LECTURE 2

Contents:
• Empirical & Molecular formula
• Oxidation Number
• Chemical Reaction
• Stoichiometry calculation
• Limiting Reagent

• Formula kimia bagi sebatian ditulis menggunakan simbol
bagi menunjukkan semua unsur dalam sebatian
mengikut nisbah tertentu.
• Formula empirikFormula empirik merupakan formula paling ringkas yang
menunjukkan nisbah terkecil bilangan atom bagi semua
unsur yang wujud dalam satu sebatian.
• Contoh: CH2O menunjukkan nisbah unsur C, H dan O
dalam sebatian ialah 1:2:1.
• Formula empirik tidak menunjukkan bilangan sebenartidak menunjukkan bilangan sebenar
atom-atom yang bergabung dalam suatu molekul.

• Formula molekul ialah formula yang menunjukkan
bilangan sebenar setiap atom yang wujud dalam satu
molekul.
• Terdapat yang mempunyai formula empirik yang sama
tetapi mempunyai formula molekul yang berbeza.
• Perkaitan antara formula empirik dan formula molekul
diringkaskan seperti berikut:
yang mana n ialah integer

Apabila dianalisis, suatu sebatian didapati mengandungi
25.5% Mg dan 74.5% Cl mengikut jisim. Apakah formula
empirik sebatian itu?
Andaikan jisim sampel sebatian = 100 g
Jisim Mg = 25.5 g dan Cl = 74.5 g
Unsur Mg Cl
Jisim (g) 25.5 74.5
Bil mol atom 25.5/24
= 1.063
74.5/35.5
= 2.098
Nisbah
terkecil
1.063/1.063
= 1
2.098/1.063
= 2
Formula empirik sebatian = MgClMgCl22

Pembakaran 0.202 g suatu sampel bahan organik yang
mengandungi unsur-unsur C, H dan O menghasilkan 0.361 g
karbon dioksida dan 0.147 g air. Jika jisim molekul relatif
sebatian ialah 148, tentukan formula molekulnya.
% O dalam sampel = 100 – (48.51 + 8.07)
= 43.42%= 43.42%
Jisim C dalam CO2 = 12
44
X 0.361 = 0.098 g
% C dalam sampel = 0.098
0.202
X 100 = 48.51%
Jisim H dalam H2O = 2
18
X 0.147 = 0.0163 g
% H dalam sampel = 0.0163
0.202
X 100 = 8.07%

Pembakaran 0.202 g suatu sampel bahan organik yang
mengandungi unsur-unsur C, H dan O menghasilkan 0.361 g
karbon dioksida dan 0.147 g air. Jika jisim molekul relatif
sebatian ialah 148, tentukan formula molekulnya.
Jisim formula molekul = n x jisim formula empirik
148 = n x [ (3x12) + (6x1) + (2x16) ]
n = 2
Formula molekul sebatianFormula molekul sebatian = 2 (C3H6O2)
= CC66HH1212OO44
Unsur C H O
Jisim (g) 48.51 8.07 43.42
Bil mol atom 48.51/12
= 4.04
8.07/1
= 8.07
43.42/16
= 2.71
Nisbah
terkecil
4.04/2.71
= 1.5 @ 3
8.07/2.71
= 3 @ 6
2.71/2.71
= 1 @ 2
Formula empirik sebatian = CC33HH66OO22

• Amaun bahan yang wujud dalam bentuk gas biasanya
diukur dalam unit isipadu.
• 1 mol gas pada suhu dan tekanan piawai, s.t.p.
(T=273.15 K, P = 1 atm) menempati isipadu 22.4 dm3
dan dinamakan isipadu molarisipadu molar.

• Larutan adalah campuran bahan larut dalam sejumlah pelarut.
Kepekatan larutan boleh dinyatakan dalam sebutan:
a.a. KemolaranKemolaran – bilangan mol bahan larut yang terlarut dalam 1
dm3
(1000 cm3
) larutan (unit = mol dm-3
).
b.b. KemolalanKemolalan – bilangan mol bahan larut yang terlarut dalam 1
kg pelarut (unit = mol kg-1
).

Hitungkan kemolaran larutan yang disediakan dengan
melarutkan 3.42 g sukrosa (C12H22O11) dalam 500 cm3
air.
Bilangan mol sukrosa = 3.42 = 0.01 mol
342
Mr C12H22O11
= (12 x 12) + (1 x 22) + (16 x 11)
= 342
Kemolaran = 0.01
500/1000
= 0.02 mol/dm3
Hitungkan kemolalan larutan yang mengandungi 6.50 g etilena
glikol (C2H6O2) dalam 200 g air.
Kemolalan C2H6O2 = 0.105
200/1000
= 0.524 mol/kg
Bilangan mol C2H6O2 = 6.5
2 (12) + 1 (6) + 2 (16)
= 0.105 mol

How to prepare 1 dm3
of 0.1 M CuSO4 starting with 2 M CuSO4 ?
50 g of benzene (C6H6) is dissolved in 80 g of toluene (C6H5CH3).
Calculate the molality of the solution.
n C6H6 = 50
12 (6) + 1 (6)
= 0.640 mol
Using M1V1 = M2V2
= 0.1 M x 1 dm3
2 M
= 0.05 dm3
= 50 cm3V1 = M2V2
M1
Molality C6H5CH3 = =
n C6H6
80/1000 kg
= 8.01 mol/kg0.640 mol
80/1000

c.c. Pecahan mol (X)Pecahan mol (X)
 Bilangan mol bahan larut dibahagikan dengan jumlah
bilangan mol semua komponen yang terdapat dalam
larutan.
 Contoh: Larutan yang mengandungi sebatian A, B dan C.
Pecahan mol bagi sebatian A dalam larutan ialah:
nA, nB dan nC adalah bilangan mol A, B dan C masing-masing
 Pecahan mol tidak mempunyai unit dan jumlah pecahan
mol semua komponen bernilai 1 iaitu:

d. Peratusan mengikut jisim atau isipadud. Peratusan mengikut jisim atau isipadu
 Kepekatan larutan dalam sebutan peratusan boleh
dinyatakan sebagai jisim bahan larut/jisim larutan (w/w)%
atau isipadu bahan larut/isipadu larutan (V/V)%
 Contohnya asid sulfurik, H2SO4 berkepekatan 36%
mengikut jisim mengandungi 36 g H2SO4 dalam 64 g air.
Bagi larutan 5% NaCl mengikut isipadu, terdapat 5 g NaCl
dalam 100 cm3
larutan.

• In solution prepared by dissolving 24 g of NaCl in 152 g of water,
• 20% w/w HCl → 20 g of HCl dissolved in 80 g of water.
Calculate the percentage by mass of NaBr in an aqueous soluting
containing 8 g of NaBr in 80 g of water.
% w/w = 8
(8 + 80)
x 100%
% w/w NaCl = 24 g x 100%
24 g + 152 g
= 14%
= 9.09%

e. ppme. ppm
 Kepekatan larutan yang sangat cair dinyatakan
dalam bahagian persejuta (ppm).
Calculate the volume of K3Fe(CN)6 used to prepare a 125 ml 85%
v/v solution.
% v/v = Volume of K3(CN)6
(125)
x 100%
= 106.25 mlVolume of K3(CN)6 = 85 x 125
100

The charge the atom would have in a molecule (or an
ionic compound) if electrons were completely transferred.
The charge the atom would have in a molecule (or an
ionic compound) if electrons were completely transferred.
1.1. Free elementsFree elements (uncombined state) have an oxidation
number of zero.
Na, Be, K, Pb, H2, O2, P4 = 0
2. In monatomic ionsmonatomic ions, the oxidation number is equal to the
charge on the ion.
Li+
, Li = +1; Fe3+
, Fe = +3; O2-
, O = -2
3. The oxidation number of oxygen is usually –2. In H2O2 and
O2
2-
it is –1.

4. The oxidation number of hydrogen is +1 except when it is
bonded to metals in binary compounds. In these cases, its
oxidation number is –1 (hydride eg: H2S).
6. The sum of the oxidation numbers of all the atoms in a
molecule or ion is equal to the charge on the molecule or
ion.
5. Group IA metals are +1, IIA metals are +2 and fluorine is
always –1.
HCO3
-
O = -2 H = +1
3x(-2) + 1 + ? = -1
C = +4
Oxidation numbers of C in
HCO3
-
?

Determine the oxidation number of the following underlined
elements:
a. Cu2O b. SO4
2-
c. ICl3 d. OF2
a. Oxidation no of O = - 2
Let oxidation no of Cu be x
2x - 2 = 0
x = +1
b. Oxidation no of O = -2
Let oxidation no of S be y
y - 8 = - 2
y = +6
c. Cl more electronegative than I
Oxidation no of Cl = - 1
Let oxidation no of I be q
q - 3 = 0
q = +3
d. F more electronegative than O
Oxidation no of F = - 1
Let oxidation no of O be r
r - 2 = 0
r = +2

• Persamaan kimia menggunakan simbol dan formula bagi
menggambarkan secara ringkas perubahan yang berlaku
dalam suatu tindak balas kimia.
• Tindak balas am:
 A dan B adalah reaktan manakala C dan D adalah hasil
tindak balas.
 Tanda ‘+’ = memisahkan setiap reaktan dan hasil tindak
balas.
 Anak panah = bermakna ‘untuk menghasilkan’

Example: 3 ways of representing the reaction of H2 with O2 to form
H2O

2 atoms Mg + 1 molecule O2 makes 2 formula units MgO
2 moles Mg + 1 mole O2 makes 2 moles MgO
48.6 grams Mg + 32.0 grams O2 makes 80.6 g MgO
IS NOT
2 grams Mg + 1 gram O2 makes 2 g MgO

1. Write the correct formula(s) for the reactants on the left
side and the correct formula(s) for the product(s) on the
right side of the equation.
Ethane reacts with oxygen to form carbon dioxide and water
C2H6 + O2 CO2 + H2O
2. Change the numbers in front of the formulas (coefficients)
to make the number of atoms of each element the same
on both sides of the equation. Do not change the
subscripts.
2C2H6
NOT C4H12

C2H6 + O2 CO2 + H2O start with C or H but not O
2 carbon
on left
1 carbon
on right
multiply CO2 by 2
C2H6 + O2 2CO2 + H2O
6 hydrogen
on left
2 hydrogen
on right
multiply H2O by 3
3. Start by balancing those elements that appear in only one
reactant and one product.

4. Balance those elements that appear in two or more
reactants or products.
multiply O2 by
7
2
2 oxygen
on left
4 oxygen
(2x2)
C2H6 + O2 2CO2 + 3H2O
+ 3 oxygen
(3x1)
= 7 oxygen
on right
C2H6 + O2 2CO2 + 3H2O
7
2
remove fraction
multiply both sides by 2
2C2H6 + 7O2 4CO2 + 6H2O2C2H6 + 7O2 4CO2 + 6H2O

5. Check to make sure that you have the same number of
each type of atom on both sides of the equation.
2C2H6 + 7O2 4CO2 + 6H2O
Reactants Products
4 C
12 H
14 O
4 C
12 H
14 O
4 C (2 x 2) 4 C12 H (2 x 6) 12 H (6 x 2)14 O (7 x 2) 14 O (4 x 2 + 6)

Ion-electron methodIon-electron method
• The ion-electron method is mainly used for
balancing the redox (reduction-oxidation)
equations.
• Reduction is a process whereby the oxidation
number of an atom decreases.
• Conversely, oxidation involves an increase in the
oxidation number.

2Mg 2Mg2+
+ 4e-
O2 + 4e-
2O2-
2Mg (s) + O2 (g) 2MgO (s)
0 0 2+ 2-
Oxidation half-reaction
- Lose e-
.
- Also a reducing agent by giving up its
electrons causing the other substance to
be reduced.
Reduction half-reaction
- Gain e-
.
- Also a oxidizing agent removes
electrons from another substance by
acquiring them itself; thus the agent is
itself reduced.

Example: Balance the equation showing oxidation of Fe2+
to Fe3+
by
Cr2O7
2-
in acid solution?
Using ion-electron method (reaction divided into two half reactions)
1. Write the unbalanced equation for the reaction ion ionic
form. Fe2+
+ Cr2O7
2-
Fe3+
+ Cr3+
2. Separate the equation into two half-reactions.
(reducing agent) Oxidation:
Cr2O7
2-
Cr3+
+6 +3
(oxidizing agent) Reduction:
Fe2+
Fe3+
+2 +3
3. Balance the atoms other than O and H in each half-reaction.
Cr2O7
2-
2Cr3+

4. For reactions in acid, add H2O to balance O atoms and H+
to
balance H atoms.
Cr2O7
2-
2Cr3+
+ 7H2O
14H+
+ Cr2O7
2-
2Cr3+
+ 7H2O
6e-
+ 14H+
+ Cr2O7
2-
2Cr3+
+ 7H2O
6. If necessary, equalize the number of electrons in the two half-
reactions by multiplying the half-reactions by appropriate
coefficients.
6Fe2+
6Fe3+
+ 6e-
5. Add electrons to one side of each half-reaction to balance the
charges on the half-reaction.
6e-
+ 14H+
+ Cr2O7
2-
2Cr3+
+ 7H2O
Fe2+
Fe3+
+ 1e-

7. Add the two half-reactions together and balance the final
equation by inspection. The number of electrons on both sidesThe number of electrons on both sides
must cancel.must cancel.
6e-
+ 14H+
+ Cr2O7
2-
2Cr3+
+ 7H2O
6Fe2+
6Fe3+
+ 6e-
Oxidation:
Reduction:
14H+
+ Cr2O7
2-
+ 6Fe2+
6Fe3+
+ 2Cr3+
+ 7H2O
8. Verify that the number of atoms and the charges are balanced.
14x1 – 2 + 6x2 = 24 = 6x3 + 2x3

Stoichiometrically Equivalent Molar ratiosStoichiometrically Equivalent Molar ratios
• Stoichiometry is the quantitative study of reactants and
products in a chemical reaction.
• Use moles to calculate the amount of product formed in
a reaction – mole methodmole method.
• Therefore, the stoichiometric coefficients in a chemical
equation can be interpreted as the number of moles of
each substance.

• Example:
N2(g) + 3H2(g) 2NH3(g)
• The equation can be read as “ 1 mole of N2 gas combines
with 3 moles of H2 gas to form 2 moles of NH3 gas”.
• In stoichiometric calculationstoichiometric calculation
– 3 moles of H2 are stoichiometrically equivalent to 2 moles of NH3
– 1 mole N2 are stoichiometrically equivalent to 2 mol NH3
– 1 mole N2 are stoichiometrically equivalent to 3 mol H2
• The conversion factorsconversion factors from this equivalence:
1 molecule 3 molecule 2 molecule
3 mol H2
2 mol NH3
2 mol NH3
3 mol H2
and

The general approach for solving stoichiometry
problems:
i. Write a balanced equation for the reaction.
ii. Convert the given amount of the reactant (gram
@ other unit) to number of moles.
iii. Use the mole ratio from the balanced equation to
calculate the number of moles of product formed.
iv. Convert the moles of product to gram ( or other
units) of product.

• Example: 16.0 g of H2 react completely with N2 to form
NH3. How many grams of NH3 will be formed?
grams H2 moles of H2 moles of NH3 grams of NH3
16 g H2
1 mol H2
2.016 g H2
x 2 mol NH3
3 mol H2
x 17.03 g NH3
1 mol NH3
x
Grams of NHGrams of NH33 ==
= 90.61 g NH= 90.61 g NH33
N2 (g) + 3H2 (g) 2NH3(g)

Methanol burns in air according to the equation
2CH3OH + 3O2 2CO2 + 4H2O
If 209 g of methanol are used up in the combustion,
what mass of water is produced?
grams CH3OH moles CH3OH moles H2O grams H2O
molar mass
CH3OH
coefficients
chemical equation
molar mass
H2O
209 g CH3OH
1 mol CH3OH
32.0 g CH3OH
x
4 mol H2O
2 mol CH3OH
x
18.0 g H2O
1 mol H2O
x
= 235 g H2O

• Goal of reaction is to produce the maximum quantity
from the starting material.
• Consequently, some reactant will be left over at the end
of reaction.
• The reactant used up first in a reaction is called the
limiting reagentlimiting reagent - the maximum amount of product
formed depends on how much of this reactant was
originally present.
• Excess reagentsExcess reagents are the reactants present in quantities
greater than necessary to react with the quantity of the
limiting reagent.

6 green used up6 red left over

Sebanyak 12 g zink dan 6.5 g sulfur bertindak balas
mengikut persamaan:
a.Apakah bahan yang merupakan reaktan penghad di dalam
tindak balas?
Daripada persamaan,
Zn(p) + S(p) ZnS(p)
Bilangan mol Zn = = 0.1836 mol12
65.37
Bilangan mol S = = 0.2031 mol6.5
32
1 mol Zn bertindak balas dengan 1 mol S
Oleh itu 0.1836 mol Zn memerlukan 0.1836 mol S
# Reaktan penghad ialah Zn# Reaktan penghad ialah Zn

Sebanyak 12 g zink dan 6.5 g sulfur bertindak balas
mengikut persamaan:
b. Berapa gramkah ZnS yang terhasil?
Daripada persamaan,
Zn(p) + S(p) ZnS(p)
Bilangan mol ZnS terbentuk = 0.1836 mol
Jisim ZnS terbentuk = 0.1836 x 97.37
= 17.88 g
Oleh kerana Zn ialah reaktan penghad maka amaun hasil
ditentukan oleh amaun Zn
1 mol Zn menghasilkan 1 mol ZnS

Sebanyak 12 g zink dan 6.5 g sulfur bertindak balas
mengikut persamaan:
c. Namakan bahan yang berlebihan dan kirakan
bilangan gram bahan tersebut yang masih belum bertindak
balas.
Bahan berlebihan ialah S
Zn(p) + S(p) ZnS(p)
Bilangan mol S berlebihan = 0.2031 - 0.1836
= 0.0195 mol
Jisim Sulfur berlebihan = 0.0195 x 32
= 0.624 g

• Example: Urea is prepares by reacting ammonia with CO2
2NH3 (g) + CO2 (g) (NH2)2CO (aq) + H2O(l)
637.2 g of NH3 are treated with 1142 g CO2. Which of the two
reactant is limiting reagent?
grams NH3 moles of NH3 moles of (NH2)2CO
637.2 g NH3
1 mol NH3
17.03 g NH3
x 1 mol (NH2)2CO
2 mol NH3
x
moles of (NH2)2CO ==
= 18.71 mol (NH= 18.71 mol (NH22)) 22COCO

• Example: Urea is prepares by reacting ammonia with CO2
2NH3 (g) + CO2 (g) (NH2)2CO (aq) + H2O(l)
637.2 g of NH3 are treated with 1142 g CO2. Which of the two
reactant is limiting reagent?
grams CO2 moles of CO2 moles of (NH2)2CO
1142 g CO2
1 mol CO2
44.01 g CO2
x 1 mol (NH2)2CO
1 mol CO2
x
moles of (NH2)2CO ==
= 25.95 mol (NH= 25.95 mol (NH22)) 22COCO
Therefore, NH3 must be the limiting reagent because it produces a
smaller amount of (NH2)2CO

• Example: Urea is prepares by reacting ammonia with CO2
2NH3 (g) + CO2 (g) (NH2)2CO (aq) + H2O(l)
Calculate the mass of (NH(NH22))22COCO formed.
18.71 mol (NH2)2CO 60.06 g (NH2)2CO
1 mol (NH2)2CO
x
Mass of (NH2)2CO ==
= 1124 g (NH= 1124 g (NH22)) 22COCO
The molar mass of (NH2)2CO is 60.06 g. Use the conversionconversion
factorfactor to convert from moles of (NH2)2CO to gram of
(NH2)2CO.

Do You Understand Limiting Reagents?
In one process, 124 g of Al are reacted with 601 g of Fe2O3
2Al + Fe2O3 Al2O3 + 2Fe
Calculate the mass of Al2O3 formed.
g Al mol Al mol Fe2O3 needed g Fe2O3 needed
or
g Fe2O3 mol Fe2O3 mol Al needed g Al needed
124 g Al
1 mol Al
27.0 g Al
x
1 mol Fe2O3
2 mol Al
x
160. g Fe2O3
1 mol Fe2O3
x = 367 g Fe2O3
Start with 124 g Al need 367 g Fe2O3
Have more Fe2O3 (601 g) so Al is limiting reagent

Use limiting reagent (Al) to calculate amount of product that
can be formed.
g Al mol Al mol Al2O3 g Al2O3
124 g Al
1 mol Al
27.0 g Al
x
1 mol Al2O3
2 mol Al
x
102. g Al2O3
1 mol Al2O3
x
= 234 g Al= 234 g Al22OO33
Do You Understand Limiting Reagents?
In one process, 124 g of Al are reacted with 601 g of Fe2O3
2Al + Fe2O3 Al2O3 + 2Fe
Calculate the mass of Al2O3 formed.

• Theoretical Yield is the amount of product that would
result if all the limiting reagent reacted (from calculation
using the balanced equation).
• Actual Yield is the amount of product actually obtained
from a reaction (almost always less than the theoretical
yield).
• Percent yield - to determine reaction efficiency, which
describe the proportion of the actual yield to the
theoretical yield.
% Yield =
Actual Yield
Theoretical Yield
x 100

• Example:
TiCl4 (g) + 2Mg(l) Ti(s) + 2MgCl2 (l)
3.54 x 107
g of TiCl4 are reacted with 1.13 x 107
g of Mg.
Calculate the theoretical yield of Ti in grams.Calculate the theoretical yield of Ti in grams.
g of TiCl4 Mol of TiCl4 Mol of Ti
First: to find limiting reagent, calculate the no. of moles of Ti
that could produces if all the TiCl4 reacted
= 1.87 x 105
mol Ti
Next: calculate the no. of moles of Ti formed from 1.13 x 107
g of Mg.
g of Mg Mol of Mg Mol of Ti = 2.32 x 105
mol Ti

• Example:
TiCl4 (g) + 2Mg(l) Ti(s) + 2MgCl2 (l)
3.54 x 107
g of TiCl4 are reacted with 1.13 x 107
g of Mg.
Calculate the theoretical yield of Ti in grams.Calculate the theoretical yield of Ti in grams.
Therefore, TiCl4 is the limiting reagent because it produces a
smaller amount of Ti
= 8.95 x 106
g Ti1.87 x 105
mol Ti
47.88 g Ti
1 mol Ti
x
Mass of Ti =Mass of Ti =

• Example:
TiCl4 (g) + 2Mg(l) Ti(s) + 2MgCl2 (l)
3.54 x 107
g of TiCl4 are reacted with 1.13 x 107
g of Mg.
Calculate the percent yield if 7.91 x 10Calculate the percent yield if 7.91 x 1066
g of Ti are actuallyg of Ti are actually
obtained.obtained.
The percent yield is given by:
% Yield = Actual Yield
Theoretical Yield
x 100
% Yield = 7.91 x 106
g
8.95 x 106
g
x 100
= 88.4%