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