
What is material balance?
A material balance equation that is related to the
law of mass conservation.
‘Mass can neither be destroyed or created’
So if what goes in must come out then,
Total Mass Input = Total Mass Output
The line that limits the system to be analyzed
SYSTEM BOUNDARY
MASS INPUT
SYSTEM
MASS OUTPUT
• A system is where physical
or chemical processes
occur.
• Any part of a process which
is considered for analysis.
Example: Process of Making Orange Juice
Oranges
Juice Extraction
System A
Evaporation
System B
Orange
Juice

Batch system
- A process whereby materials are neither added
or removed from the system during the operation.
- The product is accumulates in the system.
- The material balance equation includes a term for
accumulated material in the system.
Accumulation
within the system
Input
+
=
Generation within the system
Consumption
within the system
Output

Continuous System
- A process in which the materials constantly
enters or exists the system. Materials enters and
leaves the process without interruption.
- Also known as a steady state process when the
flow in the system remains unchanged with time.
- No material is accumulated in the system.
- The material balance equation does not include
the ‘material accumulation’ term.
=
Output
Input
+
Generation within the system
Consumption
within the system
Q: Classify the following processes as (a) batch, (b)
semi-batch, (c) neither or (d) both.
(i)
(ii)
(iii)
(iv)
A section of a river between two bridges
A reaction carried out in a beaker
Preparing a pot of curry
Water boiling in a pot on a stove.


Closed System
- Changes occur only inside the system
- No material enters or exists the system
- No mass exchange occurs with the surroundings
outside the system boundary.
Open System
- Materials can enter and exit a system
- Materials cross the system boundary
- Also known as a flow system.
QUESTION: What are examples of open and closed
systems?
Q: Discuss and catogorize the following processes
as (a) open, (b) closed, (c) neither or (d) both for
a characteristic operating level.
(i)
(ii)
(iii)
(iv)
Oil storage tank at a refinery
Flush tank on a toilet
Catalytic converter on an automobile
Gas furnace in a home

Steady State
- A system whereby all the conditions remain
constant with time. This includes temperature,
pressure and the amount of a material.

Unsteady State
- A system where one or more of its conditions
changes with time.
- Also known as a transient system.
Q: For the same processes as before, would you
consider them as a (a) steady state, (b) unsteadystate, (c) neither or (d) both?
Q: Will you save money if instead of buying
premium 89 octane gasoline at RM
1.269/gallon that has the octane you want,
you blend sufficient 93 octane supreme
gasoline at RM 1.349/gallon with 87 octane
regular gasoline at RM 1.149/gallon.
Total Mass
Input
Total Mass
Output
‘Mass can neither be destroyed or created’
So if what goes in must come out then,
Total Mass Input = Total Mass Output
What I hear, I forget;
What I see, I remember;
What I do, I understand.
Confucius

Draw and label a diagram of the process.

Select a basis for calculation purposes.

State any existing chemical equations (for
chemical processes).

Write down material balance equations.

Solve the equations to obtain the unknown
values.
Physical Process
Q. As much as 10 kgs-1 of a 10 % (mass) concentration
of NaCl solution is concentrated to 50 % in a
continuous evaporator. Calculate the production
rate of the concentrated NaCl and the rate of
water removed from the evaporator?

A.
10kgs-1
Flowrate =
NaCl = 0.1
H2O = 0.9
Flowrate Concentrated NaCl = W
NaCl = 0.5
H2O = 0.5
Flowrate H2O= D
Material Balance for NaCl :
Input rate x wt fraction = Output rate x wt fraction
(10 kgs-1) (0.1) = (W) (0.5)
W = 2 kgs-1
Material Balance for H2O:
Input rate x wt fraction = Output rate x wt fraction
(10 kgs-1) (0.9) = (2 kgs-1) (0.5) + D (1)
D = 8 kgs-1
Or
Total balance:
Input = Output
10 kgs-1 = 2 kgs-1 + D
D = 8 kgs-1
Q. A solution containing 30 % of NaOH is fed into a
reactor to produce a solution of 12 % NaOH. Distilled
water was also fed into the reactor for dilution
purposes. Calculate the feed rates of 30 % NaOH
and H2O needed to produce 2000 kg/h of 12 %
NaOH?
A.
Flowrate = W
NaOH = 0.3
H2O = 0.7
Flowrate H2O= D
Flowrate of NaOH= 2000 kg/h
NaOH = 0.12
H2O = 0.88
Material Balance for NaOH:
Input rate x wt fraction = Output rate x wt fraction
(W kgh-1) (0.3) = (2000 kgh-1) (0.12)
W = 800 kgh-1
Material Balance for H2O:
Input rate x wt fraction = Output rate x wt fraction
(800 kgh-1) (0.7) + D (1) = (2000 kgh-1) (0.88)
D = 1200 kgh-1
Or
Total balance:
Input = Output
800 kgh-1 + D = 2000 kgh-1
D = 1200 kgh-1
Q. An equimolar mixture of propane and butane is
fed into a distillation column at a rate of 67 mol s-1.
As much as 90 % of propane was recovered in the
top product which has a propane mol fraction of
0.95. Calculate the flowrates of the bottom
products and the composition of the bottom
product.
A. 1. Draw a process diagram
s-1
Flowrate = 67 mol
Propane = 0.5
Butane = 0.5
Flowrate = W
Propane = 0.95
Butane = 0.05
Flowrate = D
2. Choose a basis
Example, feed rate = 10 mol s-1.
3. Write the balance equations.
Total flowrate balance:
10 mol s-1 = W + D
(1)
Total propane balance:
10 mol s-1 (0.5) = W (0.95) + D (XP)
5 mol s-1 = 0.95 W + D (XP)
(2)
Propane balance for top product:
10 mol s-1 (0.5) (90%)= W (0.95)
W = 4.74 mol s-1
Insert W into (1):
10 mol s-1= W + D
10 mol s-1= 4.74 mol s-1 + D
D = 5.26 mol s-1
Insert W and D into (2):
5 mol s-1 = 0.95 (4.74 mol s-1 ) + 5.26 mol s-1 (XP)
XP = 0.094
4. Scale up
So,
67 mol s-1 = 6.7
10 mol s-1
W = 4.74 x 6.7 = 31.8 mol s-1
D = 5.26 x 6.7 = 35.2 mol s-1
7
95 % EtOH
Feed 10 kg s-1
1
30 % EtOH
70 % water
3
Column
1
20 %
Benzene
5
Mixer
4
2
7 % EtOH
6
Benzene
Q.
Column
2
Pure EtOH 2 kg s-1
The image shows the process for the production
of pure ethanol from an aqueous solution by
azeotropic distillation. The known quantities are
shown. What are the flow rates and compositions.
Basis = 10 kg s-1
Column 1
EtOH 10(0.3) = 0.07 (F2) + 0.95 (F3)
H2O 10(0.7) = 0.93 (F2) + 0.05 (F3)
Mixer
EtOH F3 (0.95) = X1,5 (F5)
H2O F3 (0.05) = X2,5 (F5)
Bz
F4 = 0.2 F5
Column 2
EtOH X1,5 F5 = 2 + X1,7 (F7)
H2O X2,5 F5 = X2,7 F7
Bz
0.2 F5 = X3,7 F7
Balance for Flow 7
X1,7 + X2,7 + X3,7 = 1
X1 = EtOH
X2 = Air/ H2O
X3 = Benzene
Total Balance Mixer
F3 + F4= F5
Total balance
EtOH 10(0.3) = 0.07 F2 + 2 + X1,7 (F7)
H2O 10(0.7) = 0.93 F2 + X2,7 F7
Bz
F4 = X3,7 F7
Solve the equations.
Flow
1
2
3
4
5
6
7
Flow rate
(kg s-1)
10.00
7.39
2.61
0.65
3.26
2.00
1.26
Component
EtOH
30
7
95
75
100
48
H2O
70
93
5
4
10
Bz
100
20
52
For the acetone recovery system below, calculate
the values of A, F, W, B or D in kg/hour?
Water 100 %
W kg/h
Air 99.5 %
Water 0.5 %
A kg/h
Condenser
Absorber
Column
Feed
1400 kg/h
Acetone 3.0 %
Air 95.0%
Water 2.0 %
F kg/h
Water 81 %
Acetone 19 %
Distillation
Column
Distillate
D kg/h
Water 1 %
Acetone 99 %
B kg/h
Water 96 %
Acetone 4 %
Q: In a process producing KNO3 salt, 1000 kg/h of a feed
solution containing 20 wt% KNO3 is fed to an evaporator,
which evaporates water producing 50 wt % KNO3 solution.
This is fed into a crystallizer which removes crystals that
Contain 96 wt % KNO3. The saturated solution containing
37.5 wt% KNO3 is recycled to the evaporator. Find R and P.
Water,
W kg/h
KNO3 20 %
Feed, 1000 kg/h Evaporator
422 K
KNO3 50 %
S kg/h
Crystallizer
311 K
KNO3 37.5 %
recycle, R kg/h
Water 4 %
Crystals, P kg/h
2. Choose a basis
Example, feed rate = 1000 kg h-1.
3. Write the balance equations.
Total feed balance:
1000 kg h-1 = W + P
(1)
Total KNO3 balance:
1000 kg h-1 (0.2) = W (0) + P (0.96)
P = 208.33 kg h-1
(2)
Evaporator, KNO3 balance:
1000 kg h-1(0.2) + R (0.375)= S (0.5)
(3)
Crystallizer, total balance:
S= R + P
S = R + 208.33
Insert (4) into (3):
R = 766.68 kg h-1
S = 974.9 kg h-1
(4)
The flowchart to recover crystalline potassium chromate
(K2CrO4) is shown below.The K2CrO4 accounts for 95% of
the total mass of the filter cake. Calculate the (a) rate of
water evaporation and (b) the recycle ratio.
Water
Vapour
K2CrO4 33.3 %
Feed, 4500 kg/h
Evaporator
K2CrO4
49.4 %
K2CrO4 36.4 %
Filtrate
Crystallizer
& Filter
K2CrO4 36.4 %
Solid Crystals, K2CrO4

Stoichiometry Equation
- Shows the ratios in which molecules of different species
are consumed or formed in a reaction.
C2H6 + 7/2O2 → 2CO2 + 3H2O
- Using this equation, the stoichiometric coefficient, vi, can
be defined as below:
vCO2 = 2
vC2H6 = -1
vH2O = 3
vO2 = -7/2
- vi is negative for reactants, positive for products and
zero for inerts (substance unchanged in the reaction).

Extent of Reaction,ξ
- Used to measure the amount of material consumed or
produced in a chemical reaction.
- Extent if reaction is defined as:
ξ = fi,out – fi,in
vi
where,
fi,out is the no of moles of species i leaving the reactor.
fi,in is the no of moles of i entering.
vi is the stoichiometric coefficient.

Fractional Conversion
- The maximum extent in which a reaction can occur.
- Defined as:
fractional conversion
fi, in – fi, out
=
of species, i
fi, in
Q. If you feed 10 grams of N2 gas and 10 grams of H2
gas into a reactor:
a. What is the maximum number of grams of NH3 that
can be produced?
b. What is the limiting reactant?
c. What is the excess reactant?
N2 (g)
10 g
Reactor
H2 (g)
10 g
NH3 (g)
Mass (g)
MW
Mol
N2 + 3H2
10
10
28
2
0.357
5
→ 2NH3
?
17
?
(a) 1 mol N2 produces 2 mol NH3
0.357 mol N2 produces 0.714 mol NH3
So, 0.714 x 17 = 12.2 g NH3
(b) Extent of reaction, ξN2 = - 0.357 = 0.357
-1
Extent of reaction, ξH2 = -4.960 = 1.65
-3
Limiting reactant is N2
(c) Excess reactant is H2
Q. ln the combustion of heptane, CO2 is produced.
Assume that you want to produce 500 kg of dry ice
per hour, and that 50% of the CO2, can be
converted into dry ice, as shown in the figure
below. How many kilograms of heptane must be
burned per hour?
The chemical equation is
C7H16 + 11O2 → 7CO2 + 8H2O
Other
products
CO2
Gas (50%)
C7H16 Gas
CO2 Solid
(50%)
O2 Gas
500 kg/h
50 % of y = 500 kg
y = 1000kg
Bil mol CO2 = 1000 x 103 g
44
= 22727.27 mol
1 mol heptane produces 7 mols of CO2
x mol heptane produces 22727.27 mols of CO2
x = 3246.75 mol heptane
= 324.67 kg heptane
Q: 200 mol of ethane is burned in a furnace with 50 % excess
of air. A conversion of 95 % is achieved. Calculate the
composition of the stack gasses.
A:
C2H6 + 7/2O2 → 2CO2 + 3H2O
Basis = 200 mol
Total mol ethane = 200 mol
Total mol O2= mol in stochiometric equation + excess 50%
= (7/2 x 200) + (7/2 x 200 x 0.5)
= 1050 mol
Total mol N2 = 79/21 x 1050 = 3948 mol
Stack
gasses
fC2H6, in = 200 mol
fO2, in = 1050 mol
fN2, in = 3948 mol
fC2H6, out
fO2, out
fN2, out
fCO2, out
fH2O, out
FURNACE
Conversion = fC2H6, in - fC2H6, out = 0.95
fC2H6, in
= 200 - fC2H6, out
200
= 0.95
= 10 mol
Extent of conversion = fC2H6, out - fC2H6, in = 10 – 200
vC2H6
-1
= 190 mol
Species
I
In
(mol)
fi, mas
C2H6
O2
N2
CO2
H2O
200
1050
3948
0
0
Out
fi, mas + viξ = fi,kel
Mol
fraction
200 - ξ
1050 – 7/2ξ
3948
2ξ
3ξ
0.0019
0.0727
0.7459
0.0718
0.1077
10
385
3948
380
570
The gas of a fuel which is composed of 3.1 % H2, 27.2% CO,
5.6% CO2, 0.5% O2 and 36.6%N2 mol respectively is burned
in an excess of 20% air. H2 was burned completely but only
98% of CO was burned. If the feed for the gas fuel is 100kg
mol, calculate the mol for each component that exists the
system.
U kg mol air
100 kg mol gas
3.1 % H2
27.2% CO
5.6% CO2
0.5 % O2
63.6% N2
F kg mol gas
H2O
CO
CO2
O2
N2

Reactions
H2 + ½ O2 → H2O
CO + ½ O2 → CO2

Basis = 100 kg mol gas

Composition of Air
Mol of O2 (complete combustion) =
(3.1 x ½) + (27.2 x ½) = 15.15 kg mol
Mol of O2 (given) = 0.5 kg mol
Mol of required = 15.15 – 0.5 = 14.7 kg mol
Mol of O2 in air = 14.7 + (14.7 x 0.2) = 17.6 kg mol
Mol of N2 in air = 79/21 x 17.6 = 66.1 kg mol

Conversion
Conversion CO = fi, in – fi, out = 0.98
fi, in
27.2 – fi, out
=
= 0.98
27.2
fCO, out = 0.544 kg mol
Conversion H2
fi, in – fi, out
=
= 1.00
fi, in
3.1 – fi, out
= 1.00
3.1
fH2, out = 0 kg mol

Extent of Reaction
ξCO = fi,out – fi,in = 0.544 – 27.2 = 26.65
vi
-1
ξH2 = fi,out – fi,in = 0 – 3.1 = 3.1
vi
-1
Species
I
H2
CO
O2
N2
CO2
H2O
In
Out
fi, mas
fi, mas + viξ = fi,kel
3.1
27.2
18.1
129.7
5.6
0
3.1 - ξH2
27.2 - ξCO
18.1- [½ξH2+½ξCO]
129.7
5.6 + ξCO
ξH2
0
0.5
3.2
129.7
32.3
3.1