Matter (1.2 Mole Concept)

1.2 MOLE CONCEPT

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Learning 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. 11/05/08

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(c) Determine empirical and molecular formulae from mass composition or combustion data.

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(d) Define and perform calculation for each of the following concentration measurements : i) molarity (M) ii) molality (m) iii) mole fraction, X iv) percentage by mass, % w/w v) percentage by volume, %V/V 11/05/08

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(e) Determine the oxidation number of an element in a chemical formula. (f) Write and balance : i) chemical equation by inspection method ii) redox equation by ion-electron method

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(g) Define limiting reactant and percentage yield. (h) Perform stoichiometric calculations using mole concept including reactant and percentage yield.

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1.2 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.

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 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 =

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6.02 x 1023 mol-1

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Example 1.0 mole of chlorine atom 1.0 mole of chlorine molecules

1.0 mole of NH3

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= 6.02 x 1023 chlorine atoms = 35.5 g Cl = 6.02 x 1023 chlorine molecules = 71.1 g Cl2 = 6.022 x 1023 x 2 chlorine atoms = 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 MATTER

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Molar Mass  The

mass of one mole of an element or one mole of compound is referred as molar mass.

 Unit

: g mol-1

 Example:

-

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molar mass of Mg molar mass of CH4

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= = =

24 g mol-1 (12 + 4) gmol-1 16 g mol-1 10

Number of Mole Mass (g) Number of mole = Molar Mass (g mol -1 )

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Example 1 In 14 g of N 2 , calculate; i. Number of moles molecule if molar mass = 28 g mol -1 ii. Number of molecule iii. Number of atoms

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Solution : i. Number of moles molecule if molar mass = 28 gmol -1 Mass (g) Number of mole molecules N 2 = Molar Mass (g mol -1 ) 14 g = 28 g mol -1 = 0.5 mol

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Example 1 (cont…) ii. Numberof molecule Number of moleculesN2 = Numberof mole x NA = 0.5x 6.022x 1023 = 3.011x 1023 molecules iii.Numberofatoms 1 moleculeofN2 contains2 atomsof N 3.011x 1023 moleculeofN2 contains2 x 3.011x 1023 atomsof N = 6.022x 1023 atoms 11/05/08

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Example 2 Calculate the numberH atom in 1 mole of NH3 1 mole of NH3 contains3 molesof H atoms Numberof atoms Number of moles= NA Numberof atoms= 3 x 6.022x 1023 = 1.807x 1024atoms

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Example 3 Calculate the number of bromide ions in 2 moles of CaBr2 1 mole of CaBr2 contains 2 moles of bromide ions 2 mole of CaBr2 contains 4 moles of bromide ions so, Number of bromide ions = 4 x 6.022 x 10 23 = 1.2046 x 10 24 ions

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1.2.1 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

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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 temperature At room temperature, volume of gas (dm3) = number of mole X 24.0 dm3 mol-1

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Example 1 A balloon is filled with hydrogen gas at s.t.p. If the volume of the ballon is 2.24 dm 3 , calculate the amount (mole) of hydrogen gas. Solution 1, 22.4 dm 3 consists 1 mol of hydrogen gas 1 3 2.24 dm consists x 2.24 mol of hydrogen gas 22.4 so, Number of mole = 0.1 mol 11/05/08

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Cont… from example 1 Solution 2, volume of gas (dm 3 ) Number of mole = 3 −1 22.4 dm mol 2.24dm 3 = 3 −1 22.4 dm mol = 0.1 mol

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Exercise A sample of CO2 has a volume of 56 cm3 at STP. Calculate: c.

The number of moles of gas molecules 0.0025 mol

f.

The number of molecular 1.506 x 1021 molecules

i.

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The number of oxygen atoms in the sample 3.011x1021atoms Note: 1 dm3 = 1000 cm3 MATTER 21 3 1 dm = 1L

Empirical And Molecular Formulae -

-

Empirical formula is a chemical formula that shows the simplest ratio of all elements in a molecule. Molecular formula is a formula that show the actual number of atoms of each element in a molecule.

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-

The relationship between empirical formula and molecular formula is : Molecular formula = n ( empirical formula ) Where ;

relative molecular mass n= emprical formula mass 11/05/08

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Example A sample of hydrocarbon contains 85.7% carbon and 14.3% hydrogen by mass. Its molar mass is 56. Determine the empirical formula and molecular formula of the compound.

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Solution : C

H

85.7

14.3

Number of mol

85.7 12 7.1417

14.3 1 14.3

Simplest ratio

1

2

mass

Empirical formula = CH2 11/05/08

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relative molecular mass n= emprical formula mass 56 14 =4

n = 56 14 = 4

molecular formula = C4H8 11/05/08

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1.2.2 Concentration of Solution Solution  When an amount of solute dissolved completely in a solvent and it will form a homogeneous mixture.

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Exercise A combustion of 0.202 g of an organic sample that contains carbon, hydrogen and oxygen produce 0.361g carbon dioxide and 0.147 g water. If the relative molecular mass of the sample is 148, what is the molecular formula. Ans : C6H12O4

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

Units of concentration of a solution: A. B. C. D. E.

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Molarity Molality Mole Fraction Percentage by Mass Percentage byVolume

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A. Molarity (M) 

The number of moles of solute per cubic decimetre (dm3) or litre (L) of solution.

moles of solute (mol) molarity, M = 3 volume of solution (dm ) Unit :

mol dm -3 or mol L-1 or molar Note: 1 dm3 = 1000 cm3 1L = 1000 mL

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Example Calculate the molarity of a solution of 1.71 g sucrose (C12 H 22 O 11 ) dissolved in a 0.5 L of water. [Ar H = 1, C = 12, O = 16 ] Solution, Molar mass of sucrose = (12x2) + 22 + (11x16) = 342 g mol −1

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Cont… mass Number of mole of sucrose = molar mass 1.71 g = 342 g mol −1 = 0.005 mol mole of sucrose molarity of solution sucrose = volume of solution 0.005 mol = 0.5 L = 0.01 mol L-1 11/05/08

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Exercises How many grams of potassium dichromate, K2Cr2O7 required to prepare a solution of 250 mL with 2.16 M? [Ar K = 39.1, Cr = 52, O = 16 ] Ans : 158.87 A ,matriculat ion studentprepareda solutionby dissolving 0.586g of sodiumcarbonate,Na2CO 3 in 250 cm3 of water. Calculate itsmolarity? [Ar Na = 23, C = 12, O = 16 ] Ans: 0.0221moldm-3 11/05/08

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B. Molality (m) 

Molality is the number of moles of solute dissolved in 1 kg of solvent

moles of solute (mol) molality, m = mass of solvent (kg) unit : 

mol kg

Note:  Mass of solution  Volume

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-1

or molal or m

=

of solution ≠MATTER

mass of solute + mass of solvent volume of solvent

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Example 1 Calculate the molality of sulphuric acid solution containing 24.4 g of sulphuric acid in 198 g of water? [molar mass H 2 SO 4 = 98.08 g mol ] -1

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Solution : n H SO 2

4

mass = molar mass 24.4 g = 98.08 g mol −1 = 0.2488 mol

moles of solute (mol) Molality of H 2 SO 4 = mass of solvent (kg) 0.2488 mol = 0.198 kg = 1.26 m 11/05/08

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Example 2 What is the molal concentration of a solution prepared by dissolving 0.30 mol of CuCl 2 in 40.0 mol of water? [molar mass H 2 O = 18.02 g mol ] -1

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Solution : nH O 2

mass = molar mass

mass of H 2 O

= 40.0 mol x 18.02 gmol −1 = 720.8 g or 7208 kg

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moles of solute (mol) Molality of H 2 O = mass of solvent (kg) 0.3 mol = 0.7208 kg = 0.416 m MATTER

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Exercises What is the molality of a solution containing 7.78 g of urea [(NH 2 ) 2 CO] in 203 g water? Ans : 0.639 m

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A solution containing 121.8 g of Zn(NO 3 ) 2 per litre has a density of 1.107 g mL . Calculate its molal concentration. Ans : 0.653 m - -1

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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.

moles of A mole fraction of component A, X A = total number of moles of all component nA XA = ntotal

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 It

is always smaller than 1

 The

total mol fraction in a mixture (solution) is equal to one. XA + XB + XC = 1

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Example 1 What is the mole fraction of CuCl 2 in a solution prepared by dissolving 0.30 mol of CuCl 2 in 40.0 mol of water? [molar mass H 2 O = 18.02 g mol ] -1

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Solution : nCuCl X CuCl = n total

2

2

0.3 = 0.3 + 40 = 0.007 11/05/08

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XHO = 2

nH O 2

n total

40 = 0.3 + 40 = 0.993 11/05/08

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XCuCl2 + XH2O = 1 XH2O = 1− 0.007 = 0.093 11/05/08

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Example 2 A solution is prepared by mixing 55 g of toluene, C 7 H 8 and 55 g of bromobenzene C 6 H 5 Br. What is the mole fraction of each component? [Ar C = 12.01, H = 1.01, Br = 79.9]

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Step2 :

Step1: nC7H8 =

55 7(12.01) + 8(1.01)

=

nC6H5Br =

55 92.15

=

Step4 :

Step3 : 0.5969 = 0.5969+ 0.3491

XC 6H5Br =

0.3491 0.5969+ 0.3491

= 0.37

= 0.63

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55 157.55

= 0.3491mol

= 0.5969mol

XC7H8

55 6(12.01) + 5(1.01) + 79.90

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D. Percentage by Mass (%w/w) 

Percentage by mass is defined as the percentage of the mass of solute per mass of solution.

%w

mass of solute = x100 w mass of solution

note: mass of solution = mass of solute+ mass of solvent

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Example 1 A sample of 0.892 g of potassium chloride, KCl is dissolved in 54.3 g of water. What is percentage by mass in the solution? Solution : %w

w

= =

mass of solute x100 mass of solution 0.892 x100 0.892+ 54.3

= 1.61%

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Example 2 Calculate the amount of water (in grams) that must be added to 5.00 g of urea in the preparation of a 16.2 percent by mass solution.

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Solution : =

mass of solute x100 mass of solution

16.2=

5 x100 mass of solution

%w

w

mass of solution =

5 x100 16.2

= 30.86g mass of solution = mass of solute+ mass of solvent 30.86

= 5.00+ mass of solvent

∴ mass of solvent= 30.86 - 5.00 = 25.86g

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Exercises 1. How many grams of NaOH and water are needed to prepare 250.0 g of 1.00% NaOH solution? Ans : 2.50 g ; 247.5 g

2. Hydrochloric acid can be purchased as a solution of 37% HCl. What is the mass of this solution contains 7.5 g of HCl? Ans : 20.27 g

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E. Percentage By Volume (%V / V) 

Percentage by volume is defined as the percentage of volume of solute in milliliter per volume of solution in milliliter. %V

volume of solute (mL) = x 100 V volume of solution (mL)

note : mass of solution Density of solution = volume of solution 11/05/08

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Example A 200mL of perfume contains 28 mL of alcohol. What is the % by volume of alcohol in this solution? Solution : %V

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volume of alcohol (mL) x 100 V volume of solution (mL) 28 = x 100 200 = 14 % =

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1.2.3 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.

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 Example:

xA

+

yB

zC

Reactants

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+

wD

Products

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

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 11/05/08

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Inspection Method 

Write down the unbalanced equation. Write the correct formulae for the reactants and products.



Balance the metallic element, followed by nonmetallic atoms.



Balance the hydrogen and oxygen atoms.



Check to ensure that the total number of atoms of each element is the same on both sides of equation.

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Example Balance the chemical equation by applying the inspection method.

NH3 + CuO → Cu + N2 + H2O

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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.

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ClO2 + H2O → HClO3 + HCl

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1.2.4 Redox Reaction  Redox

reaction is a reaction that involves both reduction and oxidation.

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 Oxidation The

substance loses one or more elactrons. Increase in oxidation number Act as an reducing agent (reductant)

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 Reduction The

substance gains one or more elactrons. decrease in oxidation number Act as an oxidising agent (oxidant)

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

Oxidation numbers of any atoms can be determined by applying the following rules:



In a free element , as an atom or a molecule the oxidation number is zero. Example: Na = 0 Cl2 = 0

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Br2

=

0

Mg

=

0

O2

MATTER

=

0

65



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For monoatomic ion, the oxidation number is equal to the charge on the ion. Example: Na+ = +1 Mg2+ = +2 Al3+ = +3 S2- = -2

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

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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

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

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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

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

Oxygen has an oxidation number of -2 in most of its compound. Example: Oxidation number of O in MgO Oxidation number of O in H2O

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MATTER

= =

-2 -2

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However there are two exceptional cases: - in peroxides, its oxidation number is -1 Example: Oxidation number of O in H2O2 = -1 -

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When combine with fluorine, posses a positive oxidation number Example: Oxidation number of O in OF2 =

MATTER

+2

70



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 Oxidation number of KMnO4

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MATTER

= =

0 0

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

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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

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Example : Assign the oxidation number of Cr in Cr2O72-. Solution : Cr2O7 = -2 2 Cr + 7 (-2) = -2 2 Cr = + 12 Cr = + 6

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Exercise 

Assign the oxidation number of Mn in the following chemical compounds. i. MnO2 ii. MnO4-



Assign the oxidation number of Cl in the following chemical compounds. i. KClO3 ii. Cl2O72-

7.

Assign the oxidation number of following: i. Cr in K2Cr2O7 ii. U in UO22+ iii.C in C2O42-

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1.2.4.1 Balancing Redox Reaction  Redox

reaction may occur in acidic and basic solutions.

 Follow

the steps systematically so that equations become easier to balance.

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Balancing Redox Reaction In Acidic Solution Fe2+ + MnO4- → Fe3+ + Mn2+ 3.

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Divide the equation into two half reactions, one involving oxidation and the other reduction i. Fe2+ → Fe3+ ii.MnO4- → Mn2+

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1.

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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.

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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 MATTER

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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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

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+

→

Total charge reactant = 5(+2) + (-1) + 8(+1) = + 10 - 1 + 8 = +17

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5Fe3+ + Mn2+ + 4H2O

Total charge product = 5(+3) + (+2) + 4(0) = + 15 + (+2) = +17

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Example: In Acidic Solution C2O42- + MnO4- + H+ → CO2 + Mn2+ + H2O Solution:  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. ii.

5C2O42- → 10CO2 + 10e 2MnO4- + 16H+ + 10e→ 2Mn2+ + 8H2O

_________________________________________________ 5C2O42- + 2MnO4- + 16H+ → 10CO2 + 2Mn2+ + 8H2O

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Balancing Redox Reaction In Basic Solution 

Firstly balance the equation as in acidic solution .



Then, add OH- to both sides of the equation so that it can be combined with H+ to form H2O.



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 Solution Cr(OH)3 + IO3- + OH- → CrO32- + I- + H2O Solution:  i. Oxidation

:

Cr(OH)3 → CrO32-

ii. Reduction

:

IO3- → I-

2.

i. Cr(OH)3 → CrO32- + 3H+ ii. IO3- + 6H+ → I- + 3H2O

3. 11/05/08

i. Cr(OH)3 → CrO32- + 3H+ + 1e ii. IO3- + 6H+ + 6e → I- + 3H2O MATTER

84



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 ________________________________________________ 6Cr(OH)3 + IO3- → 6CrO32- + I- + 12H+ + 3H2O

6.

6Cr(OH)3 + IO3- + 12OH- → 6CrO32- + I- + 12H+ + 3H2O + 12OH-

7.

6Cr(OH)3 + IO3- + 12OH- → 6CrO32- + I- + 15H2O

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MATTER

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Exercise Balance the following redox equations: a. In Acidic Solution i. Cu + NO3 + H+→ Cu2+ + NO2 + H2O

b.

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ii.

MnO4- + H2SO3 → Mn2+ + SO42- + H2O + H+

iii.

Zn + SO42- + H+ → Zn2+ + SO2 + H2O

In Basic Solution i. ClO- + S2O32- → Cl- + SO42ii.

Cl2 → ClO3- + Cl-

iii.

NO2 → NO3 + NO

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1.2.5 Stoichiometry  Stoichiometry

is the quantitative study of reactants and products in a chemical reaction.

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MATTER

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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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MATTER

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Example 1 How many moles of hydrochloric acid, HCl do we need to react with 0.5 moles of zinc?

Solution : Zn (s) + 2HCl (l) → ZnCl 2 (s) + H 2 (g) From the equation, 1 mole of Zn reacts with 2 mol of HCl 0.5 x 2 ∴ 0.5 mole of Zn react with mol of HCl 1 1 mol HCl

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Example 2 How many moles of H2O will be formed when 0.25 moles of C2H5OH burns in oxygen? Solution : C 2H5OH + 3O2 → 2CO2 + 3H2O Fromtheequation, 1mol of C 2H5OH gives 3 molesof H2O ∴ 0.25mol of C 2H5OH gives X molesof H2O

X=

0.25x 3 1

= 0.75 mol H2O 11/05/08

MATTER

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Exercise 1 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

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MATTER

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Exercise 2 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

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MATTER

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1.2.5.1 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.

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MATTER

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Example 1 S + 3F2 → SF6 If 4 mol of S reacts with 10 mol of F2 , which of the two reactants is the limiting reagent? Solution : From the equation, 1 mol of S reacts with 3 moles of F2 ∴ 4 mol of S reacts with X moles of F2 4x3 X= 1 = 12 mol F2 Compare the n F needed (12 mol) with the n F (10 mol) available 2

2

in the question. F2 is in limit, ∴ F2 is the limiting reactant.

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MATTER

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Example 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. d. Which is the limiting reactant? e. Calculate the number of mole(s) of C? f. How much of the excess reactant (in mol) is left at the end of the reaction?

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MATTER

95

Solution A : Fromtheequation, 1 mol of A reacts with5 molesof B ∴ 2 mol of A reacts withX molesof B

X=

2 x5 1

= 10 mol H2O Compare thenB needed (10mol) with the nB (9mol) available in the question. B is in limit,∴ Bis thelimitingreactant.

11/05/08

MATTER

96

Solution B : The amountof product formedreliesof themolesof B, thelimitingreactant. Fromtheequation, 5 mol of B produce with1 molesof C ∴ 9 mol of B produce withX molesof C

X=

9 x1 5

= 1.8 mol C

11/05/08

MATTER

97

Solution C : A is theexcess reactant. Fromtheequation, 5 mol of B produce with1 molesof A ∴ 9 mol of B produce withX molesof C 9 x1 X= 5 = 1.8 mol A The amountexcess reactant= 2- 1.8= 0.2 mol A 11/05/08

MATTER

98

Percentage yield  The

percentage yield is the ratio of the actual yield (obtained from experiment) to the theoretical yield (obtained from stoichiometry calculation) multiply by 100%

11/05/08

MATTER

99

Percentage yield

11/05/08

=

actual yield x 100% theoretical yield

MATTER

100

Exercise 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 ]

•

What was the theoretical yield of CCl2F2 in grams ?

b) What was the percentage yield of CCl2F2 ?

11/05/08

Ans : a) 11.6 g

b) 74.31 %

MATTER

101