COPPERBELT UNIVERSITY COLLEGE
CHE110: Introduction to Chemistry
MODULE 1
Copperbelt University College
Kitwe-Zambia
Science Department
Copyright
Copperbelt University College– 2009.
No part of this module may be reproduced or transmitted in any form or by any means,
electronic or mechanical, including photocopying, recording or by any information storage
and retrieval system, without permission in writing from the publisher.
P O BOX 20382
KITWE
ZAMBIA
Telephone: +260212239003
E-mail: cosetco@zamtel.zam
Acknowledgements
The Science Department in particular in wishes to thank the following people for their
positive contribution to this module:
The course material writers
Mr Mweshi E
Mr Muma E
Mukonde B
Changwe P.
Luchembe D.T.
Tumeo A.
Kayamba F
The Editors
Directorate of Distance Education DODE
Zambia College of Distance Education
The University of Zambia
CHE110: Introduction to Chemistry
Contents
The Module Overview ...................................................................................................... 1
Module outcomes .............................................................................................................. 3
Timeframe ......................................................................................................................... 3
Study skills ........................................................................................................................ 3
Need help? ........................................................................................................................ 4
Assignments ...................................................................................................................... 5
Assessments ...................................................................................................................... 5
Getting around this Module
6
Margin icons ..................................................................................................................... 6
Unit 1
7
Measurement ........................................................................................................... 7
1.0 Introduction ....................................................................................................... 7
1.1 Quantities........................................................................................................... 8
1.2 Errors in measurement................................................................................... 12
1.3 Precision and accuracy .................................................................................. 13
1.4 Significant figures .......................................................................................... 18
1.5 Units of measurements .................................................................................... 23
1.6 Conversion factors .......................................................................................... 25
Unit summary ................................................................................................................. 28
Assessment...................................................................................................................... 29
Unit 2
31
Matter .............................................................................................................................. 31
2.0 Introduction ..................................................................................................... 31
2.1 Properties of Matter ......................................................................................... 32
2.2 Types of substances. ........................................................................................ 34
2.3 Separating mixtures ......................................................................................... 36
Unit summary ................................................................................................................. 38
Assessment...................................................................................................................... 39
Unit 3
40
Atomic Theory ................................................................................................................ 40
3.0 Introduction ..................................................................................................... 40
3.1 Progression of the atomic theory ..................................................................... 41
3.1.1 Dalton’s atomic theory ................................................................................. 41
3.1.2 Avogadro’s theory ........................................................................................ 42
3.2 Composition of an atom .................................................................................. 43
3.3 Atomic number and Mass number .................................................................. 46
3.4 Isotopes ............................................................................................................ 48
3.4.1 Average atomic weight ................................................................................. 49
CHE110: Introduction to Chemistry
Unit 4
52
Laws of Chemical Combination ..................................................................................... 52
4.1 The law of conservation of matter ................................................................... 52
4. 2 Law of Constant Composition ........................................................................ 53
4.3 Law of multiple Proportions............................................................................ 54
Unit summary ................................................................................................................. 57
Assessment...................................................................................................................... 58
Unit 5
61
The Mole Concept .......................................................................................................... 61
5.0 Introduction ..................................................................................................... 61
5.1 What is a mole? ............................................................................................... 62
5.2 Calculations of moles, mass, volume and particles in a given ........................ 64
substance ............................................................................................................... 64
5.3 Chemical Formulae ......................................................................................... 69
5. 4 Calculation of Empirical and Molecular formulae ......................................... 72
5. 5 Molarity .......................................................................................................... 80
Unit summary ................................................................................................................. 86
Assessment...................................................................................................................... 87
Unit 6
88
Stoichiometry .................................................................................................................. 88
6.0 Introduction ..................................................................................................... 88
6.1Chemicl Equation ............................................................................................. 89
6. 2 Molar interpretation of a balanced chemical equation ................................... 91
6. 3 Limiting Reactants ......................................................................................... 95
6.4 Percentage yield ............................................................................................ 100
Unit summary ............................................................................................................... 103
Assessment.................................................................................................................... 104
Answers to Activities .................................................................................................... 106
Activity 1.2.1 ....................................................................................................... 106
Student A was more accurate than B because the average was close ................. 106
to the true value ................................................................................................... 106
Answers to Assessment Questions ............................................................................... 108
Readings........................................................................................................................ 110
CHE110: Introduction to Chemistry
The Module Overview
This module gives you the foundation to General Chemistry
Course. The module exposes you to measurement, the composition
of matter, and the basic atomic theory. These concepts will help
you to understand subsequent topics in chemistry. If you are a
secondary school teacher with a secondary teacher’s diploma in
science, then this is your module. It will help you to lay a very
strong foundation in chemistry as you pursue your Bachelors of
Education in Natural Sciences.
To complete this module successfully, you will need to spend three
(3) hours per week studying the module, and make sure you work
out all the activities in each unit. Don’t move to another unit before
you understand the previous unit. In case you need help contact the
course tutors.
You are expected to do all the self marked activities and one tutor
marked assignment. You are required to summit the assignment to
the nearest resource centre in your district. This module has four
units.
We strongly recommend that you read the overview carefully
before starting your study.
The module content
The module is broken down into 6 units. Each unit comprises:
An introduction to the unit content.
Unit outcomes..
Core content of the unit with a variety of learning activities.
A unit summary.
Assignments and/or assessments, as applicable.
Resources
For those interested in learning more on this subject, we provide
you with a list of additional resources at the end of this module;
these may be books, articles or web sites.
Your comments
After completing module 1 of CHE: 110 Introduction to
Chemistry we would appreciate it if you would take a few
moments to give us your feedback on any aspect of this course.
Your feedback might include comments on:
1
CHE110: Introduction to Chemistry
Course content and structure.
Course reading materials and resources.
Course assignments.
Course assessments.
Course duration.
Course support (assigned tutors, technical help, etc.)
Your constructive feedback will help us to improve and enhance
this course.
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CHE110: Introduction to Chemistry
Module outcomes
Upon completion of this module you will be able to:
Determine precision and accuracy of measurements.
Report measurements with the correct units and number of
significant figures
Outcomes
Discuss the composition of matter.
Discuss the structure of an atom
Use mole concept to perform stiochiometry calculations
Timeframe
This module is expected to be covered within a period of 100
hours. The 100 hours will include studying the actual module and
all the activities in it.
How long?
Study skills
As an adult learner, your approach to learning will be different
from that of your school days: you will choose what you want to
study, have professional and/or personal motivation for doing so
and you will most likely be fitting your study activities around
other professional or domestic responsibilities.
Essentially, you will be taking control of your learning
environment. As a consequence, you will need to consider
performance issues related to time management, goal setting, stress
management, etc. Perhaps you will also need to reacquaint yourself
in areas such as essay planning, coping with exams and use of the
web as a learning resource.
Your most significant considerations will be time and space i.e. the
time you dedicate to your learning and the environment in which
you engage in that learning.
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CHE110: Introduction to Chemistry
We recommend that you take time now, before starting self study,
to familiarize yourself with these issues. There are a number of
excellent resources on the web. A few suggested links are:
http://www.how-to-study.com/
The “How to study” web site is dedicated to study skills
resources. You will find links to study preparation (a list of nine
essentials for a good study place), taking notes, strategies for
reading text books, using reference sources, test anxiety.
http://www.ucc.vt.edu/stdysk/stdyhlp.html
This is the web site of the Virginia Tech, Division of Student
Affairs. You will find links to time scheduling (including a
“where does time go?” link), a study skill checklist, basic
concentration techniques, control of the study environment, note
taking, how to read essays for analysis, memory skills
(“remembering”).
http://www.howtostudy.org/resources.php
Another “How to study” web site with useful links to time
management, efficient reading, questioning/listening/observing
skills, getting the most out of doing (“hands-on” learning),
memory building, tips for staying motivated, developing a
learning plan.
The above links are our suggestions to start you on your way. At
the time of writing these web links were active. If you want to look
for more go to www.google.com and type “self-study basics”,
“self-study tips”, “self-study skills” or similar.
Need help?
Should you require help in the course of your studies, do not
hesitate to contact the following course tutors
Help
Mr. Mweshi E Cell: 0955-881340 / 0969-218224
E. Mail: emweshi@yahoo.com
Mr. Muma E
Cell: 0977-185244
Mr. Kayamba
Cell: 096/7 370381
E. Mail: francesbk@gmail.com
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CHE110: Introduction to Chemistry
Assignments
You will be expected to write at least two assignments in this
academic year. The first assignment will come from module one
and two..
Assignments
The assignments should be handed in to course tutors during the
residential sessions.
You will be required to submit the assignments in the order in
which they are given to you.
Assessments
Assessments
You will be expected to write two tutor- marked test which will be
written during each residential session. You are also expected to
answer the self- marked assessments in each unit of this module.
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CHE110: Introduction to Chemistry
Getting around this Module
Margin icons
While working through this module you will notice the frequent
use of margin icons. These icons serve to “signpost” a particular
piece of text, a new task or change in activity; they have been
included to help you to find your way around the module.
A complete icon set is shown below. We suggest that you
familiarize yourself with the icons and their meaning before
starting your study.
Activity
Assessment
Assignment
Summary
Outcomes
Group activity
Help
Note it!
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CHE110: Introduction to Chemistry
Unit 1
Measurement
1.0 Introduction
Welcome to unit1. In this unit we will discuss some fundamental
principles regarding measurements. Measurements are prerequisite
to practical work in chemistry. People use measurements to help to
determine relationships between quantities such as length, time,
mass, volume, and density among several others. Therefore,
knowledge of measurement will be very helpful to you in your
scientific computations. This unit will focus on the following
aspect of measurements; precision and accuracy, Systematic and
random errors and significant figures. After going through this unit,
you will realize how exciting it is to manipulate various quantities
both in your science practical activities and in real life.
During and after completion of this unit you will be able to:
Determine the precision and accuracy of a set of measurements.
Discuss the sources of systematic and random errors.
Outcomes
Express measurements to correct number of significant figures
Demonstrate understanding and use of scientific notation to
express large and very small measurements.
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CHE110: Introduction to Chemistry
1.1 Quantities
You might have had done activities involving measurements both
at school and at home. Very quickly, make a list of quantities of
measurement that you know.
………………………………………………………………………
………………………………………………………………………
………………………………………………………………………
……………………………………………………………
When you look at your list that you have made you might notice
that some measurements are exprssed using numbers while others
can be expressed using words. Mesurement is the process or the
results of determineing the magnitudeof a quantity such as lengthb
or mass, relative to a unit of measurement such as meter or a
kilogram.
Meaurable features or properties of objects are often called
physical quantities The area of a football field, the mass of a bag of
wheat and the speed of a motor car are all physical quantities.
some non phyiscal, diffficult–to-measure quantities are love, hate,
fear and hope. phyiscal quantities are expressed in terms of a
numerical valve and a units.
With the exception of a few seemily fundamental quantum
constants, units of measurement are essentially arbitrary; In other
words, people make them up and then agree to use them. Nothing
inherent in nature dictates that an inch has be a certain length, or
that a mile is a better measure of distance than a kilometer. Over
the course of history, however, first for convenience and then for
necessity, standards of measurement evolved so that communities
should have certain common benchmarks. Therefore, to be of use
in measurement, standards to define the basic units must be agreed
upon through out the world. To this effect, laws of regulating
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CHE110: Introduction to Chemistry
measurement were originally developed to prevent fraud in
commerce. Today, units of measurement are generally defined on
scientic basis.
It is now time for you to do an exercise.
Activity 1.1.1
Take a ruler and measure the length of your exercise book. Repeat
the exercise four times and record your measurement as L1, L2, L3
and L4. You should spend five minutes on this task.
L1---------------------------------------------------------------------------L2---------------------------------------------------------------------------L3---------------------------------------------------------------------------L4---------------------------------------------------------------------------Are these four measurements you have made numerically the
same? If not, can you suggest why. When you look closely at these
measurements, you will notice that the four measurements fall over
a certain range. The above observation therefore, implies that
whenever you measure a given physical quantity more than once
using the same measuring instrument, you will usually get a series
of measured values. How close these measured values are to one
another will be discussed in detail in the next section.
Now that you are acquainted with linear measurement, let’s now
look at the measurements made by a student measuring the length
of a piece of wood using rulers with two different scale divisions.
In the first measurement (measurement A) the student uses a ruler
with the scale division of 0.1cm as indicated below.
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CHE110: Introduction to Chemistry
Measurement A
Figure1.1.1
Using a ruler of this calibration, the Student is certain that the
length of the piece of wood lies between 2cm and 3cm. The
decimal place has to be estimated, which could be 0.4cm. Then the
length of a piece of wood is 2.4cm. From this measurement the first
digit is certain while the second digit is estimated (uncertain). It
makes no sense to give a second decimal place, for instance, 2.45
since the first decimal place is uncertain.
Measurement B
Figure1.1.2
In the second measurement (measurement B) the same Student
measures the same piece of wood with a ruler of scale division of
up to 0.01cm. S/he is now certain that the length of the same piece
of wood lies between 2.4 cm and 2.5 cm. The first decimal place is
now certain, but the second decimal place must be estimated, for
instance, 0.02 and the length of the piece of wood is now 2.42cm.
It is clear from the above measurements A and B that measurement
under the same conditions does not give values which are
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CHE110: Introduction to Chemistry
absolutely exact.
Instead, small differences
in successive
measurements of the same physical quantity by the same person
under the same conditions are observed. We shall return to this
point again on the next page.
Activity 1.1.2
In your view, which measurement (between A and B) is more
precise and more accurate than the other? Give reasons to your
answer. Take 15 minutes to do this activity.
………………………………………………………………………
………………………………………………………………………
………………………………………………………………………
………………………………………………………………………
………………………………………………………………………
………………………………………………………………………
………………………………………………………………………
………………………………………………………………………
………………………………………………………………………
………………………………………………………………………
………………………………………………………………………
.
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CHE110: Introduction to Chemistry
1.2 Errors in measurement
When we measure a physical quantity, there is always some degree
of error or uncertainty. These errors are caused by the limitation of
the measuring instrument, the condition under which the
measurement is made, and the different ways the operator uses the
instrument .let us now look at the errors that are encountered during
measurement.
Systematic errors are errors caused by faulty instrument, wrong
uses of instrument or wrong experimental design. These errors can
be determined and presumably be avoided; Instrumental errorbefore you carry out an experiment, always check faulty instrument
and uncalibrated glassware to avoid experiencing this type of error.
Operative error- this include personal error. This can be reduced
by experience and care of an analyst. (iii) Method error- this is the
most serious error. This includes co-precipitation of impurities,
slight solubility of a precipitate, side reaction, and impurities in
solutions. This can be reduced by running a reagent blank
Non-Systemat
ic: These are accidental or random errors
which represent experimental uncertainties that occur in any
measurement. These type of errors are indeterminate and are
unavoidable. These are reviewed by small differences in
successive measurements by the same analyst working under the
same conditions. It should also be noted that random errors
originate in the limited ability of the analyst to control
conditions.
Absolute error – this refers to the difference between the
standard (true) value and the measured value with regard to sign.
It is reported in the same unit as the measurement. For example,
if a 2.62g sample is analysed as 2.52g, the absolute error is
0.10g.
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CHE110: Introduction to Chemistry
Absolute error = measured value – True value
Mean error is when the measured value is an average of several
values,
Relative error- this is the absolute error or mean error
expressed
as
Relative error =
a
percentage
of
a
true
value.
absolute error
x100
True value
For example, if a 2.62g sample is analysed as 2.52g, the
absolute error is 0.10g, therefore its relative error will be
Relative error =
0.10
× 100 = 3.8%
2.62
1.3 Precision and accuracy
when you measure the same physical quanty more than once you
will notice that a given set of measurement of the same physical
quantity will have different values; some measured values will be
close to ane another, some set of measured values will be far
spaced. encounter an error or a degree of uncertianity in your set of
results.
Precision refers to how close the measured values of the same
quantity are to each other. The closer the experimental values are to
one another (the smaller the spread), the higher the precision of the
measured quantity. Experimental results which are precise are
usually reproducible.
1. The larger the number of repeated measurements of the
same quantity under the same conditions, the greater will be
the precision.
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CHE110: Introduction to Chemistry
2. For a single measurement, the more decimal places a
measured value (measurement) has the greater will be its
precision. In the next section we will look at another factor,
namely, significant figures, that also plays some role in the
precision of a single measurement.
While it is important to know how close measured values are to one
another, it is also important to know how close an experimental
observation is to the true value. The degree of agreement of the
measured values and the true value is known as accuracy.
Generally a more precise measurement will also be a more accurate
measurement. The value 2.45cm has a greater precision than 2.4cm
and probably also lies closer to the true length. There may be
instances when a measurement may be precise, but not particularly
accurate.
To appreciate the concepts of precision and accuracy more fully, let
us do the following task.
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CHE110: Introduction to Chemistry
Activity 1.3.1
Two students performed a titration; each of them performed the
experiment three times. Their results were as follows:
Student A
Student B
39.40 ml
39.40ml
39.57ml
39.41ml
39.51ml
39.38ml
The results of the same experiment carried by an experienced
lab Technician indicated that the correct value (volume) must
have been 39.50ml. Which student performed the experiment
with higher precision, and which student performed the
experiment with higher accuracy?
Let us work through this task together!
First, we have to find the average value for a series of
measurements made by A and B as shown below.
Average value for A =
(39.40 + 39.57 + 39.51) ml
= 39.49ml
3
Average value for B =
(39.40 + 39.41 + 39.38) ml
= 39.40ml .
3
Then, we compare each calculated average value to the correct
value (true value), 39.50ml
Which average value is closer to the true value?
…………………………………………………………………
…………………………………………………………………
…………………………………………………………………
…………………………………………………………………
…………………………………………………………………
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CHE110: Introduction to Chemistry
Which student had a much more accurate measurement than the
other? Give reason(s) for your answer.
…………………………………………………………………
…………………………………………………………………
…………………………………………………………………
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…………………………………………………………………
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…………………………………………………………………
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…………………………………………………………………
………………………………………………………………….
Next, we have to calculate the spread of the measurement for
student A and B by subtracting the lowest value from the
highest value in each set of measurements.
The spread of measurements experienced by student A is
39.40ml – 39.57ml = 0.17m
The spread of measurements experienced by student B is
39.41ml – 39.38ml = 0.03ml
Now that we have found the spread for each set of
measurements, we have to compare the two values obtained
against each other.
Looking closely at the two values that we have obtained, which
student had a higher precision of measurement than the other?
Give reason(s) for your answer-----------------------------------------------------------------------------------------------------------------.----------------------------------------------------------------------------
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CHE110: Introduction to Chemistry
The simplest measure of precision is the average deviation,
calculated by first determining the ‘best’ values (averages) of a
series of measurements. Then the difference of each individual
measurement from the average is calculated. (The differences
are treated as positives). The differences (deviations) are finally
averaged. Then the reported values is reported as the average
value plus or minus ( ± ) the average deviations. Take for
example, the measurements for students A and B illustrates
what has been discussed above.
Student A
Measurements
Student B
Deviations
Measurements
Deviations
39. 40 ml
0.01
39.40 ml
0.00
39. 57 ml
0.08
39. 41 ml
0.01
39. 51 ml
0.02
39. 38 ml
0.03
Av 39. 49 ml
0.11 ÷ 3
Av.deviation =0.04
Av 39.40ml
0.04 ÷ 3
Av deviation = 0.01
Av stands for average in the above illustration.
Student A would report his result as 39. 49 ± 0.04 and student
B as 39.40 ml ± 0.01. We will talk more about precision later
when we discuss ways of expressing precision of measured
values.
In the next section we shall consider the role of
significant figures in measurement.
Relative Accuracy – is the measured value or mean
expressed as a percentage of the true value. For a 2.62g
sample analysed as 2.52g the
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CHE110: Introduction to Chemistry
Relative accuracy =
Measured value
X100
True value
2.52
x100 = 96.18%
2.62
When you are carrying out the same experiment more than two
times, you will discover that your obtained might not be the same.
The understanding of the variation of we have noted in the previous
section that measurements are not exact; they have a degree of
uncertainty. The degree of uncertainty in the measured values of a
physical quantity is caused by various factors such as poorly
calibrated instrument as observed in the previous section. As seen
in the previous section measurement B was more precise because a
better graduated ruler was used.
Since we now know that measurements are not exact it is important
to repeat the experiment and then average out the measured values
so that we get the measurement which is close to the “true” value.
The spread of the measured values from the repeated measurements
of the same quantity tell us how precise
1.4 Significant figures
When you measure something, you obtain numbers by reading
them from a scale of a measuring device. When reading a
measurement from a device, there is nearly always some limitation
on the number of meaningful digits that can be obtained.
As a reminder, when a student in the first section of this unit
measured the length of a piece of wood using two rulers with
different scale divisions, we saw that one scale gave him/her a
reading of 2.4cm as a length of a piece of wood. While the other
scale with additional graduations gave him/her a reading of
2.45cm. The reading 2.4cm has 2 significant figures, while the
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CHE110: Introduction to Chemistry
second reading, 2.45cm has 3 significant figures. Digits obtained as
a result of measurement such as these noted above, are called
significant figures.
As a chemistry student you shall be reporting all measurements to
the correct number of significant figures in order to indicate the
degree of uncertainty in your measurement. Therefore, the
importance of significant numbers is that they indicate to us the
reliability of our measurements. Since scientific laws and theories
are derived from measured quantities, our confidence in them is
directly related to the quality of data on which they are based.
You are now familiar with some aspects of significant figures. At
this point in time, it is also important for you to note that significant
figures apply only to the following: measured quantities and
computed quantities. They do not apply to the following: counted
quantities, defined or designated numbers and mathematical
numbers such as π (3.14) or e (2.71)
In order to determine the number of significant figures in the
measurement, we shall be using the following rules:
(i)
All nonzero figures are significant
(ii)
All zeros between nonzero figures are significant
(iii)
When a decimal point is shown, zeros to the right of
nonzero figures are significant. (When a decimal point
is not shown, the number is ambiguous and the number
of significant figures cannot be determined).
(iv)
Zeros to the left of the first nonzero figure are not
significant. (If you are not sure if a zero is a significant
number, convert the measured number to standard
notation: A× 10n where 1 ≤ A < 10 and n is a whole
number).
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CHE110: Introduction to Chemistry
Now, can you put your knowledge of significant figures to
work by doing the following simple task below?
Activity 1.4.1
Use the above rules to determine the number of significant figures
in each of the following measurements.
a. 14.5g ------------------------------------------------------------b.
99.003 ------------------------------------------------------------
c.
4.750cm ----------------------------------------------------------
d.
0.45kg ------------------------------------------------------------
e.
4.065m -----------------------------------------------------------
f.
0.32g --------------------------------------------------------------
If the number of digits needed to express the magnitude of a
measurement exceeds the number of significant in the measured
value, exponential notation should be used. For example if you are
asked to express 5.2 metres in millimeters and maintain your
answer to two significant figures. To work through this problem
you will need to multiply 5.2m by 1000 in order for you to get
5200mm. In exponential notation 5200mm can be written as 5.2 x
103mm. This measurement still gives us two significant figures.
Let us now discuss how we can determine the number of
significant figures involving addition, subtraction, multiplication
and division of measured numbers.
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CHE110: Introduction to Chemistry
Addition and Subtraction of measurement
When we add and subtract measurements the measured value with
the fewest decimal places (d.p) determines the number of
significant figures (s.f) in the answer. To understand this, let us
work out the addition of the following measurements:
308.7810g
0.00034g
+ 10.31g
(4 p.d, 7 s.f)
(5 d.p, 2 s.f)
(2 d.p. 4 s.f) the measurement with the fewest
d.p
319.09134g
The answer should therefore have 2 decimal places, and answer is
319.09g. Thus the answer will have 5 significant figures. It
important that when you are adding or subtracting measurement,
you should first determine the number of decimal places in the
answer only then you can count the number of significant figures in
answer.
Multiplication and division of measurements
During multiplication and division of measurements, the value with
the fewest number of significant figures determines the number of
significant
figure
in
the
answer.
3.0 g (2 s. f ) × 4297 g (4 s. f )
= 178793.3426 g
0.0721g (3s. f )
What is the correct number of significant figures in the above
computed value?------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
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CHE110: Introduction to Chemistry
If you have observed the above rule, you should noticed that the
measurement with the fewest number of significant figures has 2
significant figures.
Your answer should therefore have two
significant figures expressed as 1.8 × 105 g.
When a number is “rounded off” non significant figures are
discarded. The last significant figure is increased by 1 if the
discarded number after rounding off is 5 or greater. The last
significant figure is unchanged if the discarded figure is less than 5.
From the above example the value 178793.3426g was rounded off
to 18000g, the last significant figure (7) increased by one to 8
because the discarde number (8) is greater than 5. To express the
answer to 2 significant figures we have to discard the zeros and
express the answer in scientific nation (standard form). The answer
is, therefore, 1.8 × 105 g
Activity1.4.2
Round off the following measurements
(a) 4.6349 to 4 s.f.---------------------------------------------------------------- ----------------------------------------------------------------------(b) 64. 629g to 2 s.f,-------------------------------------------------------------------------------------------------------------------------------------(c) 64.629 to 4 s.f----------------------------------------------------------------------------------------------------------------------------------------Since you have successfully completed the activity, let us now
move to the discussion of the types of errors that affect accuracy
and precision of our measurements
22
CHE110: Introduction to Chemistry
1.5 Units of measurements
All the measured values are expressed in different units. Each of
the quantities used by scientists and non-scientists alike are
measured in a particular unit. There is an international system of
units called the SI system. ( from the French for the system, system
International d’unites) which is most commonly used around the
world.
The seven base (or fundamental) units of the SI system are as
presented below.
Table 1.5.1: Base SI units
Quantity
Unit
Symbol
Length
metre
m
Mass
Kilogram
Kg
Time
Seconds
s
Temperature
Kelvin
K
Electric current
Ampere
A
Amount of Substance
mole
mol
Luminous intensity
candela
Cd
Using SI system is easy, partly because of the prefixes which are
used with the base units. Table 1.5.2 SI prefixes. Examples of using
prefixes with units as follows below.
23
CHE110: Introduction to Chemistry
Example 1.5.1
1 centimetre = 10-2 metres = 0.01 metre
1 Kilogram = 103 grams = 1000 grams
Table 1.5.2: SI prefixes
Factor
Prefixes
Symbol
Factor
Prefixes
Symbol
1012
Tetra
T
10-1
deci
D
109
Giga
G
10-2
Cent
C
106
Mega
M
10-3
Milli
M
103
Kilo
K
10-6
Micro
102
Hecter
H
10-9
Nano
N
101
deca
de
10-12
pico
p
In order to express very large and very small measurements, we use
the SI prefixes. These prefixes denote multiples and factors of the
SI base and derived units.
Derived units: Using base units it is possible to devise a system of
units which can be used to measure all other quantities. Quantities
formed from the basic quantities are called derived quantities.
Therefore, Base units are the basic units from which other units are
derived. Some derived quantities have been given specific names,
such as joule, while others are just made up of a combination of
base units, such as metre per second. Many quantities can be
measured in more than two or more equivalent units; for example,
one joule per metre.
24
CHE110: Introduction to Chemistry
Table 1.5.3: Derived SI Units
Quantity
Unit
Symbol
Equivalent
Volume
Cubic metre
m3
No equivalent
Force
Newtons
N
Kgm −1s −1
Pressure
Pascal
Pa
N .m −2
Energy, work
Joule
J
N.m
Power
watt
W
Js −1
Example 1.5.2
Given that the definition of area as Area = length x width,
determine the base, or fundamental unit of the area. You can work
through this problem as follows:
Area = length x width
Units of (area) = units of length (length x width)
= metre x metre
= metre2 or m2
1.6 Conversion factors
If you measure the distance between the lines marked off below,
the measurement you report might be 12.7cm, 127mm depending
on the ruler being used and the scale used. Notice that the number
reported is meaningless without reporting the unit.
It is frequently necessary to be able to convert the units of a
measurement to different units. The unit conversion method
depends on the two mathematical facts: (1) any quantity can be
used to write a fraction equal to 1; and (2) like quantities in the
25
CHE110: Introduction to Chemistry
numerators and denominators can be ‘cancelled out’. For example,
if we want to convert 1.5min to seconds we have to employ the
following steps:
Step 1: First, we start by writing a statement of equivalence as
indicated below
60 seconds = 1 min
and
60 sec onds
=1
1min
60 seconds
is called a conversion factor to change minutes
1min
to seconds by multiplying it with the given minutes.
Another conversion factor to change minutes to seconds is
1 minute
= 1 this is used to convert seconds to minutes.
60 seconds
Step 2: Since we want to convert minute to seconds we multiply the
minutes given by the appropriate conversion factor which is
60 sec onds
1min
1.5 min ×
60 sec onds
= 90 seconds
1min
In general:
Given quantity and unit × conversion factor = desired quantity and
unit
The conversion factor takes the form of ratio in which the desired
unit is in the numerator and given unit in the denominator.
Now here is a task for you
26
CHE110: Introduction to Chemistry
Activity 1.6.1
Convert the following measurements: (please take not more than 15
minute to do this task).
(a) 25cm3 to dm3
………………………………………………………………
………………………………………………………………
………………………………………………………………
……………………………………………………………....
................................................................................................
(b) 2.4 km/h to m/s
………………………………………………………………
………………………………………………………………
………………………………………………………………
………………………………………………………………
………………………………………………………………
(c) 2.6 kg/m3 to g/cm3
………………………………………………………………
………………………………………………………………
………………………………………………………………
………………………………………………………………
………………………………………………………………
Before we leave this unit, let us go through what we have been
looking at.
27
CHE110: Introduction to Chemistry
Unit summary
In this unit you have learnt that there are two types of
measurements, namely, qualitative and quantitative. Quantitative
measurements have a degree of uncertainty which is caused by
either random or systematic error. The random errors usually affect
precision while systematic errors affect the accuracy of
experimental result. When we compute the measured values it is
always right to use the SI units and express the answer to the
correct number of significant figures in order to indicate the degree
of uncertainty in the measurement.
In the multiplication and division of measurements, the number of
significant figures allowed in the answer depends on the
measurements with the fewest number of significant figures. In
addition and subtraction, the number of significant figures in the
answer depends on the measurements with the least number of
digits to the right of the decimal point. Furthermore, you have also
come to know that working with significant numbers frequently
requires ‘rounding off”. In the next unit we will discuss matter
In the next unit we shall be looking at classification of matter.
28
CHE110: Introduction to Chemistry
Assessment
Answer the following questions in the spaces provided
1. How many significant figures are there in the following
measurements?
(a) 4.990kg---------------------------------------------------------------(b) 0.067cm--------------------------------------------------------------(c) 0.0003mg-------------------------------------------------------------2. Write the following numbers in standard (scientific) notation;
(a) 105 × 104 ------------------------------------------------------------(b) 39.10 × 10 2 ---------------------------------------------------------(c) 0.000104 ------------------------------------------------------------3. Perform the following calculations and give the answer to
correct
number of significant figures
(a)
(1.014 × 105 ) × (2.34 × 102 )
---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------(b)
0.0024 + ((0.693 × 0.018) / 0.064)
---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4. Two students determined the density of the titanium. Their
results are
as follows
Student A
4.49 g / cm3 , 4.50 g / cm3 , 4.52 g / cm3 , 4.50 g / cm3
29
CHE110: Introduction to Chemistry
Student B
4.48g/ cm 3 , 4.47 g/ cm 3 , 4.44 g/ cm 3 , 4.53 g/ cm 3
(a) Which student had the more accurate results, and which student
had the more precise results? Explain your answer. (The accepted
value
for the density of titanium is 4.51 g/ cm 3 )
---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------(b) Report the results of the two students using the average
deviation.
---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
Note that the answers to the above exercise are given at the end of
the module. However, you are encouraged to work through first
before referring to them.
30
CHE110: Introduction to Chemistry
Unit 2
Matter
2.0 Introduction
Welcome to unit 2. In this unit we are going to discuss the
classification of matter. Before we discuss matter, let us relate
chemistry to matter first
Chemistry is the study of the compositions, structures and
interaction of matter. Matter is any thing that occupies space and
has mass. Matter is made of particles. These particles are either
atoms, ions or molecules.
The compositions, structures and interactions of matter will help us
understand the classification of matter.
During and after completion of this unit you will be able to:
Discuss the composition of matter.
.Describe pure and impure substances
Describe how some mixtures can be separated.
.
31
CHE110: Introduction to Chemistry
2.1 Properties of Matter
Since by now you are aware that chemistry deals with the
compositions, structure and interaction of matter, let us now look at
types of properties of particles that help us classify matter.
Chemists use the following properties to describe and classify
matter in various ways:
(i)
Intensive and extensive properties.
(ii)
Physical and chemical properties.
Intensive properties are properties that do not depend on the size
of a sample. Examples of intensive properties are; density, boiling
point, melting point and colour. Extensive properties are
properties of matter that depend on the size of a sample, for
example, mass and volume.
Let us now turn to physical and chemical properties. Physical
properties are properties of particles that change only the states of
the particle without forming a new substance. Therefore, the
identity of a substance is maintained. For example, boiling point,
melting point, density, colour and mass. Now answer the following
questions in 5 minutes.
What type of change that happens when you boil water? Does the
change affect the identity of water?
----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------In case you are not sure of the answer the following statement will
help you to confirm whether or not your response was correct.
A determination of the boiling point of water only change the state
32
CHE110: Introduction to Chemistry
of water from liquid to gas (the substance remains water) .This type
of change is called physical change.
Now let us see what chemical properties are. Chemical properties
are properties of the particles when they form a new substance. The
identity of the substance is changed since a new substance is
formed. For example, if we want to determine the chemical
properties of sodium, we can combine sodium with water and
observe what happens. Some of the observations which you note
are; sodium will react vigorously with water, heat will be given out
and the whitish solid sodium will disappear forming a colourless
solution called sodium hydroxide. The products formed, namely
hydrogen and sodium hydroxide, are different from the substances
(sodium and water) from which they are made. This type of change
is called chemical change.
Below is a table indicating some differences between a physical
property and a chemical property.
Table 2.1: Comparison of physical and chemical properties of matter
Physical property
No new substance is formed
Chemical property
A new substance is formed
The mass of the substance does The mass of the substance change
not change
since a new substance is formed
It only changes its state
state depends on the new substance
formed
The energy of the particle does The energy of the particle changes
not change
Particle mix in any proportion Particle mix in fixed proportion by
by mass
mass
33
CHE110: Introduction to Chemistry
Now let us see how these properties of substances help us to
identify and classify different types of substances in the next
section.
2.2 Types of substances.
Now we can define matter in terms of its atomic make up: matter which
is composed of ‘identical atoms’ is referred to as elements. Then, matter
which is composed of different kinds of atoms chemically combined in
simple, whole number ratios is referred to as compound. While matter
which is a physical combination of the particles of elements or
compounds is referred to as a mixture. In the rest of this section we shall
discuss elements, compounds and mixtures.
Elements and compounds are pure substances while mixtures are
impure substances. The classification of matter can be summarised
in the figure below
Figure 2.2.1 Classification of matter
An Element is a substance that is made up of atoms with the same
atomic number and cannot be split into simpler particles by
ordinary chemical means.. They are the building blocks for all
compounds and other complex substances. They are 108 elements
34
CHE110: Introduction to Chemistry
which have been organized in the periodic table of which, 91 are
natural elements, and the others are artificial elements (man-made,
with extremely short life span).
Compounds consist of atoms of different elements which have
been combined chemically to form one kind of “molecule.” The
elements making up a certain compound are always present in the
same ratio (law of constant composition or law of definite
proportions). For instance, water always occurs in the ratio of 1g
hydrogen to 8g oxygen, ammonia always occurs in a ratio of 3g
hydrogen to 14g nitrogen.
Unlike an element, a compound can be decomposed to give simpler
substances by chemical means.
For example
(i)
Use of electricity during electrolysis e.g.
2 NaCl (l ) → 2 Na ( s ) + Cl2 ( g )
(ii)
Use of heat, thermal decomposition. For example,
CaCO3 ( s ) → CaO( s ) + CO2 ( g )
(iii)
Use of light e.g. in photolysis. For example,
2 AgI ( s ) → 2 Ag ( s ) + I 2 ( s )
A mixture is made up of pure substances which are not chemically
combined. The ratio in which these substances are combined is not
fixed but can be variable, for instance a mixture of sugar and water
can be made from 1g of sugar and 100g water, 10g of sugar in 100g
of water and so on and so forth. Mixtures can be mixture of
elements, of compounds and mixtures of elements and compounds.
If we mix two substances of different states: gas and liquid, liquid
and liquid, solid and liquid, to get a mixture in one state only, the
35
CHE110: Introduction to Chemistry
mixture formed is called a solution. A solution is therefore, a
homogeneous mixture. That is, the particles are evenly distributed
or the same or uniform through out. Homogeneous matter may be
either a mixture or a pure substance.
Examples of the
homogeneous mixture include sugar solution, copper sulphate
solution and sodium chloride.
But if we mix two or more substances and we obtain a mixture
which is not of the same state through out, then we have formed a
heterogeneous mixture. That is, the particles are not evenly
distributed. For example of the heterogeneous mixture include
milk, mud and Colgate paste.
2.3 Separating mixtures
Now let us consider how different types of mixtures can be
separated. Very quickly list some of the separation techniques that
you can use to separate different mixtures.
………………………………………………………………………
………………………………………………………………………
………………………………………………………………………
………………………………………………………………………
The following answers could have been part of your list: fractional
distillation,
filtration,
evaporation,
crystallization
and
chromatography. Let us now discuss briefly each of the separation
method listed above:
Distillation: a process of separating a liquid by evaporating it out
of its mixture then condensing it to obtain a in its pure state.
Fractional Distillation is used for the separation of two or more
miscible liquid in which the components have different boiling
points. When the mixture is heated, the components which have
lower boiling points are the first to be collected as the fraction of
the mixture. For example, liquid air is separated by fractional
36
CHE110: Introduction to Chemistry
distillation into oxygen and nitrogen; separation of crude oil into
various fractions such as petrol, paraffin and lubricating oil.
Simple distillation: A process of collecting a solvent from the
solution made by dissolving a solid solute into a liquid solvent. For
example, obtaining pure water from a salt solution.
Filtration a process used to separate insoluble particle from its
liquid mixture using a filtering material such as filter paper. The
particles of a solid are held back (on account of the large size)
while the liquid is able to pass through the filtering material.
Evaporation is a process of separating out a liquid from its
solution by changing its physical state from a liquid to a gas upon
heating. The mixture is heated, the liquid turns into vapour and
becomes dispersed in the air while the solid material, if any, is left
as a residue in the basin. The solid material that remains in the
basin are called crystals and this process called crystallization.
Chromatography is a process of separating a mixture of solutes
in a solvent using a separating medium. The separating medium is a
stationary phase, and mostly used is a filter paper, and a solvent is
mobile phase Separation of the solutes depends on the solubility.
You should have noticed by now that all the separation processes
we have discussed are based on the physical properties of the
components of a mixture.
37
CHE110: Introduction to Chemistry
Unit summary
In this unit you learned that matter is composed of particles, which
are atoms, ions and molecules. These particles are classified
depending on their properties. These include intensive, extensive,
physical and chemical. You also look at how these properties are
used in separation techniques of matter in which particles are
classified as pure or impure substances. We also discovered that
pure substances are classified into elements or compounds.
.
\
38
CHE110: Introduction to Chemistry
Assessment
Answer the following questions in the spaces provided
1. On the list of substances below, classify them into mixture and
compounds and describe the composition of each substance
Water, milk, ink, crude oil, bronze, sodium chloride, sea water
Mixtures-----------------------------------------------------------------------------------------------------------------------------------------------------Compounds--------------------------------------------------------------------------------------------------------------------------------------------------2. State and describe the best technique to separate each mixture to
obtain the last pure substance?
(a) water in salt-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------(b) Alcohol in water-----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------Note that the answers to the above exercise are given at the end of
the module. However, you are encouraged to work through first
before referring to the answers.
39
CHE110: Introduction to Chemistry
Unit 3
Atomic Theory
3.0 Introduction
Welcome to this unit. In the previous unit we looked at matter and
how it is classified into elements, compounds and mixtures. Now
we shall further discuss the composition of matter. As a remainder,
matter is made of particles which are ions, atoms or molecule. It
should be noted that the ions and molecules come from atoms. So,
in this unit we will first start by discuss atom, then we shall see
how atoms are related to ions and molecules in subsequent units.
During and after completion of this unit you will be able to:
Describe the composition of an atom.
Justify the characteristics of an atom using the law of chemical
combination.
Discuss the laws of chemical combination
Calculate the relative atomic masses of isotopes.
40
CHE110: Introduction to Chemistry
In your own words can you describe an atom.
………………………………………………………………………
………………………………………………………………………
………………………………………………………………………
………………...................................................................................
In order for us to understand the atomic theory well, it is important
for us to relate an atom to matter. Well, with reference to matter, an
atom can be regarded as the smallest part of an element. The
concept of the atom is an old one. The word ‘atom’ has been in use
since the 400 B.C. With this background at hand, we can now
consider the following theories proposed by some great scientists
concerning atoms.
3.1 Progression of the atomic theory
The current atomic theory is as result of the contributions made by
different scientists. Now we shall start looking at the work of some
remarkable contributions which some scientists made to the
contemporary understanding of an atomic theory.
3.1.1 Dalton’s atomic theory
Dalton was the first scientist to suggest an atomic theory. His
theory gave a scientific basis to the existence of atoms. Dalton
stated the atomic theory in a clear manner which explained many
known scientific laws and experiments. In addition, his theory
accounted for the classification of matter into elements, compounds
and mixtures. Some of the key statements of Dalton’s theory
include the following:
41
CHE110: Introduction to Chemistry
(i)
All matter is made up of atoms which are indestructible by
ordinary chemical means. All the atoms of a given element
posses identical properties.
(ii)
Atoms of different elements have different properties and
different masses.
(iii)
Atoms are the units of chemical changes. (Chemical
changes involve the combination, separation or
rearrangement of atoms; during chemical change atoms are
not created, destroyed, changed or subdivided (it is the
smallest particle of an element that can enter into a
chemical reaction).
(iv)
When atoms combine to form “molecules”, they do so in
simple whole numbers.
(v)
All “molecules” of one compound are identical and are
different from molecules of other compounds. (A molecule
is the smallest amount of a chemical substance which can
exist in a free and separate state).
3.1.2 Avogadro’s theory
The flaw in Dalton’s atomic theory was corrected in 1811 by
Amedeo Avogadro. Avogadro had proposed that equal volumes of
any two gases under the same conditions of temperature and
pressure contain equal numbers of molecules (in other words the
mass of a gas particle does not affect its volume). Avogadro’s law
allowed him to deduce the diatomic nature of numerous gases by
studying the volumes at which they reacted .For instance, since two
litres of hydrogen will react with just one litre of oxygen to produce
two litres of water vapour (at constant pressure and temperature). It
meant that a single oxygen molecule splits in two in order to form
two particles of water .Thus, Avogadro was able to offer more
accurate estimates of the atomic mass of oxygen and various
42
CHE110: Introduction to Chemistry
elements ,and firmly establishment the distinction
between a
molecule and an atom.
Since you are now familiar with the current atomic theory, you
might have noticed as you critically read Dalton’s theory that some
areas of the original Dalton’s atomic theory do not completely
agree to the current atomic theory of matter. For instance, the
original Dalton’s theory does not address the existence of isotopes,
that is, atoms of the same element having the same properties. The
following are the modifications to Dalton’s original atomic theory:
(i)
The existence of isotopes means that all the atoms of a
particular element do not have the same mass. (Isotopes
will be explored latter in this unit).
(ii)
Atom are known to be made up of small particles called
electrons, protons and neutrons
(iii)
Atoms of different elements with same masses do exist and
are called isobars.
(iv)
Atoms are created or destroyed in special changes such as
radioactivity, nuclear fission and fusion.
3.2 Composition of an atom
Now that you are familiar with Dalton’s atomic theory, it is
important for us to look at the composition of an atom. That is, to
explore the inside of an atom. Rutherfold’s and Bohr’s models of
the atom and the Chadwick’s experimental results showed that all
atoms of different elements are made up of three fundamental subatomic particles, namely, protons, neutrons and electrons. The
hydrogen atom, the simplest atom of all, however, contains just
43
CHE110: Introduction to Chemistry
one proton and one neutron. Atoms of different elements differ in
the numbers of these particles.
The centre of the atom is occupied by a nucleus. All the protons
and neutrons, collectively called nucleons, are located in the
nucleus. The electrons are found in energy levels or shells
surrounding the nucleus. Much of the atom is empty space.
The actual mass of a proton is 1.6726 × 10−24 g but it is assigned a
relative value of 1. The mass of a neutron is virtually identical and
also has a relative mass of 1. Compared to a proton and a neutron
an electron has negligible mass with a relative mass of only
1/1836. Neutrons are neutral particles. An electron has a charge of
1.602 x 10-19 coulombs which is assigned a relative value of -1. A
proton carries the same charge as an electron but of an opposite
sign so has a relative value of +1. All atoms are neutral so must
contain equal numbers of protons and electrons.
The difference between atoms of different elements is not caused
by the type of particles they consist of, but by the number of each
sub-atomic particle in the atom contains. The properties of the
fundamental sub-atomic particles of atoms are as shown in the
table below.
44
CHE110: Introduction to Chemistry
Table 3.2.1 Summary of Relative Mass and Charge of fundamental
sub-atomic particles of an atom.
atomic mass
unit, a.m.u.
Subatomic
Particle
Mass (g)
(Relative
Relative
mass)
Charge
symbol
Proton
1.6726 × 10−24
1
+1
p
Neutron
1.6750 × 10−24
1
0
n
1
1836
-1
e
Electron
9.1095 × 10
−28
As you can see from the table above the masses of the constituent
particles of atoms are very small. Since the sub-atomic particles
are very small, their masses are expressed more convenient in
small units called atomic mass unit (a.m.u). The atomic mass unit
is defined as
1
of the mass of an atom of a
12
equal to 1.66 x 10
−24
12
6
C isotope and is
g. (almost the mass of a proton or neutron).
We shall cover the concept of isotope in the next section of this
unit.
45
CHE110: Introduction to Chemistry
The atomic mass unit of each subatomic particle is expressed as a
ratio to the mass of
1 proton =
1
of 126C as shown below.
12
1.6726 ×10−24 g
= 1.00759 = 1 a.m.u
1.66 ×10−24 g
1Neutron =
1.6750 ×10−24 g
= 1.009036 = 1 a.m.u
1.66 ×10−24 g
1Electron =
9.1095 × 10−28
= 0.000549 = 0
1.66 × 10−24
The atomic mass unit has no units because is a ratio. The atomic
mass unit for a proton and a neutron is 1 while that of an electron
is approximately equal to zero.
3.3 Atomic number and Mass number
An atom is characterised by two numbers. These being: Atomic number
and Mass number.
Atomic number is the number of protons in the nucleus of an atom.
This number is also equal to the number of electrons in the atom.
In addition, the atomic number defines which element the atom
belongs to and consequently its position in the periodic table. It is
represented by the symbol (Z). If we take ‘X’ to serve as a symbol
for an element, the atomic number is written at the bottom and on
the left of the symbol, thus: Z X
.
As for atomic mass number, it is the sum of protons and neutrons
in the nucleus of a given atom. It is represented by the symbol (A).
The mass number is indicated at the top and to the left of the
symbol, thus:
A
X
.
46
CHE110: Introduction to Chemistry
Since all atoms contain the same number of electrons and protons,
an atom is electrically neutral. Therefore, atoms have no charge.
The charge of an atom is represented by n. So, an atom has n = 0.
When charge is equal to 0, the place for charge is left blank.
However, by losing one or more electrons atoms become positive
ions, or by gaining one or more electrons atoms form negative
ions.
The shorthand notation for an atom or ion is as indicated below.
A
Z
X n +/n −
The number of electrons in an atom of an element determines the
chemical properties of a given element because the numbers of
electrons in the outermost shell are involved in a chemical change
of the element. The atoms in an element undergo chemical change
by gaining, loosing or sharing of electrons. We will look at
chemical change of elements as we go on in our course.
Atoms are conventionally represented on the periodic table. as
nuclide in this pattern, AZ X .(we shall learn more about periodic
table in the next module). For now, let us consider the nuclide of
oxygen atom that is presented as
16
8
O in the periodic table. What
do (i) the following numbers, 16 and 8 stand for? (ii) Letter ‘O’
stand for?
Given the mass number and atomic number of an atom you can
easily find the number of neutrons. How? Well, by subtracting the
atomic number of an element from its mass number. As an
example let us find together the number of neutrons contained in
16
8
O. You should arrange your work neatly as indicated below.
47
CHE110: Introduction to Chemistry
n=A–Z
= (16- 8) neutrons
= 8 neutrons
Having done the concepts of atomic number and mass number, let
us now move on and do the following task.
Activity 3.3.1
Study the table below very carefully and then, fill in the missing
values.
symbol
Atomic Mass
No. of
No. of
number number protons neutrons
9
4
No. of
electrons
Be
40
20
Ca 2+
37
17
Cl-
20
17
3.4 Isotopes
In 1913, J.J. Thomson discovered that a given element can possess
atoms with different masses, and these atoms were given the name
of isotopes. To account for their existence, Rutherford supposed
that their nuclei contained the same number of protons, but he
proposed the existence of another particle within the nucleus which
he called the neutron. This particle had the same mass as the
proton but possessed no charge. Different isotopes would have
different number of neutrons in their nuclei. Neutrons were first
observed experimentally by Chadwick in 1932.
48
CHE110: Introduction to Chemistry
From the above account, it can be deduced that isotopes are atoms
of the same element with the same number of protons but with
different number of neutrons. Isotopes of the same element have
identical chemical properties. This is due to the point that chemical
properties are determined by the valence electrons of an atom
(isotopes of the same elements have the same number of electrons).
Since their mass is different their physical properties such as
density and boiling point are different. Examples of isotopes
include the following: 11 H
35
17
2
1
Cl
H
3
1
H
37
17
Cl
Most of the elements exist as isotopes in nature with different
relative abundance. The relative abundances are quantities by
percentages of a given atom that exists in nature. Take for
example, Oxygen has the following naturally occurring isotopes;
16
8
O,
17
8
O, and
18
8
O with 99.76%, 0.04% and 0.20% respectively. All
these isotopes are atoms of oxygen because they have the same
number of protons.
3.4.1 Average atomic weight
The average atomic weight reflects the fact that elements are made
up of atoms with different masses. That is, they are isotopes. In the
case of carbon, most carbon atoms are carbon-12, mass = 12 a.m.u.
But about 1% of carbon atoms are carbon-13, mass = 13 a.m.u. and
there are a very small number of carbon-14 atoms, mass = 14
a.m.u.
The weighted average of carbon turns to be 12.01 a.m.u. The
weighted average takes into account the amount of each isotope as
well as the weight of each isotope. Notice that atomic weights are
49
CHE110: Introduction to Chemistry
not whole numbers because they are weighted averages of all
isotopes of that element.
How do we determine an average atomic weight? To do this we
need know:
(i) which isotope exist for a given element and
(ii) the percentage abundance of each isotope.
Consider, the element magnesium is made up of three isotopes:
78.7%
24
Mg , 10.1 %
25
Mg and 11.2 % 26 Mg . The weight
contributed by each isotope is it’s the percentage in fractional ( %
relative abundance) or decimal form multiplied by the isotopic
mass. Therefore, the weight contributions for magnesium are:
% relative
isotopic
Abundance x mass
Weight
=
contriibution
24
Mg Contribution: 78.7/100 x 24.0
=
18.9
25
Mg Contribution: 10.1/100 x 25
=
2.53
26
Mg Contribution: 11.2/100 x 26.0
=
2.91
24.34
In this example, the average atomic weight of Mg is 24.3 a.m.u.
Thus, the average atomic weight is the sum of the contributions.
The weighted mean molar mass is thus, 24.34 g mol-1 and the
relative atomic mass is 24.34. This is also equal to the relative
atomic mass. The relative atomic mass has the following symbol
(Ar). We shall talk more on Ar in the next unit.
Let us now look at chlorine isotopes. Chlorine atoms containing 17
35
protons, 17 electrons, and 18 neutrons ( 17
Cl) with 75.53% relative
abundance , and Chlorine atoms with x protons, 17 electrons and
50
CHE110: Introduction to Chemistry
20 neutrons ( 37
17 Cl) with 24.47% relative abundance. Let us now
calculate the relative atomic mass ( Ar ) of chlorine.
(% of
35
17
Cl × atomic mass) + (% of
37
17
Cl × atomic mass)
 75.53
  24.47

 100 × 35  +  100 × 37  = 35.45 a.m.u.
 


Activity 3.4.1
In order to consolidate your understanding of relative atomic mass
( Ar ) and relative molecular mass ( M r ), answer the following
questions
Calculate the relative molecular masses of the following substances
(a) Oxygen molecule-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------(b) Water molecule--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------(c) Calcium hydroxide----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------Now that you have learnt how the relative atomic mass and relative
molecular mass are determined, let us now discuss isotopes and
how their relative abundance relates to their relative atomic masses.
51
CHE110: Introduction to Chemistry
Unit 4
Laws of Chemical Combination
We shall start by stating the meaning of chemical combination.
Then, proceed to discuss some laws of chemical combination.
Chemical combination refers to the joining of two or more atoms to
bring about a chemical change. For example, Hydrogen and
Oxygen combine to form water. It should be noted here that some
statements of Dalton were proposed as a basis for explaining some
specific laws of chemical combination. For example, Dalton’s
statement that ‘atoms of different elements can chemically combine
in simple, whole-number ratios to form compounds’ was proposed
to explain the Law of Definite Composition.
The atomic theory is, therefore, a very important tool in chemistry
because it offers, as stated above, acceptable explanation for the
laws of chemical combination. We shall now consider the
following laws of chemical combination;
(i)
The law of conservation of matter
(ii)
The law of constant (definite proportion) composition
(iii)
Law of multiple proportion
Let us now look at each of the above stated laws much more
closely:
4.1 The law of conservation of matter
In a chemical change, the of sum of the mass of the reactants is
equals to the sum of the mass of the products. This means that
matter is neither created nor destroyed in chemical change. It
therefore, follows that the number of atoms of each element in the
52
CHE110: Introduction to Chemistry
product(s) of a reaction is the same as the number of each element
in the reactant(s). For example
2 H2
+
O2
→ 2 H2O
The total of Hydrogen atoms on the left hand side is four which, is
equals to the number of Hydrogen atoms on the right hand side of
the chemical equation. The same is true for oxygen
4. 2 Law of Constant Composition
States that all samples of a pure compound always contain the
same elements in the same ratio by mass. In other words, atoms of
an element in a compound will occur in a definite proportion by
mass. This law implies that, no matter how a certain compound is
produced; it will always have the same composition. Take for
instance, a water molecule.
This compound has general formulae H2O. Meaning it will always
contain 2 atom of Hydrogen with mass 2g and one atom of
Oxygen with mass 16 g whatever the source.
Similarly for calcium oxide, it can be produced by a direct
combination of elements calcium and oxygen, by decomposition of
calcium carbonate using heat.
2Ca(s)+O 2 (g) → 2CaO(s)
CaCO3 (s) → CaO(s)+CO 2 (g)
To illustrate that chemical compounds contain elements in the
same proportion by mass, let us work out the following example.
The following experimental figures were obtained by using
different masses of copper to yield copper oxide.
Samples: A 2.120g of copper yielded 2.653 g of copper oxide
B 3.180 g of copper yielded 3.980 g of copper oxide
53
CHE110: Introduction to Chemistry
C 4.240g of copper yielded 5.307 g of copper oxide
The law of constant composition requires that the composition by
mass of the three samples of copper oxide be constant. A
convenient way of showing this is to calculate the percentages of
copper in each of the three samples.
Sample A
2.120 g
×100 = 79.9%
2.653 g
Sample B
3.180 g
×100 = 79.9%
3.980 g
Sample C
4.240 g
×100 = 79.9%
3.980 g
Since the figures of proportionality are equal, then the law of
composition is satisfied. We will now turn to a discussion of the
law of multiple proportions.
4.3 Law of multiple Proportions
The law of multiple proportion is the third postulate of Dalton‘s
atomic theory. It states that the masses of an element that
combines with fixed masses of the second element are in the ratio
of whole numbers
Therefore, two of oxygen in the two compounds that combine
with the fixed mass of Hydrogen should be in a whole number
ratio. For example, Hydrogen and Oxygen form more than one
compound, namely, water and Hydrogen peroxide. There are 2
Hydrogen in H 2O and in H 2O2 . It means that the 2 Hydrogen in
each compound have a fixed mass. While 1 O atom in H 2O is not
the same as the mass of 2 O atoms in H 2O2 . However, the masses
of the oxygen atoms in the two compounds are related as a ratio of
whole number 1 to 2. Other examples are SO2 and SO3 , CO
and CO2 , and CuCl and CuCl2 .
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CHE110: Introduction to Chemistry
Let us now look at the illustration of how the law of multiple
proportions can be verified.
Phosphorus forms two compounds with chlorine. In one, 7.75g of
phosphorus combines with 26.60 g of chlorine. In the other, 3.10 g
of phosphorus combines with 17.73 g of chlorine. Show that these
compounds confirm the law of multiple proportions.
To answer this question we should first calculate the mass of one
element, for example, chlorine, that combines with a fixed mass,
for example, 1.00 g, of the other element.
For the first compound:
1.00 g P ×
26.60 g Cl
= 3.432g Cl
7.75 g P
For the second compound:
1.00 g P ×
17.73g Cl
= 5.719g Cl
3.10 g P
Now we find the ratio of chlorine in the two compounds:
3.432 g Cl
= 0.600
5.719g Cl
The 0.600 is the same as 3/5, a ratio of whole numbers consistent
with the law of multiple proportions. (Of course, this example
could have been answered by calculating the mass of phosphorus
that combines with 1.00 g of Cl).
Now answer the following questions
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CHE110: Introduction to Chemistry
Activity 4.1.1
1. Two samples of a substance were collected from two different
sources. The substance is composed of element A and B. One
sample contains 25.5g A and 74.5 g B. The other sample was
composed of 8.1 g of A and 23.6 g of B. Show that these data
demonstrate the law of constant composition.----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2. Hydrogen and oxygen form two different compounds as follows
Compound
Water
% Hydrogen
% Oxygen
11.21%
88.79%
Hydrogen peroxide 5.94%
94.06%
Show that these data can be used to verify the law of multiple
proportions.-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
56
CHE110: Introduction to Chemistry
Unit summary
In this unit, we have looked at the atomic theory of the great
scientists and their contribution to the current understanding of
the theory regarding the atomic structure. It was discovered that
an atom is made of three fundamental sub atomic units namely,
protons and neutrons find in the nucleus, and electrons revolving
around the nucleus in the shells. We also looked at the nuclide
that gives us how the name of the element, atomic number and
atomic mass number are express using letter and numbers. We
also discussed how to determine the relative atomic mass and
relative molecular mass of particles.
57
CHE110: Introduction to Chemistry
Assessment
Answer the following questions in the spaces provided
1. Determine the relative molecular (formula) mass of each of the
following substances;
(a) Ca ( HSO4 ) 2 --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------(b)
Na2CO3 ---------------------------------------------------------------
--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------(c) CuSO4 .5 H 2O -------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------2. Naturally occurring copper has two isotopes with relative atomic
masses 62.930 and 64.928 a.m.u. The average relative atomic mass
of copper is 63.546. What are the relative abundances of the two
isotopes in naturally occurring copper?------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
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CHE110: Introduction to Chemistry
-----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------3. A metal forms two oxides. In one, 10.0g of the metal combines
with 1.35g of oxygen. In the other, 6.50g of the metal combines
with 1.75g of oxygen. Show that these data are in conformity with
the Law of Multiple Proportions. ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------4. Two samples of DDT were analysed. In the first sample, 1.00g
of hydrogen was found to combine with 18.67g of carbon and
19.69g of chlorine. In the second sample, 10.0g of carbon were
found to combine with 0.536g of hydrogen and 10.55g of chlorine.
(i)
Calculate the mass ratio of carbon, hydrogen and chlorine
in each sample.
(ii)
Do these data illustrate the Law of Constant Composition?
(1)----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
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CHE110: Introduction to Chemistry
(ii)---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------Note that the answers to the above exercise are given at the end
of the module. However, you are encouraged to work through
first before referring to the answers.
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CHE110: Introduction to Chemistry
Unit 5
The Mole Concept
5.0 Introduction
Welcome to unit 4. In the previous unit we discussed the basic atomic theory
which gave us the insight of the composition of an atom which is the
building block of matter. In this unit we are going to discuss the mole
concept which is a fundamental unit in chemistry.
Upon completion of this unit you will be able to:
Express different quantities of substances in moles.
Calculate the number of particles ( atoms or molecules) in a given
quantity of a substances
Outcomes
Determine the percentage composition of compounds
Determine the empirical and the molecular formulae of substances
61
CHE110: Introduction to Chemistry
5.1 What is a mole?
A mole is the amount of any substance that contains as many entities as
they are atoms in exactly 12g of the carbon -12 isotope. The actual
number of atoms in 12 g of caborn-12 isotope is 6.02 × 1023 and it is
called the Avogadro’s constant.
Example 4.1.1
How many particles are in the following substances: (a) A mole of
Electrons (b) 2 moles of sugar molecule (c) 1 mole of sodium
(a) 1 mole of electrons contains 6.02 × 1023 electrons
(b) 2 moles of sugar molecules contains 2 × 6.02 × 1023 molecules
of sugar
(c) 1 mole of sodium atoms contain 6.02 × 1023 atoms of sodium.
Thus a mole is an amount of any substance that contains 6.02 × 1023
particles
Molar mass
This is the mass of one mole of a substance and it is numerically equal to the
relative atomic mass for elements and relative molecular mass for molecules
and compounds. The molar mass is calculated the same way you were
calculating the relative atomic and molecular (formula) masses in the
previous unit.
Example 5.1. 2
What is molar masses of; (a) 1mole of oxygen atom (b) 1 mole of water
molecule
(a) The relative atomic mass of an oxygen, O, atom is 16.00 a.m.u, the
mass of 1 mole of O atom is 16.00 g, and this mass contains 6.02 ×
1023 oxygen atoms
62
CHE110: Introduction to Chemistry
16.00g is not the mass of 1 oxygen atom but the mass of 6.02 × 1023 atoms of
oxygen which is 1 mole. To determine the mass of 1 oxygen atom we divide
the number of oxygen atoms (6.02 × 1023) present in 1 mole into the mass of
1 mole of oxygen atoms.
Mass of 1 O atom =
Mass of 1 mole (molar mass)
Number of particles in 1 mole (Avogadro's constant)
=
16.00g( mass of 1 mole of oxygen atoms)
6.02 × 1023 (oxygen atoms in 1 mole)
= 2.658 × 10-23 g
(b) The relative molecular mass of water ( H O ) is 18.02 a.m.u, thus
2
the mass of 1 mole of H 2 O molecule is 18.02 g
Activity 5.1.1
Now calculate the mass of 1 water molecule
…………………………………………………………………………………
…………………………………………………………………………………
…………………………………………………………………………………
…………………………………………………………………………………
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CHE110: Introduction to Chemistry
The SI unit for molar mass is grams per mol (g / mol). That is the mass of 1
mole of any substance.
Molar volume
The molar volume is the volume occupied by 1 mole of any gas at a
specified temperature and pressure. At standard temperature and pressure
(s.t,p) that is at exactly 0 o C and at 1 atmosphere (atm). The molar
volume for any gas is 22.4 dm3/mole. It means that 1 mole of any gas at
s.t.p occupies a volume of 22.4 dm3.
5.2 Calculations of moles, mass, volume and particles in a given
substance
The number of moles present in given mass of a substance is determined by
dividing the mass of 1 mole of a substance (molar mass) into the given mass.
To find the number of particles present in a given mass, the number of moles
in a given mass are multiplied by the Avogadro’s constant (number of
particles per mole)
Number of moles =
given mass (grammes)
Molar mass (grammes/mol)
=
grammes
grammes/mol
= mol
Thus Mass = moles × molar mass
= mol × grammes/ mol
= grammes
Example 5.2.1
Calculate the number of moles of sulphuric acid present in 27.0g
sample of H2 SO4.
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CHE110: Introduction to Chemistry
given mass of H 2SO 4
Number of moles of H 2SO 4 =
Molar mass of H 2SO 4
Mass of sulphuric acid given = 27.0g
Molar mass of sulphuric acid =?
The molar mass of H2 SO4 = H(1.01 × 2) +S(32.06) + O(16.00 × 4)
=
Number of moles of H2 SO4 =
98.08 g/mol
27.0g
= 0.275 mol
98.08g/mol
Number of H2 SO4 molecules present in 27.0g (0.275 mol)
0.275 mol × 6.02 × 1023 molecules/mol = 1.656 × 1023 molecules
Example 5.2.2
You are provided with 5.4g ammonia ( NH 3 ) Calculate:
(a) The number of moles of N atoms and H atoms in the sample
(b) The number of N and H atoms present in the sample
(c) The mass of N and H in the sample
(a) The number of moles of N atoms and H atoms
Mass of NH3
Number of moles of NH 3 =
=
=
Molar mass of NH3
5.4 g
14.01 + (1.01 × 3) g / mol
5.4g
17.04g/mol
=0.316mol
Moles of N atoms in 0.316 moles of NH 3
Take the mole ratio of N to NH 3
N: NH3
1: 1
This means that for 1 mole of NH 3 molecule there is 1 mole of N atoms.
Thus the number of moles of N atoms in 0.316 moles of NH 3 is
determined as follows; 1 × 0.316 mol = 0.316 moles. In other words there
is 1 mole of N atoms in every mole of NH 3
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CHE110: Introduction to Chemistry
Moles of H atoms in 0.316 mole of NH 3
Take the mole ratio of H to NH 3
H: NH 3
3: 1
This means that for 1 mole of NH 3 there is 3 moles of H atoms. Thus
the number of moles of H atoms in 0.316 moles of NH 3 is calculated as
follows; 3 × 0.316 mol = 0.948 mol
(b) The number of N and H atoms present in the sample
The number of atoms (particles) in the sample is found by
multiplying the moles of atoms (particles) by the number of atoms
per mole (6.02 × 1023)
Number of N atoms = 0.316 mol × 6.02 × 1023 atoms/mol
= 1.90 × 1023 atoms
Number of H atoms = 0.948 mol × 6.02 × 1023 atoms/mol
= 5.71 × 1023atoms
(c ) The mass of N and H in the sample
The mass of an element in a compound is calculated by multiplying
the number of moles of an element present in the compound by its
molar mass.
Mass of N = 0.316 mol × 14.01g/mol = 4.427g
Mass of H = 0.948 mol × 1.01g/mol = 0.957g
Don’t multiply the molar mass of H (1.01g/mol) by 3 because
There are 3 H atoms in NH 3 . This has been taken care off by the mole
ratio, we multiplied the moles by 3. The sum of masses of individual
elements in the compound should add up to the mass of the compound,
in this example the sum of N and H should be equal to the mass of NH 3 .
4.427g of N + 0.957g of H = 5.384g =5.4g which is the mass of NH 3
66
CHE110: Introduction to Chemistry
In order to find the number of moles in a given volume of any gas at
specified temperature and pressure we divide the volume by the
molar volume.
Number of moles =
volume (dm3 )
Molarvolume (mol/dm3 )
= mol
Volume = number of moles (mol) × molar volume (mol/dm3) = dm3
Volume should always be in decimetre cubed (dm3)
Example 5.2 .3
Calculate the number of moles of carbon dioxide present in 10.25dm3of
CO2 at s.t.p
Number of moles of CO2 =
=
volume (dm3 )
Molarvolume (mol/dm3 )
10.25 dm3
=0.458mol
22.4d m3/mol
Example 5. 2. 4
What is the volume occupied by 8.00g of oxygen gas at standard
temperature and pressure?
First we should find the number of moles of O2 (oxygen gas)
Number of moles of O2
=
=
=
Volume of by O2 at s.t.p
mass of O 2
Molar mass of O 2 Molecule
8.00g
2 × 16.00 g / mol
0.250 mol
= moles of O2 × molar volume
= 0.250mol × 22.4dm3/mol
= 5. 60dm3
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CHE110: Introduction to Chemistry
Activity 5.2.1
1. What mass of oxygen is contained in 5.5 g of KClO3 , and how many
atoms of oxygen are there in this mass?..................................................
.................................................................................................................
…………………………………………………………………………
…………………………………………………………………………
………………………………………………………………………...
………………………………………………………………………...
………………………………………………………………………….
2. Compute the mass of the following compounds
(a) 200 Fe atoms……………………………………………………......
………………………………………………………………………
……………………………………………………………………….
………………………………………………………………………..
………………………………………………………………………..
(b) 0. 050 mol of CuSO 4 .5H 2 O ………………………………………..
……………………………………………………………………….
………………………………………………………………………..
………………………………………………………………………..
……………………………………………………………………….
3. Calculate the volume occupied by 4.25 g of CO2 at s.t.p
………………………………………………………………………….
…………………………………………………………………………
…………………………………………………………………………
68
CHE110: Introduction to Chemistry
Let us see how mass and mole ratio of elements helps to determine the
chemical formula of a compound.
‘
5.3 Chemical Formulae
Before we discuss how the mass composition of elements in
compound and mole ratios help us to determine the chemical formula of
a compound, let us first briefly discuss the types of chemical formulae.
In your own words how can you define a chemical formula?
A chemical formula is a representation of the composition of the
substance qualitatively and quantitatively using symbols. It shows which
elements are present and the ratio in which they are combined. For
molecular compounds the formula shows the actual number of atoms of
each element present in a molecule of the substance. While the ionic
compounds it shows the ratio of atoms/ions of each element forming
the substance. The following are some of the types of formulae;
empirical, molecular and structural formula.
Empirical formula shows the simplest whole number of atoms of each
element in a substance.
Molecular formula shows the actual number of atoms of each element in
a molecule of a substance.
Structural formular shows the geometrical arrangement of atoms in a
molecule of a substance
To illustrate this three types of formulae let us look at the empirical,
molecular and structural formula for ethene.
Molecular formula for ethene:
C2H4 This means that for every molecule of ethene there are 2 carbon
atoms and 4 atoms of hydrogen
Emperical formula for ethene
69
CHE110: Introduction to Chemistry
CH2 This means that in every molecule of ethene C and H are present
in the simplest mole ratio 0f 1 : 2
Molecular formular for ethane
A planar molecule with angles of 1200 between all bonds.
Let us now discuss how we can determine the chemical formula of a
Compound. The first step in determining a chemical formula of a
Substance is to find out which elements are present and then find out the
ratio in which the elements combined. One way of determining the ratio
in which the elements combined is to find the mass percentage
composition of a compound.
Percentage composition is the percentage of the total mass
contributed by each element in a compound and it given by;
Percentage composition by mass =
mass of each element
mass of a compound
× 100
Example 5.3.1
What is the percentage composition of the following compounds?
(a) NO2
(b) C6H12O6
(a) Percentage composition of NO2
First we need to calculate the total relative atomic mass for each element
in the compound and then we find the relative molecular mass of the
compound by adding the masses of all the elements in the compound.
Total Relative atomic mass of N = 1 × 14.01 = 14.01
Total Relative atomic mass of O = 2 × 16.00 = 32.00
Relative molecular mass of NO2 =14.01 + 32.00 = 46.01
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CHE110: Introduction to Chemistry
Percentage of N in NO2 =
=
Percentage of O in NO2 =
=
mass of N
mass of NO 2
14.01
46.01
× 100 = 30.4%
mass of O
mass of NO 2
32.00
46.01
× 100
× 100
× 100 = 69.6%
This means that for any mass of NO2, 30.4% of that mass will be the
mass N and 69.6% will be the mass of O. If you are given 100g of NO2,
it means that, 30.4g will be the mass of NO2 and 69.6g will be the mass
of N.
Activity 5.3.1
Calculate the percentage composition of the following compounds
(a) Ca2 ( PO4 ) 2 …………………………………………………………...
……………………………………………………………………….
……………………………………………………………………….
……………………………………………………………………….
……………………………………………………………………….
……………………………………………………………………….
(b) CH 3COCH 3 …………………………………………………………
………………………………………………………………………..
………………………………………………………………………..
………………………………………………………………………..
………………………………………………………………………..
………………………………………………………………………...
(c) K 2 SO4 ……………………………………………………………….
………………………………………………………………………..
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CHE110: Introduction to Chemistry
……………………………………………………………………….
……………………………………………………………………….
………………………………………………………………………..
……………………………………………………………………….
………………………………………………………………………
(d) Al2 ( SO4 )3 .24 H 2O ……………………………………………………
………………………………………………………………………...
…………………………………………………………………………
………………………………………………………………………….
………………………………………………………………………….
………………………………………………………………………….
………………………………………………………………………….
Thus the percentage composition of a compound is used to determine the
mass of each element present in the compound. Since now you know how
the mass of each element present in a substance can be determined let us
now discuss how the masses of each element in a compound are used in
the calculation of empirical and molecular formulae of a compound in
the next section
5. 4 Calculation of Empirical and Molecular formulae
To determine the empirical formula, masses of elements present in a
compound are converted to number of moles. Then the simplest ratio of
the number of atoms of each element is obtained by dividing the moles of
each element in a compound by the least number of moles in that
compound.
Example 5.4.1
Arsenic (As) was made to react with chlorine gas (Cl2) to form 5.342 g
arsenic chloride. Calculate the empirical formular of the Chloride if it
contained 29.71% arsenic.
The empirical formular of the compound can be determined by either;
(a) using the percentages or (b) the actual masses of each element present
72
CHE110: Introduction to Chemistry
in the given mass of the product. .
(a) Using the percentages
Percentage of As in the compound = 29.71%
Percentage of Cl in the compound
100% (compound) – 29.71% (As) = 70.29% (Cl)
It follows then that in 100g of arsenic chloride we have 70.29 g of Cl and
29.71 g of As.
Mass of As
Moles of As in the compound =
Molar mass of As
=
Moles of Cl in the compound =
=
We now take the mole ratio:
29.71g
74.92g/mol
=0.3966 mol
Mass of Cl
Molar mass of Cl
70.29 g
35.45 g /mol
As
:
= 1.983 mol
Cl
0.3966 mol : 1.983 mol
Arsenic (As) has the least number of moles, 0.3966 mol, so we will
divide he least number of moles through out in order to get the simplest
ratio in which As and Cl combined, hence giving has the empirical
formular
As
:
Cl
0.3966 mol
1.983 mol
:
0.3966 mol
0.3966 mol
1.000
:
4. 999
1
:
5
Thus the empirical formula for arsenic chloride is AsCl5
(b) Using the actual masses of each element present the given mass of the
compound. .
Mass of Cl in arsenic chloride is the 70.29% of the 5.342 g of the product
Mass of Cl =
70.29
100
× 5.342 g = 3.755g Cl
Mass of As in arsenic chloride is the 29.71% of the 5.342 g of the product
Mass of As =
29.71
100
× 5.342 g = 1.587 g As
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CHE110: Introduction to Chemistry
Or 5.342 g (mass of compound) – 3. 755 g Cl = 1.587 g As
Mole ratio of
As
:
Mass of As
Molar mass of As
1.587 g
74.92g/mol
Cl
Mass of Cl
:
Molar mass of Cl
3.755 g
:
35.45 g /mol
0.02118 mol : 0.1059 mol
We divide by the least number of moles (0.02118 mol)
0.02118 mol
0.02118 mol
:
0.1059 mol
0.02118 mol
1.000
:
5.001
1
:
5
Thus the empirical formula for arsenic chloride is AsCl5
That in a compound, the ratio of the number of atoms of each
element is equal to the ratio of the number of moles of atoms of each
element.
Activity 5.4.1
Determine the simplest formulae for compounds with the following
composition.
(a) 62.05 C, 10.4% H and the remainder O
……………………………………………………………………………
……………………………………………………………………………
……………………………………………………………………………..
……………………………………………………………………………..
……………………………………………………………………………..
……………………………………………………………………………..
…………………………………………………………………………….
…………………………………………………………………………….
……………………………………………………………………………..
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CHE110: Introduction to Chemistry
(b) 26.57% K, 35.36% Cr and 38.07% O.
……………………………………………………………………………..
……………………………………………………………………………..
……………………………………………………………………………..
……………………………………………………………………………..
……………………………………………………………………………..
……………………………………………………………………………..
……………………………………………………………………………..
Let us now discuss how we can determine the empirical formula of
compounds formed as a result of combustion process.
Combustion analysis: This method is used to find the empirical
formula of a compound by using the masses of the combustion products.
It is especially employed for organic compounds, take for example a
compound containing C, H and O will give CO 2 and H 2 O upon
complete combustion. The masses of CO 2 and H 2 O are used to find
the number of moles of C, H and O atoms in the sample of the compound.
Example 5.4.2
2.00g of an organic compound gave upon complete combustion 4.86 g
CO 2 and 2.03 g H 2 O . The compound contains C, H and O only. What
is the empirical formular of the compound?
To start with we know that:
•
All C atoms from the compound after combustion are in CO 2 .
Thus the mass and the number of moles of C found in CO 2 is
equal to the mass and number of moles of C in the compound
•
All H atoms from the compound after combustion are in
H 2 O and the mass and the number of mole in H 2 O is equal to
the number moles of H in the compound.
•
The O atoms in the combustion products ( CO 2 and H 2 O ) came
from the compound as well as from the oxygen gas used to burn
the compound. It is then wrong to assume that the number of
75
CHE110: Introduction to Chemistry
moles of O atoms in H 2 O and CO 2 is equal to the number of
mole of O in the compound.
Thus we will start by finding the number of moles of C in 4.86g
of CO 2 and the moles H in 2.03g of H 2 O .
Moles of C in CO 2
Mass of CO 2
Moles of CO 2 =
Molar mass of CO 2
=
4.86g
44.01g/mol
= 0.110 mol CO 2
1 molecule of CO 2 contains 1 C atom.
The moles of C = 1 × 0.110 mol = 0.110mol of C. Thus the moles of C
in the compound = 0.110 mol
Moles of H in H 2 O
Moles of H 2 O =
Mass of H 2O
Molar mass of H 2O
=
2.03g
18.02g/mol
= 0.113 mol H 2 O
1 molecule of H 2 O of H 2 O contains 2 atoms of H
The number of moles of H = 2 × 0.113 mol = 0.226 mol H
Mole of O in the Compound
To find the mass of O present in the compound we are going
to use the masses of H and C in the sample of the compound.
Mass of C in the compound
= moles of C × molar mass of C
= 0.110 mol × 12.01g/mol = 1.32 g C
Mass of H in the compound = moles of H × molar mass of H
=0.226 mol × 1.01 g/mol = 0.228 g H
Mass of O in the compound is;
mass of compound - (mass of C + Mass of H )
2.00g – (1.32 g + 0.228 g) = 0.45 g O
Moles of O in the compound =
Mass of O
Molar mass of O
=
0.45g
16.00g/mol
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CHE110: Introduction to Chemistry
= 0.028 mol O
Mole ratio
C
:
H
:
O
0.110 mol : 0.226 mol : 0.028 mol
We divide by 0.028 mol through out
0.110 mol
0.028 mol
0.226 mol
:
0.028 mol
:
0.028 mol
0.028 mol
3.9
:
8.1
:
1.0
4
:
8
:
1
Empirical formula is C 4 H 8 O
If the ratios still gives you decimal fractions which you can not
easily round off to a whole number, for example if the ratio of C : H : O
is 3.48 : 3 : 1. it will be wrong to round off the 3.48 to 3.5 or 3. You
should round off 3.48 to 3.5 and then multiply all the ratios by 2 so that
the mole ratio becomes 7: 6: 2
Activity 5.4.2
Combustion analysis of 1.00g of an organic compound yields 2.75g
CO 2 . The compound contains C and H only calculate the empirical
formula of this compound………………………………………………...
……………………………………………………………………………..
……………………………………………………………………………..
……………………………………………………………………………..
……………………………………………………………………………..
……………………………………………………………………………..
……………………………………………………………………………..
……………………………………………………………………………..
……………………………………………………………………………..
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CHE110: Introduction to Chemistry
Molecular formular: The molecular formula is determined by
first calculating the factor n and it is given by;
n =
Molar mass of the Molecular formular
Molar mass of the Emperical formula
To find the molecular formula the mole ratios of the empirical formula
are multiplied n
Molecular Formular = n ( Emperical formular)
(n is a whole number)
Example 5.4.3
The empirical formular of an organic compound was determined to
be C 4 H 8 O . The molar mass of its molecular formular was found to be
144.22 g/mol. What is its molecular formular?
First we calculate the value of n
n
=
Molar mass of the Molecular formular
Molar mass of the Emperical formula
144.22g/mol
=
Molar mass of the C 4 H8O
The molar mass of C 4 H 8 O = (12.01 × 4) +(1.01 × 8)+ 16.00
= 72.12 g/mol
n
=
144.22 g/mol
72.12 g/mol
=2
Molecular formula = 2 (Emperical formula)
= 2 ( C4 H8O )
= C4 × 2 H 8 × 2 O1 × 2
= C8 H16 O 2
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CHE110: Introduction to Chemistry
Activity 5.4.3
Chemical analysis shows that glucose is 40.0% C, 6.71% H and
53.3% O by mass if the molar mass of glucose is 180.2g/mol, what
is its molecular formula…………………………………………………..
……………………………………………………………………………..
……………………………………………………………………………..
……………………………………………………………………………..
……………………………………………………………………………..
……………………………………………………………………………..
……………………………………………………………………………..
……………………………………………………………………………..
……………………………………………………………………………..
If we know the formula of a compound and the percentage composition
we can calculate the rative atomic mass of the unknown element in the
compound.
Example 5.4.4
A metal chloride with the formular MCl3 contains 67.2% Cl. What is the
atomic mass of the metal M?
By know you understand that in 100 g of MCl3 contains 67.2 g Cl
Thus mass of M 100 g MCl3 – 67.2 g Cl = 32.8 g
Moles of Cl =
Moles of M =
Mass of Cl
Molar mass of Cl
Mass of M
Molar mass of M
=
=
67.2 g
35.45 g/mol
32.8 g
Mole ratio according to the formula
MM
= 1.90 mol
 32.8 
 mol
 MM 
=
M : C
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CHE110: Introduction to Chemistry
1
: 3
 32.8 

 mol : 1.90 mol
 MM 
 32.8 
 mol × 3 = 1 × 1.90 mol
 MM 
The molar mass (MM) of M = 
98.4
MM
mol = 1.90 mol
MM =
98.4 mol
1.90 mol
= 51.79 g/ mol
Thus the relative atomic mass for element M is 51.79 a. m. u
Let us now discuss how the mole concept is used in the expression of
solutions.
5. 5 Molarity
There are so many ways in which the quantity of a solute in a given
quantity of solvent or solution may be expressed. One of the most
useful methods of expressing the composition of a solution is by
specifying the molarity or the concentration of the solution. The molarity
(M) of a solution is expressed as number of moles of a solute per
decimetre cubed (1Litre). When we are making a solution we first start by
determining the mass (in grammes) of substance that we want to dissolve
in a given volume which could be centimetre cubed (cm3) or decimetre
cubed (dm3). The mass of the solute should expressed in the number of
moles (mol) and the volume in decimetre cubed (dm3)
Molarity M =
moles of solute (mol)
volume of the solution (dm3 )
= mol/dm3 or mol.dm -3
Example 5.5.1
A 2.00g sample of sodium hydroxide (NaOH) was dissolved in distilled
water to produce a total volume of 200 cm3 of solution . What is the
molarity (Concentration) of this NaOH solution?
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CHE110: Introduction to Chemistry
molarity M =
moles of solute (mol)
volume of the solution (dm3 )
= mol/dm3
We start by converting 2.00g NaOH to moles
Moles of NaOH =
Mass of NaOH
Molar mass of NaOH
=
2.00g
40.00g /mol
= 0.05 mol
We then express Volume in dm3. Remember that 1000 cm3 = 1dm3
200cm3 ×
1dm3
1000 cm3
(the conversion factor) = 0.2 dm3
The concentration of NaOH solution =
0.05 mol
= 0.25 mol/ dm3
3
0.2 dm
or 0.25 M NaOH
This solution may be described in so many ways:
•
The solution is 0.25 M NaOH
•
The Molarity (concentration) of NaOH is 0.25 M
•
The molarity of the solution with respect to NaOH is
0.25 mol/dm3
The formular of the solute is essential in the description of the
solution molarity
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CHE110: Introduction to Chemistry
Activity 5.5 .1
Calculate the concentration of 49.0g of H 3 PO4 dissolved in distilled
water to make a solution of 250 cm3 ………………………………….
………………………………………………………………………….
…………………………………………………………………………..
……………………………………………………………………………
……………………………………………………………………………
……………………………………………………………………………
…………………………………………………………………………….
Let us look at how mass percentage of a solute and the density of the
solutions are used in the expression of molarity
Mass percentage, density and molarity
Mass percentage of a solute in the solution is given by
Mass of solute
Mass of solute + mass of solvent (mass of solution)
× 100
Molarity and percentage by mass are the most common ways of
expressing the composition of a solution. Molarity has moles of
solute while mass percentage has mass of a solute in the numerator.
Molarity has volume of solution in the denominator while mass
percentage has mass of the solution in the denominator. To
calculate one of these from the other , we need to know the density
(g/cm3) of the solution at a specified temperature.
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CHE110: Introduction to Chemistry
Example 5.5.2
A concentrated solution of HCl contains 37.8% HCl by mass and has a
density of 1.19 g/cm3 at 20 0 C . What is the molarity of this solution?
Density tells that 1 cm3 of the solution has a mass of 1.19g . What about
1dm3(1000cm3)?
Mass of 1 dm3 0f this solution = 1.19g/cm3 × 1000 cm3 = 1190 g
We know that 37.8 % of the mass of this solution is HCl
Thus mass of HCl in 1dcm 3 =
Moles of HCl =
37.8
100
Mass of HCl
Molar mass of HCl
The molarity of HCl solution =
× 1190 g = 449.82 g HCl
=
449.82 g
36.46 g / mol
= 12.3 mol
Moles of HCl
volume of the solution
=
12.3 mol
1dm3
= 12.3mol/ dm3
Activity 5. 5.2
Calculate the molarity of H 2SO 4 solution containing 93.2 % H 2SO 4 by
3
mass having a density of 1.83 g / cm at 20 0C .
……………………………………………………………………………..
……………………………………………………………………………..
……………………………………………………………………………..
……………………………………………………………………………..
……………………………………………………………………………..
……………………………………………………………………………..
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CHE110: Introduction to Chemistry
……………………………………………………………………………..
The HCl whose molarity we determined in example 4.5.2 is a very
Concentrated acid and normally we don’t usually use very
concentrated acids in laboratory experiments. We dilute concentrated
acids (solutions) just like the we dilute the concentrated juice we by from
shops. Now let discuss how this done.
Dilution: This is a process of converting a concentrated solution to
more diluted solution by adding a solvent. Adding the
solvent increases the volume of the solution but does not change
the quantity of the solute. It means that the number of moles of a
solute is the same before and after dilution.
Moles of solute (Concentrated solution) = Moles of solute (Diluted solution)
Number of moles of a solute in a solution is given by:
Moles of solute = molarity (M) × volume (V) of the solution
=M × V
Let us substitute this formula in the above dilution principle
M concentrated × Vconcetrated = M diluted × Vdiluted
M c × Vc = M d × Vd
The volume in this dilution equation can be in any units as long as
they are the same on both sides
Example 5.5.3
You are given a 6.00M HCl solution and asked to prepare 500 cm3
of a 1.00 M solution. What volume of 6.00 M HCl are you going to use?
M c × Vc = M d × Vd
From the question you are given
M c = 6.00 M Vd = 500 cm3 M d = 1.00 M
You need to find the volume of the concentrated ( Vc ) acid which
you are going to dilute. You need to make Vc the subject of the
formula
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CHE110: Introduction to Chemistry
V × Md
Vc = d
Mc
=
500 cm3 × 1.00M
6.00M
= 83.3cm3
Thus you need to take 83.3 cm3 of the concentrated acid
(6.00 M HCl) and dilute it to a total volume of 500 cm3
That practically it dangerous to add water to concentrated acid
instead we add acid to water. To dilute the above solution you
need to get 500 cm3 measuring cylinder and add enough distilled water
of about 200 cm3 then add 83.3 cm3 of the concentrated acid and
then add distilled water up to 500 cm3 mark.
Activity 5.5.3
A 2.00M aqueous solution of AgNO3 is available. You dilute this
20.0 cm3 of this solution with distilled water up to the mark of a
250 cm3 volumetric flask. What is the molarity of the solution you
have just made?..................................................................................
………………………………………………………………………
………………………………………………………………………
………………………………………………………………………
………………………………………………………………………
………………………………………………………………………
.
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CHE110: Introduction to Chemistry
Unit summary
In this unit we looked at the mole concept which is the measure of the
amount of substance. A mole is the amount of any substance that
Summary
contains 6.02 × 10 23 particles. The mole concept helps us to express and
calculate the following quantities of substances; mass, volume and
concentrations of substances. It also used to determine the empirical and
molecular for substances using the percentage composition of substances.
86
CHE110: Introduction to Chemistry
Assessment
Answer the following questions in the spaces provided
Assessment
1. Copper is obtained from ores containing the following minerals
Azurite, Cu3 (CO3 ) 2 (OH )2 , Chalcopyrite, CuFeS 2 and Cuprite, Cu2O
Which mineral has the highest copper content on a mass percentage
basis?.........................................................................................................
……………………………………………………………………………..
……………………………………………………………………………
……………………………………………………………………………
…………………………………………………………………………….
……………………………………………………………………………..
3
2. What mass of CuSO 4 .5H 2 O would be required to make 400 cm of a
0.250 M solution?.....................................................................................
…………………………………………………………………………….
…………………………………………………………………………….
……………………………………………………………………………..
……………………………………………………………………………..
…………………………………………………………………………….
……………………………………………………………………………..
Note that the answers to the above exercise are given at the end of the
module. However, you are encouraged to work through first before
referring to the answers.
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CHE110: Introduction to Chemistry
Unit 6
Stoichiometry
6.0 Introduction
Welcome to this unit. In the previous unit we looked at the mole
concept and now will discuss how the mole concept is used in
stoichiomentry. This is the part of chemistry that deals with
calculations relating the amounts of reactants and products in
chemical reactions using a balanced chemical equation.
Upon completion of this unit you will be able to:
Balance chemical equations.
Calculate the amounts of reactants and products in a chemical
Outcomes
reaction using a balanced chemical equation.
Determine limiting reactant in a chemical reaction to calculate
the amount of the products
Determine the percentage yield of a chemical reaction.
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CHE110: Introduction to Chemistry
6.1Chemicl Equation
A chemical equation is a short hand method of describing a
chemical
change. Below is a general representation of a chemical equation.
aA (S) + bB(aq) → cC(l) + dD (g)
The chemical equation shows the following information about a
chemical reaction;
•
The reactants and the products in a chemical change
A and B are reactants while C and D are products
•
The physical state of reactants and products: (S) for solid,
(aq)
for aqueous, (l) for liquid and (g) for gaseous state.
•
The ratios in which reactants reacted and the ratios in which
products were produced represented by a, b, c and d
During a chemical change no substances are either lost or gained,
before
and after the reaction. That is, atoms are not created or destroyed in
chemical changes but merely rearranged. This is the reason why we
need
to balance chemical equations so that the number of atoms of each
element in the reactants A and B are the same as the number of
atoms of
the same elements in the products C and D. Balancing of chemical
equations is done by placing the numbers represented by a, b, c and
d,
before each correct formular.
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CHE110: Introduction to Chemistry
Example 6.1.1
Balance the following chemical eqaution
C (s) + O 2(g) → CO( g )
This equation is not balanced because there 2 atoms of O in the reactant
but there is only 1 atom of O in the product.
To balance the equation we put 2 in front of CO and a 2 in front of C and
the equation becomes
2C (s) + O 2(g) → 2CO( g )
A number before a formular such as 2 before C and CO in known as
coefficient. A coefficient is a multiplier for the entire formular and it
represents the mole ratios of reactants and reactant to products. This will
be discussed in detail in the next section
When balancing the chemical equations never change the
subscripts in the formulae of reactants and products because it changes
the chemical identity of the substances. But changing the coefficient
changes the quantity of the substance.
Activity 6.1.1
Balance the following chemical equation
(a) Ba(OH) 2 (aq) + HCl(aq) → BaCl2 (aq) + H 2 O(l)
--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
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CHE110: Introduction to Chemistry
6. 2 Molar interpretation of a balanced chemical equation
A balanced chemical equation reveals much more than the observed
chemical change. It is also a statement of the relative numbers of moles
of reactants and products involved in a chemical reaction. Let us consider
the reaction between propane and oxygen to give carbon dioxide and
water.
C3 H 8 (g) + 5O 2(g) → 3CO 2( g ) + 4H 2 O (g)
The above balanced equation is telling us that 5 oxygen molecules
reacted with 1 propane molecule, producing 3 carbon dioxide molecules
and 4 water molecules.
The coefficients in a balanced equation give the relative number of
molecules of each reactant and the relative number of molecules of each
product. This statement may be summarised as follows;
C3 H 8 (g)
+
5O 2(g) →
(1 mol C3H8 ) ( 5 mol O 2 )
 6.02 × 10 23 C3H8   5 × 6.02 × 10 23 

 

 molecules
  O 2 molecules 
( 44.08 g C3H8 )
(5 × 32.0 g O 2 )
3CO 2( g )
+
4H 2 O (g)
(3 mol CO 2 ) ( 4 mol H 2 O )
 3 × 6.02 × 10 23   4 × 6.02 × 10 23 



 CO 2 molecules   H 2 O molecules 
(3 × 44.0g CO 2 ) ( 4
× 18.02g H 2 O
)
The quantities under each formular represent the same amount of
substance expressed in different ways. For example 5 mols of oxygen is
the same as 5 x 6.02 × 10 23 molecules and 5 x 32.0g of oxygen
Thus we may choose any mole quantity to express the amount of
substances undergoing chemical change. However it more convenient to
work with the mole ratios.
Balanced equations with fractional coefficients such as;
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CHE110: Introduction to Chemistry
H 2(g) +
1
2
O 2(g) → H 2 O (l) is read as 1 mol of H 2 combines with
1
mole of O 2 (not
2
1
2
1
2
2
molecule) to form 1 mol of H 2 O . But it is
incorrect to write O in place of
from O2 .But
1
O2 . The substance O is different
O2 and 3 O2 are not different substances from O2 .
We cannot write O for
1
2
O2 or
1
2
O2 for O, or O6
for 3 O 2
Balanced chemical equations make it possible for us to calculate
quantities of substances in a chemical reaction when the quantity of one
substance is known.
Example 6.2.1
How many moles of O 2 are required to burn 3.50 mol of liquefied
petroleum gas, propane, C3 H8
Step1: write a balanced chemical equation for the reaction
C3 H 8 (g) + 5O 2(g) → 3CO 2( g ) + 4H 2 O (g)
Step 2: Find the relationship between moles of the substance whose
quantity is given and the substance whose quantity you want to find:
Quantity given 3.50 mol C3 H8
Quantity sought: moles of O 2
Mole ratio C3 H8 to O 2 from the equation:
C3 H 8 : O 2
1
:
5
3.50 mol : moles of O 2
Moles of O 2 = 5 × 3.50 mol = 17.50 mol
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CHE110: Introduction to Chemistry
Example 6.2.2
Hydrogen is produced in large quantities by electrolysis of water for use as
rocket fuel. Find the mass in grams of hydrogen produced from 50.0g of
H 2O
Step 1: Write a balanced equation for the reaction
2H 2 O (l) → 2H 2( g ) + O 2(g)
Quantity given: 50.0g of H 2 O
Quantity sought: mass of H 2
Step 2: Convert the quantity given, 50.0g of H 2 O to the number of moles
Number of moles of H 2 O =
mass of H 2O
molar mass of H 2O
=
50.0g
18.02g/mol
= 2.77 mol
Step3: Find the moles of H 2 using the mole ratio from the equation
Mole ratio H 2 O :
1
H2
:
1
2.77 mol : moles of H 2
Moles of H 2 =
1 × 2.77 mol = 2.77 mol H 2
Step 4: Convert moles of H 2 to mass (quantity being sought)
Mass of H 2 = moles × molar mass
= 2.77 mol × 2.02 g/mol = 5.61 g
Example 6.2.3
45.50cm3 of 0.100 M NaOH reacts with 50.01cm3 of H 2SO 4 solution
Find the molarity of H 2SO 4 solution
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CHE110: Introduction to Chemistry
Step 1: Write a balanced chemical equation for this reaction
2NaOH (aq) + H 2SO 4(aq) → Na 2SO 4(aq) + 2H 2 O (l)
Quantity given: 45.50cm3 of 0.100 M NaOH
Quantity sought: molarity of 50.01cm3 of H 2SO 4 solution
Step2: Find the number of moles of 45.00 cm3 of 0.100M NaOH.
Moles of NaOH = Molarity × Volume
= 0.100 mol/dm3 × 0.0455dm3
= 4.55×10-3 mol NaOH
Step 3: Find the moles of H 2SO 4 using the mole ratio
NaOH : H 2SO 4
2
:
4.55×10-3 mol :
Moles of H 2SO 4 =
1
Moles of H 2SO 4
1 × 4.55×10-3mol
2
= 2.275 × 10−3 mol
Step 4: Find the molarity of 50.01cm3 of H 2SO 4 solution
Molarity =
=
Number of moles of H 2SO 4
Volume of the H 2SO 4 solution
2.275×10-3mol
= 0.045 mol/dm3
3
0.0501dm
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CHE110: Introduction to Chemistry
Activity 6.2.1
Calcium carbonate when heated decomposes according to the equation
CaCO3 ( s ) → CaO( s ) + CO2 ( g )
How many grams of CaCO3 must be decomposed to give 3.26 × 10 23
molecules of CO2 ?......................................................................................
……………………………………………………………………………..
……………………………………………………………………………..
……………………………………………………………………………..
…………………………………………………………………………….
Let us now discuss in the next section how we can determine the amount
of products produced in a chemical reaction when the quantities of all the
reactants are given
6. 3 Limiting Reactants
A limiting reactant is the one which limits the amount of products that
can be obtained; the other reactant or reactants are in excess. Thus when
you are given the amount of two or more reactants in a chemical reaction
you must identify the limiting reactant. The limiting reactant is totally
consumed in a chemical reaction. The reactant which is in excess is not
completely consumed.
For example if you are given 100 car bodies and 360 wheels, how many
cars are you going to assemble? And which one is going to determine the
number of cars you are going to assemble?
One car needs 4 wheels, so from 360 wheels you can only produce 90
cars. It means from 100 bodies, 10 bodies won’t be used. It means that
the number of wheels are the limiting factor, that is they determined the
number of cars to be produced while the number of car bodies were in
excess. Now let us now discuss the example below
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CHE110: Introduction to Chemistry
Example 6.3.1
Suppose we have 4 moles of Na and 3 moles of Cl and you wish to make
NaCl. The “recipe” for this reaction is the balanced chemical equation
+ Cl 2(g) → 2NaCl(s)
2Na (s)
It tells us through mole ratios the amount of products that can be obtained
from a given amount of reactants
Cl2 : NaCl
Mole ratios: Na : NaCl
1
: 1
1 :
4 mol : 4mol
2
3 mols : 6 mols
These mole ratios show that if 4 moles of Na react completely 4 moles of
NaCl will be produced. And if 3 moles of Cl2 react completely 6 moles
of NaCl will be produced.
It means that maximum number of moles that will be produced in this
reaction is 4 moles of NaCl and this will be determined by the 4 moles of
Na available. Sodium (Na) is the limiting reactant because it produces the
lesser amount of products. The Cl is in excess only 2 moles 0f Cl2 will
be consumed to produce 4 moles NaCl. Thus 1 mole of Cl will not be
consumed.
Initially
Reacted /formed
4 moles Na
- 4 mol reacted
3 moles Cl2
- 2 reacted
0 moles NaCl
+4 formed
Finally
0 mole Na
1 mole Cl
4 moles NaCl
The limiting reactants is always identified by working in moles
General methods of determining the limiting reactant
1. Convert the amounts of reactants given to moles
2. Calculate the molar amounts of products obtained based on
each reactant
3. Choose the reactant which produces the least amount of
products as a limiting reactant
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CHE110: Introduction to Chemistry
Example 6.3.2
A 50.0g sample of Mg(OH)2 is mixed with 70.0g of H2 PO4 . A
neutralised reaction takes place which is represented by the equation
3Mg (OH ) 2 + 2 H 3PO4 → Mg3 ( PO4 )2 + 6 H 2O
How many grams of Mg3 ( PO4 ) 2 are produced?
Step1: Convert the amounts of reactants given to moles
Mass of Mg(OH) 2
50.0g
=
Molar mass of Mg(OH) 2 58.3g/mol
Mole of Mg (OH ) 2 reacted =
0.858 mol Mg (OH ) 2
=
Moles of H 3 PO 4 reacted =
Mass of H3PO 4
Molar mass of H 2 PO 4
=
70.0g
98.0g/mol
0.71 mol H 3 PO 4
=
Step2: Calculate the molar amounts of products obtained based on each
reactant
Mole of Mg3 ( PO4 ) 2 produced from 0.858 mole of Mg (OH ) 2
Mole ratio
Mg (OH ) 2
3
:
:
0.858 mol
Moles of Mg3 ( PO4 ) 2 =
Mg3 ( PO4 ) 2
1
moles of Mg3 ( PO4 ) 2
0.858 mol
= 0.286 mol
3
Moles of Mg3 ( PO4 ) 2 produced from 0.71moles H 3 PO 4
Mole ratio
H 3 PO 4
2
0.71 mol
Moles of Mg3 ( PO4 ) 2 =
:
Mg3 ( PO4 ) 2
:
1
:
moles of Mg3 ( PO4 ) 2
0. 71 mol
= 0.357 mol
2
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CHE110: Introduction to Chemistry
Step3: Choose the reactant which produces the least amount of products
as a limiting reactant
0.858 mol Mg (OH ) 2 → 0.286 mol Mg3 ( PO4 ) 2
0.71 mol H 3 PO 4 → 0.357 mol Mg3 ( PO4 ) 2
Mg (OH ) 2 produced the least amount of product thus it is the limiting
reactant. 0.286 moles Mg3 ( PO4 ) 2 were produced in this reaction
The mass of Mg3 ( PO4 ) 2 produced = moles × molar mass
= 0.286 mol × 262.9 g/mol
= 75.18 g Mg3 ( PO4 ) 2
To calculate the excess mass of H 3 PO 4 we can take the mole ratio
between H 3 PO 4 and Mg (OH ) 2 to know the number of H 3 PO 4 that
reacted with 0.858 moles of Mg (OH ) 2 .
Mole ratio:
Mg (OH ) 2
3
0.858 mol
Moles of H 3 PO 4 reacted =
H 3 PO 4
:
:
2
Moles of H 3 PO 4 reacted
:
2 × 0.858 mol
3
= 0.572 mol
Excess number of moles = moles of H 3 PO 4 available – Moles of
H 3 PO 4 reacted = 0.71mol - 0.572mol = 0.14 moles
Thus mass of H 3 PO 4 in excess = 0.14mols × molar mass of H 3 PO 4
= 0.14 mol × 98.0 g/mol
=
13 .72 g of H 3 PO 4 did not react.
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CHE110: Introduction to Chemistry
Activity 6.3.1
12.0g of Zn reacted with 6.5g of sulphur to form zinc sulphide.
(a) Which is the limiting reactant?
(b) How many grams of ZnS can be formed from this particular reaction?
(c) How many grams of the excess element remained unreacted?
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When we carry out an experiment we don’t usually obtain the exact
amount the products we expected theoretically, the actual amount we
obtain is usually less than the expected. Thus there is need to compare the
two amounts and express this comparison in terms a percentage. This is
discussed in detail in the next section.
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CHE110: Introduction to Chemistry
6.4 Percentage yield
The percentage yield is the measure of the difference between the actual
(practical) yield and the theoretical yield.
Theoretical yield is the maximum amount of products that can be
obtained based on the amount of reactants given in a balanced chemical
equation. These amounts of products are calculated from the specified
amounts of reactants using a chemical equation are theoretical.
They are not the actual amounts we can only obtain the actual amount
when we perform the experiment.
Actual yield is the actual amount of products obtained by performing a
chemical reaction. The amounts obtained by practical means is less than
the theoretical yield, this is because some products are generally “lost”
This apparent lose does not mean that the law of conservation of matter is
violated, the lose of products are due to the following; non completion of
the reaction, some products may remain in the solution and some
materials may be left behind when a chemist is transferring materials
from one container to another. The actual yield can only be determine
experimentally. The percentage yield is given by:
Percentage yield =
Actual yield
Theoratical yield
× 100
Example 6.4.1
Aluminium fluoride ( AlF3 ) is obtained from the reaction of 13.49 g of
aluminium in the reaction;
2Al(s) + 3F2(g) → 2AlF3(s)
Calculate the percentage yield if 40.0g of AlF3 was actually produced.
Percentage yield =
Actual yield
Theoratical yield
× 100
Quantity given = 13.49 g of Al
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CHE110: Introduction to Chemistry
Quantity sought:
mass of AlF3 (theoretical yield) =?
Actual yield = 40.0g AlF3
Let us now calculate the theoretical mass of AlF3 that can be obtained
from 13.49 g of Al
Moles of Al reacted =
mass of Al
molar mass of Al
Moles of AlF3 produced
Al
:
1
:
0.5 mol :
=
13.49g
26.98g/mol
= 0.5 mol Al
AlF3
1 (this the same as 2 : 2)
moles of AlF3
Moles of AlF3 produced = 1 × 0.5 mol =
0.5mol AlF3
Mass of AlF3 produced = moles of AlF3 × molar mass of AlF3
= 0.5 mol × 83.98 g/ mol = 41.99g
=
Percentage yield
=
=
42.0g
Actual yield
Theoratical yield
40.0 g
42.0g
× 100
× 100 = 95 .2%
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CHE110: Introduction to Chemistry
Activity 6.4.1
How many grams of N 2 F4 can be obtained from the reaction between
NH3 and 14.00g of F2 ? And calculate the percentage yield for this
reaction if 4.80g of N 2 F4 was obtained. Below is the unbalanced for this
reaction:
NH 3 + F2 → N 2 F4 + HF
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CHE110: Introduction to Chemistry
Unit summary
In this unit we have discussed that it important to balance chemical
equations because atoms of reacting elements are not destroyed or
Summary
created, they have to be accounted for after the reaction. A balanced
chemical equation gives us the mole ratios of reactants and products in a
chemical reaction. Thus a balanced chemical equation is like a ‘recipe’ it
helps to calculate the amounts substances that reacted and the products
produced. From a balanced chemical equation we are also able to
determine the reactant(s) that reacted completely (Limiting reactants) and
the reactants that were in excess. The equation helps to calculate the
percentage yield.
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CHE110: Introduction to Chemistry
Assessment
Answer the following questions in the spaces provided
Assessment
1. Calculate the molarity of a HCl solution if 200 cm3 of it
completely neutralizes 300 cm3 of a 0.120 M barium hydroxide
solution……………………………………………………………..
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. 2. When copper is heated with sulphur, Cu2 S is formed.
(a) What mass of Cu2 S could be Obtained by reacting 80g of copper
with 45g of Sulphur?........................................................................
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CHE110: Introduction to Chemistry
(b) Calculate the mass of the excess reactant which did not take part
in a chemical Reaction…………………………………………..
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(c) Determine the amount of Cu2 S produced after carrying
out this chemical reaction if the percentage yield was 85%
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Note that the answers to the above exercise are given at the end of the
module. However, you are encouraged to work through first before
referring to the answers.
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CHE110: Introduction to Chemistry
Answers to Activities
Activity 1.2.1
Student A was more accurate than B because the average was close
to the true value
Student B was more precise because the measurements were close
together than A
Activity 1.4.1
(a) 3 s.f (b) 5 s.f (c) 5 s.f (d) 2 s. f (e) 5 s. f (f) 2 s. f
Activity 1.4.2
(a) 4.635 (b) 6.5 × 101 (c) 6.463 × 101
Activity 1.6.1
(a) 0.025 dm3 (b) 0.67 m/s (c) 2.6 × 10 −3 g / cm3
Activity 3.4.1
Relative molecular mass of the following substance
(a) Oxygen molecule (O2) = 32.00
(b) Water ( H 2O ) = 18.02
(c) Calcium hydroxide Ca (OH ) 2 = 74.10
Activity 4.1.1
1. The percentage of A in the two samples = 25.5% (this demonstrates
the law of constant composition because the percentage
composition by mass of A is constant in the two sample)
2. Mass of oxygen (O) combined with a fixed mass of 1.00g hydrogen
(H) in water ( H 2O ) = 8.00g
Mass of oxygen (O) combined with a fixed mass of 1.00g hydrogen
(H) in hydrogen peroxide ( H 2O2 ) = 16.00g
The ratio of oxygen mass in the two compounds =
mass of O in H 2O
mass of O in H 2O2
=
8.00 g
16.00 g
=
1
2
or 1: 2
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CHE110: Introduction to Chemistry
Activity 5.1.1
2.99 × 10-23 g mass of one water molecule.
Activity 5.2.1
1. 2.15g of oxygen, 7.38 × 10 22 oxygen atoms.
2. (a) 1.85 × 10-22 g (b) 12.49 g
3. 2.16 dm3
Activity 5.3.1
(a) Ca = 29.7%, P = 22.9% and O = 47.4%
(b) H = 10.4%, C = 62.1% and O = 27.5%
(c) K = 44.9% , S = 18.4% and O = 36.7%
(d) Al = 7.0% S = 12.4% H = 6.3% and O = 74.4%
Activity 5.4.1
(a) C3H 6 O (b) K 2 Cr2 O 7
Activity 5.4.2 CH 4
Activity 5.4.3 6(CH 2 O) = C6 H12 O 6
Activity 5.5.1
2.0 mol/dm 3of H 3 PO 4 or 2.0M H 3 PO 4
Activity 5.5.2 17.43 mol/dm 3of H 2SO 4 or 17.43M H 2SO 4
Activity 5.5.3 0.16 mol/dm 3 AgNO 3 or 0.16M AgNO 3
Activity 6.1.1 Ba(OH) 2(aq) + 2HCl(aq) → BCl 2(aq) + 2H 2 O (l)
Activity 6.2.1 54.0g of CaCO3
Activity 6.3.1
(a) Zn is a limiting reactant (b) 17.93g of ZnS (c) 0.61g of S
Activity 6.4.1 7.70g of N 2 F4 , 62.3%
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CHE110: Introduction to Chemistry
Answers to Assessment Questions
Unit 1
1 (a) 4 s.f (b) 1 s.f (c) 2 s.f
2. (a) 1.05 × 10
6
(b) 3.910 × 10
3
(c) 1.04 × 10
−4
7
3. (a) 2.37 × 10 (b) 0.984 (4 s.f)
4. (a) Student A was both precise and accurate with a spread of 0.03 and a
deviation of 0.01
(b) Student A results 4.50 ± 0.01 g / cm3
Student B results 4.48 ± 0.03 g / cm3
Unit 2
1. (a) Water, compound. a combination of Oxygen and Hydrogen
(b) Brass, mixture. a mixture of copper and brass
(c) Chlorine gas, element, existing as a molecule with 2 atoms of
chlorine
(d) Crude oil, mixture, a mixture of the following petroleum products;
Gas, petrol, paraffin, diesel, lubricant oil and heavy oils
(e) Sodium chloride, compound, a combination of sodium and
chlorine
2. (a) A magnet will be used to separate iron fillings from sulphur
Because sulphur is non magnetic material
(b) To obtain water distillation method is used. The mixture is put in
flask connected to a fractionating column which in turn is connected
to a condenser When the mixture is heated the alcohol with a lower
boiling point will condense in the column and drops back in the
flask. Water vapour will proceed and cooled in the condenser. This
is possible because water and alcohol have different boiling points.
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CHE110: Introduction to Chemistry
(c) Copper (II) sulphate crystals can obtained by crystallization by
making a saturated solution and then heat the solution gently to
remove water. Leave the Copper(II) sulphate with very little water
content on a dish for hours for the crystals to form. This is possible
because Cupper(II) sulphate solution is soluble in water
Unit 4
1. Relative atomic masses of the following compounds
(a) Ca ( HSO4 ) 2 = 234.22
(b) Na2CO3 =105.98
(c) CuSO4 .5 H 2O =249.71
2. Cu isotope with 62.930 a.m.u has 69.2% relative abundance
Cu isotope with 64.928 a.m.u has 30.8% relative abundance
3. The ratio of oxygen in the two compound is 1:2 it satisfies the of multiple
Composition
Unit 5
1. Cuprite has the highest percentage of copper, 88.8%.
2. 24.97g of CuSO 4 .5H 2 O will be required
Unit 6
1. (a) 0.36mol/ dm 3of HCl or 0.36M HCl
2. (a) 100.25g of Cu 2S (b) 24.81g of S was in excess
(c) 85.21g of Cu 2S was the actual yield.
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CHE110: Introduction to Chemistry
Readings
There are a number of excellent resources on the web. A few suggested links
are:
http://www.how-to-study.com/
The “How to study” web site is dedicated to study skills resources. You
will find links to study preparation (a list of nine essentials for a good
study place), taking notes, strategies for reading text books, using
reference sources, test anxiety.
http://www.ucc.vt.edu/stdysk/stdyhlp.html
This is the web site of the Virginia Tech, Division of Student Affairs.
You will find links to time scheduling (including a “where does time go?”
link), a study skill checklist, basic concentration techniques, control of the
study environment, note taking, how to read essays for analysis, memory
skills (“remembering”).
http://www.howtostudy.org/resources.php
Another “How to study” web site with useful links to time management,
efficient reading, questioning/listening/observing skills, getting the most
out of doing (“hands-on” learning), memory building, tips for staying
motivated, developing a learning plan.
The above links are our suggestions to start you on your way. At the time of
writing these web links were active. If you want to look for more go to
www.google.com and type “self-study basics”, “self-study tips”, “self-study
skills” or similar.
1. P. Mathews (1992). Advanced Chemistry, Cambridge
University Press,
Important
note : If you
have not yet
done so,
please
review the
2. Brescia F., Arent J.,Meislich H., and Turk A (1987).General
Chemistry, 5th Ed Academic press; New York
3. Atkinson A. (1983). Certificate Chemistry, 4th Ed , Longman; Hong
Kong
4. Geoff Neuss (2001) Chemistry for the IB Diploma : Standard and Higher
Level, Oxford University Press
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