Ch emis try
Transcript of Ch emis try
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
34
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
?