Process calculations

Uploaded by:
Ebooks Chemical Engineering


For More Books, softwares & tutorials Related to Chemical Engineering
Join Us

@facebook:
Engineering/238197077030
@facebook:
@facebook:

ADMIN:
I.W







<< If you like this Book, than support the author and BuY it >>
PROCESS CALCULATIONS
PROCESS CALCULATIONS
SECOND EDITION
V. VENKATARAMANI Formerly Professor
Department of Chemical Engineering National Institute of Technology Tiruchirappalli
N. ANANTHARAMAN Professor
Department of Chemical Engineering National Institute of Technology Tiruchirappalli
K.M. MEERA SHERIFFA BEGUM Associate Professor

Department of Chemical Engineering National Institute of Technology Tiruchirappalli
New Delhi-110001 2011

PROCESS CALCULATIONS, Second Edition
V. Venkataramani, N. Anantharaman and K.M. Meera Sheriffa Begum
© 2011 by PHI Learning Private Limited, New Delhi. All rights reserved. No part of this book may be reproduced in any form, by
mimeograph or any other means, without permission in writing from the publisher.
ISBN-978-81-203-4199-9
The export rights of this book are vested solely with the publisher.
Second Printing (Second Edition) … … February, 2011
Published by Asoke K. Ghosh, PHI Learning Private Limited, M-97, Connaught Circus, New Delhi-110001 and Printed by Meenakshi
Art Printers, Delhi-110006.

To My Parents
— V. Venkataramani
— K.M. Meera Sheriffa Begum
To My Mother
— N. Anantharaman
Contents
Preface xi Preface to the First
Edition xiii Acknowledgements
xv
1 UNITS AND DIMENSIONS 1–6
1.1 Introduction 1
1.2 Basic Units and Notations 1
1.3 Derived Units 2
1.4 Definitions 3
Worked Examples 3
Exercises 6
2 MASS RELATIONS 7–34

2.1 Mass Relations in Chemical Reaction 7
2.2 Conservation of Mass 8
2.3 Avogadro’s Hypothesis 9
2.4 Limiting Reactant and Excess Reactant 9
2.5 Conversion and Yield 9
2.6 Composition of Mixtures and Solutions 10
2.6.1 Weight Percent 10
2.6.2 Volume Percent 10
2.6.3 Mole Fraction and Mole Percent 11
2.6.4 Atomic Fraction and Atomic Percent 11
2.6.5 Composition of Liquid Systems 11
2.7 Density and Specific Gravity 12
2.7.1 Baume’ (°Be’) Gravity Scale 12
2.7.2 API Scale (American Petroleum Institute) 12
2.7.3 Twaddell Scale 13
2.7.4 Brix Scale 13
Worked Examples 13
Exercises 32
vii
viii
CONTENTS
3 IDEAL GASES 35–73
3.1 Relation between Mass and Volume for Gaseous Substances 35
3.1.1 Standard Conditions 35
3.1.2 Ideal Gas Law 35
3.2 Gaseous Mixture 36
3.2.1 Partial Pressure (PP) 36
3.2.2 Pure Component Volume (PCV) 36
3.2.3 Dalton’s Law 37
3.2.4 Amagat’s Law (or) Leduc’s Law 37

3.3 Average Molecular weight 38
3.4 Density of Mixture 38
Worked Examples 38
Exercises 70
4 VAPOUR PRESSURE 74–86
4.1 Effect of Temperature on Vapour Pressure 74
4.2 Hausbrand Chart 75
Worked Examples 75
Exercises 85
5 PSYCHROMETRY 87–110
5.1 Humidity 87
5.2 Definitions 87
Worked Examples 90
Exercises 106
6 CRYSTALLIZATION 111–121 Worked Examples
112 Exercises
120
7 MASS BALANCE 122–179
Worked Examples 122
Exercises 173
8 RECYCLE AND BYPASS 180–195
8.1 Recycle 180
8.2 Bypass 180
8.3 Purge 180
Worked Examples 181
Exercises 195
CONTENTS ix

9 ENERGY BALANCE 196–220
9.1 Definitions 196

9.1.1 Standard State 196
9.1.2 Heat of Formation 196
9.1.3 Heat of Combustion 197
9.1.4 The Heat of Reaction 197
9.1.5 Heat of Mixing 197
9.2 Hess’s Law 197
9.3 Kopp’s Rule 198
9.4 Adiabatic Reaction Temperature 198
9.5 Theoretical Flame Temperature 198
Worked Examples 198
Exercises 217
10 PROBLEMS ON UNSTEADY STATE OPERATIONS 221–228 Worked Examples
221 Exercises
227
T ables 229–234
I Important Conversion Factors 229
II Atomic Weights and Atomic Numbers of Elements 231
III(a) Empirical Constants for Molal Heat Capacities of Gases at Constant Pressure
234
III(b)Molal Heat Capacities of Hydrocarbon Gases 234
Answers to Exercises 235–245
Index 247–248
Preface
The objective of this book is to enrich a budding chemical engineer the techniques involved in
analyzing a process plant by introducing the concepts on units and conversions, mass and energy
balances. This will enable him to achieve a proper design of process equipment. An attempt has been
made to explain the principles involved through numerical examples. The problems are not only
confined to SI system of units but also worked out in other systems like FPS, CGS and MKS systems.
We feel that our attempt will be more rewarding if students come across data presented in FPS, CGS
and MKS systems while designing equipment, since different reference books give standard values

and data in various units.
The book covers various interesting topics such as units and dimensions, mass relations, properties of
gases, vapour pressure, psychrometry, crystallization, mass balance including recycle and bypass,
energy balance and unsteady state operations. The second edition is now enriched with additional
worked examples and exercises to give additional exposure and practice to students.
The text is designed for a one semester programme as a four credit course and takes care of the
syllabus on ‘Process Calculations’ of most of the universities in India.
V. Venkataramani N. Anantharaman K.M. Meera Sheriffa Begum
xi
Preface to the First Edition
Chemical engineers in process industries generally need to focus on design, operation, control and
management of a process plant. It is, therefore, absolutely essential for them to be conversant with the
mass and energy conservation techniques at every stage of the process to achieve economy in the
design of process equipment in various units of the plant. This book aims at imparting knowledge of
the basic chemical engineering principles and techniques used in analyzing a chemical process. By
applying the relevant techniques, a chemical engineer is able to evaluate material and energy balances
in different units and present the information in a proper form so that the data can be used by the
management in taking correct decisions. Keeping this in mind, an attempt has been made to give a
brief theory on the principle involved and more emphasis on numerical examples.
Since data are generally obtained in different units, the worked examples are not confined to SI units,
but to other systems as well, such as FPS, CGS and MKS systems of units. The examples
incorporated in the text are simple and concrete to make the book useful for self-instruction.
The text is organized into ten chapters and appends three important tables. The organization is such
that the topics are presented in order of easy comprehension rather than following a logical sequence,
e.g. the chapter on unsteady state operations has been included as the last chapter so that students can
absorb the problems easily. We strongly feel that once the student understands the topics presented in
this book, he will find other advanced courses in chemical engineering simple and easy to follow.
The topics covered in this book cater to the syllabi on ‘Process Calculations’ of most universities
offering courses in chemical engineering and its allied branches at the undergraduate level.
V. Venkataramani N. Anantharaman xiii

Acknowledgements
At the outset, we wish to thank the almighty for his blessings.
V. Venkataramani wishes to acknowledge his wife, Prof. (Mrs.) Booma Venkataramani, sons Mr. V.
Ravi Chandar, Mr. V. Hari Sundar and his daughters-in-law Mrs. Vandana Ravi Chandar and Mrs.
Ramya Hari Sundar and granddaughters Miss Vaishnavi Ravi Chandar and Miss Sadhana Hari Sundar
for their support, cooperation and patience shown during the preparation of this book.
N. Anantharaman wishes to thank his mother, wife, Dr. Usha Anantharaman, sons Master A. Srinivas
and A. Varun for all their patience, cooperation and support shown during the preparation of this
book. The encouragement received from his brothers and sisters and their family members is
gratefully acknowledged. He also wishes to place on record the support received from his brothers-
in-law and sisters-in-law and their family members.
K.M. Meera Sheriffa Begum wishes to acknowledge her mother, husband Mr. S. Malik Raj and baby
M. Rakshana Roshan for all their encouragement, support and cooperation while preparing this book.
She also wishes to place on record the support received from her parents-in-law. The support
received from her brothers, sisters, in-laws and their families is gratefully acknowledged.
We also thank Director, NIT, Tiruchirappalli for extending all the facilities and his words of
appreciation.
We wish to acknowledge the support and encouragement received from the Head of Chemical
Engineering Department and all our colleagues during the course of preparation of this book.
We also wish to place on record the suggestions received from students, especially those at NIT,
Tiruchirappalli, and also the faculty from other institutions.
xv xvi
ACKNOWLEDGEMENTS
We gratefully acknowledge all the well-wishers.
Finally, we wish to thank the publishers, PHI Learning, New Delhi for encouraging us to bring out the
second edition of the book.
V. Venkataramani N. Anantharaman K.M. Meera Sheriffa Begum
Units and Dimensions 1
1.1 INTRODUCTION
Chemical engineers are concerned with the design and development of processes which involve

changes in the bulk properties of matter. To make a quantitative estimation of these processes,
chemical equations showing the quantities of reactants and products are used. Though internationally
we follow SI system of units, a chemical engineer is expected to be familiar and conversant with all
the systems so far adopted for measuring and expressing various quantities. A review of literature and
data over the years will be available in various units. These are used to express properties, process
variables and design parameters in FPS, CGS, MKS and SI systems of Units. Hence, one has to be
conversant with their use and applications. This chapter deals with the basic notations and conversion
of a given quantity from one system of units to another.
The quantities used in our analysis are classified as fundamental quantities and derived quantities.
The fundamental quantities comprise length, mass, time and temperature. The quantities such as force,
density, pressure, mass flow rate derived from the fundamental quantities are called derived
quantities. While handling these quantities, we come across different systems of units as mentioned
earlier. Now let us see in detail these systems of units and their conversion from one unit to another.
1.2 BASIC UNITS AND NOTATIONS
English Metric Engineering System Engineering,
CGS MKS
International, FPS SI
Mass ( m) lb g kg kg Length (L)ft cmm m Time (t)ssss Temperature (T)°F °C °C K
1
Mass (m)
1 kg = 2.205 lb
Length (L)
1 ft = 30.48 cm
= 0.3048 m
Time (t)
1 h = 3600 s
Temperature (T )
°C=
°F 32
(Celsius and Centigrade are same)

1.8

1.3 DERIVED UNITS
Area: = length ´ breadth (L
2
):
1 ft
2
= 0.0929 m
2
10.76 ft
2
= 1 m
2
Force: = mass ´ acceleration (mL/t
2
):
1 dyne = 1 g cm/s
2
(Force applied on a mass of 1 g, which gives an acceleration of 1 cm/s
2
)
1 Newton (N) = 1 kg m/s
2
= (1000 g) (100 cm)/s
2
1 N = 10
5
g cm / s
2

= 10
5
dynes
Work/energy: = 1 kg m/s
2
1 erg is the amount of work done on a mass of 1 g when it is displaced by 1 cm by applying a force of
1 dyne.
1 erg = [1 dyne] ´ [1 cm]
= [1 g cm/s
2
] ´ [1 cm]
= 1 g cm
2
/s
2
1 Joule = (1 N) ´ (1 m)
= 10
5
g cm/s
2
´ 100 cm
= 10
7
g cm
2
/s
2
1 Joule = 10
7
erg

Heat Unit:
1 Btu = 0.252 kcal = 252 cal
1 cal = 4.18 J
1 J/s = 1 W
1.4 DEFINITIONS
System. This refers to a substance or group of substances under consideration, e.g. storage tank,
water in a tank, hydrogen stored in cylinder, etc.
Process. Changes taking place within the system is called process, e.g. burning of fuel, or reaction
between two substances like hydrogen and oxygen to form water.
Isolated system. Boundaries of the system are limited by a mass of material, and its energy content is
completely detached from all other matter and energy. In an isolated system, the mass of the system
remains constant, regardless of the changes taking place within the system.
Extensive property. It is a state of system, which depends on the mass under consideration, e.g.
volume.
Intensive property. This state of a system is independent of mass. An example of this property is
temperature.
WORKED EXAMPLES
1.1 The superficial mass velocity is found to be 200 lb/h.ft
2
. Find its equivalent in kg/s.m
2
G (Mass velocity) = (200) lb/h.ft
2
=
(200)
ʈ¥¥ ¥
11 1
m
2
Á˜Ë¯ (3600 s) (0.0929)

= 0.2712 kg/s.m
2
1.2 Convert the heat transfer coefficient of value 100 Btu/h.ft
2
.°F into W/m
2
°C
h (Heat transfer coefficient) = (100)È˘Í˙Î˚
(0.0929 m
2
)
(1.8 °C) [1 degree variation in Farh. scale is equivalent to 1.8 times the variation in celsius scale]
= 4.186 ¥ 10
–2
kcal/s.m
2
°C = 4.186 ¥ 10
–2
¥ 10
3
¥ 4.18 W/m
2
°C = 174.98 W/m
2
°C
1.3 The rate of heat loss per unit area is given by (0.5) [(DT)
1.25
/(D)
0.25

] Btu/h ft
2
for a process,
where, DT is in °F and D is in ft. Convert this relation to estimate the heat flux in terms of kcal/h. m
2
using DT in °C and D in m.
We know that,
9
C
1
+ 32 = F
15

9
C
2
+ 32 = F
25
Therefore,
1.8 [DC] = (DF)

qT
()1.25

0.5()
0.25
A
q
, Btu/h ft
2

= 0.5
'(°F)1.25
A (ft)
0.25
We know that,
1 Btu = 0.252 kcal
1 ft
2
= 0.0929 m
2
1 ft = 0.3048 m
For DT °F = 1.8 DT °C

ËÛ1.25
Btu/h ft
2 = 0.5 'ÌÜ
(ft)
0.25 (a)
ÌÜ
ÍÝ
(1.8
'
T
°C)

1.25

= (0.5)
(Dm/0.3048)
0.25(b)

= (0.7746)

'(°C)
1.25
0.25
(m)
Btu/h ft
2
= (0.252 kcal)/h (0.0929 m
2
)
= 2.713 kcal/h m
2
(c)
From (b) and (c) we get the expression for heat flux in units of kcal/h m
2
with temperature difference
in Celsius and diameter in metre as:
Heat flux, kcal/h m
2
=
(m)0.25
ËÛ
T °C)
1.25
'ÌÜ
´
2.713
​ÌÜ
ÍÝ

(2.101)(
'
T
°C)

1.25

=
(m)
0.25 (d)

Now let us check the conversion with the following data: D = 0.2 ft, i.e. D¢ = 0.06096 m
DT = 18 °F, i.e. DT = 10 °C
From Eq. (a), heat flux is = 0.5
(18)
1.25

0.25
= 27.72 Btu/h ft
2
= 75.2 kcal/h m
2
(0.2)
Also, from Eq. (d), heat flux = 2.101

(10)
1.25
(0.06096)
0.25
= 75.2 kcal/h m

2
Both the values agree.
1.4 If C
p
of SO
2
is 10 cal/g mole K, what is the value in FPS units? The C
p
value is the same in all
units, i.e. 10 Btu/lb mole °R.
1.5 Iron metal weighs 500 lb and occupies a volume of 29.25 litres. Find the density in kg/m
3
.
Basis: 500 lb of Iron = 500/2.2 = 227.27 kg
29.25 lit = 29.25 ´ 10
–3
m
3
227.27
´ 10
–3
= 7770 kg/m
3
Density =
29.25
1.6 Etching operation follows the relation d = 16.2 – 16.2e
–0.021t
, where t is in s. and d is in microns.
Convert this equation to evaluate d in mm with t in min.
d = 16.2 [1 – e

–0.021t
]
Let d¢ be in mm and t¢ be in min. (d = d¢ ´ 10
3
and t = t¢ ´ 60) Then, d = d¢ × 10
3
= 16.2 [1 – e
–0.021t
× 60
]
d¢ = 0.0162 [1 – e
–1.26t¢
]
1.7 The density of fluid is given by r = 70.5 exp (8.27 × 10
–7
). Convert this equation to calculate the
density in kg/m
3
with pressure in N/m
2
.
1000 kg/m
3
= 62.43 lb/ft
3
14.7 psi = 1.0133 × 10
5
N/m
2
1 kg/m

3
= 62.43 × 10
–3
lb/ft
3
Let, r¢ be in kg/m
3
and p¢ be in N/m
2
.
Then, (r¢ × 62.43 × 10
–3
) = 70.5 × exp (8.27 × 10
–7
× p¢ × 14.7/1.0133 × 10
5
) r¢ (kg/m
3
) = 1.129 ×
10
3
× exp [119.97 × 10
–12
× p¢ (N/m
2
)]
1.8 Vapour pressure of benzene in the range of 7.5 °C to 104 °C is given by log
10
(p) = 6.9057 –
1211/(T + 220.8), where T is in °C and p is in torr 1 torr = 133.3 N/m

2
. Convert it to SI units.
Let p¢ be in N/m
2
and T¢ be in K.
Then, log

ÉÙ
ÊÚ

pÈØ
= 6.9057 –
1211

(T
273)
220.8

1211
log p¢ – log 133.3 = 6.90305 –
T​ 52.2

log (p
1211
.¢) = 9.0305 –
T​52.2

EXERCISES
1.1 Convert the following quantities:
(a) 42 ft

2
/h to cm
2
/s
(b) 25 psig to psia
(c) 100 Btu to hp-h
(d) 30 N/m
2
to lbf/ft
2
(e) 100 Btu/h ft
2
°F to cal/s cm
2
°C
(f) 1000 kcal/h m°C to W/m K
1.2 The heat transfer coefficient for a stream to another is given by h = 16.6 C
p
G
0.8
/D
0.2
where h = Heat transfer coefficient in Btu/(h)(ft)
2
(°F) D = Flow diameter, inches
G = Mass velocity, lb/(s)(ft)
2
C
p
= Specific heat, Btu/(lb) (°F)

Convert this equation to express the heat transfer coefficient in kcal/ (h)(m)
2
(°C)
With D = Flow diameter in m, G = Mass velocity in kg/(s)(m)
2
and C
p
= Specific heat, kcal/(kg) (°C)
1.3 Mass flow through a nozzle as a function of gas pressure and temperature is given by m = 0.0549
p/(T)
0.5
where m is in lb/min, p is in psia and T is in °R, where T(°R) = T °F + 460. Obtain an
expression for the mass flow rate in kg/s with p in atmospheres (atm) and T in K.
1.4 The flow past a triangular notch weir can be calculated by using the following empirical formula:
q = [0.31 h
2.5
/g
0.5
] tan F
where q = Volumetric flow rate, ft
3
/s
h = Weir head, ft
g = Local acceleration due to gravity, ft/s
2
F = Angle of V-notch with horizontal plane
1.5 In the case of liquids, the local heat transfer coefficient, for long tubes and using bulk-temperature
properties, is expressed by the empirical equation,
h
= 0.023

G
0.8
k
0.67
C
0.5
/(D
0.2
m
0.47
p )
where G = Mass velocity of liquids, lb/ft
2
.s
k = Thermal conductivity, Btu/ft.h.°F
C
p
= Specific heat, Btu/lb °F
D = Diameter of tube, ft
m = Viscosity of liquid, lb/ft.s
Convert the empirical equation to SI units.
Mass Relations 2
2.1 MASS RELATIONS IN CHEMICAL REACTION
In stoichiometric calculations, the mass relations between reactants and products of a chemical
reaction are considered and are based on the atomic weight of each element involved in the reaction.
For the following reactions the material balance is established as indicated below:
(i) CaCO
3
® CaO + CO
2

(2.1) [40 + 12 + 3 ´ 16] ® [40 + 16] + [12 + 32] 100 ® 56 + 44
(ii) 3Fe + 4H
2
O ® Fe
3
O
4
+ 4H
2
(2.2) (3 ´ 55.84) + 4(2 ´ 1 + 16) ® (55.84 ´ 3 + 4 ´ 16) + 4(2 ´ 1)
167.52 + 72 ® 231.52 + 8 239.52 ® 239.52
Based on the reactions given by Eqs. (2.1) and (2.2) we conclude that when 100 parts by weight of
CaCO
3
reacts, 56 parts by weight of CaO and 44 parts by weight of CO
2
are formed. Similarly, when
167.52 parts by weight of iron reacts with 72 parts by weight of steam (water), we get 231.52 parts
by weight of magnetite and 8 parts by weight of hydrogen. Thus the total weight of reactants is always
equal to the total weight of products.
Such computations will help one to estimate the quantity of reactants needed to obtain a specified
amount of product.
gram atom (or g atom) = Mass in grams/Atomic weight katom (or kg atom) = Mass in kg/Atomic
weight
gram mole (or g mole) = Mass in grams/Molecular weight kmole (or kg mole) = Mass in
kg/Molecular weight
7
The conclusions based on reactions (2.1) and (2.2) on material balance can be expressed in other
forms too, as per the definitions given above:
1 kmole of CaCO

3
gives 1 kmole of CaO and 1 kmole of CO
2
Similarly, 3 kmoles of iron reacts with 4 kmoles of steam (water) to yield 1 kmole of oxide and 4
kmoles of hydrogen. When such balances (on molar basis) are made, the number of moles on the
reactants side need not be equal to the total numbers of moles on the product side.
One atom of oxygen weighs = 16 grams O
One atom of hydrogen weighs = 1 gram H
One molecule of oxygen weighs = 32 grams O
One molecule of hydrogen weighs = 2 grams H
In other words,
16 grams of oxygen
32 pounds of oxygen
2 g atoms of oxygen
1 gram of hydrogen
2 kg of hydrogen
2 kg atoms of hydrogen = 1 kmole of hydrogen
\ g atom or lb atom = Mass in grams or pounds/Atomic weight \ g mole or lb mole = Mass in grams or
pounds/Molecular weight = 1 g atom of oxygen = 1 lb mole of oxygen = 1 g mole of oxygen = 1 g
atom of hydrogen = 1 kmole of hydrogen
2.2 CONSERVATION OF MASS
The law of conservation of mass states that mass can neither be created nor be destroyed. It is the
basic principle adopted in solving the material balance problems in chemical process calculations,
whether a chemical reaction is involved or not. However, while applying the law of conservation of
mass, one should not apply it for the conservation of molecules. We frequently come across chemical
reactions in which the total number of moles on the reactant side is not equal to the total number of
moles on the product side. For example:
Na
2
CO

3
+ Ca(OH)
2
® CaCO
3
+ 2NaOH
The total number of moles on the reactant side is 2 and on the product side 3. Here the mass balance
is ensured but not the mole balance. Now consider the reaction:
2Cr
2
O
3
+ 3CS
2
® 2Cr
2
S
3
+ 3CO
2
Here the total number of moles both on the reactant side and the product side is 5. Hence both the
conservation of mass and the conservation of moles are observed.
2.3 AVOGADRO’S HYPOTHESIS
1 g mole of any gaseous substance at NTP occupies 22,414 cc or
22.414 litres and 1 lb mole of the same substance occupies 359 ft
3
at NTP.
1 kmole of any gaseous substance occupies 22.414 m
3
at NTP.

Note: NTP = Normal temperature (273 K) and pressure (1 atmosphere) are also referred to at times
as standard conditions (SC).
2.4 LIMITING REACTANT AND EXCESS REACTANT
For most of the chemical reactions the reactants will not be used in stoichiometric proportion or
quantities. One of the reactants will be present in excess and remain unreacted even when the other
reactant has completely reacted. The reactant thus present in excess is termed excess reactant and the
other reactant which is present in a lesser quantity and cannot react with whole of the other reactant
(excess reactant) is called limiting reactant. All calculations involved in estimating the quantity of
product and conversion are always based on the limiting reactant. The amount by which any reactant
is present in excess to that required to combine with the limiting reactant is usually expressed as
percentage excess. The percentage excess of any reactant is defined as the percentage ratio of the
excess to that theoretically required by the stoichiometric equation for combining with the limiting
reactant. A limiting reactant is the one, which will not be present in the product, whereas the excess
reactant is the one, which will always be present in the product.
Let us consider that 18 kg of carbon is burnt with 32 kg of oxygen. As per stoichiometry
C + O
2
Æ CO
2
i.e. 12 kg of carbon will burn with 32 kg of oxygen to form 44 kg of CO
2
.
Hence, for 18 kg of carbon to react fully we should have 48 kg of oxygen. Since 32 kg of oxygen
alone is available, it is called the limiting reactant and carbon is called the excess reactant. For 32 kg
of oxygen to react fully, it is sufficient to have 12 kg of carbon. However 6 kg of carbon is present in
excess.
Hence % excess of carbon is =
6
¥ 100 = 50%.
12

2.5 CONVERSION AND YIELD These terms are used for a chemical reaction where the reactants
give out new compounds or products.
Conversion is the ratio of the amount of material actually converted to that present initially, whereas
the yield is the amount of desired product actually formed compared to that which can be formed
theoretically.
Conversion for a reaction is based on the limiting reactant whereas the yield is based on the product
formed.
2.6 COMPOSITION OF MIXTURES AND
SOLUTIONS
Different methods are available for expressing composition of mixtures of gases, liquids and solids.
Conventionally the composition of solids is either expressed on weight basis or mole basis. Let us
consider a binary system comprising components A and B.
W = Total weight of the system.
W
A
, W
B
= Weight of components A and B respectively. M
A
, M
B
= Molecular weight of components A
and B respectively,
if they are compounds.
A
A,
A
B
= Atomic weight of components A and B respectively, if they are elements.

V = Volume of the system.
V
A
and V
B
= Pure component volume of components A and B respectively.
2.6.1 Weight Percent
This is defined as the ratio of weight of a particular component to the total weight of the system in
every 100 part, i.e.
Weight % of A =
WA
¥ 100
W
This method of expressing composition is generally employed in solid and liquid systems and not
used in gaseous system. One major advantage of weight percent is, its independence to changes in
temperature and pressure.
The composition of a solid mixture is to be always taken as weight % when nothing is mentioned
above its units.
2.6.2 Volume Percent
The ratio of the volume of each component and the total volume of the system, for each 100 part of the
total volume is called volume percent Volume % of A =
VA
´ 100
V
This method of expressing composition is employed always for gases, rarely for liquids and seldom
for solids. The composition of a gas mixture is to be taken as volume % when nothing is mentioning
about its units.
The volume % is also equal to mole % for ideal gases but not for liquids and solids. This is based on
Avogadro’s law.
2.6.3 Mole Fraction and Mole Percent

These concepts are generally adopted in the case of a mixture containing molecules of different
species.

W
A
MA
Mole fraction of A =
WA WB
M
A
M
B
Mole % of A = Mole fraction of A ´ 100
2.6.4 Atomic Fraction and Atomic Percent This is adopted when a mixture contains two or more
atoms. W
A
Atomic fraction of
A
=
A
A
W
A
W
B
A
A
A
B
Atomic % of A = Atomic fraction of A ´ 100

2.6.5 Composition of Liquid Systems
In the case of liquids we come across more number of methods of expressing compositions of the
liquid constituents.
(i) Weight ratio (ii) Mole ratio (iii) Molality
(iv) Molarity
= Weight of solute/weight of solvent
= g moles of solute/g moles of solvent
= g moles of solute/1 kg of solvent
= Number of g moles of solute/1 litre of solution
(v) Normality (N) = Number of gram equivalents of solute/1 litre of solution
Hence, concentration in grams per litre = Normality ( N) ´ Equivalent weight of solute
For very dilute aqueous solutions, molality = molarity
2.7 DENSITY AND SPECIFIC GRAVITY
Density is defined as mass per unit volume and it varies with temperature. Specific gravity is the ratio
of density of a liquid to that of water. However, in the case of gases, it is defined as the ratio of its
density to that of air at same conditions of temperature and pressure.
Over a narrow range of temperature, the variation in density of solids is not high. However, in the
case of liquids and gases the variation in density is significant. Similarly, the densities vary
significantly with concentration also. This property of density and specific gravity varying with
concentration is very widely used both in industries and markets as an index for finding the
composition of a system comprising a specific solute and a specific solvent.
Several scales are in use in which specific gravities are expressed in terms of a degree, which are
related to specific gravities and densities by arbitrary mathematical definitions.
2.7.1 Baume​ (
°
Be​) Gravity Scale
For liquids lighter than water
Degrees Baume’ =
140
– 130

G
G
is the specific gravity at
60
°F
15¦µ
60
15
§¶
¨·
Thus, water will have a gravity of 10° Be’ and this degree decreases with increase in specific
gravity.
For liquids heavier than water

145
Degrees Baume’ = 145 –
G
In this scale the degree increases with increase in specific gravity.
2.7.2 API Scale (American Petroleum Institute) This scale is used for expressing gravities of
petroleum products. This is similar to Baume’ scale for liquids lighter than water.
Degrees API =
141.5
– 131.5
G

2.7.3 Twaddell Scale
This scale is used for liquids heavier than water. Degrees Twaddell (Tw) = 200 (G – 1.0)
2.7.4 Brix Scale
This scale is used in sugar industry and 1 degree Brix is equal to 1% sugar in solution.
Degree Brix =

400
– 400
G

WORKED EXAMPLES 2.1 Convert 5000 ppm into weight %.
5000 ×100
= 0.5%106

2.2 The strength of H
3
PO
4
was found to be 35% P
2
O
5
. Find the weight % of the acid.
The acid can be split into
2H PO
→
P O + 3H O

34 25 2
(2 98)
142
(3 18)
×
196 units of the acid contains 142 units of the pentaoxide. The weight % of pentaoxide is (142/196)
which is 72.5% for pure acid. When the strength of pentoxide is 35%, the weight % of acid is 
=

48.3%
2.3 What is the volume of 25 kg of chlorine at standard condition?
25 kg Cl
2
=
25
kmoles of Cl
22 ×35.46

25 × 22.414
= 7.9 m
3
Volume =
2 × 35.46
2.4 How many grams of liquid propane will be formed by the liquefaction of 500 litres. of the gas at
NTP? Molecular weight of propane (C
3
H
8
) is 44
1 g mole of any gas occupies 22.414 litres at NTP
500 litres of propane at NTP =
500
= 22.31 g moles
22.414
22.31 g moles of propane weighs = 22.31 ´ 44 = 981.52 g. 2.5 Find the volume of (a) 100 kg of
hydrogen and (b) 100 lb of hydrogen at standard conditions?
(a) 100 kg of H
2
= 50 kmoles of hydrogen

volume occupied by 50 kmoles of hydrogen º 50 ´ 22.414
= 1120.7 m
3
(b) 100 lb of H
2
º 50 lb moles of hydrogen
Volume occupied by 50 lb moles of hydrogen º 50 ´ 359 = 17950 ft
3
2.6 A solution of naphthalene in benzene contains 25 mole % Naphthalene. Express the composition
in weight %. Basis: 100 g moles of solution
Component Molecular weight Weight, Actual g mole weight, g
Composition in weight percent
Naphthalene C
10
H
8
128 Benzene C
6
H
6
78 25 25 ´ 128 = 3200 75 75 ´ 78 = 5850 3200 ´ 100/9050 = 35.35 5850 ´ 100/9050 = 64.65
Total 9050 100.00
2.7 What is the weight of one litre of methane CH
4
at standard conditions? 22.414 litres of any gas at
NTP is equivalent to 1 g mole of that gas

1
= 0.0446 g mole\ 1 litre of methane º 1 ´
22.414

\ Weight of one litre methane = 0.0446 ´ 16 = 0.714 g
2.8 A compound whose molecular weight is 103 analyses C : 81.5, H : 4.9 and N : 13.6 by weight.
What is the formula?
Basis: 100 g of substance
Element Atomic weight
Carbon 12
Hydrogen 1
Nitrogen 14
Weight, g Weight, Rounding of Weight of g atom atoms each element
81.5 81.5/12 = 6.8 7 84
4.9 4.9/1 = 4.9 5 5
13.6 13.6/14 = 0.9 1 14
Total 103
Hence the formula obtained after rounding is correct. So the molecular formula is C
7
H
5
N
2.9 An analysis of a glass sample yields the following data. Find the mole %.
Na
2
O : 7.8%, MgO : 7.0%, ZnO : 9.7%, Al
2
O
3
: 2.0%, B
2
O
3
: 8.5% and rest SiO

2
.
Basis: 100 g of glass sample
Component Weight, g Molecular weight g mole mole %
Na
2
O 7.8 62.0 0.1258 7.665
MgO 7.0 40.3 0.1737 10.583
ZnO 9.7 81.4 0.1192 7.262
Al
2
O
3
2.0 102.0 0.0196 1.194
B
2
O
3
8.5 69.6 0.1221 7.439
SiO
2
65.0 60.1 1.0815 65.857
Total 100.0 — 1.6419 100.0
2.10 A gaseous mixture analyzing CH
4
: 10%, C
2
H
6
: 30% and rest H

2
at 15 °C and 1.5 atm is flowing
through an equipment at the rate of 2.5 m
3
/min. Find (a) the average molecular weight of the gas
mixture, (b) weight % and (c) the mass flow rate.
Basis: 100 g moles of the gaseous mixture.
Component Weight, Molecular Weight, Weight % g mole weight g
CH
4
10 16 160 13.56
C
2
H
6
30 30 900 76.27
H
2
60 2 120 10.17
Total 100 — 1180 100
The average molecular weight =
1180
= 11.8
100

Volumetric flow rate at standard conditions = 2.5¥ ¥ (273/288) = 3.555 m
3
/min
Moles of the gas =
3.555

= 0.156 kmole
22.414

Mass flow rate = moles ¥ average molecular weight = 0.156 ¥ 11.8 = 1.84 kg/min 2.11 In an
evaporator a dilute solution of 4% NaOH is concentrated to 25% NaOH. Calculate the evaporation of
water per kg of feed. Basis: 1 kg of feed.
NaOH present is 0.04 kg, which appears as 25% in the thick liquor formed
Weight of thick liquor formed =
0.04
= 0.16 kg
0.25
Weight of water evaporated = (1 – 0.16) = 0.84 kg
Water evaporated per kg of feed = 0.84 kg
2.12 The average molecular weight of a flue gas sample is calculated by two different engineers. One
engineer used the correct molecular weight of N
2
as 28, while the other used an incorrect value of 14.
They got the average molecular weight as 30.08 and the incorrect one as 18.74. Calculate the %
volume of N
2
in the flue gases. If the remaining gases are CO
2
and O
2
calculate their composition
also.
Basis: 100 g moles of flue gas
Component g mole I Engineer II Engineer
N
2

x 28x 14x CO
2
y 44y 44y O
2
z 32z 32z
Total 100 3008 1874
x + y + z = 100 (i)
28x + 44y + 32z = 3008 (ii)
14x + 44y + 32z = 1874 (iii) Solving Eqs. (i), (ii) and (iii), we get
x = Moles of nitrogen = 81%
y = Moles of carbon dioxide = 11%
z = Moles of oxygen = 8%
2.13 An aqueous solution contains 40% of Na
2
CO
3
by weight. Express the composition in mole
percent.
Basis: 100 g of solution
Component grams Molecular g mole Composition in weight mole % Na
2
CO
3
40 106 40/106 = 0.377
0.377 ´ 100/3.71 = 10.16 Water 60 18 60/18 = 3.333 3.333 ´ 100/3.71 = 89.84 Total 3.710 100.00
2.14 What is the weight of iron and water required for the production of 100 kg of hydrogen?

3Fe
4H O
+→

Fe O + 4H
↑234 2
(3×55.84) (4 18) (3
55.84) (4 16)
(4 2 1)×+××××
167.52 72 231.52 8

239.52 239.52
Method 1 (Based on absolute mass)
167.52 kg of Fe is required for producing 8 kg of H
2
\ For producing 100 kg of H
2
(by stoichiometry)
100 t167.52
Iron (Fe) required =
8
= 2094 kg
Similarly, for getting 100 kg of H
2
the amount of steam (H
2
O) required is
=
100 t 72
= 900 kg
8
The total weight of reactants is
2094 kg Fe and 900 kg H
2

O = 2994 kg
The weight of Fe
231.52
´ 2094 = 2894 kg
3
O
4
formed is =
167.52
\ The total weight of products is (100 kg H
2
+ 2894 kg Fe
3
O
4
) = 2994 kg
The total weight of reactants is (2094 kg Fe + 900 kg H
2
O) = 2994 kg
Method 2 (Based on moles)
100 kg of H
2
4 kmoles H
2
comes from
50 kmoles of H
2
comes from Weight of 37.5 katoms Fe
50 kmoles H
2

from
Weight of 50 kmoles H
2
O Moles of Fe
3
O
4
formed is
Weight of Fe
3
O
4
formed is
Total weight of reactants Total weight of products = 50 kmoles
= 3 katoms of Fe (by stoichiometry)
=

Ưà50
tĐả
= 37.5 katoms Fe
ăã4
= (37.5 55.84) = 2094 kg of iron
=

50Ưà
kmoles of water
4tĐả
ăã
= (50 18) = 900 kg H
2

O
=
37.5
kmoles
3
37.5
=
Ưà

Đả
231.52 = 2894 kg
ăã
= 2994 kg
= 2994 kg
2.15 How much super phosphate fertilizer can be made from one ton of calcium phosphate 93.5%
pure?
Atomic weights are: Ca : 40, P : 31, O : 16, S : 32
Ca
3
(PO
4
)
2
+ 2H
2
SO
4
ặ CaH
4
(PO

4
)
2
+ 2CaSO
4
310 (2 Ơ 98) 234 (2 Ơ 136) 506 506
One ton of raw calcium phosphate contains 0.935 tons of pure calcium phosphate
Weight of super phosphate formed is = 234 Ơ
0.935
= 0.70577 tonne\
310

2.16 SO
2
is produced by the reaction between copper and sulphuric acid. How much Cu must be used
to get 10 kg of SO
2
?
Cu +2H SO

24
→
CuSO +SO +2H O

4 2 2
63.54 64
64 kg of sulphur dioxide is obtained from 63.54 kg of copper. 10 kg of sulphur dioxide will be
obtained from 9.93 kg of copper. 2.17 How much potassium chlorate must be taken to produce the
same amount of oxygen that will be produced by 2.3 g of mercuric oxide? 2KClO
3

Æ 2KCl + 3O
2
(2¥ 122.46) (2¥ 74.46) (6¥ 16) 244.92 148.92 96
2HgO Æ 2Hg + O
2
(2¥ 216.6) (2¥ 200.6) (2¥ 16)
433.2 g HgO gives 32 g of oxygen
2.3 g of HgO will give (32 ¥ 2.3/433.2) = 0.1698 g of O
2
0.1698 g O
2
is obtained from (244.92 ¥ 0.1698)/96 = 0.4332 g of KClO
3
.
2.18 Ammonium phosphomolybdate is made up of the radicals NH
3
, H
2
O, P
2
O
5
and MoO
3
. What is %
composition of the molecule with respect to these radicals?
The formula of ammonium phosphomolybdate is
(NH
4
)

3
PO
4
◊ 12MoO
3
◊ 3H
2
O
First let us form the final product from the radicals:
3NH
3
+ 4.5H
2
O + 12MoO
3
+ ½P
2
O
5
Æ (NH
4
)
3
PO
4
12MoO
3
·3H
2
O

(3¥17 = 51) (4.5¥18 = 81) (12¥144 = 1728) (½¥142 = 71) 1931
% of NH
3
=51 ¥
100
= 2.64
1931

% of H
2
O= 81 ¥
100
= 4.19
1931

% of MoO
3
= 1728 ´
100
= 89.49
1931

% of P
2
O
5
=71
100
= 3.68´
1931

Total = 100.00
2.19 How many grams of salt are required to make 2500 g of salt cake? How much Glauber’s salt can
be obtained from this?
The molecular formula of Glauber’s salt is Na
2
SO
4
× 10H
2
O
(142 + 180 = 322)

2NaCl
H SO Na SO
+ 2HCl

(2
×
58.46 =116.92)
+→
24 24

98 142 (2×36.46)
Thus, 142 g of Na
2
SO
4
is obtained from 116.92 g NaCl. 2500 g of salt cake is obtained from 116.92 ´
2500/142 = 2058.45 g NaCl
Hence, salt needed is 2058.45 g

Glauber’s salt (Na
2
SO
4
× 10H
2
O) obtained is 2500 ´ 322/142 = 5669 g
2.20 (a) How many grams of K
2
Cr
2
O
7
are equivalent to 5 g KMnO
4
? (b) How many grams of KMnO
4
are equivalent to 5 g K
2
Cr
2
O
7
? 2KMnO
4
+ 8H
2
SO
4
+ 10FeSO

4
® 5Fe
2
(SO
4
)
3
+ K
2
SO
4
+ 2MnSO
4
+ 8H
2
O
K
2
Cr
2
O
7
+ 7H
2
SO
4
+ 6FeSO
4
® 3Fe
2

(SO
4
)
3
+ K
2
SO
4
+ (Cr
2
SO
4
)
3
+ 7H
2
O
2KMnO
4
gives 5Fe
2
(SO
4
)
3
(2 ´158 = 316) (5 ´400 = 2000)
K
2
Cr
2

O
7
gives 3Fe
2
(SO
4
)
3
(294) (3 ´400 = 1200)

1200 t 316
= 189.6 g KMnO
4
\ 294 g K
2
Cr
2
O
7
is equivalent to
2000

3 t189.6
= 1.935 g KMnO
4
\ 3 g K
2
Cr
2
O

7
º
294
Similarly, 5 g KMnO
4

5 t 294
= 7.75 g K
2
Cr
2
O
7
º
189.6

Alternatively,
53t
= 7.75 g K
2
Cr
2
O
71.935

2.21 If 45 g of iron react with H
2
SO
4
, how many litres of hydrogen are liberated at standard

condition?
There are two possible reactions in this case: (a) Case I
Fe H SO FeSO + H

(55.85)
+→

24
4 2 (i)
(2)

The weight of hydrogen formed by reaction (i) is = 45 ¥
2
= 1.611 g,
55.85

i.e.
1.611
= 0.806 g mole
2
0.806 g mole ∫ 0.806 ¥ 22.414 = 18.06 litres
(b) Case II

2Fe
3H SO
Fe (SO ) + 3H

(111.7)
+→24 2 43 2 (ii)
(6)

The moles of hydrogen formed by reaction (ii) is 45 ¥
6
= 2.418 g
111.7
2.418 g H
2
= 1.209 g moles of hydrogen ∫ 1.209 ¥ 22.414 ∫ 27.1 litres
2.22 A natural gas has the following composition by volume CH
4
: 83.5%, C
2
H
6
: 12.5%, and N
2
: 4%.
Calculate the following:
(a) composition in mole % (b) composition in weight % (c) average molecular weight (AVMWT) (d)
density at standard condition (kg/m
3
) Basis: 100 kmoles of gas mixture
Component Molecular mole % Weight, kg Weight % weight
CH
4
16 83.5 83.5¥ 16 = 1336 1336¥ 100/1823 = 73.29
C
2
H
6

30 12.5 12.5¥30 = 375 375¥ 100/1823 = 20.57
N
2
28 4.0 4.0¥28 = 112 112¥100/1823 = 6.14
Total 1823 100.0
(c) Average molecular weight =
1823
= 18.23
100

Volume at standard condition = 00 ¥ 22.414 = 2241.4 m
3