Relative atomic and molecular masses.ppt
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About This Presentation
calculating Ar and Mr
Slide Content
The Mole
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
How can we measure the mass of an
atom?
Chemists often compare masses of different atoms
with the carbon-12 atom (an isotope of carbon).
Atoms also have very small masses,
so it is not practical to use the actual masses of
atoms in calculations.
The masses of all other atoms are compared with
one-twelfth the mass of one carbon-12 atom.
How can we measure the mass of an
atom?
Chemists often compare masses of different atoms
with the carbon-12 atom (an isotope of carbon).
Atoms also have very small masses,
so it is not practical to use the actual masses of
atoms in calculations.
The masses of all other atoms are compared with
one-twelfth the mass of one carbon-12 atom.
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
Relative Atomic Mass
The relative atomic mass of any atom is
the number of times the mass of one atom
of an element is greater than 1/12 of the
mass of one carbon-12 atom.
Relative atomic mass=
Mass of 1/12 of an atom of carbon-12
Mass of one atom of the element
Relative Atomic Mass
The relative atomic mass of any atom is
the number of times the mass of one atom
of an element is greater than 1/12 of the
mass of one carbon-12 atom.
Relative atomic mass=
Mass of 1/12 of an atom of carbon-12
Mass of one atom of the element
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
Relative Atomic Mass
The symbol for relative atomic mass is A
r.
Relative atomic mass is a ratio and has no unit.
The relative atomic mass of each element is
given in the Periodic Table.
Relative Atomic Mass
The symbol for relative atomic mass is A
r.
Relative atomic mass is a ratio and has no unit.
The relative atomic mass of each element is
given in the Periodic Table.
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
Relative Atomic Mass of Some Elements
Some A
r values are not whole numbers.
35.5chlorine
16Oxygen
12Carbon
1Hydrogen
Relative atomic mass, A
r
Element
Relative Atomic Mass of Some Elements
Some A
r values are not whole numbers.
35.5chlorine
16Oxygen
12Carbon
1Hydrogen
Relative atomic mass, A
r
Element
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
Why are some A
r values not whole
numbers?
This is because such elements occur as
mixtures of isotopes.
For example, chlorine exists in two isotopic
forms: chlorine-35 and chlorine-37.
Why are some A
r values not whole
numbers?
This is because such elements occur as
mixtures of isotopes.
For example, chlorine exists in two isotopic
forms: chlorine-35 and chlorine-37.
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
How do we derive the A
r of chlorine?
Hence, relative atomic mass of chlorine
= (0.75 × 35) + (0.25 × 37)
= 26.25 + 9.25
= 35.5
A sample of chlorine is made up of
75% of chlorine-35 atoms and
25% of chlorine-37 atoms.
How do we derive the A
r of chlorine?
Hence, relative atomic mass of chlorine
= (0.75 × 35) + (0.25 × 37)
= 26.25 + 9.25
= 35.5
A sample of chlorine is made up of
75% of chlorine-35 atoms and
25% of chlorine-37 atoms.
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
Relative Molecular Mass
Many elements and compounds exist as molecules.
For example, chlorine exists as molecules.
Cl
Cl
Hence, we use relative molecular mass
instead of relative atomic mass.
Relative Molecular Mass
Many elements and compounds exist as molecules.
For example, chlorine exists as molecules.
Cl
Cl
Hence, we use relative molecular mass
instead of relative atomic mass.
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
The relative molecular mass (M
r) of an element or
compound is the mass of a molecule, compared to
1/12 the mass of one atom of carbon-12.
Like A
r, it is a ratio and therefore has no unit.
Relative Molecular Mass (M
r)
M
r
Mass of 1 molecule
Mass of 1/12 of a carbon-12 atom
=
The relative molecular mass (M
r) of an element or
compound is the mass of a molecule, compared to
1/12 the mass of one atom of carbon-12.
Like A
r, it is a ratio and therefore has no unit.
Relative Molecular Mass (M
r)
M
r
Mass of 1 molecule
Mass of 1/12 of a carbon-12 atom
=
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
How do we calculate the relative
molecular mass of ammonia (NH
3
)?
A
r
of N = 14A
r
of H = 1
M
r of NH
3
= 14 + 1 3 =
17
How do we calculate the relative
molecular mass of ammonia (NH
3
)?
A
r
of N = 14A
r
of H = 1
M
r of NH
3
= 14 + 1 3 =
17
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
How do we calculate the relative
molecular mass of ethanoic acid
(CH
3
COOH)?
A
r
of C = 12
A
r
of H = 1
M
r
of CH
3
COOH
= 12 + 1 3 + 12 + 16 + 16 + 1
= 17
A
r
of O = 16
How do we calculate the relative
molecular mass of ethanoic acid
(CH
3
COOH)?
A
r
of C = 12
A
r
of H = 1
M
r
of CH
3
COOH
= 12 + 1 3 + 12 + 16 + 16 + 1
= 17
A
r
of O = 16
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
(2 x 12) + (6 x 1) + (1 x 16) = 462 C; 6 H; 1 OC
2H
5OHEthanol
(2 x 1) + (1 x 16) = 182 H; 1 OH
2
OWater
(1 x 12) + (2 x 16) = 441 C ; 2 OCO
2
Carbon dioxide
(1 x 14) + (3 x 1) = 171 N; 3 HNH
3
Ammonia
2 x 14 = 282 NN
2
Nitrogen
Calculating M
r
Number of
atoms in one
molecule
Chemical
formula
Molecule
(2 x 12) + (6 x 1) + (1 x 16) = 462 C; 6 H; 1 OC
2H
5OHEthanol
(2 x 1) + (1 x 16) = 182 H; 1 OH
2
OWater
(1 x 12) + (2 x 16) = 441 C ; 2 OCO
2
Carbon dioxide
(1 x 14) + (3 x 1) = 171 N; 3 HNH
3
Ammonia
2 x 14 = 282 NN
2
Nitrogen
Calculating M
r
Number of
atoms in one
molecule
Chemical
formula
Molecule
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
Relative Formula Mass
Many substances such as water are covalent and exist as
molecules.
However, substances like sodium chloride are ionic and
do not exist as molecules.
The ‘relative molecular mass’ of an ionic compound
is more accurately known as relative formula mass.
Relative Formula Mass
Many substances such as water are covalent and exist as
molecules.
However, substances like sodium chloride are ionic and
do not exist as molecules.
The ‘relative molecular mass’ of an ionic compound
is more accurately known as relative formula mass.
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
Relative Formula Mass
Like relative molecular mass,
relative formula mass has the symbol M
r
and
relative formula mass has no units.
For example, the relative formula mass of sodium
chloride (NaCl) is 23 + 35.5 = 58.5.
Relative Formula Mass
Like relative molecular mass,
relative formula mass has the symbol M
r
and
relative formula mass has no units.
For example, the relative formula mass of sodium
chloride (NaCl) is 23 + 35.5 = 58.5.
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
Calculating M
r of MgSO
4
A
r of Mg = 24
A
r
of S = 32
M
r of MgSO
4 = 24 + 32 + (16 4) = 120
A
r
of O = 16
Calculating M
r of MgSO
4
A
r of Mg = 24
A
r
of S = 32
M
r of MgSO
4 = 24 + 32 + (16 4) = 120
A
r
of O = 16
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
Calculating M
r of CuSO
4 . 5H
2O
A
r of Cu = 64
A
r of S = 32
M
r
of CuSO
4
.5H
2
O
= 64 + 32 + (16 4) + 5 (2 1 + 16) = 250
A
r of O = 16
A
r
of H = 1
The ‘dot’ means ‘plus’
in calculation.
‘5’ needs to be multiplied
to entire M
r
of water.
Calculating M
r of CuSO
4 . 5H
2O
A
r of Cu = 64
A
r of S = 32
M
r
of CuSO
4
.5H
2
O
= 64 + 32 + (16 4) + 5 (2 1 + 16) = 250
A
r of O = 16
A
r
of H = 1
The ‘dot’ means ‘plus’
in calculation.
‘5’ needs to be multiplied
to entire M
r
of water.
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
Molecule
Chemical
Formula
Number of
atoms in
one
molecule
Calculating M
r
Magnesium
sulphate
MgSO
4
1 Mg; 1 S;
4 O
(1 x 24) + (1 x 32) + (4 x 16)
= 120
Calcium
carbonate
CaCO
3
1 Ca; 1 C;
3 O
(1 x 40) + (1 x 12) + (3 x 16)
= 100
Calcium nitrateCa(NO
3
)
2
1 Ca ; 2 N;
6 O
(1 x 40) + (2 x 14) + (6 x 16)
= 164
Copper(II)
sulphate
crystals
CuSO
4
. 5H
2
O 1 Cu; 1 S;
9 O; 10 H
(1 x 64) + (1 x 32) + (9 x 16) +
(10 x 1) = 250
Molecule
Chemical
Formula
Number of
atoms in
one
molecule
Calculating M
r
Magnesium
sulphate
MgSO
4
1 Mg; 1 S;
4 O
(1 x 24) + (1 x 32) + (4 x 16)
= 120
Calcium
carbonate
CaCO
3
1 Ca; 1 C;
3 O
(1 x 40) + (1 x 12) + (3 x 16)
= 100
Calcium nitrateCa(NO
3
)
2
1 Ca ; 2 N;
6 O
(1 x 40) + (2 x 14) + (6 x 16)
= 164
Copper(II)
sulphate
crystals
CuSO
4
. 5H
2
O 1 Cu; 1 S;
9 O; 10 H
(1 x 64) + (1 x 32) + (9 x 16) +
(10 x 1) = 250
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
The Mole
Atoms are too small and numerous to be counted
one at a time.
Instead, the quantity of atoms is measured by mass.
The unit of measurement for atoms and molecules is
the mole.
The symbol for the mole is mol.
The Mole
Atoms are too small and numerous to be counted
one at a time.
Instead, the quantity of atoms is measured by mass.
The unit of measurement for atoms and molecules is
the mole.
The symbol for the mole is mol.
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
The Mole
A mole of substance contains the same number of
particles as the number of atoms in 12 g of carbon-12.
12 g of carbon-12 contain approximately
6 × 10
23
carbon atoms.
The Mole
A mole of substance contains the same number of
particles as the number of atoms in 12 g of carbon-12.
12 g of carbon-12 contain approximately
6 × 10
23
carbon atoms.
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
How many particles are there in a
mole?
There are approximately 6 × 10
23
particles in
one mole of substance.
6 × 10
23
is called the Avogadro’s constant or
Avogadro’s number.
One mole of particles means 6 × 10
23
particles.
The particles could be atoms, molecules, ions or
electrons.
How many particles are there in a
mole?
There are approximately 6 × 10
23
particles in
one mole of substance.
6 × 10
23
is called the Avogadro’s constant or
Avogadro’s number.
One mole of particles means 6 × 10
23
particles.
The particles could be atoms, molecules, ions or
electrons.
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
How do we convert between number
of moles and number of particles?
Number of moles
Number of particles
6 10
23
=
How do we convert between number
of moles and number of particles?
Number of moles
Number of particles
6 10
23
=
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
Example 1: Convert 1 x 10
23
neon atoms
to moles of neon atoms
Number of moles of neon atoms
Number of moles of neon atoms
6 10
23
=
1 10
23
6 10
23
=
=0.167 mol
Example 1: Convert 1 x 10
23
neon atoms
to moles of neon atoms
Number of moles of neon atoms
Number of moles of neon atoms
6 10
23
=
1 10
23
6 10
23
=
=0.167 mol
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
Example 2: How many iron atoms are
there in 0.5 mol of iron?
Number of iron atoms
Number of moles 6 10
23
=
= 3 10
23
In 1 mole of iron, there are 6 10
23
iron atoms.
0.5 6 10
23
=
Example 2: How many iron atoms are
there in 0.5 mol of iron?
Number of iron atoms
Number of moles 6 10
23
=
= 3 10
23
In 1 mole of iron, there are 6 10
23
iron atoms.
0.5 6 10
23
=
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
Example 3: How many hydrogen atoms are
there in 3 moles of hydrogen gas?
Hydrogen gas is made up of hydrogen molecules (H
2).
In 3 moles of H
2
molecules, there are 3 2 = 6 moles of H atoms.
In 1 mole of H
2
molecules, there are 2 moles of H atoms.
Number of hydrogen atoms
Number of moles 6 10
23
=
= 3.6 x 10
24
6 6 10
23
=
Example 3: How many hydrogen atoms are
there in 3 moles of hydrogen gas?
Hydrogen gas is made up of hydrogen molecules (H
2).
In 3 moles of H
2
molecules, there are 3 2 = 6 moles of H atoms.
In 1 mole of H
2
molecules, there are 2 moles of H atoms.
Number of hydrogen atoms
Number of moles 6 10
23
=
= 3.6 x 10
24
6 6 10
23
=
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
How is the mole related to chemical
equations?
2 particles
(atoms) of
sodium
1 particle
(molecule) of
chlorine
2 moles of
sodium atoms
1 mole of chlorine
molecules
Na + Cl
2
NaCl 22
OR OR
2 units of sodium
chloride
2 moles of sodium
chloride
OR
How is the mole related to chemical
equations?
2 particles
(atoms) of
sodium
1 particle
(molecule) of
chlorine
2 moles of
sodium atoms
1 mole of chlorine
molecules
Na + Cl
2
NaCl 22
OR OR
2 units of sodium
chloride
2 moles of sodium
chloride
OR
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
The molar mass of an element is the mass
of one mole of atoms of the element.
What is the mass of one mole of
atoms of an element?
Do you notice a
relationship between the
value of A
r and the molar
mass of a substance?
Element A
r
Molar
mass
Aluminium 27 27 g
carbon 12 12 g
Neon 20 20 g
Oxygen 16 16 g
The molar mass of an element is the mass
of one mole of atoms of the element.
What is the mass of one mole of
atoms of an element?
Do you notice a
relationship between the
value of A
r and the molar
mass of a substance?
Element A
r
Molar
mass
Aluminium 27 27 g
carbon 12 12 g
Neon 20 20 g
Oxygen 16 16 g
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
What is the relationship between
mole and molar mass?
The number of moles of an element can be
calculated using the formula:
Number of moles
of an element
Mass of element (g)
A
r of element
=
What is the relationship between
mole and molar mass?
The number of moles of an element can be
calculated using the formula:
Number of moles
of an element
Mass of element (g)
A
r of element
=
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
Example 1 Determine the number of
moles in 0.196 kg of iron. (A
r
: Fe = 56)
Number of moles of iron
mass of iron in grams
A
r
of iron
=
0.196 × 1000
56
=
=3.5 mol
Example 1 Determine the number of
moles in 0.196 kg of iron. (A
r
: Fe = 56)
Number of moles of iron
mass of iron in grams
A
r
of iron
=
0.196 × 1000
56
=
=3.5 mol
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
Example 2
a) How many moles of lead are there in 1.204 ×
10
22
atoms of lead?
b) What is the mass of 1.204 × 10
22
atoms of
lead? (A
r
: Pb = 207)
a) Number of moles of lead
number of atoms of lead
Avogadro’s constant
=
1.204 × 10
22
6 × 10
23
=
=0.02 mol
Example 2
a) How many moles of lead are there in 1.204 ×
10
22
atoms of lead?
b) What is the mass of 1.204 × 10
22
atoms of
lead? (A
r
: Pb = 207)
a) Number of moles of lead
number of atoms of lead
Avogadro’s constant
=
1.204 × 10
22
6 × 10
23
=
=0.02 mol
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
Example 2 (continued)
a) How many moles of lead are there in 1.204 ×
10
22
atoms of lead?
b) What is the mass of 1.204 × 10
22
atoms of
lead? (A
r
: Pb = 207)
b) Mass of lead
= number of moles × A
r
of lead
= 0.02 × 207
= 4.14 g
Example 2 (continued)
a) How many moles of lead are there in 1.204 ×
10
22
atoms of lead?
b) What is the mass of 1.204 × 10
22
atoms of
lead? (A
r
: Pb = 207)
b) Mass of lead
= number of moles × A
r
of lead
= 0.02 × 207
= 4.14 g
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
What is the mass of one mole of
molecules or one mole of a compound?
One mole of a substance has a mass equal its M
r.
18 g(2 1) + 16
= 18
H
2
OWater
40 g24 + 16 = 40MgOMagnesium
oxide
254 g2 127 = 160I
2
Iodine
32 g2 16 = 32O
2
Oxygen
Molar massM
r
FormulaSubstance
What is the mass of one mole of
molecules or one mole of a compound?
One mole of a substance has a mass equal its M
r.
18 g(2 1) + 16
= 18
H
2
OWater
40 g24 + 16 = 40MgOMagnesium
oxide
254 g2 127 = 160I
2
Iodine
32 g2 16 = 32O
2
Oxygen
Molar massM
r
FormulaSubstance
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
The number of moles of a substance can be
calculated using the following formulae:
Number of moles
Mass of substance (g)
M
r
=
or
Number of particles
6 10
23
=
The number of moles of a substance can be
calculated using the following formulae:
Number of moles
Mass of substance (g)
M
r
=
or
Number of particles
6 10
23
=
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
Example 1
A conical flask contains 68.4 g of octane (C
8
H
18
).
How many molecules of octane are there in the
flask?
Number of moles of octane
Mass of octane
M
r
of octane
=
6.84
114
=
=0.6 mol
Example 1
A conical flask contains 68.4 g of octane (C
8
H
18
).
How many molecules of octane are there in the
flask?
Number of moles of octane
Mass of octane
M
r
of octane
=
6.84
114
=
=0.6 mol
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
Example 1 (continued)
A conical flask contains 68.4 g of octane (C
8
H
18
).
How many molecules of octane are there in the
flask?
Number of molecules of octane
= number of moles × Avogadro’s constant
= 0.6 × 6 × 10
23
= 3.6 × 10
23
Example 1 (continued)
A conical flask contains 68.4 g of octane (C
8
H
18
).
How many molecules of octane are there in the
flask?
Number of molecules of octane
= number of moles × Avogadro’s constant
= 0.6 × 6 × 10
23
= 3.6 × 10
23
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
Example 2
How many ions are there in 20 g of magnesium
oxide (MgO)?
Number of moles of MgO
Mass of magnesium oxide
M
r
of magnesium oxide
=
20
40
=
=0.5 mol
Example 2
How many ions are there in 20 g of magnesium
oxide (MgO)?
Number of moles of MgO
Mass of magnesium oxide
M
r
of magnesium oxide
=
20
40
=
=0.5 mol
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
Example 2 (continued)
How many ions are there in 20 g of magnesium
oxide (MgO)?
MgO Mg
2+
+ O
2–
1 mol of MgO 1 mol of Mg
2+
ions and 1 mol of O
2−
ions.
0.5 mol of MgO 0.5 mol of Mg
2+
ions and
0.5 mol of O
2−
ions.
Hence, 0.5 mol of MgO contains 1 mol of ions.
Number of ions = 1 × 6 × 10
23
= 6 × 10
23
Example 2 (continued)
How many ions are there in 20 g of magnesium
oxide (MgO)?
MgO Mg
2+
+ O
2–
1 mol of MgO 1 mol of Mg
2+
ions and 1 mol of O
2−
ions.
0.5 mol of MgO 0.5 mol of Mg
2+
ions and
0.5 mol of O
2−
ions.
Hence, 0.5 mol of MgO contains 1 mol of ions.
Number of ions = 1 × 6 × 10
23
= 6 × 10
23
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
Percentage Composition of
Compounds
Chemists need to conduct analysis in order to find out
how much of each element there is in a new compound.
They do so by finding out the mass of each element in
the compound.
In this way, chemists know the percentage composition
of a compound.
Percentage Composition of
Compounds
Chemists need to conduct analysis in order to find out
how much of each element there is in a new compound.
They do so by finding out the mass of each element in
the compound.
In this way, chemists know the percentage composition
of a compound.
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
How do we find the percentage
composition of element?
Percentage by mass of an element in a compound
A
r
of element × number of atoms in formula
M
r of compound
= 100%
The percentage by mass of an element in a compound
can be found using the formula:
How do we find the percentage
composition of element?
Percentage by mass of an element in a compound
A
r
of element × number of atoms in formula
M
r of compound
= 100%
The percentage by mass of an element in a compound
can be found using the formula:
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
How do we find the percentage
composition of hydrogen peroxide?
M
r
of hydrogen peroxide (H
2
O
2
)
= (2 × 1) + (2 × 16) = 34
Percentage of hydrogen in hydrogen peroxide
A
r of hydrogen × number of hydrogen atoms
M
r of hydrogen peroxide
=
1 2
34
=
=5.9 %
100%
100%
How do we find the percentage
composition of hydrogen peroxide?
M
r
of hydrogen peroxide (H
2
O
2
)
= (2 × 1) + (2 × 16) = 34
Percentage of hydrogen in hydrogen peroxide
A
r of hydrogen × number of hydrogen atoms
M
r of hydrogen peroxide
=
1 2
34
=
=5.9 %
100%
100%
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Percentage of oxygen in hydrogen peroxide
A
r
of oxygen × number of oxygen atoms
M
r
of hydrogen peroxide
=
16 2
34
=
=94.1 %
100%
100%
Percentage of oxygen in hydrogen peroxide
A
r
of oxygen × number of oxygen atoms
M
r
of hydrogen peroxide
=
16 2
34
=
=94.1 %
100%
100%
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Example 1 Calculate the percentage of water in
copper(II) sulphate crystals (CuSO
4
.5H
2
O).
M
r
of copper(II) sulphate crystals
= 64 + 32 + (4 × 16) + (5 × 18) = 250
M
r
of water = (1 × 2) + 16 = 18
Percentage of water in CuSO
4
.5H
2
O
M
r
of water × number of water molecules
M
r
of copper(II) sulphate crystals
=
18 5
250
=
=36 %
100%
100%
Example 1 Calculate the percentage of water in
copper(II) sulphate crystals (CuSO
4
.5H
2
O).
M
r
of copper(II) sulphate crystals
= 64 + 32 + (4 × 16) + (5 × 18) = 250
M
r
of water = (1 × 2) + 16 = 18
Percentage of water in CuSO
4
.5H
2
O
M
r
of water × number of water molecules
M
r
of copper(II) sulphate crystals
=
18 5
250
=
=36 %
100%
100%
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Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
How can we work out the formula of
a compound?
We can conduct experiments to find out the formula
of a compound.
1. First, we find out the mass of the reactants taking
part in the reaction.
2. Next, we work out the relative numbers of moles
of the reactants used.
How can we work out the formula of
a compound?
We can conduct experiments to find out the formula
of a compound.
1. First, we find out the mass of the reactants taking
part in the reaction.
2. Next, we work out the relative numbers of moles
of the reactants used.
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
Working Out the Formula of
Magnesium Oxide
To work out the formula of magnesium oxide produced by
the combustion of magnesium, the following apparatus is
used.
Magnesium ribbon
Clay triangle
Tripod stand
lid
crucible
Working Out the Formula of
Magnesium Oxide
To work out the formula of magnesium oxide produced by
the combustion of magnesium, the following apparatus is
used.
Magnesium ribbon
Clay triangle
Tripod stand
lid
crucible
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
Magnesium ribbon
Clay triangle
Tripod stand
lid
crucible
Procedure
1. Weigh a crucible together with the lid. Put a coil
of magnesium ribbon in it and weigh again.
2. Put the lid on the crucible and heat the crucible
gently.
When the magnesium catches fire (you will see a
white glow through the crucible), heat it more
strongly.
Magnesium ribbon
Clay triangle
Tripod stand
lid
crucible
Procedure
1. Weigh a crucible together with the lid. Put a coil
of magnesium ribbon in it and weigh again.
2. Put the lid on the crucible and heat the crucible
gently.
When the magnesium catches fire (you will see a
white glow through the crucible), heat it more
strongly.
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
Magnesium ribbon
Clay triangle
Tripod stand
lid
crucible
3. Use a pair of tongs to lift the lid slightly from time
to time to allow air in.
Quickly replace the lid to make sure that magnesium
oxide formed does not escape.
4. When the burning is complete, allow the crucible to
cool completely. Weigh the crucible together with the
lid and the magnesium oxide in it.
Magnesium ribbon
Clay triangle
Tripod stand
lid
crucible
3. Use a pair of tongs to lift the lid slightly from time
to time to allow air in.
Quickly replace the lid to make sure that magnesium
oxide formed does not escape.
4. When the burning is complete, allow the crucible to
cool completely. Weigh the crucible together with the
lid and the magnesium oxide in it.
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
Sample results
Mass of crucible + lid = 26.52 g
Mass of crucible + lid + magnesium = 27.72 g
Mass of crucible + lid + magnesium oxide = 28.52 g
Calculations
Mass of magnesium = 27.72 − 26.52 = 1.20 g
Mass of magnesium oxide produced
= 28.52 − 26.52 = 2.00 g
Mass of oxygen reacted = 2.00 − 1.20 = 0.80 g
Sample results
Mass of crucible + lid = 26.52 g
Mass of crucible + lid + magnesium = 27.72 g
Mass of crucible + lid + magnesium oxide = 28.52 g
Calculations
Mass of magnesium = 27.72 − 26.52 = 1.20 g
Mass of magnesium oxide produced
= 28.52 − 26.52 = 2.00 g
Mass of oxygen reacted = 2.00 − 1.20 = 0.80 g
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
Deriving the Formula
Element Mg O
Mass (from experiment) 1.20 g 0.80 g
Relative atomic mass 24 16
Number of moles
Molar ratio (divide by the
smallest number from
the previous row)
1.20
24
= 0.05
0.05
0.05
= 1
0.05
0.05
= 1
0.80
16
= 0.05
Step 1: List the mass of the element.
Deriving the Formula
Element Mg O
Mass (from experiment) 1.20 g 0.80 g
Relative atomic mass 24 16
Number of moles
Molar ratio (divide by the
smallest number from
the previous row)
1.20
24
= 0.05
0.05
0.05
= 1
0.05
0.05
= 1
0.80
16
= 0.05
Step 1: List the mass of the element.
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
Deriving the Formula
Molar ratio (divide by the
smallest number from
the previous row)
Number of moles
1624Relative atomic mass
0.80 g1.20 gMass (from experiment)
OMgElement
1.20
24
= 0.05
0.05
0.05
= 1
0.05
0.05
= 1
0.80
16
= 0.05Step 2: State the A
r
of the element.
Deriving the Formula
Molar ratio (divide by the
smallest number from
the previous row)
Number of moles
1624Relative atomic mass
0.80 g1.20 gMass (from experiment)
OMgElement
1.20
24
= 0.05
0.05
0.05
= 1
0.05
0.05
= 1
0.80
16
= 0.05Step 2: State the A
r
of the element.
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
Deriving the Formula
Step 3: Derive the number of moles by dividing the mass
with A
r
.
Molar ratio (divide by the
smallest number from
the previous row)
Number of moles
1624Relative atomic mass
0.80 g1.20 gMass (from experiment)
OMgElement
1.20
24
= 0.05
0.05
0.05
= 1
0.05
0.05
= 1
0.80
16
= 0.05
Deriving the Formula
Step 3: Derive the number of moles by dividing the mass
with A
r
.
Molar ratio (divide by the
smallest number from
the previous row)
Number of moles
1624Relative atomic mass
0.80 g1.20 gMass (from experiment)
OMgElement
1.20
24
= 0.05
0.05
0.05
= 1
0.05
0.05
= 1
0.80
16
= 0.05
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
Step 4: Obtain the molar ratio.
Deriving the Formula
The simplest formula is MgO.
Molar ratio (divide by the
smallest number from
the previous row)
Number of moles
1624Relative atomic mass
0.80 g1.20 gMass (from experiment)
OMgElement
1.20
24
= 0.05
0.05
0.05
= 1
0.05
0.05
= 1
0.80
16
= 0.05
Step 4: Obtain the molar ratio.
Deriving the Formula
The simplest formula is MgO.
Molar ratio (divide by the
smallest number from
the previous row)
Number of moles
1624Relative atomic mass
0.80 g1.20 gMass (from experiment)
OMgElement
1.20
24
= 0.05
0.05
0.05
= 1
0.05
0.05
= 1
0.80
16
= 0.05
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
Empirical Formula
MgO is the simplest formula that fits the
experimental results.
Other formulae such as Mg
2
O
2
or Mg
3
O
3
also fit
the results since the ratios of magnesium to
oxygen are the same.
The formula MgO is called the simplest formula
or the empirical formula of magnesium oxide.
Empirical Formula
MgO is the simplest formula that fits the
experimental results.
Other formulae such as Mg
2
O
2
or Mg
3
O
3
also fit
the results since the ratios of magnesium to
oxygen are the same.
The formula MgO is called the simplest formula
or the empirical formula of magnesium oxide.
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
Empirical Formula
The empirical formula of a compound shows
• the types of elements present in it,
• the simplest ratio of the different types of atoms
in it.
Empirical Formula
The empirical formula of a compound shows
• the types of elements present in it,
• the simplest ratio of the different types of atoms
in it.
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
To determine the empirical formula
of copper(II) oxide
Excess methane burning off
Copper(II) oxide
methane
To determine the empirical formula
of copper(II) oxide
Excess methane burning off
Copper(II) oxide
methane
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
Procedure
1.Weigh a porcelain boat. Put one spatula of copper(II)
oxide (a black powder) in the boat and weigh again.
2.Put the porcelain boat containing copper(II) oxide into
the middle of a Pyrex test tube with a small jet hole at
its end.
3.Set up the apparatus as shown. Allow methane gas
(from the gas tap) to pass through the apparatus for
about 30 seconds to remove air.
Procedure
1.Weigh a porcelain boat. Put one spatula of copper(II)
oxide (a black powder) in the boat and weigh again.
2.Put the porcelain boat containing copper(II) oxide into
the middle of a Pyrex test tube with a small jet hole at
its end.
3.Set up the apparatus as shown. Allow methane gas
(from the gas tap) to pass through the apparatus for
about 30 seconds to remove air.
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
4. Light the gas that escapes through the hole in the test
tube. Then heat the copper(II) oxide until no further
colour change is observed.
5. Turn off the Bunsen burner but allow the methane gas to
flow through the apparatus while the apparatus is cooling.
6. Turn off the gas supply when the apparatus is completely
cool. Weigh the porcelain boat and its contents.
4. Light the gas that escapes through the hole in the test
tube. Then heat the copper(II) oxide until no further
colour change is observed.
5. Turn off the Bunsen burner but allow the methane gas to
flow through the apparatus while the apparatus is cooling.
6. Turn off the gas supply when the apparatus is completely
cool. Weigh the porcelain boat and its contents.
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
Sample results
Mass of porcelain boat = 18.40 g
Mass of porcelain boat + copper(II) oxide = 35.89 g
Mass of porcelain boat + copper = 32.37 g
Calculations
Mass of copper = 32.37 − 18.40 = 13.97 g
Mass of oxygen = 35.89 − 32.37 = 3.52 g
Sample results
Mass of porcelain boat = 18.40 g
Mass of porcelain boat + copper(II) oxide = 35.89 g
Mass of porcelain boat + copper = 32.37 g
Calculations
Mass of copper = 32.37 − 18.40 = 13.97 g
Mass of oxygen = 35.89 − 32.37 = 3.52 g
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
The empirical formula of copper(II) oxide is thus CuO.
Element Cu O
Mass (from experiment) 13.97 g 3.52 g
Relative atomic mass 64 16
Number of moles
Molar ratio
(divide by the smallest
number from the
previous row)
= 13.97
13.97
13.97
= 1
0.22
0.22
= 0.22
16
3.52
= 1
0.22
0.22
The empirical formula of copper(II) oxide is thus CuO.
Element Cu O
Mass (from experiment) 13.97 g 3.52 g
Relative atomic mass 64 16
Number of moles
Molar ratio
(divide by the smallest
number from the
previous row)
= 13.97
13.97
13.97
= 1
0.22
0.22
= 0.22
16
3.52
= 1
0.22
0.22
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
Molecular Formula
Its actual formula is P
4
O
10
.
We call this the molecular
formula.
The molecular formula is
the formula that shows the
exact number of atoms of
each element in a molecule.
The empirical formula of phosphorus(V) oxide
as determined by experiment is P
2
O
5
.
Molecular Formula
Its actual formula is P
4
O
10
.
We call this the molecular
formula.
The molecular formula is
the formula that shows the
exact number of atoms of
each element in a molecule.
The empirical formula of phosphorus(V) oxide
as determined by experiment is P
2
O
5
.
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What is the relationship between
empirical formula and molecular
formula?
For many compounds, such as water and
ammonia, the empirical formula and
molecular formula are the same.
However, there are also many compounds
(especially organic compounds) whose
molecular formulae differ from their
empirical formulae.
What is the relationship between
empirical formula and molecular
formula?
For many compounds, such as water and
ammonia, the empirical formula and
molecular formula are the same.
However, there are also many compounds
(especially organic compounds) whose
molecular formulae differ from their
empirical formulae.
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
What is the relationship between
empirical formula and molecular
formula?
Where the empirical formula and molecular
formula are different, the molecular formula is
always a multiple of the empirical formula.
For example, the molecular formula and empirical
formula of phosphorus(V) oxide are P
4
O
10
and
P
2
O
5
respectively.
The multiple is 2.
What is the relationship between
empirical formula and molecular
formula?
Where the empirical formula and molecular
formula are different, the molecular formula is
always a multiple of the empirical formula.
For example, the molecular formula and empirical
formula of phosphorus(V) oxide are P
4
O
10
and
P
2
O
5
respectively.
The multiple is 2.
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
Note that:
It is possible for different compounds to have the
same empirical formula.
For example, ethene (C
2H
4) and propene (C
3H
6)
are two compounds with the same empirical
formula, CH
2.
Note that:
It is possible for different compounds to have the
same empirical formula.
For example, ethene (C
2H
4) and propene (C
3H
6)
are two compounds with the same empirical
formula, CH
2.
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How to derive molecular formula?
We can find the molecular formula of a
substance if we know two things: the
empirical formula and the relative molecular
mass of the substance. They are related by
the multiple, n.
n
Relative molecular mass
M
r
from empirical formula
=
How to derive molecular formula?
We can find the molecular formula of a
substance if we know two things: the
empirical formula and the relative molecular
mass of the substance. They are related by
the multiple, n.
n
Relative molecular mass
M
r
from empirical formula
=
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
Example 1
The empirical formula of ethane is CH
3
. Given that
the relative molecular mass of ethane is 30, what is
its molecular formula?
M
r
of ethane from empirical formula = 12 + 3 = 15
=
M
r
from empirical formula
Mass of one atom of the an element12
15
= 2
Hence, the molecular formula of ethane = C
12 H
32 = C
2H
6
Example 1
The empirical formula of ethane is CH
3
. Given that
the relative molecular mass of ethane is 30, what is
its molecular formula?
M
r
of ethane from empirical formula = 12 + 3 = 15
=
M
r
from empirical formula
Mass of one atom of the an element12
15
= 2
Hence, the molecular formula of ethane = C
12 H
32 = C
2H
6
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
Example 2
Compound X contains 40.0% carbon, 6.6%
hydrogen and 53.3% oxygen. Its relative
molecular mass is 180. What is the molecular
formula of X?
Molar ratio
Number of
moles
16112
Relative atomic
mass
53.36.640.0
Percentage in
compound
OHCElement
= 3.3
12
40.0
= 3.3
16
53.3
= 1
16
3.52
= 2
16
3.52
= 6.6
1
6.6
= 1
3.3
3.3
Example 2
Compound X contains 40.0% carbon, 6.6%
hydrogen and 53.3% oxygen. Its relative
molecular mass is 180. What is the molecular
formula of X?
Molar ratio
Number of
moles
16112
Relative atomic
mass
53.36.640.0
Percentage in
compound
OHCElement
= 3.3
12
40.0
= 3.3
16
53.3
= 1
16
3.52
= 2
16
3.52
= 6.6
1
6.6
= 1
3.3
3.3
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
Example 2 (continued)
Compound X contains 40.0% carbon, 6.6%
hydrogen and 53.3% oxygen. Its relative
molecular mass is 180. What is the molecular
formula of X?
The empirical formula of X is CH
2
O.
=
M
r
from empirical formula
Relative molecular mass 180
30
= 6
Hence, the molecular formula of X = (CH
2
O)
6
= C
6
H
12
O
6
M
r of CH
2O = 12 + (2 x 1) + 16 = 30
Example 2 (continued)
Compound X contains 40.0% carbon, 6.6%
hydrogen and 53.3% oxygen. Its relative
molecular mass is 180. What is the molecular
formula of X?
The empirical formula of X is CH
2
O.
=
M
r
from empirical formula
Relative molecular mass 180
30
= 6
Hence, the molecular formula of X = (CH
2
O)
6
= C
6
H
12
O
6
M
r of CH
2O = 12 + (2 x 1) + 16 = 30
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Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
Molar Gas Volume
Unlike solids and liquids, gases weigh very
little.
It is easier to measure the volume of a gas
than its mass.
Is there a way to relate moles to the volumes
of gases?
Molar Gas Volume
Unlike solids and liquids, gases weigh very
little.
It is easier to measure the volume of a gas
than its mass.
Is there a way to relate moles to the volumes
of gases?
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
Molar Gas Volume
One mole of any gas occupies 24 dm
3
(24 000 cm
3
)
at room temperature and pressure (r.t.p.).
24 dm
3
is called the molar volume of a gas.
Molar Gas Volume
One mole of any gas occupies 24 dm
3
(24 000 cm
3
)
at room temperature and pressure (r.t.p.).
24 dm
3
is called the molar volume of a gas.
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
This means that at r.t.p.
• 1 mol of oxygen occupies 24 dm
3
,
• 1 mol of carbon dioxide occupies 24 dm
3
,
• 2 mol of oxygen occupy 2 × 24 = 48 dm
3
,
• 2 mol of carbon dioxide occupy 2 × 24 = 48 dm
3
.
how many dm
3
?
how many dm
3
?
how many dm
3
?
how many dm
3
?2 24 = 48
dm
3
24 dm
3
2 24 = 48
dm
3
24 dm
3
This means that at r.t.p.
• 1 mol of oxygen occupies 24 dm
3
,
• 1 mol of carbon dioxide occupies 24 dm
3
,
• 2 mol of oxygen occupy 2 × 24 = 48 dm
3
,
• 2 mol of carbon dioxide occupy 2 × 24 = 48 dm
3
.
how many dm
3
?
how many dm
3
?
how many dm
3
?
how many dm
3
?2 24 = 48
dm
3
24 dm
3
2 24 = 48
dm
3
24 dm
3
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
How can we calculate the number of
moles of a gas?
The number of moles of a gas can be measured in 2 ways:
Number of moles of gas=
M
r of gas
Mass of gas in grams
Number of moles of gas=
24 000 cm
3
Volume of gas in cm
3
at r.t.p
How can we calculate the number of
moles of a gas?
The number of moles of a gas can be measured in 2 ways:
Number of moles of gas=
M
r of gas
Mass of gas in grams
Number of moles of gas=
24 000 cm
3
Volume of gas in cm
3
at r.t.p
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
Example 1 What is the volume, in dm
3
, of 8 g
of oxygen gas (O
2
) at r.t.p.?
Relative molecular mass of oxygen = 2 × 16 = 32
Volume of oxygen
= number of moles of oxygen × 24
= × 24
= 6 dm
3
32
8
Example 1 What is the volume, in dm
3
, of 8 g
of oxygen gas (O
2
) at r.t.p.?
Relative molecular mass of oxygen = 2 × 16 = 32
Volume of oxygen
= number of moles of oxygen × 24
= × 24
= 6 dm
3
32
8
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
Example 2 In an experiment, hydrochloric acid
was reacted with calcium carbonate at r.t.p.
80 cm
3
of carbon dioxide was produced. Calculate
the number of molecules of carbon dioxide given
off.
Number of moles of carbon dioxide given off
Number of molecules of carbon dioxide given off
= number of moles × Avogadro’s constant
= 3.33 × 10
–3
× 6 × 10
23
= 2.00 × 10
21
= = 3.33 × 10
–3
mol
000 24
80
Example 2 In an experiment, hydrochloric acid
was reacted with calcium carbonate at r.t.p.
80 cm
3
of carbon dioxide was produced. Calculate
the number of molecules of carbon dioxide given
off.
Number of moles of carbon dioxide given off
Number of molecules of carbon dioxide given off
= number of moles × Avogadro’s constant
= 3.33 × 10
–3
× 6 × 10
23
= 2.00 × 10
21
= = 3.33 × 10
–3
mol
000 24
80
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Example 3 Calculate the mass of oxygen gas
(O
2
) in a room that measures 4 m high, 8 m wide
and 10 m long. Assume that air contains 20%
oxygen. (1 m
3
= 10
6
cm
3
)
Volume of air in room = 4 x 8 x 10 = 320 m
3
= 320 x 10
6
cm
3
Volume of oxygen in room = 20% x 320 x 10
6
= 64 x 10
6
cm
3
Mass of oxygen=
24 000 cm
3
Volume of oxygen
Number of moles of oxygen x M
r
Molar volume
=
=
x 32
64 x 10
6
cm
3
8.53 x 10
4
g
=
x 32
Example 3 Calculate the mass of oxygen gas
(O
2
) in a room that measures 4 m high, 8 m wide
and 10 m long. Assume that air contains 20%
oxygen. (1 m
3
= 10
6
cm
3
)
Volume of air in room = 4 x 8 x 10 = 320 m
3
= 320 x 10
6
cm
3
Volume of oxygen in room = 20% x 320 x 10
6
= 64 x 10
6
cm
3
Mass of oxygen=
24 000 cm
3
Volume of oxygen
Number of moles of oxygen x M
r
Molar volume
=
=
x 32
64 x 10
6
cm
3
8.53 x 10
4
g
=
x 32
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
Do the balloons of the same volume contain
the same number of particles?
According to Avogadro’s Law, each of
these balloons contains the same number
of gaseous particles since they have the
same volume.
Yes, they do!
Do the balloons of the same volume contain
the same number of particles?
According to Avogadro’s Law, each of
these balloons contains the same number
of gaseous particles since they have the
same volume.
Yes, they do!
Copyright © 2006-2011 Marshall Cavendish International (Singapore) Pte. Ltd.
Do the balloons of the same mass
contain the same number of
particles?
No, they don’t!
Same mass but different number of moles.
Gas Mass (g) A
r
or M
r
Number of moles
= mass / A
r
or M
r
Helium (He) 0.18 4
Hydrogen
(H2)
0.18 2
Methane
(CH4)
0.18 16
= 0.011
16
0.18
= 0.090
2
0.18
= 0.045
4
0.18
Do the balloons of the same mass
contain the same number of
particles?
No, they don’t!
Same mass but different number of moles.
Gas Mass (g) A
r
or M
r
Number of moles
= mass / A
r
or M
r
Helium (He) 0.18 4
Hydrogen
(H2)
0.18 2
Methane
(CH4)
0.18 16
= 0.011
16
0.18
= 0.090
2
0.18
= 0.045
4
0.18
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