#### Transcript Mole concept

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1
The Mole Concept
2.1
The Mole
2.2
2.3
Ideal Gas Equation
2.4
Determination of Molar Mass
2.5
Dalton’s Law of Partial Pressures
2.1
The Mole
2
An undercover agent,
a counterspy,
a double agent
Mole
A burrowing mammal
with fossorial forefeet
A small congenital
pigmented spot on the skin
A breakwater
3
P. 3 / 66
A mole is the number of atoms in exactly 12.00 g
ofpure 12
isotope.
6C
This number, known as the Avogadro’s constant,
can be determined by mass spectrometry.
m
B
k
e
V
2
4
B is the magnetic field strength.
V is the accelerating potential.
k is a constant of the instrument.
m
B
k
e
V
2
At fixed e, k, B and V
m can be determined.
5
Q.5
Mass of one mole of
C
12
6
atoms = 12.00 g mol1
= Mass of an Avogadro’s number of
C
12
6
atoms
= Avogadro’s number  1.992648  10-23 g
12.00 g mol1
23
1.992648 10 g

= 6.022  1023 mol1
6
2.1 The mole (SB p.18)
What is “mole”?
Item
Unit used to
count
Shoes
pairs
Eggs
dozens
Paper
reams
No. of
items per
unit
for
2
counting
common
12
objects
500
Particles in
Chemistry
moles
6.022  1023
for counting particles like atoms, ions, molecules
7
sextillion
billion
~602,200,000,000,000,000,000,000 mol1
quintillion
trillion
million

1 mole ~ 602.2 sextillions
8
P. 8 / 66
The fastest supercomputer can count 1.7591015 atoms per
second.
Calculate the time taken for the superconductor to count
1 mole of carbon-12 atoms.
6.022 10
8

3.424

10
s
15
1.759 10
23
 10.85 years
9
We can count the number of coins by
weighing if the mass of one coin is known.
Similarly, we can count the number of 12C by
weighing if the mass of one 12C is known.
no. of particles
no. of moles 
6.02 1023
mass

molar mass
10
P. 10 / 66
Molar mass is the mass, in grams, of 1 mole of
a substance
11
Q.6
12
13
C
C
Relative
isotopic mass
Relative
intensity

12.000
100.00

13.003
1.12
Relative atomic mass
 100.00
 1.12 
 12.000
  13.003
  12.01
 101.12 
 101.12
12
Q.6
12
13
C
C
Relative
isotopic mass
Relative
intensity

12.000
100.00

13.003
1.12
Molar mass of carbon
= 12.01 g mol1
13
Q.6
12
13
C
C
Relative
isotopic mass
Relative
intensity

12.000
100.00

13.003
1.12
Relative isotopic mass is not exactly
equal to mass number of the isotope
14
Number of moles of a substance
number of particles
=
6.022 
1023
mol1
mass
=
molar mass
Q.7
Number of moles of oxygen atoms
=
15
number of oxygen atoms
6.022  1023 mol1
=
2g
16 g mol1
Q.7
Number of moles of oxygen atoms
=
number of oxygen atoms
6.022  1023 mol1
=
2g
16. g mol1


2
23
6
.
022

10
Number of oxygen atoms =  
 16 
2
17
23
Number of
atoms



6
.
022

10
 0.04%
O
 
 16 
= 3.011  1019
16
2.1
The mole (SB p.20)
Molar mass is the same as the relative
atomic mass in grams.
Molar mass is the same as the relative
molecular mass in grams.
Molar mass is the same as the formula
mass in grams.
Example 2-1A
Example 2-1B
Example 2-1C
Example 2-1D
Example 2-1E
Check Point 2-1
17
2.2
Molar Volume and
18
2.2
Molar volume and Avogadro’s law (SB p.24)
What is molar volume of gases?
Volume occupied by one mole of
molecules of a gas.
19
2.2
Molar volume and Avogadro’s law (SB p.24)
What is molar volume of gases?
Depends on T & P
Two sets of conditions
20
2.2
Molar volume and Avogadro’s law (SB p.24)
What is molar volume of gases?
at 298 K & 1 atm
(Room temp & pressure / R.T.P.)
21
2.2
Molar volume and Avogadro’s law (SB p.24)
What is molar volume of gases?
22.4 dm3
22.4 dm3
22.4 dm3
at 273 K & 1 atm
(Standard temp & pressure / S.T.P.)
22
Gas
Molar
mass/g
Molar volume Molar volume
at R.T.P./dm3 at S.T.P./dm3
O2
32
24.0
22.397
N2
28
24.0
22.402
H2
2
24.1
22.433
He
4
24.1
22.434
CO2
44
24.3
22.260
NH3
17
24.1
22.079
~ 24
~ 22.4
Not constant
23
2.2
Molar volume and Avogadro’s law (SB p.24)
Equal volumes of ALL gases at the same
temperature and pressure contain the
same number of moles of molecules.
At fixed T & P,
Vn
If n = 1, V = molar volume
24
2.2
Molar volume and Avogadro’s law (SB p.24)
no. of moles of gas molecules
volume of gas (dm )

molar volume (dm3 mol-1 )
3
volume of gas (dm )

at R.T.P.
3
-1
24 dm mol
3
volume of gas (dm3 )

at S.T.P.
3
-1
22.4 dm mol
25
Vn
V = Vm  n
2.2
Molar volume and Avogadro’s law (SB p.24)
Interconversions involving number
of moles
Example 2-2A
Example 2-2B
Example 2-2D
26
Example 2-2C
Check Point 2-2
2.3
27
Ideal Gas
Equation
2.3 Ideal gas equation (SB p.27)
Boyle’s law
At fixed n and T,
PV = constant or
1
V
P
n = number of moles of gas molecules
28
2.3 Ideal gas equation (SB p.28)
Schematic diagrams explaining Boyle’s law
29
1/P
30
2.3 Ideal gas equation (SB p.28)
A graph of volume against the reciprocal of
pressure for a gas at constant temperature
31
2.3 Ideal gas equation (SB p.28)
Charles’ law
At fixed n and P,
V T
T is the absolute temperature in Kelvin, K
32
2.3 Ideal gas equation (SB p.28)
Schematic diagrams explaining Charles’ law
33
2.3 Ideal gas equation (SB p.28)
Volume
-273.15 oC
0o C
Temperature / oC
A graph of volume against temperature for a gas at
constant pressure
34
2.3 Ideal gas equation (SB p.28)
/K
A graph of volume against absolute temperature
for a gas at constant pressure
35
2.3 Ideal gas equation (SB p.27)
Ideal gas equation
Vn
1
V
P
VT
Charles’ law
RnT
V
P
PV = nRT
36
Boyle’s law
R is the same for all gases
R is known as the universal gas constant
Ideal gas equation
2.3 Ideal gas equation (SB p.29)
Relationship between the ideal gas equation and the
individual gas laws
37
At fixed n,
PV
 nR  a constant
T
P1V1 P2V2 P3V3


 ...... = a constant
T1
T2
T3
Ideal gas behaviour is assumed in all gas
laws
38
2.3 Ideal gas equation (SB p.27)
Gas laws
Vn
1
V
P
VT
Charles’ law
PV = nRT
39
Boyle’s law
Ideal gas equation
2.2 Molar volume and Avogadro’s law (SB p.24)
Gas laws vs kinetic theory of gases
What is the difference between a theory and a law?
A law describes what happens under a given
set of circumstances.
A theory attempts to explain why that
behaviour occurs.
40
Ideal gas behaviour
Four assumptions as stated in kinetic
theory of gases
1. Gas particles are in a state of constant
and random motion in all directions,
undergoing frequent collisions with one
another and with walls of the container.
2. Gas particles are treated as point
masses, i.e. they do not occupy volume.
Volume of a gas = capacity of the vessel
41
Ideal gas behaviour
3. There is no interaction among gas
particles.
4. Collisions between gas particles are
perfectly elastic, i.e. kinetic energy is
conserved.
42
The ideal gas equation is obeyed by
real gases only at
(i) low pressure
(ii) high temperature
(less deviation from 24 dm3 at R.T.P.)
43
(i) At low pressure, gas particles are so far
apart that
(1) any interaction among them becomes
negligible (assumption 3)
(2) the volume occupied by the gaseous
molecules becomes negligible when
compared with that of the container
(assumption 2)
44
At high temperature,
gaseous molecules possess sufficient
energy to overcome intermolecular
45
2.3 Ideal gas equation (SB p.31)
(b) A reaction vessel is filled with a gas at 20 oC and 5 atm. If
the vessel can withstand a maximum internal pressure of
10 atm, what is the highest temperature it can be safely
heated to?
P1V P2V

T1
T2
5 atm
10 atm

(273 20)K
T2
46
T2 = 586 K
2.3 Ideal gas equation (SB p.31)
(c) A balloon is filled with helium at 25 oC. The pressure
exerted and the volume of balloon are found to be 1.5 atm
and 450 cm3 respectively. How many moles of helium
have been introduced into the balloon?
PV  nRT
1.5 atm  0.450dm3  n  0.082atm dm3 K -1 mol-1  298K
n = 0.0276 mol
Or
1.5  101325Nm-2  450  10-6 m3  n  8.314J K -1 mol-1  298K
n = 0.0276 mol
47
2.3 Ideal gas equation (SB p.31)
(d) 25.8 cm3 sample of a gas has a pressure of 690 mmHg and
a temperature of 17 oC. What is the volume of the gas if
the pressure is changed to 1.85 atm and the temperature to
345 K?
(1 atm = 760 mmHg)
P1V1 P2V2

T1
T2
690 mmHg
760 mmHg
 25.8 cm3
(273 17) K
48
V2 = 15.1 cm3
1.85 atm  V2

345K
2.3 Ideal gas equation (SB p.29)
Q.8
Calculate the universal gas constant at S.T.P.
For one mole of an ideal gas at S.T.P.,
P = 1 atm or 101,325 Nm-2 (Pa)
V = 22.4 dm3 or 0.0224 m3
n = 1 mol
T = 273K
49
2.3 Ideal gas equation (SB p.29)
3



PV 1 atm 22.4 dm 
R

 0.082 atm dm3 K1 mol1
1 mol273K 
nT
Or,

101325Nm 0.0224m 
R
2
1 mol273K 
= 8.314 Nm K1 mol1
= 8.314 J K1 mol1
50
3
Q.9
PV = nRT
m = mass of the gas
m
 PV  RT
M = molar mass of the gas
M
 m  RT
P 
V  M
ρRT
ρRT
M

M
P
51
2.4
Determination
of Molar Mass
52
Determination of Molar Mass
1.
Mass Spectrometry
2.
Density Measurement
ρRT
M
P
53
m
ρ
V
2.4 Determination of molar mass (SB p.32)
(Mass of syringe + liquid) before injection (m1)
= 38.545 g
54
m
ρ
V
2.4 Determination of molar mass (SB p.32)
(Mass of syringe + liquid) after injection (m2)
= 38.260 g
55
m
ρ
V
2.4 Determination of molar mass (SB p.32)
Mass of liquid injected (m1 – m2)
= 38.545 g – 38.260 g = 0.285 g
56
m
ρ
V
2.4 Determination of molar mass (SB p.32)
Volume of air in syringe before injection (V1)
= 10.5 cm3
57
m
ρ
V
2.4 Determination of molar mass (SB p.32)
Volume of air + vapour in syringe after injection (V2)
= 146.6 cm3
58
m
ρ
V
2.4 Determination of molar mass (SB p.32)
Volume of vapour in syringe (V2 – V1)
= 146.6 cm3 - 10.5 cm3 = 136.1 cm3
59
2.4 Determination of molar mass (SB p.32)
Once m and V of the vapour are known,
m
density( ρ 
)can be determined
V
60
2.4 Determination of molar mass (SB p.32)
Temperature = 273 + 65 = 338 K
61
2.4 Determination of molar mass (SB p.32)
1 atm
Pressure = 1 atm
62
2.4 Determination of molar mass (SB p.32)
Q.10
ρRT  m  RT  m2  m1   RT 
  
Molar mass 



P
 V  P  V2  V1   P 


0.285g

 0.082atm dm K mol (338K)

3
3 
1 atm
 136.1 10 dm 
3
1
= 58.0 g mol1
Relative molecular mass = 58.0
63
1
2.5 Dalton’s law of partial pressures (SB p.35)
Unit conversions : R = 8.314 J K1mol1 = 0.082 atm dm3 K1mol1
1m3 = 103 dm3 = 106 cm3
1 atm = 760 mmHg = 101325 Nm2 = 101325 Pa
64
2.4 Determination of molar mass (SB p.34)
(a) 0.204 g of phosphorus vapour occupies a volume of
81.0 cm3 at 327 oC and 1.00 atm. Determine the molar mass
of phosphorus.
ρRT
M
P
0.204g

3
-1
-1

0.082
atm
dm
K
mol
 (273 327)K
0.0810dm3
= 124 g mol-1
65
1.00 atm
2.4 Determination of molar mass (SB p.34)
(b) A sample of gas has a mass of 12.0 g and occupies a
volume of 4.16 dm3 measured at 97 oC and 1.62 atm.
Calculate the molar mass of the gas.
(1 atm = 101325 Nm-2; ideal gas constant = 8.314 J K-1 mol-1)
ρRT
M
P
12.0 g

4.1610-3 m3
Nm
 8.314 J K mol  (273 97) K
-1
1.62  101325Nm-2
= 54.1 g mol-1
66
-1
2.4 Determination of molar mass (SB p.34)
(c) A sample of 0.037 g magnesium reacted with hydrochloric
acid to give 38.2 cm3 of hydrogen gas measured at 25 oC
and 740 mmHg. Use this information to calculate the
relative atomic mass of magnesium.
PV
n
RT

740 mmHg
760 mmHg
0.082atm dm K mol  (273 25) K
= 1.52103 mol
67
 0.0382dm
3
3
-1
-1
2.4 Determination of molar mass (SB p.34)
(c) A sample of 0.037 g magnesium reacted with hydrochloric
acid to give 38.2 cm3 of hydrogen gas measured at 25 oC
and 740 mmHg. Use this information to calculate the
relative atomic mass of magnesium.
Mg(s) + 2HCl(aq)  MgCl2(aq) + H2(g)
1.52103 mol
1.52103 mol
0.037 g
mass
1.52 10 mol 

molar mass of Mg molar mass of Mg
3
0.037 g
-1
molar mass of Mg 

24.3
g
mol
1.52  10-3 mol
68
2.5
Dalton’s Law of
Partial Pressures
69
Experiment 1
Tap opened
empty
Gas A
At fixed T & n,
PV = constant
(15 atm)(5 dm3) = (PA)(15 dm3)
 PA = 5 atm
70
Experiment 2
Tap opened
12 atm
empty
Gas B
Gas B
At fixed T & n
PV = constant
(12 atm)(10 dm3) = (PB)(15 dm3)
 PB = 8 atm
71
Experiment 3
Tap opened
12 atm
Gas B
Gas A + Gas B
The total pressure PT = 13 atm
= 5 atm + 8 atm
= PA + PB
Partial pressures of gases A & B
72
Tap opened
12 atm
Gas B
PA = 5 atm
PB = 8 atm
Partial pressure of a constituent gas in a
mixture is the pressure that the gas would
exert if it were present alone under the
same conditions
73
2.5 Dalton’s law of partial pressures (SB p.35)
Dalton’s Law of Partial Pressures
In a mixture of gases which do not react chemically,
the total pressure of the mixture is the sum of the
partial pressures of the component gases (the sum
of the pressure that each gas would exert if it were
present alone under the same conditions).
PT
74
=
PA
+
PB
+
PC
2.5 Dalton’s law of partial pressures (SB p.35)
Derivation from ideal gas equation
Consider a mixture of gases A, B and C at fixed T & V.
nA, nB and nC are the numbers of moles of each gas.
The total number of moles of gases in the mixture
n T = nA + n B + n C
Multiply by the constant RT/V
nT(RT/V) = nA (RT/V) + nB (RT/V) + nC (RT/V)
If gases A, B and C obey ideal gas behaviour
Ptotal = PA + PB + PC
 RT  nT RT
 (nA  nB  nC )

V
 V 
75
Partial Pressures and Mole Fractions
Consider a mixture of two gases A and B in a
container of capacity V at temperature T
PAV  nART
PBV  nBRT
PTV  nTRT
PAV nART
PA
nA

 
 XA
PTV nT RT
PT nA  nB
Mole fractions
of A & B
PB
nB
PBV nBRT
 
 XB

PT nA  nB
PTV nTRT
PA = PTXA
76
PB = PTXB
XA  XB  1
Consider a mixture of gases A, B, C, D,…
XA  XB  XC  XD  ...  1
PA = PTXA
PB = PTXB
PC = PTXC
PD = PTXD
…
77
Q.11
At fixed T & n, PV = constant
For N2, P1V1 = P2V2
(0.20 Pa)(1.0 dm3) = P2(4.0 dm3)  P2 = 0.05 Pa
For O2, P1’V1’ = P2’V2’
(0.40 Pa)(2.0 dm3) = P2’(4.0 dm3)  P2’ = 0.2 Pa
By Dalton’s law of partial pressures
PT  PN2  PO2 = 0.05 Pa + 0.2 Pa = 0.25 Pa
78
Q.12
At –40oC, only N2 exists as a gas in the mixture
For a given amount of N2 at fixed V, P  T
P1 T1
P1
(273 200)K



1.50 atm (273 40) K
P2 T2
At 200oC
P1  3.05 atm  PN2
PT  4.50 atm  PN2  Ppropane
Ppropane  PT - PN2  (4.50- 3.05)atm  1.45 atm
79
At fixed T & V,
Ppropane  PTXpropane
Xpropane 
80
Ppropane
PT
1.45 atm

 0.322
4.50 atm
Q.13(a)
PT  PNH3  PH2  PN2  9.810 Nm
4
2
At fixed P & T, V  n
XNH3 
nNH3
nT

VNH3
VT
= 20%
PNH3  PT XNH3  (9.810 Nm )(20%)
4
2
= 1.96  104 Nm2
81
Q.13(a)
XH2 
nH2
nT

VH2
VT
 55%
PH2  PT XH2  (9.810 Nm )(55%)
4
2
= 5.39  104 Nm2
82
Q.13(a)
XN2 
nN2
nT

VN2
VT
 25%
PN2  PT XN2  (9.810 Nm )(25%)
4
2
= 2.45  104 Nm2
83
Q.13(b)
NH3 is removed
PT  PNH3  PH2  PN2
 PH2  PN2
= 5.39  104 Nm2 + 2.45  104 Nm2
= 7.84  104 Nm2
Note :
PH2 & PN2 remain unchanged,
but PT changes
84
2.5 Dalton’s law of partial pressures (SB p.39)
(c) The valve between a 6 dm3 vessel containing gas A at a
pressure of 7 atm and an 8 dm3 vessel containing gas B at
a pressure of 9 atm is opened. Assuming that the
temperature of the system remains constant and there is
no reaction between the gases, what is the final pressure
of the system?
85
2.5 Dalton’s law of partial pressures (SB p.39)
P1V1  P2V2
P1V1 7 atm  6 dm3
Partialpressure of gas A  P2 

 3 atm
3
V2
(6  8) dm
P1V1 9 atm  8 dm3
Partialpressure of gas B  P2 

 5.1 atm
3
V2
(6  8) dm
Total pressure = PA + PB = (3 + 5.1) atm = 8.1 atm
86
2.5 Dalton’s law of partial pressures (SB p.39)
(d) 2.0 g of helium, 3.0 g of nitrogen and 4.0 g of argon are
introduced into a 15 dm3 vessel at 100 oC.
(i) What are the mole fractions of helium, nitrogen and
argon in the system?
2.0 g
mass
no. of moles of He 

 0.50 mol
-1
molar mass 4.0 g mol
3.0 g
mass
no. of moles of N2 

 0.11 mol
-1
molar mass 28.0 g mol
4.0 g
mass
no. of moles of Ar 

 0.10 mol
-1
molar mass 39.9 g mol
Total no. of moles = (0.50 + 0.11 + 0.10) mol = 0.71 mol
XHe
87
0.50

 0.70
0.71
XN2
0.11

 0.15
0.71
0.10
XAr 
 0.14
0.71
2.5 Dalton’s law of partial pressures (SB p.39)
(d) 2.0 g of helium, 3.0 g of nitrogen and 4.0 g of argon are
introduced into a 15 dm3 vessel at 100 oC.
(ii) Calculate the total pressure of the system, and hence
the partial pressures of helium, nitrogen and argon.
nT RT 0.71 mol  0.082atm dm3 K -1 mol-1  373K
PT 

 1.45 atm
3
V
15 dm
PHe  PT XHe  1.45 atm 
PN2  PT XN2
0.50
 1.0 atm
0.71
0.11
 1.45 atm 
 0.22 atm
0.71
0.10
PAr  PT XAr  1.45 atm 
 0.20 atm
0.71
88
The END
89
2.1 The mole (SB p.20)
Back
What is the mass of 0.2 mol of calcium carbonate?
The chemical formula of calcium carbonate is CaCO3.
Molar mass of calcium carbonate = (40.1 + 12.0 + 16.0  3) g mol-1
= 100.1 g mol-1
Mass of calcium carbonate = Number of moles  Molar mass
= 0.2 mol  100.1 g mol-1
= 20.02 g
90
2.1 The mole (SB p.21)
Back

14gold pendant.
Calculate the number of gold atoms in a 片
20 g
Molar mass of gold = 197.0 g mol-1
20 g
Number of moles =
197.0 g mol 1
= 0.1015 mol
Number of gold atoms
= 0.1015 mol  6.02  1023 mol-1
= 6.11  1022
91
2.1 The mole (SB p.21)
It is given that the molar mass of water is 18.0 g mol-1.
(a) What is the mass of 4 moles of water molecules?
(b) How many molecules are there?
(c) How many atoms are there?
92
2.1 The mole (SB p.21)
(a) Mass of water = Number of moles  Molar mass
= 4 mol  18.0 g mol-1
= 72.0 g
(b) There are 4 moles of water molecules.
Number of water molecules
= Number of moles  Avogadro constant
= 4 mol  6.02  1023 mol-1
= 2.408  1024
93
2.1 The mole (SB p.21)
Back
(c) 1 water molecule has 3 atoms (i.e. 2 hydrogen atoms and 1 oxygen
atom).
1 mole of water molecules has 3 moles of atoms.
Thus, 4 moles of water molecules have 12 moles of atoms.
Number of atoms = 12 mol  6.02  1023 mol-1
= 7.224  1024
94
2.1 The mole (SB p.22)
A magnesium chloride solution contains 10 g of magnesium
chloride solid.
(a) Calculate the number of moles of magnesium chloride
in the solution.
(a) The chemical formula of magnesium chloride is MgCl2.
Molar mass of MgCl2 = (24.3 + 35.5  2) g mol-1 = 95.3 g mol-1
10 g
Number of moles of MgCl2 =
95.3 g mol 1
= 0.105 mol
95
2.1 The mole (SB p.22)
(b) Calculate the number of magnesium ions in the solution.
(b) 1 mole of MgCl2 contains 1 mole of Mg2+ ions and 2 moles of Clions.
Therefore, 0.105 mol of MgCl2 contains 0.105 mol of Mg2+ ions.
Number of Mg2+ ions
= Number of moles of Mg2+ ions  Avogadro constant
= 0.105 mol  6.02  1023 mol-1
= 6.321  1022
96
2.1 The mole (SB p.22)
(c) Calculate the number of chloride ions in the solution.
(c) 0.105 mol of MgCl2 contains 0.21 mol of Cl- ions.
Number of Cl- ions
= Number of moles of Cl- ions  Avogadro constant
= 0.21 mol  6.02  1023 mol-1
= 1.264  1023
97
2.1 The mole (SB p.22)
Back

14solution.
(d) Calculate the total number of ions in
(d) Total number of ions
= 6.321  1022 + 1.264  1023
= 1.896  1023
98
2.1 The mole (SB p.23)
Back
What is the mass of a carbon dioxide molecule?
The chemical formula of carbon dioxide is CO2.
Molar mass of CO2 = (12.0 + 16.0  2) g mol-1 = 44.0 g mol-1
Mass
Number of molecules
Number of moles =
=
Molar mass
Mass of a CO2 molecule
1
=
44.0 g mol -1
6.02  10 23 mol -1
44.0 g mol -1
Mass of a CO2 molecule =
6.02  10 23 mol -1
= 7.31  10-23 g
99
2.1 The mole (SB p.23)
(a) Find the mass in grams of 0.01 mol of zinc sulphide.
(a) Mass = No. of moles  Molar mass
Mass of ZnS = 0.01 mol  (65.4 + 32.1) g mol-1
= 0.01 mol  97.5 g mol-1
= 0.975 g
100
2.1 The mole (SB p.23)
(b) Find the number of ions in 5.61 g of calcium oxide.
5.61 g
(b) No. of moles of CaO =
(40.1  16.0) g mol 1
= 0.1 mol
1 CaO formula unit contains 1 Ca2+ ion and 1 O2- ion.
No. of moles of ions = 0.1 mol  2
= 0.2 mol
No. of ions = 0.2 mol  6.02  1023 mol-1
= 1.204  1023
101
2.1 The mole (SB p.23)
(c) Find the number of atoms in 32.05 g of sulphur dioxide.
32.05 g
(c) Number of moles of SO2 =
(32.1  16.0  2) g mol -1
= 0.5 mol
1 SO2 molecule contains 1 S atom and 2 O atoms.
No. of moles of atoms = 0.5 mol  3
= 1.5 mol
No. of atoms = 1.5 mol  6.02  1023 mol-1
= 9.03  1023
102
2.1 The mole (SB p.23)
(d) There is 4.80 g of ammonium carbonate. Find the
(i) number of moles of the compound,
(ii) number of moles of ammonium ions,
(iii) number of moles of carbonate ions,
(iv) number of moles of hydrogen atoms, and
(v) number of hydrogen atoms.
103
2.1 The mole (SB p.23)
Back
(d) Molar mass of (NH4)2CO3 = 96.0 g mol-1
4.80 g
(i) No. of moles of (NH4)2CO3 =
= 0.05 mol
96.0 g mol 1
(ii) 1 mole (NH4)2CO3 gives 2 moles of NH4+ ions.
No. of moles of NH4+ ions = 0.05 mol  2 = 0.1 mol
(iii) 1 mole (NH4)2CO3 gives 1 mole of CO32- ions.
No. of moles CO32- ions = 0.05 mol
(iv) 1 (NH4)2CO3 formula unit contains 8 H atoms.
No. of moles of H atoms = 0.05 mol  8
= 0.4 mol
(v) No. of H atoms = 0.4 mol  6.02  1023 mol-1 = 2.408  1023
104
2.2 Molar volume and Avogadro’s law (SB p.24)
What is the difference between a theory and a law?
A law tells what happens under a given set of
circumstances while a theory attempts to
explain why that behaviour occurs.
Back
105
2.2 Molar volume and Avogadro’s law (SB p.25)
Find the volume occupied by 3.55 g of chlorine gas at room
temperature and pressure.
(Molar volume of gas at R.T.P. = 24.0 dm3 mol-1)
Molar mass of chlorine gas (Cl2) = (35.5  2) g mol-1 = 71.0 g mol-1
3.55 g
Number of moles of Cl2 =
71.0 g mol 1
= 0.05 mol
Volume of Cl2 = Number of moles of Cl2  Molar volume
= 0.05 mol  24.0 dm3 mol-1
= 1.2 dm3
106
Back
2.2 Molar volume and Avogadro’s law (SB p.25)
Find the number of molecules in 4.48 cm3 of carbon dioxide
gas at standard temperature and pressure.
(Molar volume of gas at S.T.P. = 22.4 dm3 mol-1; Avogadro
constant = 6.02  1023 mol-1)
Molar volume of carbon dioxide at S.T.P. = 22.4 dm3 mol-1
= 22400 cm3 mol-1
4.48 cm3
Number of moles of CO2 =
22400 cm3 mol 1
= 2  10-4 mol
Number of CO2 molecules = 2  10-4 mol  6.02  1023 mol-1
= 1.204  1020
107
Back
2.2 Molar volume and Avogadro’s law (SB p.26)
The molar volume of nitrogen gas is found to be
24.0 dm3 mol-1 at room temperature and pressure. Find the
density of nitrogen gas.
Molar mass of nitrogen gas (N2) = (14.0 + 14.0) g mol-1 = 28.0 g mol-1
Molar mass
Mass
Density =
=
Molar volume
Volume
Density of N2
28.0 g mol -1
=
24.0 dm3 mol -1
=1.167 g dm-3
Back
108
2.2 Molar volume and Avogadro’s law (SB p.26)
1.6 g of a gas occupies 1.2 dm3 at room temperature and
pressure. What is the relative molecular mass of the gas?
(Molar volume of gas at R.T.P. = 24.0 dm3 mol-1)
1.2 dm3
Number of moles of the gas =
24.0 dm3 mol 1
= 0.05 mol
1.6 g
Molar mass of the gas =
0.05 mol
= 32 g mol-1
Relative molecular mass of the gas = 32 (no unit)
Back
109
2.2 Molar volume and Avogadro’s law (SB p.27)
(a) Find the volume occupied by 0.6 g of hydrogen gas at
room temperature and pressure.
(Molar volume of gas at R.T.P. = 24.0 dm3 mol-1)
0.6 g
(a) No. of moles of H2 =
= 0.3 mol
(1.0  2) g mol 1
Volume = No. of moles  Molar volume
= 0.3 mol  24.0 dm3 mol-1
= 7.2 dm3
110
2.2 Molar volume and Avogadro’s law (SB p.27)
(b) Calculate the number of molecules in 4.48 dm3 of
hydrogen gas at standard temperature and pressure.
(Molar volume of gas at S.T.P. = 22.4 dm3 mol-1)
4.48 dm3
(b) No. of moles of H2 =
= 0.2 mol
3
1
22.4 dm mol
No. of H2 molecules = 0.2 mol  6.02  1023 mol-1
= 1.204  1023
111
2.2 Molar volume and Avogadro’s law (SB p.27)
(c) The molar volume of oxygen gas is 22.4 dm3 mol-1 at
standard temperature and pressure. Find the density of
oxygen gas in g cm-3 at S.T.P.
Mass
Molar mass
(c) Density =
=
Volume
Molar volume
Molar mass of O2 = (16.0  2) g mol-1 = 32.0 g mol-1
Molar volume of O2 = 22.4 dm3 mol-1 = 22400 cm3 mol-1
32.0 g mol -1
Density =
22400 cm3 mol 1
= 1.43  10-3 g cm-3
112
2.2 Molar volume and Avogadro’s law (SB p.27)
(d) What mass of oxygen has the same number of moles as
that in 3.2 g of sulphur dioxide?
3.2 g
(d) No. of moles of SO2 =
(32.1  16.0  2) g mol 1
No. of moles of O2 = 0.05 mol
Mass = No. of moles  Molar mass
Mass of O2 = 0.05 mol  (16.0  2) g mol-1
= 1.6 g
Back
113
2.3 Ideal gas equation (SB p.30)
Back
A 500 cm3 sample of a gas in a sealed container at 700 mmHg
and 25 oC is heater to 100 oC. What is the final pressure of the
gas?
As the number of moles of the gas is fixed,
P1V1 P2 V2
=
T2
T1
PV should be a constant.
T
700 mmHg  500 cm3
P2  500 cm3
=
(273  25) K
(273  100) K
P2 = 876.17 mmHg
The final pressure of the gas at 100 oC is 876.17 mmHg.
Note: All temperature values used in gas laws are on the Kelvin scale.
114
2.3 Ideal gas equation (SB p.30)
Back
A reaction vessel of 500 cm3 is filled with oxygen gas at 25 oC
and the final pressure exerted on it is 101 325 Nm-2. How
many moles of oxygen gas are there?
(Ideal gas constant = 8.314 J K-1 mol-1)
PV = nRT
101325 Nm-2  500  10-6 m3 = n  8.314 J K-1 mol-1  (273 + 25) K
n = 0.02 mol
There is 0.02 mol of oxygen gas in the reaction vessel.
115
2.3 Ideal gas equation (SB p.30)
Back
A 5 dm3 vessel can withstand a maximum internal pressure
of 50 atm. If 2 moles of nitrogen gas are pumped into the
vessel, what is the highest temperature it can be safely
heated to?
Applying the equation,
PV
50  101325 Nm -2  5  10 -3 m3
T=
=
= 1523.4 K
-1
-1
nR
2 mol  8.314 J K mol
The highest temperature it can be safely heated to is 1250.4 oC.
116
2.3 Ideal gas equation (SB p.31)
(b) A reaction vessel is filled with a gas at 20 oC and 5 atm. If
the vessel can withstand a maximum internal pressure of
10 atm, what is the highest temperature it can be safely
heated to?
P1V P2V

T1
T2
5 atm
10 atm

(273 20)K
T2
117
T2 = 586 K
2.3 Ideal gas equation (SB p.31)
(c) A balloon is filled with helium at 25 oC. The pressure
exerted and the volume of balloon are found to be 1.5 atm
and 450 cm3 respectively. How many moles of helium
have been introduced into the balloon?
PV  nRT
1.5 atm  0.450dm3  n  0.082atm dm3 K -1 mol-1  298K
n = 0.0276 mol
Or
1.5  101325Nm-2  450  10-6 m3  n  8.314J K -1 mol-1  298K
n = 0.0276 mol
118
2.3 Ideal gas equation (SB p.31)
(d) 25.8 cm3 sample of a gas has a pressure of 690 mmHg and
a temperature of 17 oC. What is the volume of the gas if
the pressure is changed to 1.85 atm and the temperature to
345 K?
(1 atm = 760 mmHg)
P1V1 P2V2

T1
T2
690 mmHg
760 mmHg
 25.8 cm3
(273 17) K
119
V2 = 15.1 cm3
1.85 atm  V2

347 K
2.4 Determination of molar mass (SB p.33)
Back
A sample of gas occupying a volume of 50 cm3 at 1 atm and
25 oC is found to have a mass of 0.0286 g. Find the molar mass
of the gas.
(Ideal gas constant = 8.314 J K-1 mol-1; 1 atm = 101325 Nm-2)
PV 
m
RT
M
0.0286 g
 8.314 J K 1 mol 1  (273  25) K
M
M = 13.99 g mol-1
101325 Nm - 2  50  10  6 m3 
Therefore, the molar mass of the gas is 13.99 g mol-1.
120
2.4 Determination of molar mass (SB p.34)
Back
The density of a gas at 450 oC and 380 mmHg is
0.0337 g dm-3. What is its molar mass? (1 atm = 760 mmHg =
101325 Nm-2; ideal gas constant = 8.314 J K-1 mol-1)
The unit of density of the gas has to be converted to g
calculation.
for the
m-3
0.0337 g dm-3 = 0.0337  103 g m-3 = 33.7 g m-3
PM = RT
M = RT
P
33.7 g m -3  8.314 J K -1 mol -1  (273  450) K
=
= 4.0 g mol-1
380
 101325 Nm  2
760
Therefore, the molar mass of the gas is 4.0 g mol-1.
121
2.4 Determination of molar mass (SB p.34)
(a) 0.204 g of phosphorus vapour occupies a volume of
81.0 cm3 at 327 oC and 1 atm. Determine the molar mass of
phosphorus.
(1 atm = 101325 Nm-2; ideal gas constant = 8.314 J K-1 mol-1)
(a) PV = m RT
M
101325 Nm-2  81.0  10-6 m3
0.204 g
=
 8.314 J K-1 mol-1  (273 + 327) K
M
M = 123.99 g mol-1
The molar mass of phosphorus is 123.99 g mol-1.
122
2.4 Determination of molar mass (SB p.34)
(b) A sample of gas has a mass of 12.0 g and occupies a
volume of 4.16 dm3 measured at 97 oC and 1.62 atm.
Calculate the molar mass of the gas.
(1 atm = 101325 Nm-2; ideal gas constant = 8.314 J K-1 mol-1)
(b) PV = m RT
M
1.62  101325 Nm-2  4.16  10-3 m3
12.0 g
=
 8.314 J K-1 mol-1  (273 + 97) K
M
M = 54.06 g mol-1
The molar mass of the gas is 54.06 g mol-1.
123
2.4 Determination of molar mass (SB p.34)
(c) A sample of 0.037 g magnesium reacted with hydrochloric
acid to give 38.2 cm3 of hydrogen gas measured at 25 oC
and 740 mmHg. Use this information to calculate the
relative atomic mass of magnesium.
(1 atm = 760 mmHg = 101325 Nm-2;
ideal gas constant = 8.314 J K-1 mol-1)
124
2.4 Determination of molar mass (SB p.34)
Back
(c) Mg(s) + 2HCl(aq)  MgCl2(aq) + H2(g)
PV = nRT
740
 101325 Nm-2  38.2  10-6 m3
760
= n  8.314 J K-1 mol-1  (273 + 25) K
n = 1.52  10-3 mol
No. of moles of H2 produced = 1.52  10-3 mol
No. of moles of Mg reacted = No. of moles of H2 produced
= 1.52  10-3 mol
0.037 g
Mass
Molar mass of Mg =
=
= 24.34 g mol-1
-3
No. of moles 1.52  10 mol
The relative atomic mass of Mg is 24.34.
125
2.5 Dalton’s law of partial pressures (SB p.36)
Back
Air is composed of 80 % nitrogen and 20 % oxygen by
volume. What are the partial pressures of nitrogen and
oxygen in air at a pressure of 1 atm and a temperature of
25 oC?
80
Mole fraction of N2 =
100
Mole fraction of O2 = 20
100 80
Partial pressure of N2 =
 101325 Nm  2
100
= 81060 Nm-2
20
 101325 Nm  2
Partial pressure of O2 =
100
= 20265 Nm-2
126
2.5 Dalton’s law of partial pressures (SB p.36)
The valve between a 5 dm3 vessel containing gas A at a
pressure of 15 atm and a 10 dm3 vessel containing gas B at a
pressure of 12 atm is opened.
(a)
Assuming that the temperature of the system remains
constant, what is the final pressure of the system?
(b)
What are the mole fractions of gas A and gas B?
127
2.5 Dalton’s law of partial pressures (SB p.36)
(a) Total volume of the system = (5 + 10) dm3 = 15 dm
By Boyle’s law, P1V1 = P2V2
Partial pressure of gas A (PA)
15 atm  5 dm3
=
15 dm3
= 5 atm
Partial pressure of gas B (PB)
12 atm  10 dm3
=
15 dm3
= 8 atm
By Dalton’s law of partial pressures, Ptotal = PA + PB
Final pressure of the system = (5 + 8) atm = 13 atm
128
2.5 Dalton’s law of partial pressures (SB p.37)
Back
PA
(b) Mole fraction of gas A =
Ptotal
5 atm
=
13 atm
= 0.385
P
Mole fraction of gas B = B
Ptotal
8 atm
=
13 atm
= 0.615
129
2.5 Dalton’s law of partial pressures (SB p.37)
0.25 mol of nitrogen and 0.30 mol of oxygen are introduced
into a vessel of 12 dm3 at 50 oC. Calculate the partial
pressures of nitrogen and oxygen and hence the total
pressure exerted by the gases.
(1 atm = 101325 Nm-2; ideal gas constant = 8.314 J K-1 mol-1)
130
2.5 Dalton’s law of partial pressures (SB p.37)
Let the partial pressure of nitrogen be PA.
Using the ideal gas equation PV = nRT,
PA  12  10-3 m3 = 0.25 mol  8.314 J K-1 mol-1  (273 + 50) K
PA = 55946 Nm-2 (or 0.552 atm)
Let the partial pressure of oxygen be PB.
Using the ideal gas equation PV = nRT,
PB  12  10-3 m3 = 0.30 mol  8.314 J K-1 mol-1  (273 + 50) K
PB = 67136 Nm-2 (or 0.663 atm)
131
2.5 Dalton’s law of partial pressures (SB p.37)
Back
Total pressure of gases
= (55946 + 67136) Nm-2
= 123082 Nm-2
Or
Total pressure of gases
= (0.552 + 0.663) atm
= 1.215 atm
Hence, the partial pressures of nitrogen and oxygen are 0.552 atm and
0.663 atm respectively, and the total pressure exerted by the gases is
1.215 atm.
132
2.5 Dalton’s law of partial pressures (SB p.38)
4.0 g of oxygen and 6.0 g of nitrogen are introduced into a
5 dm3 vessel at 27 oC.
(a)
What are the mole fraction of oxygen and nitrogen in
the gas mixture?
(b) What is the final pressure of the system?
(1 atm = 101325 Nm-2;
ideal gas constant = 8.314 J K-1 mol-1)
133
2.5 Dalton’s law of partial pressures (SB p.38)
4.0 g
(a) Number of moles of oxygen =
32.0 gmol 1
= 0.125 mol
6.0 g
Number of moles of nitrogen =
28.0 gmol 1
= 0.214 mol
Total number of moles of gases = (0.125 + 0.214) mol
= 0.339 mol
0.125 mol
Mole fraction of oxygen =
= 0.369
0.339 mol
0.214 mol
Mole fraction of nitrogen = 0.339 mol = 0.631
134
2.5 Dalton’s law of partial pressures (SB p.38)
(b) Let P be the final pressure of the system.
Using the ideal gas equation PV = nRT,
P  5  10-3 m3 = 0.339 mol  8.314 J K-1 mol-1  (273 + 27) K
P = 169 107 Nm-2 (or 1.67 atm)
Back
135
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