1.0 Matter

d) i. 1 mol S Ξ 6.02 x 1023 atoms
5.10 mol S Ξ 5.10 x 6.02 x 1023 atoms
= 3.07 x 1024 atoms of S

ii. nCu = 3.14 g = 4.9371 x 10-2 mol
63.6 gmol-1

1 mol Cu Ξ 6.02 x 1023 atoms
4.9371 x 10-2 mol Cu Ξ 4.9371 x 10-2 x 6.02 x 1023 atoms

= 2.97 X 1022 atoms of Cu

e) i. nCa = 77.4 g = 1.9302 mol

40.1 gmol-1

ii. 6.02 x 1023 Co atoms Ξ 1 mol Co atoms
6.00 x 109 Co atoms Ξ 6.00 x 109 atoms x 1 mol Co atoms
6.02 x 1023 atoms
= 9.97 x 10-15 mol Co atoms

EMPIRICAL MOLECULAR
FORMULA FORMULA

The simplest ratio of The actual number of
all elements in a atoms of each element
molecule.
in a molecule.

Molecular formula = n ( empirical formula )

Where ; n = relative molecular mass
empirical formula mass

52

QUESTION 13

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

53

Assume mass of hydrocarbon sample = 100g

Mass of C = 85.7 x 100 g Mass of H = 14.3 x 100 g
100 100

= 85.7 g = 14.3 g

Element C H

Mass (g) 85.7 14.3

Moles 85.7 = 7.1417 14.3 = 14.3
(mol) 12.0 1.0

Simplest ratio 7.1417 = 1 14.3 = 2
7.1417 7.1417

Empirical formula : CH2

[ 12 + 2(1) ] n = 56
n=4

Molecular formula = (CH2)n = (CH2)4 = C4H8

54

QUESTION 14

A 0.5438g sample of liquid contains C, H and O burned in
pure O2 and obtained 1.039g CO2 and 0.6369g H2O
The molar mass of compound is 376 g/mol.
Find the empirical and molecular formula.

Ans : C2H6O, C16H48O8

55

Mass of C CxHyOz + O2 CO2 + H2O

= 1.039 g x 12g Mass of H = 0.6369 g x 2g
44 g 18 g

= 0.2834 g = 0.07077 g

Mass of O = 0.5438 –( 0.2834 + 0.07077) = 0.1896 g

Element C HO

Mass (g) 0.2834 0.07077 0.1896

Moles 0.2834 = 0.02362 0.07077 = 0.07077 0.1896 =0.01185
(mol)
12.0 1.0 16.0

Simplest 0.02362 = 1.99≈ 2 0.07077 = 5.97 ≈ 6 0.01185 = 1

ratio 0.01185 0.01185 0.01185

Empirical formula : C2H6O

[ 12(2) + 1(6) + 16(1) ] n = 376
n=8

Molecular formula = (C2H6O)8 = C16H48O8

56

A SOLUTION

SOLUTION
When an amount of solute dissolved completely in a

solvent and form a homogeneous mixture.

SOLUTE SOLVENT

SOLUTE + SOLVENT = SOLUTION

57

A SOLUTION

SOLVENT

SOLUTE SOLUTION

58

SOLUTE + SOLVENT = SOLUTION
NaCl + H2O = NaCl SOLUTION

ZnCl2 + H2O = ZnCl2 SOLUTION
NaOH + H2O = NaOH SOLUTION

59

CONCENTRATION MEASUREMENT

1. Molarity (M)
2. Molality (m)
3. Mole fraction (X)
4. Percentage by volume (% v/v)
5. Percentage by mass (% w/w)

60

MOLARITY (M)

The number of moles of solute dissolved
in 1 L/ 1dm3 of solution

moMlamroiostfyloe(,lmsutoel
voolsfuom(ldu3)emtio

U:nmdi-3tomomlrL-1omlr ola

Note:
1 dm3 = 1000 cm3
1 L = 1000 mL

61

QUESTION 15.

Calculate the molarity of a solution of 1.71 g sucrose ( C12H12011) dissolved
in a 0.5 L of water. [ Ar H = 1.01 C = 12.01 O = 16.00 ]

Mr of C12H12011 = 332.24

n of C12H12011 = mass
molar mass

= 1.71 g .
332.24 g/mol

= 5.1469 x 10-3 mol

molarity, M = mole of solute (mol)
volume of solution (L)

M of C12H12011 = 5.1469 x 10-3 mol
0.5 L

= 0.0103 mol L-1

62

QUESTION 16

Calculate the molarity of a solution made by dissolving 5.0 g of
C6H12O6 in water to form 100 mL solution

Mr of C6H12O6 = 180

n of C6H12O6 = mass
molar mass

= 5.0 g .

180 g/mol

= 0.027778 mol

mole of solute (mol)
molarity, M = volume of solution (L)

M of C6H12O6 = 0.027778 mol
0.10 L

= 0.278 M

63

MOLALITY (m)

Number of moles of solute dissolved in 1 kg of solvent

momlamliotsyfloe,(slmutoel)
moasfsosl(vkegn) t

u:nmitk-o1golmr olmral

Note:
Mass of solution = mass of solute + mass of solvent
Volume of solution ≠ volume of solvent

64

QUESTION 17

Calculate the molality of sulphuric acid solution containing 24.4 g of
sulphuric acid in 198 g of water . [ Mr H2SO4 = 98.08 ]

n of H2SO4 = mass
molar mass
= 24.4 g .

98.08 g/mol

= 0.24878 mol

mole of solute (mol)
molality, m = mass of solvent (kg)

Molality of H2SO4 = 0.24878 mol
0.198 kg

= 1.26 m

65

QUESTION 18

A solution containing 121.8 g of Zn(NO3)2 per litre has a density of
1.107 g mL-1. Calculate its molal concentration.
(Ans: 0.653 m)

Assume volume of solution = 1 L

n of Zn(NO3)2 = 121.8 g = 0.64308 mol

189.4 g/mol

density, = mass
volume

Mass solution = 1000 mL x 1.107 g/mL = 1107 g

Mass solvent, H2O = Mass solution – mass solute

= 1107g – 121.8 g
= 985.2 g

m = n of solute = 0.64308 mol = 0.653 m

mass solvent 0.9852 kg

66

MOL FRACTION (X)

Number of moles of a component divided by total
number of moles of all components

mofrlaecoctfioomn ApXo,AntoenntmatuloomolAfemfbseorl
oaf lclompo

XAnntAotal

It is always smaller than 1

XA + XB + XC = 1

67

BBB A
A
BB A
A BA
B A

XA = nA
nA + nB

XB = nB
nA + nB

XA + XB = 1

68

QUESTION 19

What is the mole fraction of CuCl2 in a solution
prepared by dissolving 0.30 mol of CuCl2 in 40.0 mol of
water? (Ans: 0.00744)

n of CuCl2 = 0.3 mol
n of H2O = 40 mol

X of CuCl2 = n CuCl2 .

n CuCl2 + n H2O

= 0.3
40.3

= 0.00744

69

QUESTION 20
How many grams MgCl2 needed to be added to 1000 g of water to
prepare a 0.1 mole fraction solution of MgCl2 ? ( 588.27 g)

X of MgCl2 = n MgCl2 = 0.1

n MgCl2 + n H2O

n of H2O = mass = 1000 g = 55.5556 mol
molar mass 18 g/mol

0.1 = n MgCl2 .

n MgCl2 + 55.5556

n of MgCl2 = 6.1728 mol

Mass MgCl2 = 6.1728 mol x 95.3 g/mol
= 588.27 g

70

PERCENTAGE BY MASS (% w/w)

Mass of solute divided by total mass of solution
in percentages

%wwmmoosasaffoosxs1lsls%uu0tt0ieon
n:omtoesafosmlsuosatfioosmlsnuosatfoes

71

QUESTION 21

A sample of 0.892 g of potassium chloride, KCl is dissolved in 54.3
g of water. What is the percentage by mass of this solution ?

(Ans: 1.62 % )

% w/w of KCl solution = mass of KCl l x 100 %

mass of water + mass of KCl

= 0.892 g x 100 %

0.892 g + 54.3 g

= 1.62 %

72

QUESTION 22

How many grams of HNO3 and water needed to prepare
250 g of 10% HNO3 solution ?

% wൗw = mass of solute × 100 %
mass of solution

10 % = mass of HNO3 solute × 100 %
250 g

Mass of HNO3 solute = 10 x 250 g = 25 g
100

Mass of water = mass of solution – mass of solute
= 250 g – 25 g
= 225 g

73

PERCENTAGE BY VOLUME (% V/V)

Volume of solute divided by volume of solution in
percentage

%VVvvoolluuoommfsfsoeoelluu((tmtimeoLxnL1))0% 0

no:te
Densoifstyolutimonaosfssolution
voluomfsoelution

74

QUESTION 23
Calculate the volume of antifreeze required to make 10 dm3 of a
solution of antifreeze which is 40% by volume. ( 4 dm3)

% v/v antifreeze solution = volume of antifreeze x 100 %
volume of antifreeze solution

40 % = volume of antifreezep x 100 %100
10 dm3

volume of antifreeze = 40 × 10 dm3
100

= 4 dm3

75

QUESTION 24
A solution contains glycerol, C3H8O3 and water. This solution is usually
used as coolant in vehicles such as cars. If the solution consists of
66.0 g of glycerol and 46.0 g of water and its density is 1.10 g cm-3,
calculate;

(a) The percent by mass of glycerol
(b) The concentration of glycerol

( Ans: 58.93 %, 7.05 M)

76

a) % wൗw of glycerol = mass of solute × 100 %
mass of solution

= 66 g x 100 %px 100
46 + 66 g

= 58.93 %

b) of solution = mass of solution
volume of solution

Volume of solution = 112 g .
1.10 g/cm3

= 101.8182 cm3

n glycerol = 66 g = 0.71739 mol
92 g/mol

M of glycerol = n glycerol______

volume of solution

= 0.71739 mol
0.10182 dm3

= 7.05 M

1.3 STOICHIOMETRY

LEARNING OUTCOMES

At the end of this topic, I should be able to:

(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

(c) Define limiting reactant and percentage yield.

(d) Perform stoichiometric calculations using mole concept
including limiting reactant and percentage yield.

79

OXIDATION NUMBER

Can be determined using these rules:

1. In a free element, an atom or a molecule, the
oxidation number is zero.

Eg: Cl2 = 0
Na = 0

80

2. For monoatomic ion, the oxidation number
is equal to the charge on the ion.

Eg: S2- = -2
Al3+ = +3

81

3. Fluorine and other halogens always have
oxidation number of -1 in its compound. Only
have a positive oxidation number when combine
with oxygen.
Eg:
Oxidation number of F in NaF = -1
Oxidation number of Cl in Cl2O7 = +7

82

4. Hydrogen has an oxidation number of +1 in its
compound except in metal hydrides which
hydrogen has an oxidation number of -1

Eg: -1
Oxidation number of H in NaH = -1
Oxidation number of H in MgH2 =

83

5. Oxygen has an oxidation number of -2 in most
of its compound.

Eg:
Oxidation number of O in MgO = -2
Oxidation number of O in H2O = -2

84

However there are two exceptional cases:
a) in peroxides, its oxidation number is -1

Eg:
Oxidation number of O in H2O2 = -1

b) when combine with fluorine, it has positive
oxidation number
Eg:
Oxidation number of O in OF2 = +2

85

6. In neutral molecule, the total oxidation number
is equal to zero.

Eg:
Oxidation number of H2O = 0
Oxidation number of KMnO4 = 0

86

7. For polyatomic ions, the total oxidation number
is equal to the net charge of the ion.

Eg:
Oxidation number of MnO4- = -1
Oxidation number of Cr2O72- = -2

87

EXAMPLE.

Assign the oxidation number of Cr in Cr2O72-.

Solution :

Cr2O7 = -2
2 Cr + 7(-2) = -2

2 Cr = + 12

Cr = + 6

88

CHEMICAL EQUATION

Shows a chemical reaction using symbols for the
reactants and products.

Ex: zC + wD
xA + yB

Reactants Products
CH4 + 2O2 CO2 + 2H2O

89

BALANCING CHEMICAL EQUATION

INSPECTION METHOD

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

2. Balance the metallic element, followed by non-metallic
atoms.

3. Balance the hydrogen and oxygen atoms.
4. Check to ensure that the total number of atoms of

each element is the same on both sides of equation.

90

EXAMPLE

Balance the chemical equation by applying the inspection
method.

NH3 + CuO → Cu + N2 + H2O

Answer:
2NH3 + 3CuO → 3Cu + N2 + 3H2O

91

REDOX REACTION

Oxidation Reduction

a) The substance loses a) The substance gains
one or more electrons. one or more electrons.

b) Increase in oxidation b) Decrease in oxidation
number number

c) Losing of hydrogen c) Losing of oxygen atoms
atoms
d) Gain of hydrogen atom
d) Gain of oxygen atoms

92

BALANCING REDOX EQUATION

ION-ELECTRON METHOD
IN ACIDIC MEDIUM

Cu + NO3- → Cu2+ + NO

1. Divide the equation into two half reactions, one
involving oxidation and the other reduction

i. Cu Cu2+ (oxidation)
ii. NO3- NO (reduction)

93

2. Balance the element other than oxygen and hydrogen

i. Cu Cu2+
ii. NO3- NO

3. Balance the O atom by adding H2O to side that needs O

i. Cu Cu2+

ii. NO3- NO + 2H2O

94

4. Balance the H atom by adding H+ to side that need H

i. Cu Cu2+
ii. 4H+ + NO3- NO + 2H2O

5. Balance the charge by adding electrons to the
side with the greater overall positive charge.

i. Cu Cu2+ + 2e-
ii. 4H+ + NO3- + 3e- NO + 2H2O

95

6. Make number of electron at both side equal

i. ( Cu Cu2+ + 2e- ) x 3
3Cu 3Cu2+ + 6e-

ii. (4H+ + NO3- + 3e- NO + 2H2O ) x 2
8H+ + 2NO3- + 6e- 2NO + 4H2O

96

7. Add the two half-reactions and cancel anything that
same on the both side

i. 3Cu 3Cu2+ + 6e-

ii. 8H+ + 2NO3- + 6e- 2NO + 4H2O

__________________________________

3Cu + 2NO3- +8H+ 3Cu2+ + 2NO + 4H2O

97

8. Check the equation, make sure same number of atoms
of each element and same total charge on both sides.

3Cu + 2NO3- +8H+ 3Cu2+ + 2NO + 4H2O
Total charge reactant Total charge product
= 3(0) + 2(-1) + 8(+1) = 3(+2) + 2(0) + 4(0)
= +6 =+6

98

QUESTION 25

P4 → PH3 + H2PO2-
(acidic medium)

Solution: :
1. i. Oxidation :

ii. Reduction

2. i.
ii.

3. i.
ii.

Answer: P4 + 6H2O → PH3 + 3H2PO2- + 3H+

99

IN BASIC MEDIUM

Cu + NO3- Cu2+ + NO

1. Balance the equation as in acidic solution

3Cu + 2NO3- +8H+ 3Cu2+ + 2NO + 4H2O

100