Chapter 4 1CHEM ENG 1007 Reading Materials: Chapter 4 LECTURE 9.
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Transcript of Chapter 4 1CHEM ENG 1007 Reading Materials: Chapter 4 LECTURE 9.
Chapter 4 1CHEM ENG 1007
Reading Materials: Chapter 4
LECTURE 9
Chapter 4 2CHEM ENG 1007
Objectives
At the end of these lectures you should be able to: Explain in your own words the meaning of Explain in your own words the meaning of gram-mole, lb-mole gram-mole, lb-mole
and kilogram-mole (Lecture 7)and kilogram-mole (Lecture 7) Calculate two of the quantities mass (or mass flow rate), Calculate two of the quantities mass (or mass flow rate),
volume (or volumetric flow rate), and moles (or molars flow volume (or volumetric flow rate), and moles (or molars flow rate) from a knowledge of the third quantity for any species of rate) from a knowledge of the third quantity for any species of known density and molecular weight. (Lecture 8)known density and molecular weight. (Lecture 8)
Transform a material from one measure of concentration to Transform a material from one measure of concentration to another including mass/volume, moles/volume, ppm, ppb and another including mass/volume, moles/volume, ppm, ppb and molarity (Lecture 8)molarity (Lecture 8)
Calculate the average molecular weight of the mixture Convert the composition of a mixture from mass fraction (or
mass percent) to mole fraction (or mole percent) and vice versa.
Chapter 4 3CHEM ENG 1007
In the calculation of the mass fraction and mole fraction, the same units for mass and moles must be used in the numerator as in the denominator so that the calculated fraction is without units.
§4.2.3 Mixture Concentration
A A T A A A A T A Am m MW n MW n MW V= x y c
Summary:
x = =
y
A A A mA
T T T m
A A AA
T T T
m (kg) m (g) m (lb )
m (kg) m (g) m (lb )
n (kmol) n (mol) n (lbmol)
n (kmol) n (mol) n (lbmol)
Chapter 4 4CHEM ENG 1007
Two compounds, one with a high molecular weight and one with a low molecular weight, each comprise 50 mole% of the same mixture. Which has the greater mass fraction in the mixtureSolution:
Learning CheckLearning Check
Answer: the stream that has higher molecular weight will have greater mass fraction.
A A T A A A A T A Am m MW n MW n MW V= x y c
A A T B B T
A BT T
A B A B
MW n MW n
m m
MW MW
y yx ; x
x x
Chapter 4 5CHEM ENG 1007
A solution of salt dissolved in water is diluted with additional water. With each of the following variables, indicate whether the dilution process will cause the value of the variable to increase, decrease, or stay the same. Support your answerSolution:
Learning CheckLearning Check
salt
salt
salt
a) x
b) V
c) c
d) m
saltsalt T salt
T
ma) x V m x
m
T water salt water Tb) m m m m m V
salt salt saltn
c) c V c V
saltd) m cons tant
Chapter 4 6CHEM ENG 1007
From mole fraction:
From mass fraction:
ave i i
1 1 2 2 3 3
MW y MW
y MW y MW y MW ....
i
ave i
31 2
1 2 3
x1
MW MW
xx x ...
MW MW MW
Average/Mean Molecular Weight
Chapter 4 7CHEM ENG 1007
Illustration 7Calculate the average molecular weight of air
(1) From its approximate molar composition of 79% N2, 21% O2
Solution:
2 2
ave 1 1 2 2
ave
MW(O ) 2(16) 32 g /mol; MW(N ) 2(14) 28 g /mol
MW y MW y MW
MW (0.79)(28) (0.21)(32) 29 g /mol
Chapter 4 8CHEM ENG 1007
Illustration 7Calculate the average molecular weight of air
(2) From its approximate mass composition of 76.7 wt% N2, 23.3 wt% O2
Solution:
2 2
1 2
ave 1 2
ave
ave
MW(O ) 2(16) 32 g /mol; MW(N ) 2(14) 28 g /mol
x x1
MW MW MW
1 0.767 0.2330.0347
MW 28 32
MW 28.8 29 g /mol
Chapter 4 9CHEM ENG 1007
Illustration 7
2
2
ave
m(air) n(air)xMW(N ) 100x28 2800 g (wrong)
m(air) n(air)xMW(O ) 100x32 3200 g (wrong)
m(air) n(air)xMW 100x29 2900 g (closed enough)
(3) Assume we have 100 mol of air mixture, what is this amount in grams?
From (1) molar composition of 79% N2, 21% O2
Solution:
Chapter 4 10CHEM ENG 1007
Illustration 7
2 2 T
2 2 T
T 2
2 2 2
2
2
2 2
n(O ) y(O ) x n 0.21 100 21 mol
n(N ) y(N ) x n 0.
m(O ) n(O ) x MW(O ) 21 32 672 g
m(N ) n(N ) x MW(N )
79 100 79 mol
m =m(O ) m(N ) 672 22
79 28 221
12 28
2 g
84 g
(3) Assume we have 100 mol of air mixture, what is this amount in grams?
From (1) molar composition of 79% N2, 21% O2
Solution: (alternative method)
Chapter 4 11CHEM ENG 1007
System Properties
Quantities necessary to describe the state or condition of a system.
Extensive properties: depend on the size of system
eg. weight, force, energy, flow rate
Intensive properties: are independent of the mass or size of system,
eg. temperature, pressure, density, viscosity, concentration, mass fraction, mole fraction, etc.
Chapter 4 12CHEM ENG 1007
A solution of NaOH in water flows in a stream, and the mass flow rate of the stream suddenly increased. For each of the following properties of the stream, indicate whether the increase of the flow rate will cause the property to increase, decrease, or remain the same each case, explain your answer.Solution:
Learning CheckLearning Check
NaOH
NaOH
NaOH
a)
b) c
c) m
d) y
a) intensive property no change
b) intensive property no change
T NaOHc) extensive property m m
d) intensive property no change
Chapter 4 13CHEM ENG 1007
A solution of NaOH in water flows in a stream, and the mass flow rate of the stream suddenly increased. For each of the following properties of the stream, indicate whether the increase of the flow rate will cause the property to increase, decrease, or remain the same each case, explain your answer.Solution:
Learning CheckLearning Check
NaOH
e) n
f) V
g) MW g) intensive property no change
Te) extensive property m n
Tf ) extensive property m V
Chapter 4 14CHEM ENG 1007
Example 4.5From the acid-neutralization problem, the volumetric flow rate of the HCl solution coming from our manufacturing process is 11,600 L/hr, and the average molarity of HCl in that stream is 0.014 M.
a) How many mol of HCl are in 88 m3 of the solution?
V = 88 m3; cHCl = 0.014 M = 0.014 mol/L; n = ? mol
HClHCl
3
HCl HCl 3
nc
V
0.014 mol 88 m 1000 Ln c V 1232 mol HCl
L 1 m
Chapter 4 15CHEM ENG 1007
Example 4.5
b) How many mole of HCl are flowing from the process per minute when the volumetric flow rate of the solution is 11,600 L/hr?
HClHCl
HCl HCl
nc
V
0.014 mol 11600 L 1 hrn c V 2.71 mol /min HCl
L hr 60 min
HCl HClV 11,600 L /hr; c 0.014 M 0.014 mol /L; n ?mol /min
Chapter 4 16CHEM ENG 1007
Example 4.5
c) What is the mass fraction of HCl in the solution?
V and c were given; xHCl=?
A A T A A A A T A Am m MW n MW n MW V= x y c
x
x
x
HCl T HCl HCl
HCl HCl HCl HClHCl
T
HCl
HCl
HCl HCl
m MW n
MW n MW n
m V
MW 1 35.5 36.5 g /mol
36.5 g 2.71 mol L 1 kg hr 60 min
mol min 1 kg 1000 g 11600 L 1 hr
MW n
x
4
HCl
V
0.00051 5.1x10
Chapter 4 17CHEM ENG 1007
Illustration 8
A 2.25 L gas cylinder holds a mixture of hydrocarbons gases: methane, ethane and propane. The mass of the gas mixture in the cylinder is measured at 2.0 g. The mixture is at 25oC & 1 atm. Individual masses are given in the below table.
im
Constituent mi(g of i)
CH4 1.0
C2H6 0.6
C3H8 0.4
Total 2.0
Chapter 4 18CHEM ENG 1007
a) Mass fraction of methane?
4
4
CH 4CH
i
m g of CH1.0x 0.5
g totalm 1.0 0.6 0.4
Illustration 8
Same procedure for ethane and propane. Constituent (g of i/g total)
CH4 0.5 (i.e., 1/2)
C2H6 0.3 (i.e., 0.6/2)
C3H8 0.2 (i.e., 0.4/2)
Total 1.0
ix
Chapter 4 19CHEM ENG 1007
b) Molecular weight of Propane?
3 8MW(C H ) 12 . 3 1 . 8 44 g /mol
Constituent MW (g/mol)
CH4 16 (=12+4)
C2H6 30 (=2 x12+6x1)
C3H8 44 (=12x3+8x1)
Total -
Average 21.84
Illustration 8
Ave
16 30 44MW 30 g/mol
3
Very wrong!!!
Chapter 4 20CHEM ENG 1007
c) Mole fraction of ethane?
2 6
2 6
2 6
C HC H
C H
m 0.6n 0.02 mol
MW 30
Same procedure for methane and propane.
Total number of moles is
2 6
2 6
C HC H
T
n 0.02y 0.218
n 0.0916
Tn 0.0625 0.02 0.0091 0.0916 mol
Illustration 8
Chapter 4 21CHEM ENG 1007
Constituent yi
CH4 0.682
C2H6 0.218
C3H8 0.099
Check result by summing mole fractions?
iy 0.999 1 Thought Problem: Why don’t the mole fractions sum to 1?
Illustration 8
Chapter 4 22CHEM ENG 1007
d) Average Molecular Weight?
By mole fraction:
ave i iMW y MW
0.682 16 0.218 30 0.099 44
21.81 g /mol
Illustration 8
By mass fraction:
ave
1MW 22.0 g/mol
0.5 0.3 0.216 30 48
Chapter 4 23CHEM ENG 1007
Constituent Ci
(kg of i/m3)CH4 0.444
C2H6 0.267
C3H8 0.178
e) Mass concentration of propane:
3 8
3 8
3C H
C H
m 0.4 10c
V 32.25 10 30.178 kg/m
Apply same procedure for methane and ethane
Illustration 8
Chapter 4 24CHEM ENG 1007
§4.2.3 Conversion between mole fraction and mass fraction
Strategy for converting mass fractions/percentage to mole fractions/percentages
Assume a basis of 100 mass units
From the known mass fractions, calculate the mass of each species
Using the molecular weights, convert the mass into moles for each species
Compute the desired moles fractions or percentages.
Chapter 4 25CHEM ENG 1007
Conversion between mole fraction and mass fraction
Assume a basis of 100 moles
From the known mole fractions, calculate the number of moles of each species
Using the molecular weights, convert the moles into mass for each species
Compute the desired mass fractions or percentages.
Note: When performing such conversions, the best way is to tabulate your data
§4.2.3 Conversion between mole fraction and mass fraction
Chapter 4 26CHEM ENG 1007
Example 4.6Air has the following approximate mole%:
N2: 78.03 mole%; O2: 20.99 mole%; Ar: 0.94 mole%
Solution: Assume 100 mol of Air
2 2
2 2
2 N N T 2
2 O O T 2
Ar Ar T
N : n y n 0.7803 100 78.03 mol N
O : n y n 0.2099 100 20.99 mol O
Ar : n y n 0.0094 100 0.94 mol Ar
Convert from mole to mass
2 2 2
2 2 2
2 N N N 2
2 O O O 2
Ar Ar Ar
N : m MW n 28 78.03 2185 g N
O : m MW n 32 20.99 672 g O
Ar : m MW n 39.95 0.94 38 g Ar
Chapter 4 27CHEM ENG 1007
Example 4.6
Hence, the total number of grams is
mT = 2186 + 672 + 38 = 2896 g
2
2
2
2
N 22 N
T
O 22 O
T
ArAr
T
m 2186 g NN : 0.7548
m 2896 g
m 672g OO : 0.2320
m 2896 g
m 38 g ArAr : 0.0131
m 2895 g
x
x
x
Chapter 4 28CHEM ENG 1007
Component yi ni Mwi mi (ni.Mwi) xi
N2
0.7807 78.07 28 2186 0.7548
O2
0.2099 20.99 32 672 0.2320
Ar0.0094 0.94 39.95 38 0.0131
Total 1 100 2896 0.9999
Example 4.6
Alternatively, we can set up our work as shown in the below table
Basis: 100 mol