1

Chemistry

SK016

CHAPTER 1

MATTER

2

1.1 Atoms & Molecules

LEARNING OUTCOMES:

a) Describe proton, electron and

neutron in terms of the relative mass

and relative charge

b) Define proton number, Z, nucleon

number, A and isotope.

c) Write isotope notation.

LECTURE 1

3

Model of an Atom

Nucleus atom:

Electron, e (-ve)

Surrounding nucleus atom

i. proton, p (+ve) ii. neutron, n (neutral)

4

Points to Ponder

All the elements, except hydrogen atom, possess neutrons in the atomic nuclei.

What does a neutron do in an atomic nucleus?

Distinguish the nucleus, neutron and nucleon.

5

The properties of the particles

Name Charge Mass (amu) Mass (grams)

Electron (e) -1 5.4 x 10-4 9.1095 x 10-28

Proton (p) +1 1.00 1.675 x 10-24

Neutron (n) 0 1.00 1.675 x 10-24

6

Isotope Notation

(notation for Nuclides)

X

Symbol of element

A

Nucleon number

Z Proton number

c Charge on

an ion

Example:

327

13Al 216

8O

A = Z + neutron

e = p atom

e > p anion (-ve)

e < p cation (+ve)

7

Proton number ,Z: The number of proton in the

nucleus of an atom.

The particles that are found in the nucleus of an

atom is termed as nucleon. It consists of proton

and neutron.

Nucleon number, A: The total sum of protons

and neutrons found in the nucleus of the atom.

Note:

* Proton number is also known as Atomic Number

* Nucleon Number is also known as Mass Number

8

X A Z

c

1) ____ is the proton number.

It is the total number of _________ in

an atomic nucleus.

2) A is the ______________ of the

nuclide X.

3) Nucleon number is defined as the total

number of _________ and _________

in an atomic nucleus.

Exercise

9

The nucleon number of Kr = _______

The proton number of Co3+ = _______

The number of neutrons in Kr = _____

_____ contains 10 electrons.

Co3+ consists of ___ protons, ___ electrons

and ______ neutrons.

Exercise:

36 84 Kr

27 59 Co 3+

8 16 O 2-

10

Exercise: complete the following table

Species Proton

no.

Nucleon

no.

No.

n

No.

p

No.

e

Charge

on the

species

M 24 28 3+

Q -- 14 13 0

F 9 19 -- 10

T -- -- 10 8 10

Write the notation for nuclide T.

52

27

24 21

13

10

2-

18 8 T 2-

11

Points to Remember:

Writing the charge on an ion

An atom lost electrons – cation

An atom gained electrons – anion

An atom CANNOT donate protons!!!

Na+

Cl

correct incorrect Na1+

Cl1

Na+1

Cl1 Ca2+ Ca+2

S2 S2

12

Isotopes

Isotopes are two or more atoms of the

same element that have the same number

of protons in their nucleus but differ in

number of neutrons.

13

Isotopes

Hydrogen atom:

1 1 1

1 2 3 H H H

protium deuterium tritium

Proton number: 1 1 1

Number of neutrons: 0 1 2

H +

Hydrogen ion @ proton

_ H

Hydride ion

14

The nuclides are , , and :

Exercise

13 6 T 14

6 R 15 7 Z

a) Write the nuclides that are isotopes.

b) State the atoms that have the same number

of neutrons.

15

Compare and contrast between the

isotopes of an element by referring to

Ne-20, Ne-21 and Ne-22

Homework

16

LECTURE 2

1.1 Atoms & Molecules

17

LEARNING OUTCOMES:

(d) Define relative atomic mass, Ar and relative molecular mass, Mr based on the C-12 scale.

(e) Calculate the average atomic mass of an element given the relative abundances of isotopes or a mass spectrum.

18

Definition

Relative atomic mass, Ar of an element: mass of an

atom in comparison to 1/12 of mass of carbon-12

atom

Ar =

average mass of one atom of the element

mass of one atom C-12 1

12 x

19

Definition

Relative molecular mass, Mr of a molecular

substance: mass of a molecule in comparison to

1/12 of mass of a carbon-12 atom.

Mr =

average mass of one molecule of the substance

mass of one atom C-12 1

12 x

20

Mass spectrum of Rubidium

Interpretation

The mass spectrum of rubidium shows that naturally occurring rubidium consists of two isotopes (two peaks): 85Rb and 87Rb.

The height of each line is proportional to the abundance of each isotope. In this example, Rb-85 is more abundance than Rb-87.

Relative

Abundance

m/e

18

85 87

7

21

Average atomic mass

Average atomic mass

= ∑ (isotopic mass x abundance) ∑ abundance

= (mi x Qi)

Qi

22

Example: A sample of Rubidium is analyzed in a spectrometer. i) Calculate

the relative atomic mass of Rubidium. ii) What is the percentage

abundance of each of the isotopes?

Calculation:

i) Average atomic mass of Rb

=

Relative

Abundance

m/e

18

85 87

7

(mi x Qi)

Qi

85x18 + 87x7

(18 + 7) =

85.56 a.m.u =

23

ii) Percentage abundance of each of the isotopes:

% Rb-85 = 18

25 x 100 = 72 % % Rb-85 = 100% - 72%

= 28 %

Relative atomic

mass of Rb = 85.56 a.m.u

1/12 x 12.01 a.m.u

= 85.56 no unit!

24

Exercise:

The ratio of relative abundance of naturally

occurring of copper isotopes is as follow:

Based on the carbon-12 scale, the relative atomic

mass of 63Cu=62.9396 and 65Cu=64.9278.

Calculate the Ar of copper.

63

65

Cu

Cu = 2.333

25

Solution: 63

65

Cu

Cu = 2.333

63

65

Cu

Cu

2.333

1 =

The abundance of Cu-63 = 2.333 and

The abundance of Cu-65 = 1

Ar Cu =

26

The atomic masses of 6Li and 7Li are 6.0151u and

7.0160u respectively.

What is the relative abundance of each isotope if

the relative atomic mass of lithium is 6.941?

Exercise:

w = (the abundance of Li-6)

Solution: Let the abundance of 6Li = w and

the abundance of 7Li = 1-w

6.941 =

The abundance of Li-7 =

27

1.2 Mole Concept LEARNING OUTCOMES:

a) Define mole in terms of mass of carbon-12 and

Avogadro’s constant, NA.

b) Interconvert between moles, mass, number of

particles, molar volume of gas at s.t.p. and room

temperature.

c) Define the terms empirical and molecular formulae

d) Determine empirical and molecular formulae from

mass composition or combustion

LECTURE 3

28

Chemists have adopted the mole concept as a convenient way to deal with the enormous numbers of atoms, molecules or ions in the samples they work with.

Mole (unit mol):

The amount of substance that contains as many

elementary particles as there are atoms in exactly

12.000 g of carbon-12.

Mole Concept

29

The term mole, just like a dozen or gross, refers to

a particular number of things.

one dozen = 12

one gross = 144

one mole = 6.02 x 1023

Avogadro’s constant (NA)

The number of atoms in a 12 gram sample of carbon-

12 is called Avogadro’s constant (symbol NA or

L)

30

Molar mass

The molar mass of a substance is the mass of one

mole of the substance

For all substances, the molar mass in grams per

mole is numerically equal to the formula weight in

atomic mass units.

Molar mass = Ar or Mr (g mol1)

31

Mole Concept

1 mol = 6.02 x 1023 particles

(Avogadro’s constant = 6.02 x 1023 mol1 )

• 1 mol = Ar or Mr in unit gram

Ar of an element obtained from the Periodic Table

Mr of compound = Ar of each constituent atom

• 1 mol of any gas = 22.4 L at STP (0oC, 1 atm)

= 24 L at room temperature,RTP

(25oC, 1 atm)

32

Case 1: mole mass

1 mol copper, Cu = ____ g

Ar for

1 mol monatomic element = Ar g

1 mol silver ion, Ag+ = ____ g

0.25 mol silver, Ag = 0.25 x ____ g = g

a) Monatomic element

Ag = Cu = 108 63.5

0.1 mol copper(II) ion, Cu2+ = 0.1 x ____ g = g

63.5 63.5

6.36

108 108

27

33

Case 1: mole mass

1 mol chlorine, Cl2 =

Ar for

1 mol Polyatomic compound = Mr g

1 mol carbonate ion, CO32

0.4 mol CO32

b) Polyatomic substance

C = Cl = 12 35.5

2 x Ar (Cl) g

= 2 x _____ g = 71 g

= [Ar (C) + 3xAr (O)] g

= [___+ 3x____] g = 60 g

O = 16 35.5 12 16

= 0.4 x __ g

60

= 24 g

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Points to remember

1 mole 1 molecule

Units:

the atomic mass = Ar amu

the molecular mass = Mr amu

the molar mass = Ar or Mr g mol1

the mass of 1 mol of X = Ar or Mr g

Ar, Mr, the weight of X --- dimensionless

35

Exercise:

A sample of iron, Fe weighed 1.00 kg. What is the amount (mole) of Fe?

Solution:

Mole, n

Note:

= Mass, g

Molar mass, g mol1

Remember that mass must be expressed in grams here!

36

1 mol substance = 6.02 x 1023 particles

Case 2: mole particles

1 mol sodium, Na = 6.02 x 1023 atoms

1 mol fluorine, F2 = 6.02 x 1023 molecules F2

1 mol ammonium ion, NH4+ = 6.02 x 1023 cations

= 2 x 6.02 x 1023 atoms F

(I molecule of fluorine is made up of 2 atoms)

= 5 x 6.02 x 1023 atoms

(I particle of NH4+ consists of 5 atoms)

37

Exercise:

Calculate the number of atoms in 0.20 mol of

magnesium.

The number of Mg atoms

Solution:

= 0.20 mol x mol

atoms

=

Mole, n

Note:

= No. particles

Avogadro’s constant

38

1 mol Ar = 22.4 L at STP

1 mol gas = 22.4 L at STP

Case 3: mole volume gas

1 mol N2 = 22.4 L at STP

1 mol C2H2 = 22.4 L at STP

0.6 mol C2H2 = 0.6 x 22.4 L = 13.4 L

39

Exercise:

How many moles are there in 6.5 L oxygen at STP

?

Mole O2

Solution:

= 6.5 L x L

mol

= mol

40

2X + 3Y M + 2Q

a) Utilise stoichiometric proportions Case 4: mole X mole Y

Mole X

Mole Y =

2

3

a) 0.2 mole of X reacts with _______ mole of Y

b) 0.1 mole of Y needs _________ mole of X

to react completely

0.3

0.067

41

An antacid tablet contained 450 mg Na2CO3. When

swallowed, the Na2CO3 reacted with stomach acid

(HCl), according to the reaction equation,

Na2CO3 + 2HCl 2NaCl + CO2 + H2O

How many grams of HCl were neutralized by the

tablet?

Exercise:

42

Strategy:

Since the Na2CO3 is in units of mg, we need to

convert to g by a metric conversion before changing

to moles:

Solution:

= = mol

Mole of Na2CO3

1 mole Na2CO3 reacts with 2 moles of HCl

If mol of Na2CO3 Ξ x 2/1 mol HCl

= mol HCl

43

Therefore,

Mass of HCl = mol x 36.5 gmol-1

= g

= g

44

1 mole of C2H4 consists of

___ moles of C and ___ moles of H

b) Ratio by atoms in a molecular formula Case 4: mole X mole Y

3 moles of C2H4 give ____ moles of C

give ____ moles of H

0.5 mole of C2H4 gives ___ mole of C

gives ___ moles of H

2 4

6

12

1

2

45

Exercise:

Calculate the mass of (NH4)2CO3 that contains

a) 0.300 mol NH4+ b) 6.02 x 1023 H atoms

Solution (a):

Ξ 1 mol (NH4)2CO3

1 unit of mass of (NH4)2CO3 contains 2 units of NH4+

= 14.4 g

mol NH4+ 2

If 0.300 mol NH4+ Ξ 0.300 mol x ½ mol (NH4)2CO3

= 0.15 mol (NH4)2CO3

1 mol of (NH4)2CO3 = 96 g

If 0.15 mol of (NH4)2CO3 = 0.15 mol x 96 g mol-1

46

Exercise:

Calculate the mass of (NH4)2CO3 that contains

a) 0.300 mol NH4+ b) 6.02 x 1023 H atoms

Solution (b):

From (NH4)2CO3:

= 12.0 g

8 mol mol of H atoms = 1 mol of (NH4)2CO3

If 1 mol of H atom = 1 x 1/8 mol of (NH4)2CO3

= 0.125 mol of (NH4)2CO3

Then, 1 mol of (NH4)2CO3 = 96 g

Therefore, 0.125 mol (NH4)2CO3 = 0.125 mol x 96g/1 mol

6.02 x 1023 H atoms = 1 mol of H atoms

47

HOMEWORK

1. Sulphur is nonmetallic element. Its presence in coal gives rise to the acid rain phenomenon. How many atoms are in 16.3 g of sulphur.

(3.06 x 1023 atoms) 2. How many hydrogen atoms are present in

25.6 g of urea [(NH2)2CO], which is used as a fertilizer, in animal feed, and in the manufacture of polymers? The molar mass of urea is 60.06 g.

(1.03 x 1024 atoms)

48

Empirical Formula:

indicates which elements are present and the

simplest whole-number ratio of their atoms in a

molecule.

Molecular Formula shows the exact number of atoms of each element

in a molecule.

n x empirical formula = molecular formula

49

Compound EF MF n

formaldehyde CH2O CH2O 1

acetic acid CH2O C2H4O2 2

glucose CH2O C6H12O6 6

All 3 compounds are 40.00 % C

6.714 % H

53.27 % O

Compounds with different molecular formulae

can have the same empirical formula, and such

substances will have the same percentage

composition.

50

Consider the following flow-diagram:

Per cent composition

Mass Composition

Number of moles of

each element

Divide by smallest number of

moles to find the molar ratios

Multiply by appropriate number to

get whole number subscripts

51

Ascorbic acid (vitamin C) cures scurvy and may help

prevent the common cold. It is composed of 40.92% carbon,

4.58% hydrogen and 54.50% oxygen by mass.

The molar mass of ascorbic acid is 176 g mol1.

Determine its empirical formula and molecular formula.

Exercise

52

Solution

Element C H O

Mass/g

Amount/mol

Simplest ratio

40.92 4.58 54.50

40.92

12 1 16

4.58 54.50

= 3.41 = 4.58 = 3.41

3.41 3.41

4.58 3.41

3.41 3.41

= 1 = 1.33 = 1

1 x 3 1.33 x 3 1 x 3

= 3 = 3 = 4

53

Since the simplest ratio of C:H:O is 3:4:3

Therefore, the empirical formula is C3H4O3

Point to remember

Never round off values close to

whole number in order to get a

simple ratio, but multiply the

value by a factor until we get a

whole number.

54

Given Mr (C3H4O3)n = 176

Calculating Molecular Formula

[(12x3) + (4x1) + (16x3)] x n = 176

88 x n = 176

n = 2

Molecular formula = (C3H4O3)2

= C6H8O6

55

Homework

Define the terms relative atomic mass of an

element, relative molecular mass of a molecule,

empirical formula, and molecular formula.

Composition analysis of acyclic compound, Y with

molecular mass of 82 containing 87.8 % carbon and

12.2 % hydrogen.

Calculate the empirical formula and molecular

formula for the compound Y.

56

EXERCISE

A colourless liquid used in rocket engines, whose empirical formula is NO2, has a molecular mass of 92.0. What is its molecular formula?

57

Homework

A compound Y with chemical formula as

shown below:

CH2=CHCOOCH3

a) Write the empirical formula and molecular

formula of the compound

b) Calculate the percentage composition of

carbon in the compound Y

58

1.2 Mole Concept

LEARNING OUTCOMES:

e) Define and perform calculations for each of the following concentration measurements:

i) Molarity, M

ii) Molality, m

iii) Mole fraction, X

iv) Precentage by mass, % w/w

v) Percentage by volume, %v/v

LECTURE 4

59

Concentration Units

The concentration of solutions is the quantity of dissolved substance per unit quantity of solvent in a solution.

Concentration is measured in various ways:

a) concentration (formerly molarity), c

b) molal concentration (or molality), m

c) mole fraction, X

d) percentage by mass, %w/w

e) Percentage by volume, %v/v

60

Molarity, M or Amount Concentration

The number of mole of dissolved solute divided by the

volume of the mixture.

Alternative name: molarity

Symbol: M or c

Unit: mol L1 or mol dm3 or M

Molarity = Mole solute (mol)

Volume solution (dm3)

61

Example 1

A matriculation student prepared a solution by dissolving

0.586 g of sodium carbonate, Na2CO3 in 250.0 cm3 of

water. Calculate its concentration.

Solution:

Mole of Na2CO3(aq) = = 5.5283 x 10-3mol 0.586 g

g 106

Molarity of Na2CO3 (aq) = 5.5283 x 10-3 mol = 0.0221 mol dm-3

250.0 x 10-3 dm3

Molarity = mole solute (mol)

volume solution (dm3)

62

Molal Concentration or molality

The number of mole of dissolved solute divided by the mass (in kg) of the solvent.

Alternative name: Molality

Symbol: m

Unit: mol kg1 or molal or m

molality = Mole solute (mol)

mass solvent (kg)

63

Calculate the molal concentration of ethylene glycol

(C2H6O2) solution containing 8.40 g of ethylene glycol in

200 g of water. The molar mass of ethylene glycol is 62

g/mol.

Example 2

Solution:

= 0.677 mol kg1.

Mass of solute , ethylene glycol = 8.40g

Mole of solute = 8.40 g / 62.0 gmol-1 = 013548 mol

Mass of solvent = 200g

= 200 x 10-3 kg

Therefore, molality = 0.13548 mol

200 x 10-3 kg

64

Mole Fraction, X The mole fraction of component A is given by

XA =

where

nA = the number of mole of one component in a

mixture, A (for a given entity) and

nT = the total number of mole of all substances present in

the mixture (for the same entity).

n A

n T

65

What is the mole fraction of CuCl2 in a solution prepared

by dissolving 0.30 mol of CuCl2 in 40.0 mol of H2O.

Example 3

X =

Solution:

= 0.0074

0.30 mol

(0.30 + 40.0) mol CuCl2

66

Percentage by Mass, %w/w Alternative name: weight per cent

Symbol: %w/w

%w/w =

10% w/w NaOH means 10 g NaOH dissolve in 90 g water

Mass solute

mass solution X 100 %

67

A sample of 0.892 g of potassium chloride, KCl is

dissolved in 54.3 g of water. What is the per cent by mass

of KCl in this solution?

Example 4

% w/w KCl =

Solution:

= 1.62 %

0.892 g

(0.892 + 54.3) g X 100 %

Mass of solute = 0.892 g

Mass of solvent = 54.3 g

Mass of solution = mass of solute + mass of solvent

68

Percentage by volume, %v/v

% v/v =

5% v/v of KCl 5 mL of KCl dissolved in 100 mL

of solution

The units of volume used in the ratio must be same.

Example, both in mililiters or both in liters

volume solute, mL

volume solution, mL X 100 %

69

What volume of NaCl is needed to prepare 250

mL of 0.9% v/v solution.

Example 5

0.9 % v/v NaCl =

Solution:

= 2.25 mL

volume NaCl

250 mL solution X 100 %

Volume of NaCl

70

Calculation of mass from density

Example 6

Given the density of a 2.50 M aqueous solution of

methanol (CH3OH) is 0.954gcm-3 . Calculate the

molality of the solution.

71

Solution:

Assume that V solution = 1L @ 1000cm3

from density,0.954 g/cm3

so, mass of solution = 954 g

mole of solute = 2.5 mol/L x 1L= 2.5 mol

mass of solute = 2.5 mol x 32g/mol = 80 g

mass of solvent = 954 g – 80 g = 874 g

molality = mol solute (mol)

mass of solvent (kg)

= 2.5 mol

0.874 kg

= 2.86 mol/kg @ m

72

1.3 Stoichiometry

LEARNING OUTCOMES:

a) Determine the oxidation number of an element in a

chemical formula

b) Write and balance :

i) chemical equation by inspection method

ii) redox equation by ion-electron method

LECTURE 5

73

Oxidation Number / State

The oxidation number of an atom

is the charge assigned to the atom

according to a set of rules.

INTRODUCTION

74

Example (Free elements)

Ox. no. of Cl in Cl2 =

• Ox. no. of O in O3 =

• Ox. no. of P in P4 =

• Ox. no. of S in S8 =

0

If an element not combined chemically with

a different element, each atom has an oxidation

number of zero

• Ox. no. of Ne =

0

0

0

0

75

Example (Monatomic ion)

Ox. no. of Al3+ =

• Ox. no. of K+ =

• Ox. no. of S2 =

• Ox. no. of Cl =

+3

+1

2

1

For ions composed of only one atom, the oxidation

number is equal to the charge on the ion

76

Example (Polyatomic ion or molecule)

Ox. no. of Cl in Cl2O7

2 x (ox.no. Cl) + 7 x (ox.no. O) = 0

2 x (ox.no. Cl) + 7 x (-2) = 0

2 x (ox.no. Cl) = + 14

(ox.no. Cl) = + 14

2 = + 7

77

Example (Polyatomic ion or molecule)

Ox. no. of S in S4O62

4 x (ox.no. S) + 6 x (ox.no. O) = -2

4 x (ox.no. S) + 6 x (-2) = -2

4 x (ox.no. S) = -2 + 12

(ox.no. S) = + 10

4 = + 2.5

78

Redox Reaction

H2 Cl2 2HCl + 0 0 +1 -1

Oxidising reagent

or oxidant Reducing reagent

or Reductant

The hydrogen is oxidised (increased in oxidation number) and the

chlorine is reduced (decreased in oxidation number).

Thus, the reducing agent is H2 and the oxidising reagent is Cl2

79

Chemical Equations

A chemical equation is a way of denoting a chemical

reaction using the symbols (chemical formulae) for

the participating particles (atoms, molecules, ions or

free radicals).

CaCO3(s) + 2HCl(aq) CaCl2(aq) + CO2(g) + H2O(l)

A reactant is a starting substance in a chemical

reaction (on the left)

A product is a substance that results from a reaction

(on the right).

80

The state or phase of a substance may be indicated

with one of the following labels:

(g) ; (I) ; (s) ; (aq)

You can also indicate the conditions under which a

reaction takes place, as well as the presence of a

catalyst.

eg. 2H2O2(aq) Pt 2H2O(l) + O2(g)

Stoichiometric coefficients

81

81

General rules:

1) Write correct formulae for all reactants and products on the correct side of the “reaction arrow”

2) The equation can be balanced only by adjusting the coefficients of the formulae, as necessary.

3) Never introduce extraneous formulae (spectator ions) that are not involved in the reaction.

4) The total numbers of the atoms of each element is the same on both sides.

Balancing a Chemical Equation

82

Balancing a

Chemical Equation

Method 1: By Inspection Method

Method 2: Ion-Electron Method

83

Write correct formulae for all reactants and products:

ammonia reacts with copper(II) oxide to yield copper

metal, nitrogen and water.

Single arrow ( ) : irreversible reaction

double arrows ( ): reversible reaction

By Inspection

Cu Cu O + + + N H 3

N 2

H O 2

84

By Inspection

N N H H Cu Cu O O + + + 3 2 2

2 3 3 3

Start with the compound which has the most complex

formula or

Balance the element that occurs in only one compound on

each side of the equation or

Balance the atoms of each element except H and O

(on trial and error basis)

(g) (s) (g) (l) (s)

85

EXERCISES

1. By using inspection method, balance these

equations:

• I2 + Na2S2O3 NaI + Na2S4O6

• C3H8 + O2 CO2 + H2O

86

86

Ion-Electron Method 1. Write the half-equation

2. Balance all the elements except H and O

3. Balance O by adding H2O to the appropriate side of

the eq.

4. Balance H by adding H+ to the appropriate side.

5. Balance the charge by adding e-s to the side which

is more +ve.

6. Combine the two half-equations such that the e-s

cancel each other.

7. *In basic medium, add OH- to the both side of the

equation equal to the no. of H+ present.

87

Ion-Electron Method

+ + S O 2-

4 ClO

-

Apply to redox equations with the presence of ions

O S 2 3

2- Cl

-

Separate the equation into two half-equation:

Oxidation and reduction reactions

ClO -

ClO -

Cl -

Cl -

O S 2 3

2-

O S 2 3

2-

S O 2-

4

S O 2-

4

reduction

oxidation

88

Ion-Electron Method

+ + S O 2-

4 ClO

-

Apply to the redox equations with the presence of ions

O S 2 3

2- Cl

-

ClO -

Cl -

H + + +

O S 2 3

2-

O H 2

O H 2

S O 2-

4 + + H +

2

2 5 10

Balance the atoms of the elements undergoing changes in

oxidation number

Balance oxygen atoms by using H2O

Balance hydrogen atoms by using H+

89

Ion-Electron Method

+ + S O 2-

4 ClO

-

Apply to the redox equations with the presence of ions

O S 2 3

2- Cl

-

ClO -

Cl -

H + + +

O S 2 3

2-

O H 2

O H 2

S O 2-

4 + + H +

2 + e 2

2 5 10 + e 8

Balance the charge in each half-reaction by adding electrons to

equalise the ionic charges. Multiply the half-reactions by appropriate integer to ensure that

the number of electrons lost in oxidation is equal to the number

of electrons gained in reduction

x4 4 8 8 4 4

90

Ion-Electron Method

+ + S O 2-

4 ClO

-

Apply to the redox equations with the presence of ions

O S 2 3

2- Cl

-

O S 2 3

2- O H 2

S O 2-

4 + + H +

2 5 10 + e 8

ClO - Cl

- H + + + O H

2 + e 4 8 8 4 4

Sum the half-reactions. Simplify the overall equation

algebraically.

ClO - 4 O S

2 3

2- O H 2

+ + Cl -

+ 4 + 2 2 S O 2-

4 H +

4 O H 2

+ 4 2 + 2 H +

4 O H 2

+ 4 2 + 2 H +

For Acidic Medium

91

Ion-Electron Method

+ + S O 2-

4 ClO

-

Apply to the redox equations with the presence of ions

O S 2 3

2- Cl

- 4 O H

2 + 4 2 + 2 H +

For Basic Medium

Add enough OH- to both sides of the equation. The number of OH-

added is equal to the number of H+ in the equation so that OH-

combines with H+ to form H2O.

OH -

OH - + 2 + 2

O H 2

2

+ ClO -

O S 2 3

2- 4 + OH -

2 + S O 2-

4 Cl

- 4 2 + O H

2

92

Exercises

1. Balance the following equation in basic

solution by ion electron method.

2. Balance the redox equation by ion electron

method in acidic medium

MnO4- + NO2

- MnO2 + NO3-

MnO4- + C2O4

2- Mn2+ + CO2

93

1.3 Stoichiometry

LEARNING OUTCOMES

c) Define limiting reactant and

percentage yield.

d) Perform stoichiometric calculations

using mole concept including

limiting reactant and percentage

yield.

LECTURE 6

94

Stoichiometry

Theoretical Yield: The maximum obtainable product

calculated from the balanced equation when all of the

limiting reagents have reacted.

Actual Yield: The amount of product determined

experimentally from a reaction.

Stoichiometry is the calculation of the quantities of

reactants and products involved in a chemical

reaction

95

% yield = actual yield x 100%

theoretical yield

Percentage yield:

describes the proportion of the actual yield

to the theoretical yield.

it is calculated as follows:

96

Quantitative Information from

Balanced Equations

Grams and/or moles of

any two reactants or

products in a chemical

reaction are related by

coefficients in the

balanced chemical

equation relating the

two species.

Mass A

Moles A

Moles B

Mass B

Use molar mass of A

Use molar mass of B

Use coefficients of A & B

in balanced eqn

97

Stoichiometry

xA + yB zC + wD

The numbers x, y, z, and w, showing the relative numbers of molecules reacting, are called the stoichiometric coefficients.

x moles of A react with y moles of B to yield z moles of C and w moles of D.

98

Molecular Interpretation:

Molar interpretation:

1 molecule N2 + 3 molecules H2 2 molecules NH3

1 mol N2 + 3 mol H2 2 mol NH3

N2(g) + 3H2(g) 2NH3(g)

Consider the reaction:

Point to remember:

The number of moles involved in a reaction is proportional to

the coefficients in the balanced chemical equation.

Mole H2

Mole NH3 =

3

2

99

Stoichiometric Calculations

How much hydrogen would be needed to produce

153 g of ammonia?

N2(g) + 3H2(g) 2NH3(g)

Mole of NH3 = 153g = 9.0 mol

(14.0) + 3 (1.0) gmol-1

2 mol of NH3 = 3 mol of H2

If 9.0 mol of NH3 = 9.0 x 3/2 = 13.5 mol H2

SOLUTION:

100

a) What is the maximum number of grams of Ag that could

have obtained?

Cu + 2AgNO3 Cu(NO3)2 + 2Ag.

When 10.0 g of copper was reacted with excess silver nitrate

solution, 30.0 g of silver was obtained.

Exercise

Solution Mole of Cu = 10.0 g / 63.5 gmol-1 = 0.15748 mol

1mol Cu = 2 mol Ag

If 0.15748 mol Cu = 0.15748 x 2/1 = 0.31496 mol Ag

Therefore, mass of Ag = 0.31496 mol x 108 gmol-1 = 34.02 g

(theoritical yeild)

101

b) What was actual yield of Ag in grams?

Cu + 2AgNO3 Cu(NO3)2 + 2Ag.

When 10.0 g of copper was reacted with excess silver nitrate

solution, 30.0 g of silver was obtained.

Solution = 30.0 g Ag Actual yield of Ag

c) Calculate the percentage yield for this reaction.

Solution % yield = actual yield theoretical yield

X 100

= 30.0 g

34.0 g X 100

= 88.2 %

102

Limiting Reactant

The limiting reactant is the reactant which is

entirely consumed and limits the amount of

products that can be formed when a reaction goes

to completion.

The limiting reactant in a chemical reaction is

present in insufficient quantity to consume the other

reactant (s).

This situation arises when reactants are mixed in

non-stoichiometric ratios.

103

Excess Reactant

Notes:

1) The first step in a stoichiometric calculation is to write a balanced equation.

2) The coefficients in the balanced equation tell you only the molar ratios in which the species combine or are formed.

3) Always base the calculation of the maximum yield of product on the stoichiometric equivalency (molar ratio) between it and the limiting reactant.

Excess reactant are the reactants in quantities greater than necessary to react with the quantity of the limiting reactant.

104

9 slices of Bread

3 slices of cheese

product

Excess

Reactant:

bread reactants Limiting reactant:

cheese

The Cheese Sandwich Analogy

105

Given: X2(g) + 2Y (g) 2XY(g)

If X2 and Y were mixed in the quantities shown in the container

as follows and allowed to react, which of the following is the

correct representation of the contents of the container after the

reaction has occurred?

Point to Ponder:

Before reaction After reaction:

Option #1 Option #2 Option #3 = atom X

= atom Y

106

Exercise 1

Given: 2Al + Fe2O3 Al2O3 + 2Fe

In one process 124 g of Al are reacted with 601 g of Fe2O3.

(a) Identify the limiting reagent.

Solution: method 1 Mole reactant

Stoichiometric coefficient

The reactant that gives the smaller ratio is the limiting reactant---

Mole Al = = 4.59 mol 124

27

Mole Fe2O3 = = 3.77 mol 601

160

For Al : = 2.30 4.59

2

For Fe2O3 : = 3.77 3.77

1

Al

107

Exercise 1

Given: 2Al + Fe2O3 Al2O3 + 2Fe

In one process 124 g of Al are reacted with 601 g of Fe2O3.

(a) Identify the limiting reagent.

Solution: method 2

Stoichiometric equivalency: 2 moles Al 1 mole Fe2O3

Mole Al = = 4.59 mol 124

27 mol Mole Fe2O3 = =

601

160

4.59 moles Al need moles Fe2O3 2.295

3.77

4.59 2

1 x moles Fe2O3 =

The limiting reactant is Al (all Al will be consumed/used up)

The excess reactant is Fe2O3 (still available)

108

(b) Calculate the mass (in grams) of Al2O3 formed.

Exercise 1

Given: 2Al + Fe2O3 Al2O3 + 2Fe

In one process 124 g of Al are reacted with 601 g of Fe2O3.

Solution: limiting reactant = Al

Stoichiometric equivalency: 2 moles Al 1 mole Al2O3

If 4.59 mol of Al = 4.59 x ½ Al2O3

= 2.295 mol Al2O3

Therefore, mass of Al2O3 = 2.295 mol x 102 gmol-1

= 234g

109

(c) How much of the excess reagent (in grams) is left over at

the end of the reaction?

Exercise Given: 2Al + Fe2O3 Al2O3 + 2Fe

In one process 124 g of Al are reacted with 601 g of Fe2O3.

Solution: limiting reactant = Al

From (a): Mole Fe2O3 (left over) = 3.77 – 2.295

= 1.475 mol

Mass Fe2O3 (left over) = 1.475 mol x mol

g

1

160

= 236 g

110

Exercise:

1. 2.00 g of sodium is reacted with 2.45 g chlorine to produce sodium chloride.

i. Balance the chemical equation

ii. Name the limiting reactant

iii Determine the mass of sodium chloride produced in this reaction.

iv. Calculate the excess amount of other reactant remaining after the reaction is completed.

111

2. 10.0g of Zn are added into a 50.0mL HCl

solution with a molarity of 0.20M to form H2

and ZnCl2. Calculate

i. the volume of H2 gas evolved at s.t.p

ii. the amount of excess reactant remains

after the reaction has completed.

112

Urea [(NH2)2CO] is prepared by reacting ammonia

with carbon dioxide:

2NH3(g) + CO2(g) (NH2)2CO(aq) + H2O(l)

In one process, 637.2 g of NH3 are allowed to

react with 1142 g of CO2

a) Which of the two reactants is the limiting reagent?

b) Calculate the mass of (NH2)2CO formed.

c) How much of the excess reagent (in grams) is left at the end of the reaction?

3)

113

Industrially, vanadium metal, which is used in steel

alloys, can be obtained by reacting vanadium (V)

oxide with calcium at high temperatures:

5Ca + V2O5 5CaO + 2V

In process 1.54 x 103 g of V2O5 react with

1.96 x 103 g of Ca

i) Calculate the theoretical yield of V

ii) Calculate the percent yield if 803 g of V are obtained.

4)

114

5. 10 g of Zn is added to a beaker containing

0.18 mole of hydrochloric acid to form ZnCl2

and hydrogen gas. Determine;

a) The limiting reactant

b) The mass in gram for H2 produced

115

Homework

Aspirin is produced from salicylic acid (C7H6O3) and

acetic anhydride (C4H6O3) according to the following

balanced equation:

C7H6O3 + C4H6O3 C9H8O4 + C2H4O2

salicylic acetic aspirin acetic

acid anhydride acid

How many grams of aspirin would you obtain

from 4.5 g salicylic acid if the percentage yield of the

reaction is 75%?

116 116

Stoichiometry of Reactions

in Solution

Learning Outcomes:

Perform stoichiometric calculations using

mole concept including limiting reactant

and percentage yield.

LECTURE 7

117

c1

V1 Volume of water or

solvent added

V2

c2

Dilution

Initial moles of solute, n1 = final moles of solute, n2

c1 V1 = c2 V2 c@M = n (mol)

V (L)

(concentrated) (dilute)

M1 V1 = M2 V2 Or

118

Exercise A particular analytical chemistry procedure

requires 0.0500 M K2CrO4. What volume of

0.250 M K2CrO4 must be diluted with water to

prepare 100 mL of 0.0500 M K2CrO4?

Solution

M1 = 0.250 M M2 = 0.0500 M V1 = ? V2 = 100 mL

M1 V1 = M2 V2

(0.250 M) V1 = (0.0500 M) (100 mL)

V1 = 20.0 mL

119

The laboratory procedure for preparing a solution by dilution is as follows:

A pipette is used to withdraw a 20.0-mL sample of 0.250 M K2CrO4(aq).

The pipetteful of 0.250 M K2CrO4 is discharged into a 100.0-mL volumetric flask.

Following this, water is added to bring the level of the solution to the calibration mark etched on the neck of the flask. At this point the solution is 0.0500 M K2CrO4.

120

Titrant:

Mt : concentration of titrant

Vt: volume of added

titrant or titre value

Analyte:

Ma: concentration of analyte

Va: volume of pippeted analyte

50

40

30

20

10

0

Mt Vt

Ma Va = stoichiometric equivalency

121

Exercise A 25.0-mL sample of HCl solution is titrated

against Na2CO3 solution of 0.150 M. It requires

21.2 mL of Na2CO3 for complete neutralisation.

Calculate the concentration of HCl solution.

2HCl(aq) + Na2CO3(aq) 2NaCl(aq) + CO2(g) + H2O(l)

Solution

M

V

M(HCl) = ? M(Na2CO3) = 0.150 M

V(HCl) = 25.0 mL V(Na2CO3) = 21.2 mL

(Na2CO3) M V

(HCl) =

2

1

(21.2) (0.150)

M (HCl) =

2

1

(25.0)

M (HCl) = 0.253 M

122

Exercise

Solution

A 16.42-mL volume of 0.1327 M KMnO4 solution

is needed to oxidise 20.00 mL of a FeSO4

solution in an acidic medium. What is the

concentration of the FeSO4?

5Fe2+ + MnO4 + 8H+ Mn2+ + 5Fe3+ + 4H2O

M

V M V =

5

1

(Fe2+)

(MnO4 )

M (Fe2+)

(16.42) (0.1327) =

(20.00) 5

1

M (Fe2+) = 0.5447 M

M

(MnO4 )

(MnO4 ) M (Fe2+)

(Fe2+) V V

=

=

=

=

0.1327 M

20.00 mL 16.42 mL

?