11/23 MATTERMATTER 2222
Empirical And Molecular FormulaeEmpirical And Molecular Formulae
- Empirical formulaEmpirical formula is a chemical formula is a chemical formula that shows the simplest ratio of all that shows the simplest ratio of all elements in a molecule.elements in a molecule.
- Molecular formulaMolecular formula is a formula that show is a formula that show the actual number of atoms of each the actual number of atoms of each element in a molecule.element in a molecule.
04/11/2304/11/23 MATTERMATTER 2323
- The relationship between empirical formula and The relationship between empirical formula and molecular formula is :molecular formula is :
Molecular formula = n ( empirical formula )Molecular formula = n ( empirical formula ) Where ;Where ;
mass formula emprical
mass molecular relativen
04/11/2304/11/23 MATTERMATTER 2424
Example Example
A sample of hydrocarbon contains 85.7% A sample of hydrocarbon contains 85.7% carbon and 14.3% hydrogen by mass. Its carbon and 14.3% hydrogen by mass. Its molar mass is 56. Determine the empirical molar mass is 56. Determine the empirical formula and molecular formula of the formula and molecular formula of the compound. compound.
04/11/2304/11/23 MATTERMATTER 2525
Solution :Solution :
Empirical formula = Empirical formula = CHCH22
CC HH
massmass 85.785.7 14.314.3
Number of molNumber of mol 85.7 85.7
1212
7.14177.1417
14.3 14.3
11
14.314.3
Simplest ratioSimplest ratio 11 22
04/11/2304/11/23 MATTERMATTER 2626
mass formula emprical
massmolecular relativen
4
14
56
n = 56
14
= 4
molecular formula = C4H8
04/11/2304/11/23 MATTERMATTER 2727
1.2.2 Concentration of Solution1.2.2 Concentration of SolutionSolution When an amount of solute dissolved completely in a solvent and
it will form a homogeneous mixture.
04/11/2304/11/23 MATTERMATTER 2828
ExerciseExercise
A combustion of 0.202 g of an organic sample A combustion of 0.202 g of an organic sample that contains carbon, hydrogen and oxygen that contains carbon, hydrogen and oxygen produce 0.361g carbon dioxide and 0.147 g produce 0.361g carbon dioxide and 0.147 g water. If the relative molecular mass of the water. If the relative molecular mass of the sample is 148, what is the molecular formula.sample is 148, what is the molecular formula.
Ans : Ans : CC66HH1212OO44
04/11/2304/11/23 MATTERMATTER 2929
Units of concentration of a solution:
A. Molarity
B. Molality
C. Mole Fraction
D. Percentage by Mass
E. Percentage byVolume
04/11/2304/11/23 MATTERMATTER 3030
A.A. Molarity (M)Molarity (M) The number of moles of solute per cubic decimetre
(dm3) or litre (L) of solution.
Note:1 dm3 = 1000 cm3
1 L = 1000 mL
molar or L molor dm mol :Unit
)(dmsolution of volume
(mol) solute of molesM molarity,
1-3-
3
molar or L molor dm mol :Unit
)(dmsolution of volume
(mol) solute of molesM molarity,
1-3-
3
04/11/2304/11/23 MATTERMATTER 3131
ExampleExample
1
112212
342
)1611(22(12x2)sucrose of massMolar
Solution,
] 16O 12,C 1,H[Ar
water.of L 0.5 ain issolved
)OH(C sucrose g 1.71 ofsolution a ofmolarity theCalculate
molg
x
d
1
112212
342
)1611(22(12x2)sucrose of massMolar
Solution,
] 16O 12,C 1,H[Ar
water.of L 0.5 ain issolved
)OH(C sucrose g 1.71 ofsolution a ofmolarity theCalculate
molg
x
d
04/11/2304/11/23 MATTERMATTER 3232
Cont…Cont…
L mol 0.01 L 5.0
mol 0.005
solution of volume
sucrose of mole sucrosesolution ofmolarity
mol 0.005
342
g 1.71
massmolar
mass sucrose of mole of
1-
1
molg
Number
04/11/2304/11/23 MATTERMATTER 3333
ExercisesExercises
87.158:
] 16O 52,Cr 1,.39K[Ar
M? 2.16 with mL 250 ofsolution a prepare to
required O7CrK ,dichromate potassium of gramsmany How 22
Ans
87.158:
] 16O 52,Cr 1,.39K[Ar
M? 2.16 with mL 250 ofsolution a prepare to
required O7CrK ,dichromate potassium of gramsmany How 22
Ans
3-
332
moldm 0221.0:Ans
] 16O 12,C ,23Na[Ar
molarity? its Calculate
water.of cm 250 in CONa carbonate, sodium of g 0.586
dissolving by solution a prepared student ionmatriculat, A
3-
332
moldm 0221.0:Ans
] 16O 12,C ,23Na[Ar
molarity? its Calculate
water.of cm 250 in CONa carbonate, sodium of g 0.586
dissolving by solution a prepared student ionmatriculat, A
04/11/2304/11/23 MATTERMATTER 3434
B.B. Molality (Molality (mm)) Molality is the number of moles of solute dissolved in 1
kg of solvent
Note: Mass of solution = mass of solute + mass of solvent Volume of solution ≠ volume of solvent
mor molalor kg mol :
(kg)solvent of
(mol) solute of moles molality,
1-unit
massm
mor molalor kg mol :
(kg)solvent of
(mol) solute of moles molality,
1-unit
massm
04/11/2304/11/23 MATTERMATTER 3535
Example 1Example 1
] mol g 98.08SOH mass[molar
water?of g 198in
acid sulphuric of g 24.4 containingsolution
acid sulphuric ofmolality the
1-
42
Calculate
04/11/2304/11/23 MATTERMATTER 3636
m 1.26
kg 198.0
mol 2488.0
(kg)solvent of
(mol) solute of SOH ofMolality
mol 0.2488
08.98
4.24
mass
:
42
1
42
mass
moles
molg
gmolar
massn
Solution
SOH
04/11/2304/11/23 MATTERMATTER 3737
Example 2Example 2
] mol g 18.02OH mass[molar
water?of mol 40.0
in CuCl of mol 0.30 dissolvingby prepared
solution a ofion concentrat molal theishat
1-
2
2
W
04/11/2304/11/23 MATTERMATTER 3838m 0.416
kg 7208.0
mol 3.0
(kg)solvent of
(mol) solute of OH ofMolality
kg 7208or g 720.8
gmol 18.02 x mol 0.40 OH of mass
mass
:
2
1
2
2
mass
moles
molar
massn
Solution
OH
04/11/2304/11/23 MATTERMATTER 3939
ExercisesExercises
mAns 639.0:
water?g 203in CO])[(NH urea of
g 7.78 containingsolution a ofmolality theisWhat
22
mAns 639.0:
water?g 203in CO])[(NH urea of
g 7.78 containingsolution a ofmolality theisWhat
22
04/11/2304/11/23 MATTERMATTER 4040
mAns 653.0:
ion.concentrat
molal its Calculate .mL g 1.107 ofdensity a has
litreper ) Zn(NOof g 121.8 containingsolution A -1-
23
mAns 653.0:
ion.concentrat
molal its Calculate .mL g 1.107 ofdensity a has
litreper ) Zn(NOof g 121.8 containingsolution A -1-
23
04/11/2304/11/23 MATTERMATTER 4141
C. Mole Fraction (X)C. Mole Fraction (X)
Mole fraction is the ratio of the number of moles of one component to the total number of moles of all component present.
totaln
numbertotal
AA
A
nX
component all of
moles of
A of molesX A, component offraction mole
totaln
numbertotal
AA
A
nX
component all of
moles of
A of molesX A, component offraction mole
04/11/2304/11/23 MATTERMATTER 4242
It is always smaller than 1
The total mol fraction in a mixture (solution) is equal to one.
XA + XB + XC = 1
04/11/2304/11/23 MATTERMATTER 4343
Example 1Example 1
] mol g 18.02OH mass[molar
water?of mol 40.0
in CuCl of mol 0.30 dissolvingby prepared
solution ain CuCl offraction mole theishat
1-
2
2
2
W
04/11/2304/11/23 MATTERMATTER 4444
0.007
40 0.3
3.0
n
:
total
2
2
CuCl
CuCl
nX
Solution
04/11/2304/11/23 MATTERMATTER 4545
0.993
40 0.3
40
n total
2
2
OH
OH
nX
04/11/2304/11/23 MATTERMATTER 4646
0.093
007.01X
1XX
O2H
O2H2CuCl
04/11/2304/11/23 MATTERMATTER 4747
Example 2Example 2
79.9]Br 1.01,H 12.01,C[Ar
component?each offraction mole theisWhat
Br.HC nebromobenze of g 55 and HC
toluene,of g 55 mixingby prepared issolution
5687
A
04/11/2304/11/23 MATTERMATTER 4848
mol 0.5969
92.1555
8(1.01)7(12.01)55
n
:1 Step
8H7C
mol 0.3491
157.5555
79.905(1.01)6(12.01)55
n
:2 Step
Br5H6C
63.0
3491.05969.05969.0
X
:3 Step
8H7C
37.0
3491.05969.03491.0
X
:4 Step
Br5H6C
04/11/2304/11/23 MATTERMATTER 4949
D. Percentage by Mass (%w/w)D. Percentage by Mass (%w/w)
Percentage by mass is defined as the percentage of the mass of solute per mass of solution.
solvent of masssolute of masssolution of mass :note
100xsolution of masssolute of mass
ww%
solvent of masssolute of masssolution of mass :note
100xsolution of masssolute of mass
ww%
04/11/2304/11/23 MATTERMATTER 5050
Example 1Example 1
solution? in the mass
by percentage is What water.of g 54.3in dissolved
is KCl chloride, potassium of g 0.892 of sampleA
1.61%
100x3.54892.0
892.0
100xsolution of masssolute of mass
ww%
:Solution
1.61%
100x3.54892.0
892.0
100xsolution of masssolute of mass
ww%
:Solution
04/11/2304/11/23 MATTERMATTER 5151
Example 2Example 2
solution. massby percent
16.2 a ofn preparatio in the urea of g 5.00 toadded be
must that grams)(in water ofamount theCalculate
04/11/2304/11/23 MATTERMATTER 5252
g 25.86
5.00 - 30.86 solvent of mass
solvent of mass 5.00 30.86
solvent of masssolute of masssolution of mass
g 30.86
100x16.2
5 solution of mass
100xsolution of mass
5 16.2
100xsolution of masssolute of mass
ww%
:Solution
g 25.86
5.00 - 30.86 solvent of mass
solvent of mass 5.00 30.86
solvent of masssolute of masssolution of mass
g 30.86
100x16.2
5 solution of mass
100xsolution of mass
5 16.2
100xsolution of masssolute of mass
ww%
:Solution
04/11/2304/11/23 MATTERMATTER 5353
ExercisesExercises
g 247.5 ;50.2:
solution? NaOH 1.00% of g 250.0 prepare to
needed are water and NaOH of gramsmany How 1.
gAns g 247.5 ;50.2:
solution? NaOH 1.00% of g 250.0 prepare to
needed are water and NaOH of gramsmany How 1.
gAns
g 27.20:
HCl? of g 7.5 containssolution thisof mass theis What HCl.
37% ofsolution a as purchased becan acid icHydrochlor 2.
Ans g 27.20:
HCl? of g 7.5 containssolution thisof mass theis What HCl.
37% ofsolution a as purchased becan acid icHydrochlor 2.
Ans
04/11/2304/11/23 MATTERMATTER 5454
E. Percentage By Volume (%E. Percentage By Volume (%V / VV / V))
Percentage by volume is defined as the percentage of volume of solute in milliliter per volume of solution in milliliter.
solution of volume
solution of masssolution ofDensity
:
100x (mL)solution of volume
(mL) solute of volumeV
V%
note
solution of volume
solution of masssolution ofDensity
:
100x (mL)solution of volume
(mL) solute of volumeV
V%
note
04/11/2304/11/23 MATTERMATTER 5555
Example Example
solution? in this alcohol of by volume % theisWhat
alcohol. of mL 28 contains perfume of 200mLA
% 14
100x 200
28
100x (mL) solution of volume
(mL) alcohol of volumeV
V%
:Solution
% 14
100x 200
28
100x (mL) solution of volume
(mL) alcohol of volumeV
V%
:Solution
04/11/2304/11/23 MATTERMATTER 5656
1.2.3 Balancing Chemical Equation1.2.3 Balancing Chemical Equation
A chemical equation shows a chemical reaction using symbols for the reactants and products.
The formulae of the reactants are written on the left side of the equation while the products are on the right.
04/11/2304/11/23 MATTERMATTER 5757
Example:
x A + y B z C + w D
Reactants Products
04/11/2304/11/23 MATTERMATTER 5858
The total number of atoms of each element is the same on both sides in a balanced equation.
The number x, y, z and w, showing the relative number of molecules reacting, are called the stoichiometric coefficients.
The methods to balance an equation: Inspection Method
04/11/2304/11/23 MATTERMATTER 5959
Inspection MethodInspection Method
a. Write down the unbalanced equation. Write the correct formulae for the reactants and products.
b. Balance the metallic element, followed by non-metallic atoms.
c. Balance the hydrogen and oxygen atoms.
d. Check to ensure that the total number of atoms of each element is the same on both sides of equation.
04/11/2304/11/23 MATTERMATTER 6060
Example Example
Balance the chemical equation by applying the
inspection method.
NH3 + CuO → Cu + N2 + H2O
04/11/2304/11/23 MATTERMATTER 6161
Exercise Exercise
1. Balance the chemical equation below by applying inspection method.
a. Fe(OH)3 + H2SO4 → Fe2(SO4)3 + H2O
b. C6H6 + O2 → CO2 + H2O
c. N2H4 + H2O2 → HNO3 + H2O
d. ClO2 + H2O → HClO3 + HCl
04/11/2304/11/23 MATTERMATTER 6262
1.2.4 Redox Reaction1.2.4 Redox Reaction
Redox reaction is a reaction that involves both reduction and oxidation.
04/11/2304/11/23 MATTERMATTER 6363
Oxidation The substance loses one or more
elactrons.Increase in oxidation numberAct as an reducing agent (reductant)
04/11/2304/11/23 MATTERMATTER 6464
ReductionThe substance gains one or more
elactrons.decrease in oxidation numberAct as an oxidising agent (oxidant)
04/11/2304/11/23 MATTERMATTER 6565
Oxidation numbers of any atoms can be determined by applying the following rules:
1. In a free element , as an atom or a molecule the oxidation number is zero.
Example:
Na = 0 Cl2 = 0
Br2 = 0 O2 = 0
Mg = 0
04/11/2304/11/23 MATTERMATTER 6666
2. For monoatomic ion, the oxidation number is equal to the charge on the ion.
Example:
Na+ = +1 Mg2+ = +2
Al3+ = +3 S2- = -2
04/11/2304/11/23 MATTERMATTER 6767
3. Fluorine and other halogens always have oxidation number of -1 in its compound. Only have a positive number when combine with oxygen.
Example:
Oxidation number of F in NaF = -1
Oxidation number of Cl in HCl = -1
Oxidation number of Cl in Cl2O7 =+7
04/11/2304/11/23 MATTERMATTER 6868
4. Hydrogen has an oxidation number of +1 in its compound except in metal hydrides which hydrogen has an oxidation number of -1
Example:
Oxidation number of H in HCl = +1
Oxidation number of H in NaH = -1
Oxidation number of H in MgH2 = -1
04/11/2304/11/23 MATTERMATTER 6969
5. Oxygen has an oxidation number of -2 in most of its compound.
Example:
Oxidation number of O in MgO = -2
Oxidation number of O in H2O = -2
04/11/2304/11/23 MATTERMATTER 7070
However there are two exceptional cases:
- in peroxides, its oxidation number is -1
Example:
Oxidation number of O in H2O2 = -1
- When combine with fluorine, posses a
positive oxidation number
Example:
Oxidation number of O in OF2 =+2
04/11/2304/11/23 MATTERMATTER 7171
6. In neutral molecule, the sum of the oxidation number of all atoms that made up the molecule is equal to zero.
Example:
Oxidation number of H2O = 0
Oxidation number of HCl = 0
Oxidation number of KMnO4 = 0
04/11/2304/11/23 MATTERMATTER 7272
7. For polyatomic ions, the total oxidation number of all atoms that made up the polyatomic ion must be equal to the nett charge of the ion.
Example:
Oxidation number of KMnO4- = -1
Oxidation number of Cr2O72- = -2
Oxidation number of NO3- = -1
04/11/2304/11/23 MATTERMATTER 7373
Example :Example :
Assign the Assign the oxidation number of Cr in Cr in Cr22OO772-2-..
Solution :Solution :
CrCr22OO7 7 = -2 = -2
2 Cr + 7 (-2) = -22 Cr + 7 (-2) = -2
2 Cr = + 122 Cr = + 12
Cr = + 6Cr = + 6
04/11/2304/11/23 MATTERMATTER 7474
ExerciseExercise
1. Assign the oxidation number of Mn in the following chemical compounds.i. MnO2 ii. MnO4
-
2. Assign the oxidation number of Cl in the following chemical compounds.i. KClO3 ii. Cl2O7
2-
3. Assign the oxidation number of following:i. Cr in K2Cr2O7
ii. U in UO22+
iii. C in C2O42-
04/11/2304/11/23 MATTERMATTER 7575
1.2.4.1 Balancing Redox Reaction1.2.4.1 Balancing Redox Reaction
Redox reaction may occur in acidic and basic solutions.
Follow the steps systematically so that equations become easier to balance.
04/11/2304/11/23 MATTERMATTER 7676
Balancing Redox Reaction In Acidic Balancing Redox Reaction In Acidic SolutionSolution
Fe2+ + MnO4- → Fe3+ + Mn2+
1. Divide the equation into two half reactions, one involving oxidation and the other reductioni. Fe2+ → Fe3+ ii. MnO4
- → Mn2+
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2. Balance each half-reaction
a. first, balance the element other than oxygen and hydrogen
i. Fe2+ → Fe3+
ii. MnO4- → Mn2+
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b. second, balance the oxygen atom by adding H2O
and hydrogen by adding H+
i. Fe2+ → Fe3+
ii. MnO4- + 8H+ → Mn2+ + 4H2O
c. then, balance the charge by adding electrons to the
side with the greater overall positive charge.
i. Fe2+ → Fe3+ + 1e
ii. MnO4- + 8H+ + 5e → Mn2+ + 4H2O
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3. Multiply each half-reaction by an interger, so that number of electron lost in one half-reaction equals the number gained in the other.
i. 5 x (Fe2+ → Fe3+ + 1e)
5Fe2+ → 5Fe3+ + 5e
ii. MnO4- + 8H+ + 5e → Mn2+ + 4H2O
4. Add the two half-reactions and simplify where possible by canceling species appearing on both sides of the equation.
i. 5Fe2+ → 5Fe3+ + 5e
ii. MnO4- + 8H+ + 5e → Mn2+ + 4H2O
____________________________________________
5Fe2+ + MnO4- + 8H+ → 5Fe3+ + Mn2+ + 4H2O
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5. Check the equation to make sure that there are the same number of atoms of each kind and the same total charge on both sides.
5Fe2+ + MnO4- + 8H+ → 5Fe3+ + Mn2+ + 4H2O
Total charge reactant
= 5(+2) + (-1) + 8(+1)
= + 10 - 1 + 8
= +17
Total charge product
= 5(+3) + (+2) + 4(0)
= + 15 + (+2)
= +17
04/11/2304/11/23 MATTERMATTER 8181
Example: In Acidic Solution Example: In Acidic Solution
C2O42- + MnO4
- + H+ → CO2 + Mn2+ + H2O
Solution:
1. i. Oxidation : C2O42- → CO2
ii. Reduction : MnO4- → Mn2+
2. i. C2O42- → 2CO2
ii. MnO4- + 8H+ → Mn2+ + 4H2O
3. i. C2O42- → 2CO2 + 2e
ii. MnO4- + 8H+ + 5e→ Mn2+ + 4H2O
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4. i. 5 x (C2O42- → 2CO2 + 2e)
→ 5C2O42- → 10CO2 + 10e
ii. 2 x (MnO4- + 8H+ + 5e→ Mn2+ + 4H2O)
→ 2MnO4- + 16H+ + 10e→ 2Mn2+ + 8H2O
5. i. 5C2O42- → 10CO2 + 10e
ii. 2MnO4- + 16H+ + 10e→ 2Mn2+ + 8H2O
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5C2O42- + 2MnO4
- + 16H+ → 10CO2 + 2Mn2+ + 8H2O
04/11/2304/11/23 MATTERMATTER 8383
Balancing Redox Reaction In Basic Balancing Redox Reaction In Basic SolutionSolution
1. Firstly balance the equation as in acidic solution .
2. Then, add OH- to both sides of the equation so that it can be combined with H+ to form H2O.
3. The number of hydroxide ions (OH-) added is equal to the number of hydrogen ions (H+) in the equation.
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Example: In Basic SolutionExample: In Basic Solution
Cr(OH)3 + IO3- + OH- → CrO3
2- + I- + H2O
Solution:
1. i. Oxidation : Cr(OH)3 → CrO32-
ii. Reduction : IO3- → I-
2. i. Cr(OH)3 → CrO32- + 3H+
ii. IO3- + 6H+ → I- + 3H2O
3. i. Cr(OH)3 → CrO32- + 3H+ + 1e
ii. IO3- + 6H+ + 6e → I- + 3H2O
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4. i. 6 x (Cr(OH)3 → CrO32- + 3H+ + 1e)
→ 6Cr(OH)3 → 6CrO32- + 18H+ + 6e
ii. IO3- + 6H+ + 6e → I- + 3H2O
5. i. 6Cr(OH)3 → 6CrO32- + 18H+ + 6e
ii. IO3- + 6H+ + 6e → I- + 3H2O
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6Cr(OH)3 + IO3- → 6CrO3
2- + I- + 12H+ + 3H2O
6. 6Cr(OH)3 + IO3- + 12OH- → 6CrO3
2- + I- + 12H+ + 3H2O + 12OH-
7. 6Cr(OH)3 + IO3- + 12OH- → 6CrO3
2- + I- + 15H2O
04/11/2304/11/23 MATTERMATTER 8686
ExerciseExercise
Balance the following redox equations:
a. In Acidic Solution
i. Cu + NO3 + H+→ Cu2+ + NO2 + H2O
ii. MnO4- + H2SO3 → Mn2+ + SO4
2- + H2O + H+
iii. Zn + SO42- + H+ → Zn2+ + SO2 + H2O
b. In Basic Solution
i. ClO- + S2O32- → Cl- + SO4
2-
ii. Cl2 → ClO3- + Cl-
iii. NO2 → NO3 + NO
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1.2.5 Stoichiometry1.2.5 Stoichiometry Stoichiometry is the quantitative study of reactants and products in a chemical
reaction.
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Example:CaCO3 (s) + 2HCl (aq) → CaCl2 (aq) + CO2 (g) + H2O (l)
1 mole of CaCO3 reacts with 2 moles of HCl to yield 1 mole of CaCl2, 1 mole of CO2 and 1 mole of H2O.
Stoichiometry can be used for calculating the species we are interested in during a reaction.
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Example 1Example 1
How many moles of hydrochloric acid, HCl do we need to react with 0.5 moles of zinc?
HCl mol 1
HCl of mol 1
2 x 0.5h react wit Zn of mole 0.5
HCl of mol 2 with reacts Zn of mole 1
equation, theFrom
(g) H (s) ZnCl (l) 2HCl (s) Zn :Solution 22
04/11/2304/11/23 MATTERMATTER 9090
Example 2Example 2
How many moles of H2O will be formed when 0.25 moles of C2H5OH burns in oxygen?
OH mol 75.0
13 x 0.25
X
OH of moles X gives OHHC of mol 0.25
OH of moles 3 gives OHHC of mol 1
equation, theFrom
O3H 2CO 3O OHHC
:Solution
2
252
252
22252
04/11/2304/11/23 MATTERMATTER 9191
A 16.50 mL 0.1327 M KMnO4 solution is needed to oxidise 20.00mL of a FeSO4 solution in an acidic medium. What is the concentration of the FeSO4 solution? The net ionic equation is:
5Fe 2+ + MnO4- +8H+ Mn 2+ +5Fe 3+ +4H2O
Answer : 0.5474 M
Exercise 1Exercise 1
04/11/2304/11/23 MATTERMATTER 9292
How many mililitres of 0.112 M HCl will react exactly with the sodium carbonate in 21.2 mL of 0.150 M Na2CO3 according to the following equation?
2HCl(aq)+Na2CO3(aq) 2NaCl(aq)+CO2(g)+H2O(l)
Answer : 56.8 mL
Exercise 2 Exercise 2
04/11/2304/11/23 MATTERMATTER 9393
1.2.5.1 Limiting Reactant1.2.5.1 Limiting Reactant A limiting reactant is the reactant that is
completely consumed in a reaction and limits the amount of products formed.
An excess reactant is the reactant that is not completely consumed in a reaction and remains at the end of the reaction.
04/11/2304/11/23 MATTERMATTER 9494
Example 1Example 1S + 3F2 → SF6
If 4 mol of S reacts with 10 mol of F2 , which of the two reactants is the limiting reagent?
reactant. limiting theis F limit,in is F question. in the
available mol) (10 n he with tmol) (12 needed n theCompare
F mol 12 1
3 x 4X
F of moles X with reacts S of mol 4
F of moles 3 with reacts S of mol 1
equation, theFrom
:Solution
22
FF
2
2
2
22
04/11/2304/11/23 MATTERMATTER 9595
Example 2Example 2
C is prepared by reacting A and B :
A + 5B → C
In one process, 2 mol of A react with 9 mol of B.
a. Which is the limiting reactant?
b. Calculate the number of mole(s) of C?
c. How much of the excess reactant (in mol) is left at the end of the reaction?
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reactant. limiting theis B limit, in is B question. thein
available mol) (9 n the withmol) (10 needed n theCompare
OH mol 10
15 x 2
X
B of moles X withreacts A of mol 2
B of moles 5 withreacts A of mol 1
equation, theFrom
:A Solution
BB
2
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C mol 8.1
51 x 9
X
C of moles X withproduce B of mol 9
C of moles 1 withproduce B of mol 5
equation, theFrom
reactant. limiting the
B, of moles theof relies formedproduct of amount The
:B Solution
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A mol 0.21.8-2 reactant excess amount The
A mol 8.1
51 x 9
X
C of moles X withproduce B of mol 9
A of moles 1 withproduce B of mol 5
equation, theFrom
. reactant excess theis A
:C Solution
04/11/2304/11/23 MATTERMATTER 9999
Percentage yieldPercentage yieldThe percentage yield is the ratio of the The percentage yield is the ratio of the
actual yield (obtained from experiment) to actual yield (obtained from experiment) to the theoretical yield (obtained from the theoretical yield (obtained from stoichiometry calculation) multiply by stoichiometry calculation) multiply by 100% 100%
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Percentage yield = actual yield x 100% theoretical yield
04/11/2304/11/23 MATTERMATTER 101101
ExerciseExercise
In a certain experiment, 14.6g of SbF3 was allowed to react with CCl4 in excess. After the reaction was finished, 8.62g of CCl2F2 was obtained.
3 CCl4 + 2 SbF3 3 CCl2F2 + 2 SbCl3
[ Ar Sb = 122, F = 19, C= 12, Cl = 35.5 ]
a) What was the theoretical yield of CCl2F2 in grams ?
b) What was the percentage yield of CCl2F2 ?
Ans : a) 11.6 g b) 74.31 %
