1.0 MATTER

Chapter 1.0

1.1 Atoms and Molecules

COURSE LEARNING OUTCOMES

CLO OUTCOMES

1 Explain basic concepts and principles of physical chemistry in novel
and real life situations.
C2, PLO1, MQF LOD 1
Teaching & Learning Strategy : LECTURE

2 Demonstrate the correct techniques in handling laboratory apparatus Jens Martensson
and chemicals when carrying out experiments.
P4, PLO2, MQF LOD 2
Teaching & Learning Strategy : PRACTICAL, DEMONSTRATION

3 Solve chemistry related problems by applying basic concepts and 2
principle in physical chemistry.
C4, PLO4, CTPS3, MQF LOD 6
Teaching & Learning Strategy : TUTORIAL

CHEMISTRY
SK015

MODE : LECTURE (CLO 1) Jens Martensson

Student Learning Time (SLT) for lecture :

Non Face-to-face Face-to-face
(Preparation) (During Class)

1 hour 1 hour

3

Assessment of learning outcomes :

CLO T&L Strategy Assessment

1 Lecture Final Examination Jens Martensson

2 Practical Practical Test

Demonstration

3 Tutorial Assignment

4

1.1 Atoms and Molecules Jens Martensson

a) Write isotopic notation
b) Interpret mass spectrum

5

Isotopic Notation

Nucleon number A X atomic symbol
Proton number Z
Jens Martensson
EXAMPLE:

35 Cl 52 Cr Number of proton = Z = 24
17 24 Number of neutron = A - Z = 52 - 24 = 28
Number of elctrons = 24

6

Mass Spectrometer Jens Martensson

An analytical instrument
used to measure atomic
and molecular masses
directly

Mass Spectrum Jens Martensson

Mass spectrum of naturally occurring chlorine

Atomic mass

Isotopic Atomic Mass Average Atomic Mass Jens Martensson

The atomic mass of an isotope The average of mass of its
naturally occurring isotopes
Unit: a.m.u. weighted according to their
abundances
Atomic mass unit (amu)
One twelfth (1/12) the mass of a Unit: a.m.u.
carbon-12
atom .

Atomic mass

Relative Atomic Mass (Ar) Relative Molecular Mass (Mr) Jens Martensson

The mass of one atom of an The mass of one molecule of a
element compared to one compound compared to one
twelfth the mass of one atom of twelfth the mass of one atom of
carbon-12. carbon-12.

Unit: no unit Unit: no unit

Atomic mass Jens Martensson

Average Atomic Mass

( )


Relative Atomic Mass (Ar)

NO
unit!

Average atomic mass (6.94) Jens Martensson

12

Jens Martensson

13

Question 1 Jens Martensson

Naturally occurring chlorine is a mixture of two isotopes. In every
sample of this element, 75.77 % of the atoms are 35Cl and 24.23 %
are atoms of 37Cl. The accurately measured atomic mass of 35Cl is
34.9689 amu and that of 37Cl is 36.9659 amu.

From these data, calculate
i) Average atomic mass of chlorine.
ii) Relative atomic mass of chlorine.

Answer Q1

i) Average atomic mass of chlorine = % abund. of 35Cl x atomic mass of 35Cl +
% abund. of 37Cl x atomic mass of 37Cl

= 75.77 x 34.9689 amu + 24.23 x 36.9659 amu Jens Martensson
100 100

= 35.45 amu

ii) Relative atomic mass of chlorine = mass of one atom of chlorine (amu)
1x mass of one 12C atom (amu)
12

= 35.45 = 35.45

1x 12

12

Question 2 Jens Martensson

Nitrogen (N, Z = 7) has two naturally occurring isotopes. Calculate the
percentage abundances for 14N and 15N from the following:

Average atomic mass of N = 14.0067 amu;
Isotopic mass of 14N = 14.0031 amu;
Isotopic mass of 15N = 15.0001 amu.

Answer Q2

Let x = % of 14N and y = % of 15N ①
x + y = 100

Average atomic mass of nitrogen Jens Martensson
= ( % of 14N x isotopic mass of 14N ) + (% of 15N x isotopic mass of 15N )
=

= 14.0067 amu ②
0.140031x + 0.150001 y = 14.0067

x = 99.64 y = 0.36
% of 14N = 99.64 % % of 15N = 0.36 %

Chapter 1.0

1.2 Mole Concepts

1.2 Mole Concepts Jens Martensson

a) Define the terms empirical and molecular formulae
b) Define each of the following concentration measurements

i. Molarity (M)
ii. Molality (molal)
iii. Mole fraction (X)
iv. Percentage by mass (%w/w)
v. Percentage by volume (%v/v)

19

Chemical Formula

Emphirical Formula Molecular Formula Jens Martensson

Shows the relative number of Shows the actual number of
atoms of each element in the atoms of each element of a
compound molecule in the compound

The simplest whole number ratio Eg.: Hydrogen peroxide
of all element in a molecule. Emphirical formula  H2O2

Eg.: Hydrogen peroxide
Emphirical formula  HO

Solution

Solvent Solution Jens Martensson

Substance present Homogenous mixture of
In the larger amount. two or more substances
To dissolve solute

Solute

Substance present in the smaller amount.
Dissolve in solvent.

Concentration Jens Martensson

Amount of solute present in a given quantity of solution or solvent

Molarity (M)
Molality (molal)
Mole fraction (X)
Percentage by mass ( % w/w)
Percentage by volume ( % v/v)

Concentration

Molarity (M) Molality (molal) Mole Fraction (X)

Number of moles of solute The number of moles of The ratio of the number of Jens Martensson
solute dissolved in 1 kg of moles of one component to
in 1 L of solution the total number of moles
solvent present in the solution

M m No unit
mol/ L
mol/ dm3 molal
mol kg-1

Concentration

Percentage by mass Percentage by volume
(% w/w) (% v/v)

The ratio of the mass of a solute to the The ratio of the volume of a solute to Jens Martensson
mass of the solution, multiplied by 100 the volume of the solution, multiplied
% by 100 %

For questions that give concentration and density,
assumptions can be made based on concentration.

Example: Unit Formula Assumption
Concentration ( Suggestion )
Mole solute Volume solution = 1 L Jens Martensson
Molarity V solution (L)
Solvent = 1 Kg = 1000 g
Molality Mole solute
Kg solvent Mass solution = 100 g
% w/w
Mass solute x 100% Volume solution = 100 ml
% v/v Mass solution

Volume solute x 100%
Volume solution

25

Dilution Jens Martensson

Procedure for preparing less concentrated solution from a more
concentrated solution

The number of moles of solute does not change when more solvent is
added to the solution.
Concentration, however, does change with the added amount of solvent

YTohuank Prepared by Miss Eva
Kolej Matrikulasi Johor

Question 1 Jens Martensson

A sample of commercial concentrated hydrochloric acid is 11.8 M HCl
and has a density of 1.190 g/mL.

Calculate the % by mass, molality, and mole fraction of HCl.

Jens Martensson

Answer Q1

Jens Martensson

Answer Q1

Jens Martensson

Answer Q1

Chapter 1.0

1.2 Stoichiometry

1.3 Stoichiometry Jens Martensson

a) Write and balance:

i. Chemical equation by inspection method
ii. Redox equation by ion-electron method

b) Define limiting reactant and percentage yield

33

Stoichiometry Jens Martensson

The numerical relationship between chemical quantities in a balanced
chemical equation.

Allows us to predict
• the amounts of product form in a chemical reaction
• the amount of reactant necessary to form a given product

Chemical Equation

A statement that uses chemical formulas to express the identities and
quantities of the substances involved in a chemical or physical change

There are TWO methods to balance a chemical equation
• Inspection method
• Algebraic method

i) Inspection method Jens Martensson

Write the unbalanced equation using correct
chemical formulas for all substances

Find coefficients to balance the equation.
Start with the most complex substance and deal
with one element at a time.

Adjust the coefficients to obtain whole numbers :

Specify the states of matter : (s), (l), (g) or (aq)

i) Inspection method

C8H18 + O2 CO2 + H2O unbalanced

1 C8H18 + O2 8 CO2 + H2O Balance for
8CO2 + 9H2O C
1 C8H18 + O2 8CO2 + 9H2O Jens Martensson
Balance for
1 C8H18 + 25 O2 H
2
Balance for
O

1 C8H18 + 25 O2  8CO2 + 9 H2O x2
2

2C8H18(l) + 25O2(g)  16CO2(g) + 18H2O(g)

ii) Redox Reaction Jens Martensson

Oxidation–reduction reaction
Net movement of electrons from one reactant to another

E.g.: Fe2+ + Cr2O72–  Fe3+ + Cr3+

What is the oxidation state of the underlined element?

Cr2O72–

2(Cr) + 7(-2) = -2 Jens Martensson
2(Cr) = -2+14 = +12
Cr = +6

MnO4-

Mn + 4(-2) = -1

38

Step 1 ii) Redox Reaction Jens Martensson
Step 2
Step 3 Write the unbalanced equation (ionic form)
Step 4 Separate into two half-reactions.
Balance the elements other than O and H in each half- reaction.
Step 5 Add H2O to balance O atoms
Step 6 Add H+ to balance H atoms.
Step 7 Add e- to balance the charges
Step 8 Balance the e- in both half- reaction.
Add the two half-reactions
CHECK !!

STEPS: Jens Martensson
1. Write the unbalanced equation for the reaction in ionic form.

Fe2+ + Cr2O72–  Fe3+ + Cr3+

2. Separate the equation into two half-reactions.
Oxidation : Fe2+  Fe3+
Reduction : Cr2O72–  Cr3+

3. Balance the atoms other than O and H in each half- Jens Martensson
reaction.
Fe2+  Fe3+ (no change)
Cr2O72–  2Cr3+

4. Add H2O to balance O atoms and H+ to balance H atoms.

Cr2O72–  2Cr3+ + 7H2O
14H+ + Cr2O72–  2Cr3+ + 7H2O

5. Add e- to one side of each half-reaction to balance the
charges on the half-reaction.

+2 +3 –1 Jens Martensson

Fe2+  Fe3+ + 1e-

6e- + 14H+ + Cr2O72–  2Cr3+ + 7H2O

–6 + 14 –2 +6

6. Balance the e- in the two half-reactions by multiplying the
half-reactions by appropriate coefficients.

Fe2+  Fe3+ + 1e- X 6 Jens Martensson

6e- + 14H+ + 6Fe2+  6Fe3+ + 6e-
Cr2O72–  2Cr3+ + 7H2O

7. Add the two half-reactions together and balance the final
equation by inspection. The number of e- on both sides
must be equal.

Oxidation: 6Fe2+  6Fe3+ + 6e-
2Cr3+ + 7H2O
Reduction: 6e- + 14H+ + Cr2O72–  Jens Martensson

14H+ + Cr2O72– + 6Fe2+  6Fe3+ + 2Cr3+ + 7H2O

8. Verify that the number of atoms and the charges are
balanced.

(14 x 1) + ( – 2) + (6 x 2) = 24 = (6 x 3) + (2 x 3)

9. For reactions in basic solutions, add OH- to both sides of Jens Martensson
the equation for every H+ that appears in the final
equation.

14H+ + Cr2O72- + 6Fe2+  6Fe3+ + 2Cr3+ + 7H2O

14OH– + 14H+ + Cr2O72- + 6Fe2+ 6Fe3+ + 2Cr3+ + 7H2O + 14OH–

14H2O + Cr2O72- + 6Fe2+  6Fe3+ + 2Cr3+ + 7H2O + 14OH–

7H2O + Cr2O72– + 6Fe2+  6Fe3+ + 2Cr3+ + 14OH–

Limiting Reactant Jens Martensson

Reactant that is completely consumed in a reaction and limit the amount
of products formed.

• will be used up first in a reaction
• to determine the amount of products formed

1 Car = 4 Wheels + 1 Engine + 1 Body + 2 Bumpers Jens Martensson

How many cars can be produced if given :

20 wheels The body limits number of
2 bodies car that can be produced
4 engines
Body : Limiting reactant

Wheels and engines : Excess reactant

Answer : 2 cars

Question 1 Jens Martensson

N2(g) + 3H2(g)  2NH3(g)

If 10.0 g of nitrogen gas and 30.0 L of hydrogen gas is used for the
reaction at STP, calculate the volume of the ammonia gas produced at
STP?( C3 )

Answer Q1

Given,

Mol of N2 = 10.0 = 0.3571 mol Mol of H2 = 30.0 = 1.3393 mol
28.0 22.4

N2(g) + 3H2(g)  2NH3(g) Jens Martensson

From the equation, 1 mol of N2 needs 3 mol of H2

0.3571 mol of N2 needs (3 x 0.3571) mol of H2
= 1.0713 mol of H2
(needed) < 1.3393 mol (provided)

Therefore, H2 is in excess (excess reactant)
Limiting reactant = N2

Answer Q1 Jens Martensson

Based on the limiting reactant, 1 mol of N2 produces 2 mol of NH3
0.3571 mol of N2 produces (2 x 0.3571) mol of NH3
= 0.7142 mol NH3

Volume of NH3 at STP = 0.7142 mol x 22.4 dm3 mol-1
= 16 dm3