Postsecondary lesson for pre-u students or form 6th students on mole concept, stoichiometry, limiting reagent, spectrometry and percent yield and percent purity.

1. CHAPTER 1:
ATOMS, MOLECULES &
STOICHIOMETRY
CHEMISTRY 962/1

2. How to study Chemistry?
 Always Pay attention during the lecture hours
 Always question
 Always revise the lessons on the same day it was taught
 Always study one steP ahead of the TEACHER
 Always test your understanding by doing
exercises/discussions

3. LEARNING OUTCOMES ;
 (a) describe the properties of protons, neutrons
and electrons in terms of their relative charges
and relative masses;
 (b) predict the behaviour of beams of protons,
neutrons and electrons in both electric and
magnetic fields;
 (c) describe the distribution of mass and charges
within an atom;

4. Fundamental Particles of An Atom
Particle Relative
Mass
Relative
Charge
Deflection
in electric
& magnetic
field
Electron (e) 1/1834 -1 deflected
Proton (p) 1 +1 deflected
Neutrons (n) 1 0 not

5. Behaviour of p, e- & n in Electric
Field
-
+
Beams
of P, e-
& n
P
n
e-
Deflection greater
due to lighter mass
a > b
a
b

6. Behaviour of P, e- & n in Magnetic
Field
N
S
Beams
of P, e-
& n
n = not
affected
P = deflectd
to south Pole
e- = deflectd
to north Pole
•Magnetic field is acting PerPendicularly to the Plane of
PaPer
a > b
a
b

7. LEARNING OUTCOMES ;
 (d) determine the number of protons, neutrons and
electrons present in both neutral and charged
species of a given proton number and nucleon
number;
 (e) describe the contribution of protons and
neutrons to atomic nuclei in terms of proton
number and nucleon number;
 (f) distinguish isotopes based on the number of
neutrons present, and state examples of both
stable and unstable isotopes.

8. Nuclide symbol
(Notation for nuclides)
nucleon number A C± charge on an ion
x
Proton number Z
A = Z(P.number) + neutron
e = P atom
Example: e > P anion(-ve charge ion)
27 3+ 16 2- e < P cation(+ve charge ion)
AI O
13 8

9. A Positive charge ion
formed when a neutral
atom loses an electrons(s).
11 Protons 11 Protons
11 electrons 10 electrons
A negative charge ion
formed when a neutral
atom gains an electron (s).
17 Protons 17 Protons
17 electrons 18 electrons
Two types of ions:
Na Na+ CI CI-

10. Exercise 1
84 59 16
Kr Co3+ O2-
36 27 8
 The nucleon number of Kr =
 The Proton number in O2- =
 The number of neutrons in O2- =
 contains 10 electrons.
 Co3+ consist of Protons, electrons and
neutrons.
84
8
8
Oxide ion
32
24
27

11. Isotopes
• Def : Isotopes are two or more
atoms of the same elements that
have the same number of Protons
but different number of neutrons.

12. • Most elemts exist as isotopes. The
abundance of each isotope in the
mixture is called isotopic
abundance.
• Same chemical Properties but
different Physical Props.
• Written as, for eg : chlorine-35 ;
chlorine-37 @ 35Cl ; 37Cl

13. a. Type of Isotope –
1) stable – depends on P & n
eg : 1H, 12C, 14N, 127I
2) unstable – too many P or n
unstable nucleus emits
radiation = radioactive
eg : 3H, 14C, 15N, 131I

14. b) Types of radiation
1) alpha (α) => α Particle ( He2+)
2) Beta (β) => β Particle γ ( e-)
3) Gamma (γ) => γ rays with α or β
Particle
4
2
0
-1

15. Isotopes undergo spontaneous
radioactive decay = radioisotopes
Eg : Uranium
Thorium
Radium
235
92
U
231
90
Th
223
88
Ra

16. LEARNING OUTCOMES ;
 (a) define the terms relative atomic mass, Ar,
relative isotopic mass, relative molecular
mass, Mr, and relative formula mass based
on 12C;
 (b) interpret mass spectra in terms of relative
abundance of isotopes and molecular
fragments;
 (c) calculate relative atomic mass of an element
from the relative abundance of its isotopes or
its mass spectrum.

17.  Relative mass of an atom is expressed in atomic
mass unit (a.m.u)
 Indication of how heavy one atom of an element
compared to another element
 C-12 = as standard for measurement
= the most abundant isotope
= solid and easily available
mass of a C-12 atom = 12 a.m.u
1 a.m.u = 1/12 x the mass of a C-12 atom
RAM, RIM, RMM & RFM

18. RAM,Ar, of an element =
Average mass of one atom of the element
1 x Mass of one atom C-12
12
Relative Atomic &
Molecular Mass

19.  RIM (on 12C scale) =
Mass of one atom of the isotope
1 X Mass of one atom of 12C
12

20. RMM, Mr, of a molecular substance =
Average mass of one molecule of the substance
1 x Mass of one atom C-12
12
RFM = RMM,
Mass of one formula of ionic compound
1 x Mass of one atom C-12
12

21. Mass Spectrometry
A mass spectrometer is used to determine
1. Relative atomic mass of an elements
2.Relative molecular mass of a compound
3.Types of isotopes, the abundance and its
relative isotopic mass
4.Recognize the structure of the compound in an
unknown sample

22. Mass Spectrometer Diagram
VAPORISATION

23.  Sample of element is placed in the vaporisation
chamber → converted to gaseous atoms
 Gaseous atoms ionised by the bombardment of high
energy e- emitted by a hot cathode to become +ve ions
M(g) → M+(g) + e
 The +ve ions then accelerated to a high and constant
velocity by two –vely charged plates
 The +ve ions then deflected by magnetic field
 Ion with smaller mass will deflect more than heavier
ones
 These ions are detected by ion detector
 Mass spectrum is produced

24. Mass spectrum of magnesium
24 25 26 m/e
 The mass of spectrum of Mg show that Mg
consist of three isotopes: 24Mg, 25Mg, 26Mg
 The height of each line is Proportional to
the abundance of each isotopes
 24Mg is the most abundant of three isotopes
63
8.1 9.1
%
abundance

25. How to calculate average atomic mass
from mass spectrum?
RAM=∑Qi Mi = (m1 x a) + (m2 x b) + (m3 x c)
∑ Qi a + b + c
Q= the relative abundance / Percentage abundance
of an isotope of the element
M= the relative isotopic mass (m/e) of the elements
= m
a = b = c = relative @ % abundance @ intensity of
each isotope

26. Example 1
Calculate the relative atomic mass of Mg
Relative intensity
63
9.1
8.1
24 25 26 m/e

27. Solution :
RAM Mg = (24 x 63) + (25 x 8.1) + (26 x 9.1)
63 + 8.1 + 9.1
= 24.3 #

28. Example 2
Neon has 3 isotopes , 20Ne, 21Ne, & 22Ne
in the ratio of 9.1 : 0.02 : 0.88
a) Sketch the mass spectrum of Ne
b) Calculate RAM of Ne

29. Solution :
a) % abundance : 20Ne = 9.1 = 0.91 = 91%
10
21Ne = 0.02 = 0.2%
10
22Ne = 0.88 = 8.8%
10

30.  Mass sPectrum ;
20 21 22 m/e
91.0
8.8
0.2
% abundance

31. b) RAM Ne =(20 x 91.0) + (21 x 0.2) + (22 x 8.8)
100
= 20.178 @ 20.2
= 20.2 #

32. Mass Spectra of Molecular Species
Molecule breaks into fragments ions
when bombarded ĉ high E electrons
Eg : a water sample is analysed :-
H2O(g) + ↑E e-  [H2O]+(g) + e-
║
Molecular ion @ Parent ion

33. H2O can break to form fragments ions
H2O(g) + ↑E e-  H+(g) + OH●(g) +e-
or H2O(g) + ↑E e-  H●(g) + OH+(g) +e-
H2O(g) + ↑E e-  2H+(g) + O+(g) + 3e-
H H
O
H H
O

34. The mass spectrum of water would
consists of 2 set of lines
Relative
abundance
1 16 17 18 m/e
H+
O+
OH+
[H2O]+
RMM of a
molecular sps is
given by the Peak
/ line ĉ highest m/e
ratio
But not
necessarily ĉ
highest
abundance

35. Eg 1 : Simplified mass spectrum of pentane,
C5H12
 What are the ions responsible for the peaks
at 29, 43, 57 and 72?
 Explain the peak at 44.

36. Eg. 2 : On analysis, organic
compound Q is found to be
Phenylethanone, C6H5COCH3. m/e
ratio of mass spectrum Q gives
Peaks at 15, 28, 43, 77, 105 & 120.
What are the ions responsible for
these Peaks?

37. RMM = 77 RMM = 43
RMM = 105
RMM = 15

38. RMM
= 77
RMM
= 28
RMM
= 15

39. Solutn :
m/e Peak Ion responsible
15
28
43
77
105
120
[C6H5]+
[CO]+
[C6H5COCH3]+
[CH3]+
[CH3CO]+
[C6H5CO]+

40. Eg 3 : The mass spectrum (not to scale)
of an alcohol with the formula of C3H8O
is shown below :
15 17 43 45 60 61

41. a) Draw the Possible structures of the
alcohol
b) Give the ions responsible for the peaks
ĉ mass 15, 17, 43, 45 & 60.
c) Suggest a structural formula of the
alcohol
d) What sps responsible for the Peak 61?

42. Solution :
H H H H H H
a) H – C – C – C – OH , H – C – C – C –H
H H H H OH H
b) 15 : [CH3]+
17 : [OH]+

43. 43 : [CH3CH2CH2]+ @ CH3C+H/CH3CH +
CH3 CH3
45 : [CH2CH2OH]+ @ CH3C+H / CH3CH +
OH OH
60 : [CH3CH2CH2OH]+ @ CH3CHCH3
+
OH

44. c) Since there is no Peak corresponding
to the fllwg ‘signature’ fragments :
CH3CH2CH2OH  29 : CH3CH2
+ @
31 : CH2OH+,
alcohol cannot be CH3CH2CH2OH,
:. Structure is H H H
H – C – C – C – H
H OH H

45. d) Caused by isotope of carbon, C-13 in
the molecule 13CH3CHCH3
+
OH

46.  No. of an atom in 12 g of C-12 is 6.02 x 10²³
 Called Avogadro’s number @ constant
 Symbol = NA @ L
 Based on C-12 scale, One mole of any
substance which contains the same number
of particles (atoms, molecule or ion) as the
no. of atoms in 12.000 grams of C-12
THE MOLE AND
AVOGADRO CONSTANT

47.  1 mol of He = 6.02 x 10²³
 1 mol of CO2 = 6.02 x 10²³ molecules
= 6.02 x 10²³ C atoms
= 6.02 x 10²³ x 2 O atoms
 1.0 mole of Cu(OH)2 = 6.02 x 10²³ units
formula
= 6.02 x 10²³ x 1 Cu2+ ions
= 6.02 x 10²³ x 2 OH- ions
= 6.02 x 10²³ x 3 ions
EXAMPLE

48. Mole & Concentration of Solution
[solutn] in moldm-3
= n solute
V solution
[solutn] in gdm-3
= [moldm-3] x molar mass

49. 
MOLE AND CONCENTRATION

50.  Conc in mol𝒅𝒎−𝟑
= MOLAR CONCENTRATION @
MOLARITY (M)
 MOLARITY & CONCENTRATION
MOLARITY (mol𝒅𝒎−𝟑)
= Conc (g 𝐝𝐦−𝟑)
molar mass of solute (g𝒎𝒐𝒍−𝟏
)
 No. of mole, n = MV/1000

51. What is the conc. of a solution containing 2.46 g of
sodium chloride in 250 𝒄𝒎𝟑
of water?
[NaCl] in g 𝒅𝒎−𝟑 = 2.46 x 1000/250
= 9.84 g𝒅𝒎−𝟑
[NaCl] in mol𝒅𝒎−𝟑 = 9.84/58.5
= 0.168 mol𝒅𝒎−𝟑
EXAMPLE 1

52. EXAMPLE 2
A sample of 0.40 g of anhydrous NaOH
was dissolved in water & made up to 100
mL of solutn. 25 mL of the solutn was
then neutralised ĉ dilute H2SO4
a) Find the molarity of NaOH solution
b) How many moles of H2SO4 were
required for the neutralisation

53. Solutn :
a) n NaOH = 0.01 mole ???
M = n x 1000 = 0.01 x 1000
V 100
= 0.1 moldm-3

54. b) 2NaOH + H2SO4  Na2SO4 + H2O
MaVa = a
MbVb b
n NaOH = 0.1 x 25 = 2.5 x 10-3 mole
1000
2 moles NaOH ≡ 1 mole H2SO4
:. 2.5 x 10-3 NaOH = 2.5 x 10-3 x 1
2
= 1.25 x 10-3 mole H2SO4#

55. A solutn containing an unknown [Sn2+]
was titrated ĉ a solutn containing Ce4+
ions, which oxidise Sn2+ to Sn4+ ions. In
one titration, 1.00 L of the unknown
solutn required 46.45 mL of a 0.105 M
solutn to reach equivalent Point. Calculate
[Sn2+] in the unknown solutn which
reduce Ce4+ ions to Ce3+ ions.
EXAMPLE 3

56. Solutn :
S1 – balance eq≡n :
2Ce4+(aq) + Sn2+(aq)  2Ce3+(aq) +
Sn4+(aq)
S2 – n Ce4+ ions = 0.105 x 46.45
1000
= 4.877 x 10-3 mole

57. S3 – Stoichiometry :
2 moles Ce4+ ≡ 1 mole Sn2+
:. 4.877 x 10-3 Ce4+ ≡ 4.877 x 10-3 x 1mole
2
= 2.439 x 10-3 mole Sn2+
S4 – Answer qstn
[Sn2+] = 2.439 x 10-3
1.00
= 2.439 x 10-3 moldm-3 @ molar#

58. EXAMPLE 2
Acidified KMnO4 oxidises Fe2+ ions to
Fe3+ ions as represented by the
equation :
MnO4(aq)+5Fe2+(aq)+8H+(aq)
Mn2+(aq)+4H2O(l)+5Fe3+(aq)
What volume of 0.02 M KMnO4 solution
will oxidise 25 cm3 of 0.1 M FeSO4?

59.  Solution :
S1 – n…
S2 – stoichio…
S3 – answer qstn; V KMnO4 solution
@ alnernative method
=> MMnO4V = 1
MFe2+V 5
ANS: 25 cm3

60. MOLE AND GASES
STP; T = 273K, P = 101 kPa (1 atm)
RtP; T = 298K, P = 101 kPa
1 mole of any gas = 22.4 L or 22.4 dm3
aka molar volume at STP
1 mole of any gas = 24 L or 24 dm3
aka molar volume at room condition
* Equal no. of moles o any gas, under the
same conditions, would occupy the same
volume. It does not depends on the nature
o the gas.

61. A sample of carbon dioxide contains 0.011 g of the gas.
Calculate the volume occupied by the sample at s.t.p.
n, CO2 = 0.011/44
= 2.50 x10−4
mol
V of CO2 at s.t.p = 2.50 x10−4 x 22.4
= 0.0056 dm3
= 5.60 cm3
EXAMPLE 1

62. Summary Of Relationship Of Mole With
Mass, Volume & NA
mole X 24L
X 22.4L
No of
Particles
Volume of
gas at STP
Volume of gas
room
temperature
Mass
X NA
X Ar or Mr

63. LIMITING
REACTANT
 When 2 or more reactants are combined in non-
stoichiometric ratios, the amount of product
produced is limited by the reactant that is not in
excess.
 This reactant is referred to as limiting reactant.
 When doing stoichiometric problems of this
type, the limiting reactant must be determined
first before proceeding with the calculations.

64. LIMITING
REACTANT
 When solving limiting reactant problems,
assume each reactant is limiting reactant, and
calculate the desired quantity based on that
assumption.
A + B  C
A is LR
B is LR
Calculate
amount of C
Calculate
amount of C
 Compare your answers for each assumption;
the lower value is the correct assumption.
OR

65. A fuel mixture used in the early days of rocketry
was a mixture of N2H4 and N2O4, as shown below.
How many grams of N2 gas is produced when 100.0
g of N2H4 and 200.0 g of N2O4 are mixed?
Example 1:
2N2H4 (l) + N2O4 (l)  3N2 (g) + 4H2O (g)
Limiting
reactant

66. Solution:
S1 :100.0 g N2H4 ≡ 3.125
mole
S1 : calc no of mole reactants given
S2 : stoichiometry of required amount
S3 : identify LR
200.0 g N2O4 = 2.174 mole
S2 : 2 mole N2H4 ≡ 1 mole N2O4
3.125 mole N2H4 ≡ 1.563 mole N2O4 REQUIRED
S3 : compare n given & required; n given > n required
:. N2O4 is in EXCESS so LR is N2H4
2N2H4 (l) + N2O4 (l)  3N2 (g) + 4H2O (g)

67. SOLUTION:
2 N2H4 (l) + 1 N2O4 (l)  3 N2 (g) + 4 H2O (g)
S1 :100 g N2H4 ≡ 3.125 mole 200 g N2O4 = 2.174 mole
S2 : 1 mole N2O4 ≡ 2 mole N2H4
2.174 mole N2O4 ≡ 4.348 mole N2H4 REQUIRED
S3 : compare n given & required; n given < n required
:. LR is N2H4 since NOT ENOUGH TO REACT WITH
GIVEN N2O4

68. Solution :
2N2H4 (l) + N2O4 (l)  3N2 (g) + 4H2O (g)
n product follows n LR
2 mol N2H4 ≡ 3 mol N2
3.125 mol N2H4 = 4.688 mol N2
:. Mass N2 = 131.3 g #
Calculate mass of N2

69. How many grams of AgBr can be produced when
50.0 g of MgBr2 is mixed with 100.0 g of AgNO3, as
shown below:
Example 2:
MgBr2 + 2AgNO3  2AgBr + Mg(NO3)2
Limiting
Reactant

70. Solution 2:
MgBr2 is LR
MgBr2 + 2AgNO3  2AgBr + Mg(NO3)2
102 g AgBr

71. PERCENT YIELD
 The amount of product calculated through
stoichiometric ratios are the maximum amount
product that can be produced during the
reaction, and is thus called theoretical yield.
 The actual yield of a product in a chemical
reaction is the actual amount obtained from the
reaction.

72. PERCENT YIELD
 The percent yield of a reaction is obtained as
follows:
Actual yield
x100 = Percent yield
Theoretical yield

73. In an experiment forming ethanol, the theoretical
yield is 50.0 g and the actual yield is 46.8 g. What is
the percent yield for this reaction?
Example 1:
Actual yield
% yield = x100
Theoretical yield
46.8 g
= x100 =
50.0 g
= 92.7 %

74. Silicon carbide can be formed from the reaction of
sand (SiO2) with carbon as shown below:
Example 2:
SiO2 (s) + 3C (s)  SiC (s) + 2CO (g)
When 100 g of sand are processed, 51.4g of SiC is
produced. What is the percent yield of SiC in this
reaction?
Actual
yield

75. Solution 2:
100 g SiO2
66.7 g SiC
Calculate theoretical yield
SiO2 (s) + 3C (s)  SiC (s) + 2CO (g)

76. Solution 2:
Actual yield
% yield = x100
Theoretical yield
= 77.1 %
Calculate percent yield

77.  Chlorine gas exists as diatomic molecule Cl2. In
an experiment, 326.6 g of Cl2 was used to react
with excess iron to produce iron (III) chloride,
FeCl3.
 Write a balanced equation for this reaction
 Calculate the theoretical yield of FeCl3
 If the percentage of the actual yield of FeCl3
obtained in the laboratory is 82.0%, calculate
the mass of FeCl3 produced
 Explain why there is a difference between the
calculated theoretical yield of FeCl3, with that
of the actual yield obtained in the laboratory
EXERCISES

78. 2Fe(s) + 3Cl2(g) → 2FeCl3(s)
 (3x71) g Cl2 will produce (2x162.3) g of FeCl3
326.6 g of Cl2 would produce
326.6/(3x71) x (2x162.3) g of FeCl3
=497.72
 Actual mass of FeCl3 produced = 497.72 x 0.82
= 408.0 g
 Some of the clorine gas escapes from the reacting
vessel. Formation of some iron (II) chloride, FeCl2.
some of FeCl3 formed decomposed to FeCl2.
ANSWER