Chapter 4 1CHEM ENG 1007 Reading Materials: Chapter 4 LECTURE 9

Chapter 4 1CHEM ENG 1007

Reading Materials: Chapter 4

LECTURE 9


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.


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)


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 B B T

A BT T

A B A B

MW n MW n

m m

MW MW

y yx ; x

x x


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:


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


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


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


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


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:


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)


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.


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:


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


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:


NaOH

e) n

f) V

g) MW g) intensive property no change

Te) extensive property m n

Tf ) extensive property m V


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


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


Example 4.5

c) What is the mass fraction of HCl in the solution?

V and c were given; xHCl=?


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


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


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


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!!!


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


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


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


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


§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.


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


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


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


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

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Chapter 4 1CHEM ENG 1007 Reading Materials: Chapter 4 LECTURE 9