Transcript 2.3 Relative Molecular Mass Determination

Chapter 2
The Mole Concept
2.1
The Mole
2.2
Ideal Gas Equation
2.3
Determination of Relative Molecular Mass
2.4 The Faraday and the Mole
1
New Way Chemistry for Hong Kong A-Level Book 1
2.1 The Mole (SB p.28)
What is “Mole”?
Item
Unit used to count
No. of
items per
for counting unit
common
2
Shoes
pairs
Eggs
dozens
12
Paper
reams
500
Particles in
Chemistry
moles
6.02 x 1023
objects
for counting particles like atoms, ions, molecules
2
New Way Chemistry for Hong Kong A-Level Book 1
2.1
The Mole (SB p.28)
How large is the amount in 1 mole?
6.02 x 1023
= 602000000000000000000000
Avogadro constant
(the amount in
1 mole)
3
New Way Chemistry for Hong Kong A-Level Book 1
2.1
The Mole (SB p.28)
$ 6.02 x 1023
All the people in the
world
so that each
get:
$ 1000 note
?
2000 years
4
count at a
rate of 2
notes/sec
New Way Chemistry for Hong Kong A-Level Book 1
2.1 The Mole (SB p.29)
How to find the number of moles?
Number of moles =
Number of particles
Avogadro consant
or
number of particles
= number of moles x (6.02 x 1023)
5
New Way Chemistry for Hong Kong A-Level Book 1
1
2.1
The Mole (SB p.29)
Why defining 6.02 x 1023 as the amount for one mole?
12 g carbon contains 6.02 x 1023 12C atoms
The mole is the amount of substance containing as
many particles as the number of atoms in 12 g of
carbon-12.
6
New Way Chemistry for Hong Kong A-Level Book 1
2.1
Molar
mass
The Mole (SB p.29)
Relative
mass
…….
12
C atom
6.02 x 1023
…….
1
H atom
1g
6.02 x 1023
Relative atomic
masses
Molar mass is the mass, in grams, of 1 mole of a substance,
e.g. the molar mass of H atom is 1 g.
7
New Way Chemistry for Hong Kong A-Level Book 1
2.1
The Mole (SB p.29)
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.
8
New Way Chemistry for Hong Kong A-Level Book 1
2.1
The Mole (SB p.30)
Calculations using molar mass
Question
We have a sample of 2 g of hydrogen atoms.
What is the number of moles?
(Relative atomic mass of H = 1)
…….
…….
1 g (1 mole)
2g
Number of moles =
1g
1 g (1 mole)
mass
= 2
9
New Way Chemistry for Hong Kong A-Level Book 1
molar mass
2.1
The Mole (SB p.30)
Number of moles =
mass (in g)
molar mass (in g)
or
Mass
= number of moles x molar mass
10
New Way Chemistry for Hong Kong A-Level Book 1
2
2.1
The Mole (SB p.30)
Example 2-1
What is the mass of 0.2 mole of calcium carbonate?
(R.a.m.* : C = 12.0, O = 16.0, Ca = 40.1)
Solution:
The chemical formula of calcium carbonate
is CaCO3.
Molar mass of calcium carbonate
= (40.1 + 12.0 + 3 x 16.0) g mol-1
= 100.1 g
11
mol-1
New Way Chemistry for Hong Kong A-Level Book 1
Answer
2.1
The Mole (SB p.30)
Solution: (cont’d)
Mass of calcium carbonate
= Number of moles x Molar mass
= 0.2 mol x 100.1 g mol-1
= 20.02g
12
New Way Chemistry for Hong Kong A-Level Book 1
2.1
The Mole (SB p.31)
Example 2-2
Calculate the number of
gold atoms in 20g of gold
atom. (R.a.m. : Au = 197.0)
Solution:
Number of gold atoms in 20g of gold coin
=
20g
x 6.02 x 1023 mol-1
197.0 g mol - 1
= 6.11 x 1023
13
New Way Chemistry for Hong Kong A-Level Book 1
Answer
2.1
The Mole (SB p.31)
Example 2-3
It is given that the molar mass of water is 18.0g mol-1.
(a)What is the mass of 4 moles of water molecule?
(b) How many molecules are there?
(c) How many atoms are there?
Solution:
(a)Mass of water
= Number of moles x Molar mass
= 4 mol x 18.0 g mol-1
Answer
= 72.0 g
14
New Way Chemistry for Hong Kong A-Level Book 1
2.1
The Mole (SB p.31)
Solution: (cont’d)
(b) There are 4 moles of water molecules.
Number of water molecules
= Number of moles x Avogadro constant
= 4 mol x 6.02 x 1023 mol-1
= 2.408 x 1024
15
New Way Chemistry for Hong Kong A-Level Book 1
2.1
The Mole (SB p.31)
Solution: (cont’d)
(c)
1 water molecule has 3 atoms (including
2 hydrogen atoms and 1 oxygen atoms).
1 mole of water molecule has 3 moles of atoms.
Thus, 4 moles of water molecules have 12 moles
of atoms.
Number of atoms
= 12 mol x 6.02 x 1023 mol-1
= 7.224 x 1024
16
New Way Chemistry for Hong Kong A-Level Book 1
2.1
The Mole (SB p.31)
Solution:
Example
2-4 formula of magnesium chloride
(a)The chemical
is MgCl2.chloride solution contains 10 g of
A magnesium
magnesium
chloride
solid
Molar mass
of MgCl
2
(a)= Calculate
the number
of moles
of magnesium
-1
(24.3 + 35.5
x2) g mol
chloride in the solution.
= 95.3 g mol-1
(b) Calculate the number of magnesium ions in the
Number
solution.of moles of MgCl2
(c)= Calculate
the number of chloride ions in the solution.
10g
1 number of ions in the solution.
(d) Calculate
the
total
95.3 g mol
=(R.a.m.:
0.105 mol
Mg = 24.3, Cl = 35.5)
17
New Way Chemistry for Hong Kong A-Level Book 1
Answer
2.1
The Mole (SB p.31)
Solution: (cont’d)
(b) 1 mole of MgCl2 contains 1 mole of Mg2+ and 2
moles of Cl- . Therefore, 0.105 mole of MgCl2
contains 0.105 mol x 6.02 x 1023 mol-1.
Number of Mg2+ ions
= Number of moles of Mg2+ x Avogadro constant
= 0.105 mol x 6.02 x 1023 mol-1
= 6.321 x 1022
18
New Way Chemistry for Hong Kong A-Level Book 1
2.1
The Mole (SB p.31)
Solution: (cont’d)
(c) 0.105 mole of MgCl2 contains 0.21 mole of Cl- .
Number of Cl- ions
= Number of moles of Cl- x Avogadro constant
= 0.21 mol x 6.02 x 1023 mol-1
= 1.264 x 1023
(d) Total number of ions
= 6.321 x 1022 + 1.264 x 1023
= 1.896 x 1023
19
New Way Chemistry for Hong Kong A-Level Book 1
Solution:
2.1 The Mole (SB p.32)
The chemical
Example
2-5formula of carbon dioxide is CO2.
Molar mass of CO2
What is the mass of carbon dioxide
molecule?
-1
= (12.0 + 16.0 x 2) g mol
(R.a.m. : C = 12.0,
O = 16.0)
-1
= 44.0 g mol
∵ Number of mole
Number of molecules
Mass
=
=
Avogadro constant
Molar Mass
∴
Mass of CO2 molecule =
1
44.0 g mol -1
6.02 10 23
-1
44.0
g
mol
Mass of a CO2 molecule=
6.02 x 10
= 7.31 x
20
10-23
New Way Chemistry for Hong Kong A-Level Book 1
23
g
Answer
2.1 ThePoint
Mole (SB2-1
p.32)
Check
(a) Find the mass in grams of 0.01 mole of zinc
sulphide.(R.a.m. : S = 32.1, Zn = 65.4)
(b) Find the number of ions in 5.61 g of calcium oxide.
(R.a.m. : O = 16.0, Ca = 40.1)
(a)
Mass = No. of moles x Molar mass
(c) Find the number of atoms in 32.05g of sulphur
dioxide. (R.a.m.
: Oof16.0,
Mass
ZnS S = 32.1)
(d) There is 4.80 g of ammonium carbonate. Find-1the
= 0.01 mol x (65.0 + 32.1) g mol
(i) number of moles of the compound,
-1
=
0.01
mol
x
95.7
g
mol
(ii) number of moles of ammonium ions,
(iii) number of
moles of hydrogen atoms, and
= 0.975g
(v) number of hydrogen atoms.
Answer
21
New Way Chemistry for Hong Kong A-Level Book 1
2.1
The Mole (SB p.32)
5.61 g
(b) No. of moles of CaO =
(40.1  16.0) g mol
= 0.1 mol
1 CaO formula unit contains 1 Ca2+ ion and
1 O2- ion.
No. of moles of ions = 0.1 mole x 2 = 0.2 mol
No of ions = 0.2 mol x 6.02 x 1023 mol-1
= 1.204 x 1023
22
New Way Chemistry for Hong Kong A-Level Book 1
-1
2.1
The Mole (SB p.32)
32.05 g
(c) No. of moles of SO2 =
(32.1  16.0  2) g mol
= 0.5 mol
1 SO2 molecule contains 1 S atom and 2O
atoms.
No. of mole of atoms = 0.5 mole x 3
= 1.5 mol
No of atoms = 1.5 mol x 6.02 x 1023 mol-1
= 9.03 x 1023
23
New Way Chemistry for Hong Kong A-Level Book 1
-1
2.1
The Mole (SB p.32)
(d) Molar mass of (NH4)2CO3 = 96.0 g mol1
4.80g
(i) No. of mole of (NH4)2CO3 = 96.0 g mol 1
(ii) ∵ 1 mole (NH4)2CO3 gives 2 moles
NH4+.
No. of moles of (NH4)2CO3 = 0.05 mol x 2
= 0.1 mol
24
New Way Chemistry for Hong Kong A-Level Book 1
2.1
The Mole (SB p.32)
(iii) ∵ 1 mole (NH4)2CO3 gives 1 mole CO32- .
No. of moles of CO32- = 0.05 mol
(iv) 1 (NH4)2CO3 formula unit contains 8H
atoms.
No. of moles of H atoms = 0.05 mol x 8
= 0.4 mol
(v) No. of H atoms
= 0.4mol x 6.02 x 1023 mol-1 = 2.408 x 1023
25
New Way Chemistry for Hong Kong A-Level Book 1
2.1
The Mole (SB p.33)
What is Molar Volume of Gases?
at 250C & 1 atm
(Room temp & pressure / rtp)
26
New Way Chemistry for Hong Kong A-Level Book 1
2.1
The Mole (SB p.33)
22.4 dm3
22.4
22.4 dm3
22.4
22.4 dm3
22.4
at 00C & 1 atm
(Standard temp & pressure / stp)
27
New Way Chemistry for Hong Kong A-Level Book 1
2.1
The Mole (SB p.33)
V
V
v
V
v
V
v
at any other fixed temp & pressure
28
New Way Chemistry for Hong Kong A-Level Book 1
2.1
The Mole (SB p.33)
Avogadro’s Law
Equal volumes of all gases at the same
temperature and pressure contain the same
number of molecules.
Equal volumes of all gases at the same
temperature and pressure contain the same
number of moles of gases.
So 1 mole of gases should have the same volume
at the same temperature and pressure.
Vn
29
where n is the no. of moles of gas
New Way Chemistry for Hong Kong A-Level Book 1
2.1
The Mole (SB p.34)
Solution:
Example 2-6
Molar mass of chloride gas (Cl2) = 35.5 x 2 g mol-1
Find the volume occupied by 3.55 g of chloride gas
at
-1
= 71.0
g mol
room temperature and pressure (Molar
volume
of gas at
R.T.P. =of
24.0
dm3 of
mol
R.a.m. :3.55
Cl =g35.5)
Number
moles
Cl-12; =
71.0 g mol 1
= 0.05 mol
Volume of Cl2
= Number of moles of Cl2 x Molar volume
= 0.05 mol x 24.0 dm3mol-1
=1.2
30
Answer
dm3
New Way Chemistry for Hong Kong A-Level Book 1
2.1
The Mole (SB p.34)
Solution:
Example 2-7
Find the
number
of molecules
in 4.48
of carbon
Molar
volume
of carbon
dioxide
gascm
at 3S.T.P.
dioxide gas at standard temperature and pressure.
4.48 cm3 3 -1
= 22.4 dm3 mol-1
(Molar volume of gas at S.T.P. = 22.4 dm3 mol1 ;
23400
-1
22
cm
Avogrado
constant
=
6.02
x
10
mol
) mol
Number of moles of CO2 =
= 2 x 10 –4 mol
Number of CO2 molecules
= 2 x 10-4 mol x 6.02 x 1023 mol-1
=1.204 x 1020
31
Answer
New Way Chemistry for Hong Kong A-Level Book 1
2.1
The Mole (SB p.35)
Solution:
Example 2-8
Molar
massvolume
of nitrogen
gas (N
The molar
of nitrogen
gas
2) is found to be 24.0
dm3 mol-1 at room temperature
and pressure. Find the
-1
= density
(14.0 +of14.0)
g mol
nitrogen
gas. (R.a.m. : N = 14.0)
= 28.0 g mol-1
∵
Density =
Mass
=
Volume
Molar Mass
Molar Volume
∴ Density of N2
28.0 g mol -1
=
24.0 dm 3 mol -1
= 1.167 g dm-3
Answer
32
New Way Chemistry for Hong Kong A-Level Book 1
2.1
The Mole (SB p.35)
Solution:
Example 2-9
Number
of moles
of the
gas3 at room temperature and
1.6
g of a gas
occupies
1.2
dm
3
1.2
dm
pressure. What is the relative molecular mass of the gas?
=
3
-1
24.0
dm
mol
(Molar Volume of gas at R.T.P. = 24.0 dm3 mol-1)
0.05 mol
Molar mass of the gas
1.6 g
=
0.05 mol
= 32 g mol-1
Relative molecular mass of the gas = 32 (no unit)
Answer
33
New Way Chemistry for Hong Kong A-Level Book 1
2.1
The Mole (SB p.35)
Check Point 2-2
(a) Find the volume of 0.6 g of hydrogen gas at room
temperature and pressure. (R.a.m.
0.6: H
g = 1.0; molar
(a) No.
of moles
of H
2 =3 mol-1)
volume
at R.T.P.
= 24.0
dm
1.0  2 g mol -1
= 0.3 molin 4.48 dm3 of
(b) Calculate the number of molecules
hydrogen at standard temperature and pressure.
Volume
= No.
of moles
x Molar
( Molar
volume
at S.T.P.
= 22.4
dm3 volume
mol-1)
3 mol
-1 3 mol-1 at
(c) The molar volume
of oxygen
dm
= 0.3 mol
x 24.0 is
dm22.4
standard temperature and pressure. Find the density
-3 at3 S.T.P.. (R.a.m. : O = 16.0)
of oxygen in=g7.2
cmdm
(d) What mass of oxygen has the same number of
moles as that in 3.2 g of sulphur dioxide? (R.a.m. :
O = 16.0, S = 32.1)
Answer
34
New Way Chemistry for Hong Kong A-Level Book 1
2.1
The Mole (SB p.35)
4.48 g
(b) No. of moles of H2 =
22.4 dm 3 mol
= 0.2 mol
-1
No of H2 molecules = 0.2 mol x 6.02 x 1023 mol-1
= 1.204 x 1023
35
New Way Chemistry for Hong Kong A-Level Book 1
2.1
The Mole (SB p.35)
( c) Density =
Mass = Molar mass
Volume Molar volume
-1
Molar mass of O2 = 16.0 x 2 g mol
=32.0 g mol-1
Molar volume of O2 = 22..4 dm3 mol-1
= 22 400 cm3 mol-1
1
32.0
g
mol
Density =
22 400 cm3 mol -1
= 1.43 x 10-3 g cm-3
36
New Way Chemistry for Hong Kong A-Level Book 1
2.1
The Mole (SB p.35)
3.2 g
(d) No. of moles of SO2 =
(32.1  16.0 x2) g mol
= 0.05 mol
No of moles of O2 = 0.05 mol
Mass = No. of moles x Molar mass
Mass of O2 = 0.05 mol x 16.0g mol-1
= 1.6g
37
New Way Chemistry for Hong Kong A-Level Book 1
-1
2.2 Ideal Gas Equation (SB p.36)
Gas Laws
Boyle’s law states that:
At constant temperature,
the volume of a given
mass of a gas is inversely
proportional to the
pressure exerted on it
PV = constant
38
New Way Chemistry for Hong Kong A-Level Book 1
2.2 Ideal Gas Equation (SB p.36)
39
New Way Chemistry for Hong Kong A-Level Book 1
2.2 Ideal Gas Equation (SB p.37)
Charles’ Law states that:
At a constant pressure, the volume of a given mass
of a gas is directly proportional to the absolute
temperature.
40
New Way Chemistry for Hong Kong A-Level Book 1
2.2 Ideal Gas Equation (SB p.37)
41
New Way Chemistry for Hong Kong A-Level Book 1
2.2 Ideal Gas Equation (SB p.37)
The Boyle’s law and Charles’ law
gives the ideal gas equation
Combining:
Vn
(Avogadro’s Law)
V  1/P
(Boyle’s Law)
VT
(Charles Law)
V  nT/P
V = RnT/P
where R is a constant
(called the Universal Gas Constant)
PV = nRT
42
New Way Chemistry for Hong Kong A-Level Book 1
2.2 Ideal Gas Equation (SB p.37)
For one mole of an ideal gas at standard temperature and
pressure,
P = 760 mmHg, 1 atm or 101 325 Nm-2 (Pa)
V = 22.4 dm3 mol-1 or 22.4 x 10-3 m3 mol-1
T = 0 oC or 273K
By substituting the values of P, V and T in S.I. Units into
the equation, the value of ideal gas constant can be found.
R = PV/T
101325 Nm 2  22.4 10 -3 m3 mol 1
=
273K
= 8.314 JK-1mol-1
43
New Way Chemistry for Hong Kong A-Level Book 1
2.2
Ideal gas equation (SB p.38)
Example 2-10
3 sample of a gas in a sealed container at 700
A
500
cm
Solution:
mm Hg and 25oC is heated to 100oC. What is the final
pressure
of the gas?
As the number
of moles of the gas is fixed, PV/T
should be a constant
P1V1
P2V2

T1 3 T2
700 mmHg  500 cm = P2  500 cm3
(273  25)K
(273  100) K
P2 = 876.17 mmHg.
Note: All temperature values used in gas laws
are on the Kelvin scale.
Answer
44
New Way Chemistry for Hong Kong A-Level Book 1
2.2
Ideal gas equation (SB p.38)
Example 2-11
Solution:
A
reaction vessel of 500 cm3 is filled with oxygen at 25oC
and the final pressure exerted on it is 101 325 Nm-2. How
= nRTare there? (Ideal gas constant =
many moles ofPV
oxygen
-1)
8.314 J K-1 mol
101 325 Nm-2 x 500 x 10-6 m3
= n x 8.314 J K-1 mol-1 x (273 + 25) K
n = 0.02 mol
There is 0.02 mole of oxygen in the reaction vessel.
.
Answer
45
New Way Chemistry for Hong Kong A-Level Book 1
2.2
Ideal gas equation (SB p.39)
Example 2-12
A 5 dm3 vessel can withstand a maximum internal pressure
of 50 atm. If 2 moles of nitrogen gas is pumped into the
vessel,
what is the highest temperature it can be safely
Solution:
heated to? (1 atm = 101 325 Nm-2, ideal gas constant =
-1)
Applying
8.314 J K-1the
molequation,
PV
T=
nR
=
50 101325 Nm -2  5  10 -3
2 mol  8.314 JK -1 mol -1
= 1523.4 K
The highest temperature it can be safely heated to is
1250.4oC.
Answer
46
New Way Chemistry for Hong Kong A-Level Book 1
2.2
Ideal gas equation (SB p.39)
Check Point 2-3
(a) A reaction vessel is filled with a gas at 20oC and 5atm.
If the vessel can withstand a maximum internal
pressure of 10 atm, what is the highest temperature it
can be(a)
safely heated to?
P1 P2
(b) A balloon is filled with
 helium at 25oC. The pressure
T2of balloon are found to be 1.5
T1
exerted and the volume
atm and 450 cm3 respectively. How many moles of
5 atm
10 atm
helium have
been introduced
into the balloon? ( 1 atm

-2; ideal gas constant = 8.314 J K-1 mol-1)
= 101 (273
325 Nm
T2
 50)
K
(c) 25.8 cm3 sample of a gas has a pressure of 690 mm Hg
586
oC.KWhat is the volume if the
and a temperatureT2of= 17
pressure is changed to 1.85 atm and the temperature to
345 K? ( 1 atm = 760 mmHg)
Answer
47
New Way Chemistry for Hong Kong A-Level Book 1
2.2
Ideal gas equation (SB p.39)
(b) PV= nRT
1.5 x 101 325 Nm-2 x 450 x 10-6 m3
= n x 8.314 J K-1 mol-1 x (273+25) K
n = 0.0276 mol
48
New Way Chemistry for Hong Kong A-Level Book 1
2.2
Ideal gas equation (SB p.39)
(c )
P1V1
P2 V2

T1
T2
690
atm  25.8 cm3
1.85 atm  V2
760

(273  17)K
345K
V=15.06cm3
49
New Way Chemistry for Hong Kong A-Level Book 1
2.3 Determination of Relative Molecular Mass (SB p.39)
Mass of volatile liquid injected = 26.590 - 26.330 = 0.260 g
50
New Way Chemistry for Hong Kong A-Level Book 1
2.3 Determination of Relative Molecular Mass (SB p.39)
Volume of trichloromethane vapour = 74.4 - 8.2 = 66.2 cm3
51
New Way Chemistry for Hong Kong A-Level Book 1
2.3 Determination of Relative Molecular Mass (SB p.39)
Temperature = 273 + 99 = 372 K
52
New Way Chemistry for Hong Kong A-Level Book 1
2.3 Determination of Relative Molecular Mass (SB p.39)
Pressure = 101325 Nm-2
53
New Way Chemistry for Hong Kong A-Level Book 1
2.3 Determination of Relative Molecular Mass (SB p.40)
PV  nRT
101325 x 66.2 x 10  n x 8.314 x 372
-6
n  2.16 x 10 mol
-3
54
New Way Chemistry for Hong Kong A-Level Book 1
2.3 Determination of Relative Molecular Mass (SB p.40)
mass
molar mass 
number of moles

0.260
-3
2.169 x 10
 119.88 g mol
55
New Way Chemistry for Hong Kong A-Level Book 1
-1
2.3 Determination of Relative Molecular Mass (SB p.40)
PV = nRT………..(1)
n = m / M………(2)
Where
m is the mass of the volatile substance
M is the molar mass of the volatile
substance
Combing (1) and (2), we obtain
PV = (m/M) RT
M = (mRT)/PV
56
New Way Chemistry for Hong Kong A-Level Book 1
2.3 Determination of Relative Molecular Mass (SB p.41)
Example 2-13
A sample of gas occupying a volume of 50 cm3 at 1 atm
oC is found to have a mass of 0.0286 g. Find the
and
25
Solution:
relative molecular mass of the gas. ( Ideal gas constant =
-1; 1 atm = 101 325 Nm-2)
8.314 J K-1molPV
= m/M RT
101 325 Nm-2 x 50 x 10-6 m3
0.0286 g
=
x 8.314 JK-1 mol-1 x (273 + 25) K
M
∴ M = 13.99 g mol-1
Therefore, the relative molecular mass of the gas
is 13. 99.
57
Answer
New Way Chemistry for Hong Kong A-Level Book 1
2.3 Determination of Relative Molecular Mass (SB p.42)
Solution:
Example 2-14
The unit of density of the gas has to be converted
The
a gas
at 450 oC and 380 mmHg is 0.033 7
to gdensity
m-3 forofthe
calculation.
g dm-3. What is-3its relative molecular
mass? ( 1 atm-3=
3
3
0.0337
g dm
0.0337
10 gm
= 33.7 g=m
760
mmHg
= 101=325
Nm-2x; ideal
gas constant
8.314 J
K-1mol-1) PM = RT
M = RT/ P
=
33.7gm 3  8.314 JK -1 mol -1  (273  450)K
380
101325Nm 2
760
= 4.0 g mol-1
Therefore, the relative molecular mass of the gas
Answer
is 4.0.
58
New Way Chemistry for Hong Kong A-Level Book 1
2.3 Determination of Relative Molecular Mass (SB p.42)
Check Point 2-4
(a) 0.204 g of phosphorus vapour occupies a volume of
81.0 cm3 at 327oC and 1 atm. Determine the relative
molecular
= 101 325
(a)-2PV = mass
m/M of
RTphosphorus. ( 1 atm
Nm ; ideal gas constant = 8.314 J K-1 mol-1)
-2 x 81.0 x 10-6m3
101 325
(b) A sample
of Nm
gas has
a mass of 12.0g and occupies a
volume of 4.16dm3 measured at-1 97oC-1 and 1.62 atm.
= (0.204
M) x 8.314J
K mass
mol of
x the gas.
Calculate
the g/
relative
molecular
-2; ideal gas constant = 8.314 J
( 1 atm =(273
101+325
Nm
327) K
-1
-1
K mol )
-1
M
=
123.99
g
mol
(c) A sample of 0.037g magnesium reacted with
hydrochloric acid to give 38.2 cm3 of hydrogen gas
∴ The relative
of phosphorus
measured
at 25omolecular
C and 740 mass
mmHg.
Use this
information
to calculate the relative atomic mass of
is 123.99.
-2;
magnesium. (1 atm = 760 mmHg = 101 325 Nm
Answer
ideal gas constant = 8.314 J K-1 mol-1)
59
New Way Chemistry for Hong Kong A-Level Book 1
2.3 Determination of Relative Molecular Mass (SB p.42)
(b) PV= (m/M)RT
1.62x 101 325 Nm-2 x 4.16 x 10-3m3
= 12.0g/ M x 8.314 J K-1 mol-1 x
(273+97)K
M = 54.06 g mol-1
∴ The relative molecular mass of the gas is
54.06.
60
New Way Chemistry for Hong Kong A-Level Book 1
2.3 Determination of Relative Molecular Mass (SB p.42)
(c) Mg(s) + 2 HCl(aq)
MgCl2(aq) + H2(g)
PV = nRT
740/760 x 101 325 Nm-2 x 38.2 x 10-6 m3
= n x 8.314 J K-1 mol-1 x (273 + 25 ) K
n = 1.52 x 10-3 mol
∴ No. of moles of H2 produced = 1.52 x 10-3.
61
New Way Chemistry for Hong Kong A-Level Book 1
2.3 Determination of Relative Molecular Mass (SB p.42)
No. of mole of Mg reacted
= No. of moles of H2 produced = 1.52 x 10-3 mol
Molar mass of Mg = Mass/ No. of moles
= 0.037g / 1.52 x 10-3 mol-1= 24.3 g mol-1
∴ The relative atomic mass of Mg is 24.3.
62
New Way Chemistry for Hong Kong A-Level Book 1
2.4 Dalton’s Law of Partial Pressures (SB p. 43)
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 that each gas
would exert as if it is present alone under the same
conditions).
PT
63
=
PA
+
PB
New Way Chemistry for Hong Kong A-Level Book 1
+
PC
2.4 Dalton’s Law of Partial Pressures (SB p. 43)
Consider a mixture of gases A, B and C occupying a
volume V. It consists of nA, nB and nC moles of each gas.
The total number of moles of gases in the mixture
ntotal = nA + nB + nC
If the equation is multiplied by RT/V, then
ntotal (RT/V) = nA (RT/V) + nB (RT/V) + nC (RT/V)
i.e. Ptotal = PA + PB + PC
(so Dalton’s Law is a direct consequence of the
Ideal Gas Equation)
64
New Way Chemistry for Hong Kong A-Level Book 1
2.4 Dalton’s Law of Partial Pressures (SB p. 43)
Besides, the partial pressure of each component gas
can be calculated from the Ideal gas law.
PA = nA(RT/V) and Ptotal = ntotal(RT/V)
i.e. PA= (nA/ntotal) Ptotal
PA = xA Ptotal
65
New Way Chemistry for Hong Kong A-Level Book 1
2.4 Dalton’s Law of Partial Pressures (SB p.43)
Example 2-15
Solution:
Air is composed of 80 % nitrogen and 20% oxygen by
volume. What are the partial pressures of nitrogen and
Mole
of fraction
N2 = 80/100
oxygen
in air at of
a pressure
of 1 atm and a temperature of
oC? ( 1 atm = 101 325 Nm-2)
25
Mole of fraction of O2 = 20/100
Partial pressure of N2 = 80/100 x 101 325 Nm-2
= 81 060 Nm-2
Partial pressure of O2 = 20/100 x 101 325 Nm-2
= 20 265 Nm-2
Answer
66
New Way Chemistry for Hong Kong A-Level Book 1
2.4 Dalton’s Law of Partial Pressures (SB p.44)
Solution:
Example 2-16
The value
between
a 5 system
dm3 vessel containing gas A at a
(a)Total
volume
of the
pressure of 15 atm and a 10 dm3 vessel containing gas B
at +
a pressure
= (5
10) dm3of
= 12
15 atm
dm3is opened.
that
temperature
of the system remains
By(a)Assuming
Boyle’s Law:
P1the
V1 =
P2V2
constant, what is the final pressure in the vessel?
Partial
pressure
ofmole
gasfractions
A
3
(b)
What
are
the
of gas A and gas B?
15 atm  5 dm
=
= 5 atm
15dm 3
Partial pressure of gas B
3
= 12 atm 10 dm = 8 atm
3
15 dm
Answer
67
New Way Chemistry for Hong Kong A-Level Book 1
2.4 Dalton’s Law of Partial Pressures (SB p.44)
Solution: (cont’d)
By Dalton’s law: Ptotal = PA + PB
Total pressure of the system= (5 + 8) atm = 13 atm
(b) Mole fraction of gas A
= PA/Ptotal = 5 atm/13 atm = 0.385
Mole fraction of gas B
= PB/Ptotal = 8atm/13 atm = 0.615
68
New Way Chemistry for Hong Kong A-Level Book 1
2.4 Dalton’s Law of Partial Pressures (SB p. 44)
Solution:
Example 2-17
Let 0.25
the partial
of nitrogen
beofPoxygen
mole ofpressure
nitrogen and
0.30 mole
are
A.
3 at 50 oC. Calculate
introduced
into
a
vessel
of
12
dm
Using
the
ideal
gas
equation
PV
=
nRT,
the partial pressures of nitrogen and oxygen and
PA xhence
12 x the
10-3total
m3 pressure exerted by the gases.
-1 -2
-1 x gas
( 1 atm
325JK
Nm
; ideal
constant
= 0.25
mol=x101
8.314
mol
(273
+ 50) K
-1)
8.314
K-1-2mol
PA ==55
946JNm
( or
0.552 atm)
Let the partial pressure of oxygen be PB.
Using the ideal gas equation PV = nRT,
PB = 0.30 mol x 8.314 JK-1 mol-1 x (273 + 50) K
PB = 67 136 Nm-2 ( or 0.663 atm)
69
New Way Chemistry for Hong Kong A-Level Book 1
Answer
2.4 Dalton’s Law of Partial Pressures (SB p. 44)
Solution: (cont’d)
Total pressure
= (55 946 + 67 136) Nm-2 = 123 082 Nm-2
Hence, the partial pressures of nitrogen and oxygen
are 0.552 atm and 0.663 atm respectively, and the
total pressure of the mixture is 1.215 atm.
70
New Way Chemistry for Hong Kong A-Level Book 1
2.4 Dalton’s Law of Partial Pressures (SB p. 45)
Solution:
Example 2-18
g oxygen
and 6.0ofgoxygen
of nitrogen are introduced
(a)4.0
Number
of moles
into a4.0g
5 dm3 vessel at 27oC.
=
1
(a)What
are the
mole fractions of oxygen and
32.0 g mol
nitrogen
in the mixture?
= 0.125
mol
(b)What
the final
pressure of the system? ( 1 atm =
Number
ofismoles
of
nitrogen
-2; ideal gas constant = 8.314 J K-1 mol-1;
101 325
Nm
6.0g
: N = 14.0,
O = 16.0)
= R.a.m.
1
28.0 g mol
= 0.214 mol
Total number of moles of gases
= ( 0.125 + 0.214) mol = 0.339 mol
71
New Way Chemistry for Hong Kong A-Level Book 1
Answer
2.4 Dalton’s Law of Partial Pressures (SB p. 45)
Solution: (cont’d)
Mole fraction of oxygen = 0.125 mol = 0.369
0.339 mol
0. 214 mol
Mole fraction of nitrogen =
= 0.631
0.339 mol
(b) Let P be the total pressure of the system.
By ideal gas equation PV = nRT,
P x 5 x 10-3 m3
= 0.339 mol x 8.314 JK-1mol-1 x (273 + 27)K
P = 169 107 Nm-2 ( or 1.67 atm)
72
New Way Chemistry for Hong Kong A-Level Book 1
2.4 Dalton’s Law of Partial Pressures (SB p. 45)
Check Point 2-5
By Boyle’s law: P1V1 = P2V2
(a) The valve between a 6 dm3 vessel containing gas A at
Partial
of an
gas8 A
a pressire
ofpressure
7 atm and
dm3 vessel containing
3
6
dm
gas B=at7 aatm
pressure
of 9 atm is opened. Assuming that
x
3
(6

8)
dm
the temperature of the system remains constant and
atm
there=is3.00
no reaction
between the gases, what is the
-2;
final Partial
pressure
of
the
system?
(
1
atm
=
101
325
Nm
pressure of gas B
3
ideal gas constant 8=dm
8.314
J K-1 mol-1)
= 9 atm x (6  8) dm 3
= 5.14 atm
Final pressure = 3.00 atm + 5.14 atm
= 8.14 atm
73
New Way Chemistry for Hong Kong A-Level Book 1
Answer
2.4 Dalton’s Law of Partial Pressures (SB p. 45)
Check Point 2-5
2g
(i)No. of moles of He =
-1 = 0.05 mol
4.0g mol
3 g4 g of argon are
(b) 2 g of helium, 3 g of nitrogen and
No. of moles of N23 =
-1 = 0.11 mol
o
14.04
g mol
introduced into 15 dm vessel
atg2100
C.
No. of moles of Ar =
-1 = 0.10 mol
39.9ofghelium,
mol
(i) What are the mole fractions
nitrogen and
Total
no. system?
of moles of gases
argon
in the
= 0.05 the
moltotal
+ 0.11
mol +of
0.10
= 0.71
(ii) Calculate
pressure
themol
system,
andmol
hence
0.05 mol
the partial pressures of helium,
nitrogen and argon.
Mole faction of He = 0.71 mol = 0.704
0.11
mol
(1 atm = 101 325 Nm-2; ideal gas
constant
Mole-1 fraction
of N2 = 0.71 mol = 0.155
-1
= 8.314 J K mol ; R.a.m. : He0.10
= 4.0,
N = 14.0, Ar = 39.9)
mol
Mole fraction of Ar = 0.71 mol = 0.141
Answer
74
New Way Chemistry for Hong Kong A-Level Book 1
2.4 Dalton’s Law of Partial Pressures (SB p. 45)
(ii) Let the total pressure of the system be P.
PV = nRT
P x 15 x 10-3m3
= 0.71 mol x 8.314 JK-1mol-1 x ( 273 + 100 ) K
P = 146 786 Nm-2
Partial pressure of He
= 146 786 Nm-2 x 0.704 = 103 337 Nm-2
Partial pressure of N2
= 146 786 Nm-2 x 0.155 = 22 752 Nm-2
Partial pressure of Ar
= 146 786 Nm-2 x 0.141 = 20 697 Nm-2
75
New Way Chemistry for Hong Kong A-Level Book 1
2.5 The Faraday and the Mole (SB p. 46)
The Faraday and the Mole?
Quantity of electricity = current x time
(coulombs, C)
(amperes, A) (seconds, s)
76
New Way Chemistry for Hong Kong A-Level Book 1
2.5 The Faraday and the Mole (SB p. 46)
Faraday’s first law of electrolysis
Faraday’s First Law of Electrolysis
The mass of a substance liberated at or dissolved from
an electrode during electrolysis is directly proportional to
the quantity of electricity passing through the electrolyte.
Cu2+(aq) + 2e-  Cu(s)
vary directly
77
New Way Chemistry for Hong Kong A-Level Book 1
2.5 The Faraday and the Mole (SB p. 47)
Faraday’s second law of electrolysis
78
New Way Chemistry for Hong Kong A-Level Book 1
2.5 The Faraday and the Mole (SB p. 47)
Faraday’s Second Law of Electrolysis
The no. of moles of different ions discharged by
the same quantity of electricity is inversely
proportional to their respective charge.
2Ag
Ag+(aq) + 2e
e-  2Ag(s)
Ag(s)
Cu2+(aq) + 2e-  Cu(s)
79
New Way Chemistry for Hong Kong A-Level Book 1
2.5 The Faraday and the Mole (SB p. 48)
Faraday Constant
Element
Electrolyte
No. of coulombs to liberates
1 mole of metal atom
Iron
Copper
Iron
Zinc
Silver
Fe2(SO4)3(aq)
CuSO4(aq)
Fe(NO3)2(aq)
ZnCl2(aq)
AgNO3(aq)
289 500
193 000
193 000
193 000
96 500
80
New Way Chemistry for Hong Kong A-Level Book 1
2.5 The Faraday and the Mole (SB p. 48)
Charge of 1 mole of electrons
= 1 Faraday
= 96 500 coulombs
Number of moles of product formed
= It / nF
81
New Way Chemistry for Hong Kong A-Level Book 1
2.5 The Faraday and the Mole (SB p. 48)
Example 2-19
How many Faradays have been pssing through a
resistance in a circuit carrying a current of 5 A for 1
hour? ( 1 F = 96 500 C)
Solution:
Q = It = 5A x ( 60 x 60) s = 18 000 C
18000 C
Number of moles of electrons =
96 500 C mol 1
= 0.187 mol
The number of Faradays passed is 0.187.
Answer
82
New Way Chemistry for Hong Kong A-Level Book 1
2.5 The Faraday and the Mole (SB p. 49)
Example 2-20
What is the number of moles of silver formed when a
current of 0.3 A is passed through a silver nitrate
solution for 30 minutes?
Solution:
Ag+(aq) + e-
Ag(s)
To form 1 mole of Ag, 1 mole of electrons (i.e. 1 F)
is required.
Number of moles of Ag formed = It / nF
0.3 A x (30x60)s
=
1 x 96 500 C mol 1
Answer
= 5.60 x 10-3 mol
83
New Way Chemistry for Hong Kong A-Level Book 1
2.5 The Faraday and the Mole (SB p. 49)
Solution:
Example 2-21
Cu2+
(aq)is+the
2emass
Cu(s) at the cathode
What
of copper formed
when a current of 0.25 A is
passed through a
2+
To discharge
1 mole of
Cu , for
2 moles
electrons
copper(II) sulphate
solution
1 hourof( R.a.m.
: Cu
= 63.5)?
(i.e. 2F) are required.
Number of moles of Cu formed = It/nF
0.25 A x (60x60)s
= 2 x 96 500 C mol 1
= 4.66 x 10-3 mol
Mass of Cu formed = 4.66 x 10-3 mol x 63.5g mol-1
= 0.296 g
84
New Way Chemistry for Hong Kong A-Level Book 1
Answer
2.5 The Faraday and the Mole (SB p. 50)
Solution:
Example
2-22
When
a dilute sulphuric
acid is electrolysed,
hydrogen is formed at the cathode and oxygen is
Find the
masses
of products
formed when
a dilute
formed
at the
anode
of the electrolytic
cell.
sulphuric acid+ solution is- electrolysed with a current
At cathode:
2H90(aq)
+ 2e
H2(g)
of 0.6 A for
minutes.
To give
1 mole
of
hydrogen
gas, 2 moles of
(R.a.m.
:
H
=
1.0,
O
=
16.0)
electrons (i.e. 2F) are required.
Number of moles of H2 (g) formed= It/nF
0.6 A x (90x60)s
= 2x96500 C mol 1
= 0.016 8 mol
Mass of H2(g) formed
= 0.016 8 mol x 1.0 x 2 g mol-1 = 0.033 6g
85
New Way Chemistry for Hong Kong A-Level Book 1
Answer
2.5 The Faraday and the Mole (SB p. 50)
Solution: (cont’d).
At anode: 4OH-(aq)
O2(g) + 2H2O(l) + 4e-
To give 1 mole of oxygen gas, 4 moles of electrons
(i.e. 4F) are given out by the hydroxide ions.
Number of moles of O2 (g) formed= It/nF
0.6 A x (90x60)s
=
4x96500 C mol 1
= 8.394 x 10-3 mol
Mass of O2(g) formed
= 8.394 x 10-3 mol x 16.0 x 2 g mol-1 = 0.268 6g
86
New Way Chemistry for Hong Kong A-Level Book 1
2.5 The Faraday and the Mole (SB p. 50)
Example 2-23
Solution:
What
mass of copperOwould
be2H
desposited
by
-(aq)
- the
4OH
(g)
+
O(l)
+
4e
2
2
quantity of electricity that
liberates
2.4 dm3 of oxygen
at room
temperature
pressure?
Tomeasured
give 1 mole
of oxygen
gas,and
4 mole
of electrons
-1;
(i.e. 4 volume
F)are given
by the= hydroxide
ions.
(Molar
of gasout
at R.T.P.
24.0 dm3 mol
R.a.m. :ofOmoles
=316.0,ofCu
63.5)
Number
O=
2 given out
2.4 dm
= 24.0dm 3 mol 1
= 0.1 mol
Number of moles of electrons given out
= 0.1 mol x 4 = 0.4 mol
Answer
87
New Way Chemistry for Hong Kong A-Level Book 1
2.5 The Faraday and the Mole (SB p. 51)
Solution: (cont’d)
Cu2+(aq) + 2e-
Cu(s)
To discharge 1 mole of Cu2+, 2 moles of electrons
(i.e. 2F) are required.
Number of moles of Cu formed = 0.4 mol / 2
= 0.2 mol
Mass of Cu deposited = 0.2 mol x 63.5 g mol-1
= 12.7 g
88
New Way Chemistry for Hong Kong A-Level Book 1
2.5 The Faraday and the Mole (SB p. 51)
Check Point 2-6
(a) current
Q =inItamperes is required to deposit 6.35g
(a) What
of copper in 50 minutes from a copper(II) sulphate
96 500 C = 0.35 A x t
solution? (1 F = 96 500; R.a.m. : Cu = 63.5)
t = 275 714 s
(b) Cu2+(aq) + 2eCu(s)
(b) What is the time required to pass 1 Faraday of
No. of moles
of Cu
= It/2F
electricity
through
anformed
electrolyte
with a current of
0.35A? ( 1F
= 96
I x 50 x 60s
6.35
g 500C)
=
1
1
63.5 g mol
2 x 96 500 C mol
I = 6.43 A
89
New Way Chemistry for Hong Kong A-Level Book 1
Answer
2.5 The Faraday and the Mole (SB p. 51)
Check Point 2-6
(c)
(c) Calculate the mass of aluminium that would be
3+(l) + 3eAl
Al(s)of a molten
deposited during the electrolysis
aluminum
salt byofa Al(s)
current
of 10 A
for/ 3F
5 hours. ( 1F
No. of moles
formed
= It
= 96 500C; R.a.m. : Al = 27.0) = 10A x 5 x 60 x 60s
3 x 96 500Cmol 1
(d) A current of 0.37A flowing for 15 =minutes
through an
0.622 mol
electrolyte liberates 0.20 g of metal X. what mass of
Mass of Al formed
X would be liberated by a current of 0.30 A for
= 0.622 mol x 27.0g mol-1
minutes?
= 16.794
Answer
90
New Way Chemistry for Hong Kong A-Level Book 1
2.5 The Faraday and the Mole (SB p. 51)
(d) No. of moles
of Xof
formed
Mass
X =
Molar Mass of X
It
nF
Since molar mass, n and F are constants,
It /( Mass of X) is a constant.
0.30 A x 30 x 60s
0.37 A x 15 x 60s
=
Mass of X
0.20 g
Mass of X = 0.324 g
91
New Way Chemistry for Hong Kong A-Level Book 1
The END
92
New Way Chemistry for Hong Kong A-Level Book 1