Chapter
12
Reacting masses
12.1 The mole, Avogadro constant and molar mass
12.2 Percentage by mass of an element in a compound
12.3 Chemical formulae of compounds
12.4 Empirical formulae and molecular formulae derived
from experimental data
12.5 Reacting masses from chemical equations
Learning goal
After studying this chapter, you should be able to:
12.1
• perform calculations related to moles, Avogadro constant and molar masses
12.2
• calculate the percentage by mass of an element in a compound using appropriate information
12.3-12.4
• determine empirical formulae and molecular formulae from compositions by mass and molar
masses
12.5
• understand and use the quantitative information provided by a balanced chemical equation
�
• calculate masses of reactants and products in a reaction from the relevant equation and state the
interrelationship between them
�
• perform calculations related to moles and reacting masses
�
• solve problems involving limiting reactants
Chapter
12
Reacting masses
Nitroglycerin (C3H5N3O9) is a powerful explosive. It is used for mining, construction and military. It
decomposes to generate a large amount of gases and heat. The equation of the chemical reaction is:
4C3H5N3O9(,)
6N2(g) + 12CO2(g) + 10H2O(g) + O2(g)
For controlling the effect of explosion, the right amount of gases and heat should be released.
Therefore, scientists must determine the amount of nitroglycerin used in an explosion accurately.
Think about...
Nitroglycerin molecules are too tiny to be seen. How can we count their number in an explosive?
We can count the number of nitroglycerin molecules by weighing.
What is the quantitative relationship between the reactant (nitroglycerin) and products (nitrogen,
carbon dioxide, water vapour and oxygen) in the above chemical reaction?
The reactant and any of the products in the above reaction are related by a ratio, as given by the stoichiometric coefficients in the chemical
equation.
After studying this chapter, you should be able to answer the above questions.
nitroglycerin 硝化甘油
explosive 炸藥
military 軍事
12
Reacting masses
12.1 The mole, Avogadro constant and
molar mass
PowerPoint
Mole and Avogadro constant
Learning tip
In daily life, some special units are used to describe the quantity of items.
1 ream of paper
refers to 500
identical sheets of
paper.
For example, socks are in pairs, eggs are often packed in dozens, papers
are packed in reams. See Figure 12.1.
Note 1
The word ‘mole’ is derived from the
Latin word meaning a collection or
pile.
Note 2
Remind
students
that
the
abbreviation of mole is mol (NOT
m); that of gram is g (NOT gm).
Note 3
Determination of L using different
methods leads to values which are
very close to each other. The most
recent value proposed is 6.022 141
23
–1
79 × 10 mol . By approximation,
23
–1
we take L = 6.02 × 10 mol .
Figure 12.1 Socks are in pairs, eggs are often packed in dozens, and papers are often
packed in reams.
Unlike socks, eggs and papers, particles (i.e. atoms, ions or molecules)
are too small to be seen. It is very difficult to count particles one by one.
Chemists use a special unit called mole (abbreviation: mol), to describe
N1,
the quantity of particles in a substance.
N2
CE2007(II)14
Chemists have chosen the number of atoms in exactly 12.0 g of
Learning tip
A pure substance has a
formula. The simplest unit
of a substance is its
formula unit. For example,
Substance
Formula
unit
Water
H2O
Copper
Cu
Carbon
C
Sodium
chloride
NaCl
carbon-12 as the reference unit for the mole. The number of atoms in
23
exactly 12.0 g of carbon-12 is 6.02 × 10 . This number is called the
Avogadro constant (symbol: L; unit: mol ). Hence, we can also say
that, one mole of any substance contains 6.02 × 10
23
formula units
(Figure 12.2).
1 mole of
water contains
23
6.02 × 10
H2O molecules
Figure 12.2 One mole of
each of the four substances.
23
They all contain 6.02 × 10
formula units.
Avogadro constant 亞佛加德羅常數
formula unit 式單位
N3
–1
mole 摩爾
1 mole of copper
metal contains
23
6.02 × 10 Cu
atoms
1 mole of
carbon contains
23
6.02 × 10 C
atoms
ream 令 (紙張計算單位)
1 mole of sodium chloride
23
contains 6.02 × 10 NaCl
formula units (or contains
+
1 mole of Na ions and 1
–
mole of Cl ions)
3
12
III
Metals
S
XTRA
E
Historical note
Avogadro constant
The Avogadro constant
is named after the Italian
scientist Amedeo
Avogadro. He
suggested that the
volume of a gas is
related to the number of
its molecules present.
Chemists define mole in the following way:
Key point
One mole is the amount of a substance that contains the same
number of formula units as the number of atoms in exactly 12.0 g of
carbon-12.
Class practice 12.1
1.
N4
Note 4
Tell students that the term ‘formula unit’ is preferred to ‘particle’ as the former
is related to the formula of a species.
CE1999(II)2
DSE2015(IA)36
0.2 pt
A gas jar contains 0.5 mole of oxygen molecules.
(a) Calculate the number of oxygen molecules in the gas jar.
(b) Hence, calculate the number of oxygen atoms in the gas jar.
2.
Amedeo Avogadro
(1776–1856)
24
Given that a beaker contains 1.204 × 10 sodium atoms. How many moles
of sodium atoms are there in the beaker?
A12.1
23
23
1. (a) Number of oxygen molecules = 0.5 × 6.02 × 10 = 3.01 × 10
(b) As there are two oxygen atoms in each oxygen molecule, number of oxygen atoms
23
23
= 3.01 × 10 × 2 = 6.02 × 10
24
1.204 × 10
2. Number of moles of sodium atoms =
23 mol = 2 mol
6.02 × 10
Mole and molar mass
*formula
*Book 1, Section 8.5, p.17
One mole of a substance has a mass equal to its
Note 5
Remind students again that
formula mass has no unit
while molar mass has the unit
–1
of g mol .
molar mass. The unit of molar mass is gram per mol (g mol ).
mass
expressed in gram unit. The mass of one mole of a substance is called its
–1
N5
Substances consisting of atoms
In Figure 12.2, one mole of carbon and one mole of copper both contain
23
6.02 × 10 atoms. However, one mole of carbon weighs 12.0 g while one
mole of copper weighs 63.5 g. The relative atomic masses of carbon and
copper are 12.0 and 63.5 respectively. Thus, the molar masses of carbon
–1
Note 6
Tell students NOT to use ‘•’ as
the multiplication sign ‘×’. Thus,
it is 1.0 × 2, NOT 1.0 • 2.
Think about
Why does one mole
of Cl2 correspond
to a mass of 71.0 g?
Think about
Relative molecular mass of
chlorine = 35.5 × 2 = 71.0
Hence, the mass of one
mole of chlorine is 71.0 g.
12
4
formula mass 式量
molar mass 摩爾質量
–1
and copper are 12.0 g mol and 63.5 g mol respectively.
Substances consisting of molecules
CE2008(II)46
One mole of water weighs 18.0 g. The relative molecular mass of water
–1
= 1.0 × 2 + 16.0 = 18.0. Thus, the molar mass of water is 18.0 g mol .
Substances consisting of ions
One mole of sodium chloride weighs 58.5 g. The formula mass of sodium
chloride = 23.0 + 35.5 = 58.5. Thus, the molar mass of sodium chloride is
–1
58.5 g mol .
relative atomic mass 相對原子質量
relative molecular mass 相對分子質量
N6
12
Reacting masses
Key point
The mass of one mole of a substance is called its molar mass. (Unit of
–1
molar mass: g mol )
Class practice 12.2
A12.2
1.
–1
1. (a) Molar mass of Ag = 107.9 g mol
(b) Molar mass of F2
–1
–1
= (19.0 × 2) g mol = 38.0 g mol
(c) Molar mass of NH3
–1
–1
= (14.0 + 1.0 × 3) g mol = 17.0 g mol
(d) Molar mass of C2H5OH
–1
= (12.0 × 2 + 1.0 × 6 + 16.0) g mol
–1
= 46.0 g mol
2.
(e) Molar mass of Fe2(SO4)3
–1
= 55.8 × 2 + 3 × (32.1 + 16.0 × 4) g mol
–1
= 399.9 g mol
2. (a) Mass of 1 mole of Na2SO4
= (23.0 × 2 + 32.1 + 16.0 × 4) g = 142.1 g
(b) Mass of 0.5 mole of CCl4
= 0.5 × (12.0 + 35.5 × 4) g = 77.0 g
What is the molar mass of each of the following substances?
(a) Silver
(b) Fluorine
(c) Ammonia
(d) Ethanol (C2H5OH)
(e) Iron(III) sulphate
(Relative atomic masses: H = 1.0, C = 12.0, N = 14.0, O = 16.0, F = 19.0,
S = 32.1, Fe = 55.8, Ag = 107.9)
What is the mass of each of the following substances?
(a) 1 mole of sodium sulphate
(b) 0.5 mole of tetrachloromethane (CCl4)
(Relative atomic masses: C = 12.0, O = 16.0, Na = 23.0, S = 32.1,
Cl = 35.5)
Relationship between number of formula units,
number of moles and mass
The diagram below summarizes the relationship between ‘number of
formula units’, ‘number of moles’ and ‘mass’ in a substance.
÷ Avogadro constant
number of
formula units
CE2005(II)23
CE2010(II)4
× molar mass
number of
moles
× Avogadro constant
mass
÷ molar mass
We can calculate one item from another by using the following
equations:
Key point
Number of moles (mol) =
Number of moles (mol) =
ethanol 乙醇
tetrachloromethane 四氯甲烷
mass (g)
–1
molar mass (g mol )
number of formula units
–1
Avogadro constant (mol )
5
12
III
Metals
H20
Example 12.1
CE2004(II)3
CE2006(II)18
CE2008(II)10
DSE2018(IA)4
Calculating the mass and number of molecules in a substance using the number of moles
Note 7
A gas jar contains 1.85 moles of methane (CH4).
(a) Remind students to write the ‘subject’
(a) Calculate the mass of methane in the gas jar.
of an expression clearly. Thus, it is
‘Molar mass of methane = (12.0 + 1.0
(b) Hence, calculate the number of methane molecules in the gas jar.
–1
–1
× 4) g mol = 16.0 g mol ’, NOT
–1
(Relative atomic masses: H = 1.0, C = 12.0)
‘Methane = (12.0 + 1.0 × 4) g mol =
–1
Solution
–1
–1
(a) Molar mass of methane = (12.0 + 1.0 × 4) g mol = 16.0 g mol
N7
Mass of methane (g)
16.0 g mol ’.
(b) Do not write ‘gm’ for ‘g’; ‘m’ or ‘M’ for
‘mol’.
(c) It is a good practice to write a unit for
each separate line (if applicable).
–1
= number of moles of methane (mol) × molar mass of methane (g mol )
–1
= 1.85 mol × 16.0 g mol
= 29.6 g
(b) Number of methane molecules
–1
= number of moles of methane (mol) × Avogadro constant (mol )
23
–1
= 1.85 mol × 6.02 × 10 mol
24
= 1.11 × 10
Self-test 12.1
A beaker contains 10.21 g of magnesium hydroxide.
Self-test 12.1
(a) Molar mass of magnesium hydroxide (Mg(OH)2)
–1
–1
= (24.3 + 16.0 × 2 + 1.0 × 2) g mol = 58.3 g mol
Number of moles of Mg(OH)2
10.21 g
=
–1 = 0.175 mol
58.3 g mol
(a) Calculate the number of moles of magnesium hydroxide in the beaker.
(b) Hence, calculate the number of hydroxide ions in the beaker.
(Relative atomic masses: H = 1.0, O = 16.0, Mg = 24.3)(b)
Try Chapter Exercise Q9
H20
Example 12.2
–
Since 1 formula unit of Mg(OH)2 contains 2 OH ions,
–
number of moles of OH ions = 0.175 × 2 mol = 0.350 mol
–
23
–1
23
Number of OH ions = 0.350 mol × 6.02 × 10 mol = 2.11 × 10
DSE2019(IB)2(b)(i)
Calculating the mass of a particle of a substance
Calculate the mass of
(a) 1 Na atom
(b) 1 H2O molecule
(c) 1 formula unit of NaCl
(Relative atomic masses: H = 1.0, O = 16.0, Na = 23.0, Cl = 35.5)
Solution
One mole of a substance corresponds to its molar mass and contains the Avogadro constant of
formula units.
–1
molar mass (g mol )
 mass of 1 formula unit =
–1
Avogadro constant (mol )
–1
(a) Mass of 1 Na atom =
23.0 g mol
23
–1
6.02 × 10 mol
–23
= 3.82 × 10 g
–1
(1.0 × 2 + 16.0) g mol
(b) Mass of 1 H2O molecule =
23
–1
6.02 × 10 mol
–23
= 2.99 × 10
12
6
g
cont’d
12
–1
(c) Mass of 1 formula unit of NaCl =
(23.0 + 35.5) g mol
23
–1
6.02 × 10 mol
–23
= 9.72 × 10 g
Self-test 12.2
Calculate the mass of
(a) 1 Mg atom
(b) 1 I2 molecule
Self-test 12.2
(a) Mass of 1 Mg atom =
Reacting masses
–1
24.3 g mol
–23
= 4.04 × 10 g
23
–1
6.02 × 10 mol
–1
126.9 × 2 g mol
–22
(b) Mass of 1 I2 molecule =
= 4.22 × 10 g
23
–1
6.02 × 10 mol
(c) Mass of 1 formula unit of CaCO3
–1
(40.1 + 12.0 + 16.0 × 3) g mol
=
23
–1
6.02 × 10 mol
–22
= 1.66 × 10 g
(c) 1 formula unit of calcium carbonate
(Relative atomic masses: C = 12.0, O = 16.0, Mg = 24.3, Ca = 40.1, I = 126.9)
Class practice 12.3
1.
Calculate the mass of
(a) 0.200 mole of chlorine atoms.
(b) 0.200 mole of chlorine molecules.
A12.3
1. (a) Mass of 0.200 mole of Cl atoms = 0.200 × 35.5 g = 7.1 g
(b) Mass of 0.200 mole of Cl2 molecules
= 0.200 × (35.5 × 2) g = 14.2 g
(c) Mass of Cl2 = 1.20 × (35.5 × 2) g = 85.2 g
(c) chlorine which contains the same number of molecules as there are in 1.20 mole of water.
(Relative atomic masses: H = 1.0, O = 16.0, Cl = 35.5)
2.
Complete the following table.
Substance
Molar mass
–1
(g mol )
Mass (g)
No. of moles
(mol)
No. of molecules/
formula units
(a)
Sodium hydroxide
40.0
10.0
0.250
1.51 × 10
23
(b)
Helium
4.0
0.20
0.05
3.01 × 10
22
(c)
Sulphur dioxide
64.1
320.5
5
3.01 × 10
(d)
Compound X
46.0
23.0
0.5
3.01 × 10
24
23
(Relative atomic masses: H = 1.0, He = 4.0, O = 16.0, Na = 23.0, S = 32.1)
12.2 Percentage by mass of an element in
PowerPoint
a compound
Calculating percentage by mass of an element in
the compound
From the formula of a compound, we can work out the percentage by
mass of each element in the compound.
Key point
Percentage by mass of element A in a compound
relative atomic mass of A × number of atoms of A in the formula
× 100%
=
formula mass of the compound
percentage by mass 質量百分比
7
12
III
Metals
H20
Example 12.3
CE2001(II)26
CE2002(I)1(c)(i)
CE2002(I)7(a)(v)
CE2007(I)5(e)
CE2009(II)5
DSE2012(IA)9
Calculate the percentage by mass of an element in a compound
Bauxite is the main ore of aluminium. It contains mainly aluminium oxide (Al2O3). Calculate the
percentage by mass of aluminium in aluminium oxide.
(Relative atomic masses: O = 16.0, Al = 27.0)
Solution
Formula mass of Al2O3
Al2
O3
= 27.0 × 2 + 16.0 × 3
= 102.0
Percentage by mass of Al in Al2O3
relative atomic mass of Al × number of atoms of Al in the formula
× 100%
=
formula mass of Al2O3
27.0 × 2
× 100%
=
102.0
= 52.9%
Self-test 12.3
Sodium hydroxide is the main ingredient of drain cleaners. Calculate the percentage by mass of sodium
Self-test 12.3
in sodium hydroxide.
–1
(Relative atomic masses: H = 1.0, O = 16.0, Na = 23.0)
Try Chapter Exercise Q10
Calculating the mass of an element in the
compound
The mass of an element in a compound can be calculated from the
formula of the compound and the percentage by mass of that element in
the compound.
H20
Example 12.4
CE2011(II)8
Calculate the mass of an element in a compound
Calculate the mass of copper in 15.0 g of copper(II) sulphate-5-water (CuSO4․5H2O).
(Relative atomic masses: H = 1.0, O = 16.0, S = 32.1, Cu = 63.5)
Solution
Formula mass of CuSO4․5H2O
= 63.5 + 32.1 + 16.0 × 4 + 5 × (1.0 × 2 +16.0)
= 249.6
cont’d
12
8
bauxite 鋁土礦
–1
Formula mass of NaOH = (23.0 + 16.0 + 1.0) g mol = 40.0 g mol
Percentage by mass of Na in NaOH
23.0
× 100%
=
40.0
= 57.5%
drain cleaner 通渠劑
12
Reacting masses
Percentage by mass of Cu in CuSO4․5H2O
relative atomic mass of Cu × number of atoms of Cu in the formula
× 100%
=
formula mass of CuSO4․5H2O
63.5
× 100%
=
249.6
= 25.4%
That means for every gram of CuSO4․5H2O, there is 25.4% (or 0.254 gram) of Cu in it.
 mass of Cu in 15.0 g of CuSO4․5H2O
Self-test 12.4
Formula mass of K2Cr2O7
–1
–1
= (39.1 × 2 + 52.0 × 2 + 16.0 × 7) g mol = 294.2 g mol
Percentage by mass of K in K2Cr2O7
39.1 × 2
× 100%
=
294.2
= 26.6%
= 15.0 g × 25.4%
= 3.81 g
Self-test 12.4
Calculate the mass of potassium in 7.91 g of potassium dichromate (K2Cr2O7).
(Relative atomic masses: O = 16.0, K = 39.1, Cr = 52.0)
Try Chapter Exercise Q11
Mass of K in 7.91 g of K2Cr2O7
= 7.91 g × 26.6%
= 2.10 g
Calculating the relative atomic mass of an
element
The relative atomic mass of an element may be calculated from the
formula of the compound and percentage by mass of that element in the
compound.
H20
Example 12.5
CE1999(II)17
CE2002(II)3
Calculating the relative atomic mass of an element
The chloride of a metal M has the formula of MCl3 and contains 34.4% by mass of M. Calculate the
relative atomic mass of M.
(Relative atomic mass: Cl = 35.5)
Solution
Let the relative atomic mass of M be a.
Percentage by mass of M in MCl3
=
relative atomic mass of M × number of atoms of M in the formula
formula mass of MCl3
a
× 100%
a + 35.5 × 3
a = 55.8
34.4% =
 the relative atomic mass of M is 55.8.
Self-test 12.5
× 100%
Self-test 12.5
Let the relative atomic mass of X be a.
a
× 100%
25.6% =
a + 79.9 × 2
100a
25.6 =
a + 159.8
a = 55.0
 the relative atomic mass of X is 55.0.
The bromide of a metal X has the formula of XBr2 and contains 25.6% by mass of X. Calculate the
relative atomic mass of X.
(Relative atomic mass: Br = 79.9)
9
12
III
Metals
2. Let the relative atomic mass of M be a.
5.68
35.5
=
26.88 a + 35.5
a = 132.5
 the relative atomic mass of M is 132.5.
Class practice 12.4
1.
100
mol = 4.35 mol
23.0
Since 1 formula unit of NaNO3 contains 1 Na,
number of moles of NaNO3 = 4.35 mol
Mass of NaNO3
2.
= 4.35 × (23.0 + 14.0 + 16.0 × 3) g
= 369.75 g
Percentage by mass of N in NaNO3
14.0
× 100% = 16.5% 3.
=
23.0 + 14.0 + 16.0 × 3
Mass of N in the NaNO3 sample
= 369.75 g × 16.5% = 61.0 g
3. Number of moles of Na =
4.6
4. Number of moles of Na =
mol
23.0
= 0.2 mol
Since 1 formula unit of Na2CO3․10H2O
contains 2 Na,
number of moles of Na2CO3․10H2O
0.2
=
mol = 0.1 mol
2
Molar mass of Na2CO3․10H2O
= [23.0 × 2 + 12.0 + 16.0 × 3 + 10 ×
–1
(1.0 × 2 + 16.0)] g mol
–1
= 286.0 g mol
Mass of Na2CO3․10H2O
= 0.1 × 286.0 g
= 28.6 g
4.
A12.4
1. Let the relative atomic mass of M be a.
A metal oxide MO contains 79.87% by mass of the metal M. Find the
a
× 100%
79.87% =
relative atomic mass of M.
a + 16.0
100a
(Relative atomic mass: O = 16.0)
79.87 =
a + 16.0
26.88 g of a metal chloride MCl contains 5.68 g of chlorine. Find the
a = 63.5
relative atomic mass of the metal M.
 the relative atomic mass of M is 63.5.
(Relative atomic mass: Cl = 35.5)
What is the mass of nitrogen present in the sample of sodium nitrate
(NaNO3) which contains 100 g of sodium?
(Relative atomic masses: N = 14.0, O = 16.0, Na = 23.0)
What is the mass of water of crystallization present in the sample of sodium
carbonate-10-water (Na2CO3․10H2O) which contains 4.6 g of sodium?
(Relative atomic masses: H = 1.0, C = 12.0, O = 16.0, Na = 23.0)
PowerPoint
12.3 Chemical formulae of compounds
Chemical formulae are part of the language of chemistry. In this section,
we are going to learn three types of chemical formulae.
Percentage by mass of H2O in Na2CO3․10H2O
10 × (1.0 × 2 + 16.0)
× 100% = 62.9%
=
286.0
Mass of H2O in the Na2CO3․10H2O
sample = 28.6 g × 62.9% = 18.0 g
Empirical formula
e-Learning
Flipped classroom
Determining the
empirical formula of a
compound from the
percentage by mass
DSE2012(IA)29
DSE2013(IA)29
The empirical formula of a compound is the formula which shows the
simplest whole number ratio of the atoms or ions present. It is
applicable to all compounds.
Molecular formula
DSE2012(IA)28
The molecular formula of a substance shows the actual number of each
kind of atoms in one molecule of the substance. It is only applicable to
https://e-aristo.hk/r/
cm2fc12i01.e
molecular compounds and elements consisting of molecules.
Structural formula
The structural formula of a molecular substance is the formula which
shows how the constituent atoms are joined up in one molecule of the
substance.
The empirical, molecular and structural formulae of some substances
are given in Table 12.1.
12
10
empirical formula 實驗式
molecular formula 分子式
structural formula 結構式
water of crystallization 結晶水
12
Substance
Reacting masses
Empirical formula
Molecular formula
Structural formula
—
N2
N≡N
Carbon dioxide
CO2
CO2
O=C=O
Ethene
CH2
C2H4
Nitrogen
H H
H C
C
H
H H H
CH2
Propene
H C
C3H6
C
C
H
H
CE2010(II)13
H H
C2H6O
Ethanol
H C
C2H6O
O H
C
H H
C2H6O
Methoxymethane
C2H6O
SiO2
Quartz
H
H
H C
O C
H
H
—
H
—
Table 12.1 The different formulae of some substances.
We should note that:
•
The empirical and molecular formulae of a compound may be the
same (e.g. carbon dioxide) or different (e.g. ethene). The molecular
formula is the empirical formula multiplied by a whole number (1, 2,
3, etc.).
•
Although different compounds may have the same empirical formula
and the same molecular formula (e.g. ethanol and methoxymethane),
they have different structural formulae.
Class practice 12.5
H
Complete the following table:
H
C
C
H
H
H
Substance
Empirical formula
Molecular formula
Structural formula
Oxygen
/
O2
O=O
Water
H2O
H2O
H–O–H
Ethane
CH3
C2H6
But-1-ene
but-1-ene 丁-1-烯
ethane 乙烷
ethene 乙烯
A12.5
H
methoxymethane 甲氧基甲烷
propene 丙烯
CH2
C4H8
H
H
H
H
H
C
C
C
C
H
H
H
11
12
III
Metals
12.4 Empirical formulae and molecular
PowerPoint
formulae derived from experimental
data
In Table 7.4 of Chapter 7, we learnt that copper may form copper(I) oxide
(Cu2O) or copper(II) oxide (CuO) with oxygen. In this section, we will
study how the empirical formula of an oxide of copper is determined.
Determination of empirical formulae
The empirical formula of a compound can be calculated from its
composition by mass i.e. the mass of each element present in a known
mass of the compound. The composition of a compound has to be
determined by experiment. Let us consider the following two examples.
Example 1: Determining the empirical formula of an oxide
of copper
To determine the empirical formula of an oxide of copper, we have to find
the ratio by mass of copper and oxygen in the compound.
Learning tip
Hydrogen and carbon
monoxide reduce the
oxide by removing
oxygen from it.
Pass town gas into a combustion tube. Then heat a known mass of
oxide of copper (black) in the combustion tube. The hydrogen and carbon
monoxide in the town gas reduce the oxide to reddish brown copper. Then
find the mass of copper. A set-up for conducting the experiment is shown
in Figure 12.3.
SBA note
• At the beginning of
the experiment, town
gas is passed into the
combustion tube.
This is to expel the air
inside the tube.
• The hot copper
formed may react
with the oxygen in air
again. Therefore, it is
necessary to pass the
town gas through the
combustion tube,
even after heating has
stopped.
12
12
combustion tube 燃燒管
composition by mass 質量組成
oxide of copper
hole
excess town gas
burns here
town gas
supply
heat
combustion tube
Figure 12.3 To determine the empirical formula of an oxide of copper by passing town gas
over the heated oxide.
Table 12.2 shows the specimen results of the experiment.
reddish brown 紅棕色
specimen result 樣本結果
town gas 煤氣
12
Reacting masses
Specimen results
Item
Table 12.2 The specimen
results of the experiment.
Mass (g)
Combustion tube
18.100
Combustion tube + oxide of copper
18.701
Combustion tube + copper
18.579
Mass of copper in oxide
18.579 – 18.100 = 0.479
Mass of oxygen in oxide
18.701 – 18.579 = 0.122
From the experimental results, we can work out the empirical formula
of the oxide of copper as shown in ‘Problem-solving strategy 12.1’.
Problem-solving strategy 12.1
CE2006(II)37
DSE2015(IB)3(b)(i)
Determining the empirical formula of a compound
The steps of how to determine the empirical formula of an oxide of copper are as follows:
1 Write down the mass of
each element
2 Write down the relative
atomic mass of each element
(After some practice, you will
be able to skip this step.)
3 Calculate the number of
moles of each type of atoms
Cu
O
Mass (g)
0.479
0.122
Relative atomic mass
63.5
16.0
0.479
= 0.00754
63.5
0.122
= 0.00763
16.0
0.00754
=1
0.00754
0.00763
= 1.01  1
0.00754
Number of moles of atoms (mol)
mass (g)
(=
)
–1
molar mass (g mol )
4 Calculate the simplest whole Simplest whole number mole ratio
number mole ratio of atoms of atoms
(divided by the smallest number of
moles)
(multiplied by the smallest possible
whole number to turn all the values
into whole number if necessary)
 the empirical formula of the oxide of copper is CuO.
Note:
Due to experimental error, the number(s) obtained for the simplest mole ratio may have a small difference
from a whole number. Hence, it is an acceptable practice to ‘round off’ the value(s) to the nearest whole
number(s). However, we must be very careful when doing so. For example, 1.01 can be rounded off to 1, but
1.2 is usually NOT rounded off to 1.
mole ratio 摩爾比
round off 四捨五入
13
12
III
Metals
Extension
magnesium
ribbon
crucible
pipe-clay
triangle
rocksil
Heat a known mass of magnesium strongly in a crucible (also of known
mass) until it catches fire (Figure 12.4). Lift the crucible lid slightly from
time to time. This lets in air to react with magnesium.
heat very
strongly
tripod
Example 2: Determining the empirical formula of an oxide
of magnesium
From the experimental results, the empirical formula of the oxide of
Figure 12.4 To find the empirical
formula of an oxide of magnesium
by heating magnesium in air.
PowerPoint
Experiment
Video
Experiment
Video
Extension
PowerPoint
magnesium can be worked out to be MgO. Try the experiment yourselves.
Experiment 12.1
Experiment Workbook 1
Determining the empirical formula of an oxide of copper
In this experiment, you are going to determine the empirical formula of an
oxide of copper.
Experiment 12.2
Experiment Workbook 1
Determining the empirical formula of an oxide of magnesium
In this experiment, you are going to determine the empirical formula of an
oxide of magnesium.
The empirical formula of a compound can also be determined if the
percentage by mass of each element in the compound is known. See
Example 12.6.
H20
Example 12.6
CE2000(II)4
CE2004(II)12
CE2005(I)8(a)
CE2010(I)4(b)
CE2010(II)33
CE2011(II)38
DSEPP2012(IB)5(a)
DSE2012(IA)3
DSE2013(IA)13
DSE2019(IA)8
Determining the empirical formula of a compound from the percentage by mass
Compound Y was found to contain iron and oxygen only. Experiments showed that it contains 70%
iron and 30% oxygen by mass. Calculate the empirical formula of Y.
(Relative atomic masses: O = 16.0, Fe = 55.8)
Solution
Assume that there are 100 g of Y. Then there are 70 g of iron and 30 g of oxygen.
Mass (g)
Relative atomic mass
Number of moles of atoms (mol)
mass (g)
(=
–1 )
molar mass (g mol )
Fe
O
70
30
55.8
16.0
70
= 1.25
55.8
30
= 1.88
16.0
cont’d
12
14
12
Fe
O
Simplest whole number mole ratio of atoms
(divided by the smallest number of moles)
1.25
=1
1.25
1.88
= 1.5
1.25
(multiplied by the smallest possible whole
number (2 here) to turn all the values into
whole numbers)
1×2=2
1.5 × 2 = 3
Reacting masses
 the empirical formula of Y is Fe2O3.
Try Chapter Exercise Q23
Class practice 12.6
An experiment was performed to determine the empirical formula of an oxide of magnesium. The
experimental results are tabulated below.
Item
Mass (g)
Crucible + lid
28.092
Crucible + lid + magnesium
28.698
Crucible + lid + oxide of magnesium
29.103
Extension
1.
(Answers on p.12-27.)
Determine the empirical formula of the oxide of magnesium using the above data.
(Relative atomic masses: O = 16.0, Mg = 24.3)
2.
1.200 g of a compound containing only carbon, hydrogen and oxygen gave 1.173 g of carbon dioxide and
0.240 g of water on complete combustion. Find the empirical formula of the compound.
(Relative atomic masses: H = 1.0, C = 12.0, O = 16.0)
3.
A compound has the empirical formula CxHy. On analysis, 1.000 g of the compound was found to
contain 0.857 g of carbon. Find the values of x and y.
(Relative atomic masses: H = 1.0, C = 12.0)
4.
Compound X contains 26.95% sulphur, 13.44% oxygen and 59.61% chlorine by mass. Find the empirical
formula of X.
(Relative atomic masses: O = 16.0, S = 32.1, Cl = 35.5)
CE2009(II)46
Determination of molecular formulae
Once the empirical formula and the relative molecular mass of a
compound are known, we can determine the molecular formula of the
compound. This is because molecular formula is a whole number
multiple of empirical formula.
complete combustion 完全燃燒
15
12
III
Metals
H20
Example 12.7
Determining molecular formula using empirical formula and relative molecular mass
A compound has an empirical formula CH2 and a relative molecular mass of 42.0. Determine its
molecular formula.
(Relative atomic masses: H = 1.0, C = 12.0)
Solution
Let the molecular formula of the compound be (CH2)n, where n is a whole number.
Relative molecular mass of (CH2)n = 42.0
n(12.0 + 1.0 × 2) = 42.0
n=3
 the molecular formula of the compound is (CH2)3, i.e. C3H6.
H20
Example 12.8
CE2003(I)3(b)(i)
CE2008(II)31
AS2010(II)4(a)
DSE2016(IB)1(c)(i)
Determining empirical formula and molecular formula using percentage by mass
Compound X was found to contain carbon and hydrogen only. Experiments showed that it contained
80% carbon and 20% hydrogen by mass. If its relative molecular mass is 30.0, calculate the empirical
formula and molecular formula of X.
(Relative atomic masses: H = 1.0, C = 12.0)
Solution
Assume that there are 100 g of X. Then there are 80 g of carbon and 20 g of hydrogen.
Mass (g)
Relative atomic mass
Number of moles of atoms (mol)
mass (g)
(=
–1 )
molar mass (g mol )
Simplest whole number mole ratio of atoms
(divided by the smallest number of moles)
C
H
80
20
12.0
1.0
80
= 6.67
12.0
20
= 20
1.0
6.67
=1
6.67
20
=3
6.67
 the empirical formula of X is CH3.
Let the molecular formula of X be (CH3)n, where n is the whole number.
Relative molecular mass of (CH3)n = 30.0
n(12.0 + 1.0 × 3) = 30.0
15.0n = 30.0
n=2
 the molecular formula of X is C2H6.
Note: 2.99 can be rounded off to 3, but 2.8 is usually NOT rounded off to 3.
Try Chapter Exercise Q24
12
16
12
H20
Example 12.9
CE2007(I)12(a)(b)
DSE2013(IB)3(a)
Reacting masses
DSE2017(IA)3
Determining empirical formula and molecular formula using masses of combustion products
Compound Z containing only carbon, hydrogen and oxygen burnt completely in air to form carbon
dioxide and water as the only products. 2.43 g of Z gave 3.96 g of carbon dioxide and 1.35 g of water.
Determine the empirical formula of Z. If its relative molecular mass is 162.0, determine the
molecular formula of Z.
(Relative atomic masses: H = 1.0, C = 12.0, O = 16.0)
Solution
Since all the C in CO2 and H in H2O came from Z,
12.0
mass of C in Z = 3.96 g ×
= 1.08 g;
12.0 + 16.0 × 2
1.0 × 2
= 0.15 g
1.0 × 2 + 16.0
The rest of mass of Z must come from oxygen.
mass of H in Z = 1.35 g ×
 mass of O in Z = (2.43 – 1.08 – 0.15) g = 1.20 g
Now go on to find the empirical formula of Z as follows:
C
H
O
Mass (g)
1.08
0.15
1.20
Relative atomic mass
12.0
1.0
16.0
Number of moles of atoms (mol)
mass (g)
(=
–1 )
molar mass (g mol )
1.08
= 0.090
12.0
0.15
= 0.15
1.0
1.20
= 0.075
16.0
Simplest whole number mole ratio of atoms
(divided by the smallest number of moles)
0.090
= 1.2
0.075
0.15
=2
0.075
0.075
=1
0.075
(multiplied by the smallest possible whole
number (5 here) to turn all values into
whole numbers)
1.2 × 5 = 6
2 × 5 = 10
1×5=5
 the empirical formula of Z is C6H10O5.
Let the molecular formula of Z be (C6H10O5)n, where n is the whole number.
Relative molecular mass of (C6H10O5)n = 162.0
n(12.0 × 6 + 1.0 × 10 + 16.0 × 5) = 162.0
162.0n = 162.0
n=1
 the molecular formula of Z is C6H10O5.
17
12
III
Metals
H20
Example 12.10
CE2003(II)11
DSE2014(IA)5
Determining the number of water of crystallization in a hydrated salt
5.60 g of hydrated copper(II) sulphate CuSO4․nH2O was heated in a crucible to drive off the water of
crystallization. The white residue was anhydrous copper(II) sulphate, which was found to have a mass
of 3.59 g.
hydrated copper(II) sulphate
anhydrous copper(II) sulphate
(a) Deduce a reasonable value for n.
(b) Explain why the answer you gave in (a) differs a bit from the value actually calculated.
(Relative atomic masses: H = 1.0, O = 16.0, S = 32.1, Cu = 63.5)
Solution
(a) Mass of water of crystallization = (5.60 – 3.59) g
= 2.01 g
CuSO4
H2O
Mass (g)
3.59
2.01
Formula mass
159.6
18.0
3.59
= 0.0225
159.6
2.01
= 0.112
18.0
0.0225
=1
0.0225
0.112
= 4.98
0.0225
Number of moles of formula units (mol)
mass (g)
(=
–1 )
molar mass (g mol )
Simplest whole number mole ratio of formula units
(divided by the smallest number of moles)
Since n should be a whole number, a reasonable value of n would be 5.
(b) The experimental value of n (4.98) is lower than 5. This might be due to two reasons:
(1) Not all water of crystallization has been removed in the heating process.
(2) The anhydrous salt has absorbed some moisture from the atmosphere during weighing.
Try Chapter Exercise Q25
Experiment 12.2
PowerPoint
12
18
anhydrous 無水的
hydrated salt 水合鹽
Experiment
Video
Experiment Workbook 1
Determining the formula of hydrated copper(II) sulphate
In this experiment, you are going to determine the formula of hydrated
copper(II) sulphate.
12
Class practice 12.7
Reacting masses
(Answers on p.12-28.)
1.
A compound contains only carbon, hydrogen and oxygen. 0.81 g of the compound gave 1.32 g of carbon
dioxide and 0.45 g of water on complete combustion. Find the empirical formula of the compound. If
the relative molecular mass of the compound is 320.0, find its molecular formula.
(Relative atomic masses: H = 1.0, C = 12.0, O = 16.0)
2.
A compound was found to contain 40.00% by mass of carbon, 6.67% by mass of hydrogen and 53.33%
by mass of oxygen. It has a relative molecular mass of 60.0. Calculate its molecular formula.
(Relative atomic masses: H = 1.0, C = 12.0, O = 16.0)
3.
Epsom salts are used as bath salts to relieve aches and pains. They are hydrated salts of magnesium sulphate
with formula MgSO4․nH2O.
Experiments were carried out to find the formula of the salt. It was found that it contained 51.22% by mass
of water of crystallization. Find the value of n.
(Relative atomic masses: H = 1.0, O = 16.0, Mg = 24.3, S = 32.1)
12.5 Reacting
PowerPoint
masses
from
chemical
equations
Chemical equations and reacting masses
In Chapter 11, we have learnt that a balanced equation provides us useful
information about the reaction it represents. Consider the equation
representing the reaction between magnesium and oxygen:
2Mg(s) + O2(g)
2MgO(s)
The equation tells us that,
2Mg(s)
+
2 magnesium atoms
23
The molar masses of
Mg, O2 and MgO are
–1
24.3 g mol , (16.0 × 2 =
–1
32.0) g mol , and (24.3
–1
+ 16.0 = 40.3) g mol
respectively.
2 moles of
magnesium atoms
–1
2 mol × 24.3 g mol
= 48.6 g of
magnesium atoms
2MgO(s)
1 oxygen molecule
2 formula units of
magnesium oxide
23
2 × 6.02 × 10
magnesium atoms
Learning tip
O2(g)
23
1 × 6.02 × 10
oxygen molecules
react
with
to
form
1 mole of oxygen
molecules
–1
1 mol × 32.0 g mol
= 32.0 g of oxygen
molecules
Epsom salt 瀉鹽
2 × 6.02 × 10 formula
units of magnesium oxide
2 moles of formula units
of magnesium oxide
–1
2 mol × 40.3 g mol
= 80.6 g of formula units
of magnesium oxide
19
12
III
Metals
Therefore, a balanced equation shows the quantitative relationship
Learning tip
of the reactant(s) and the product(s) in a reaction. The stoichiometric
The quantitative study
of reactants and
products in a reaction
is called stoichiometry.
coefficients in the equation indicate the relative number of moles (i.e.
mole ratio) of reactants and products involved in the reaction. Besides, the
total mass of reactants is equal to the total mass of products (i.e.
conservation of mass).
Calculations from chemical equations — reacting
masses
When the mass of one of the substances in the reaction is known, the
masses of other substances reacted or formed can be calculated based on
the balanced chemical equation. See ‘Problem-solving strategy 12.2’.
Problem-solving strategy 12.2
CE1999(II)8
CE2005(I)10(b)(ii)
CE2011(II)8
DSE2018(IB)1(b)(ii)
Calculating reacting masses from chemical equations
Calculate the mass of magnesium oxide produced when 2.43 g of
magnesium burns completely in air.
1
Write down the balanced chemical equation for the reaction.
2Mg(s) + O2(g)
2
Try it now
Calculate the mass of copper
produced when 15.9 g of
copper(II)
oxide
reacts
completely with hydrogen.
➊ CuO(s) + H2(g)
Cu(s) + H2O()
2MgO(s)
Convert the mass(es) of the given substance(s) into number
of moles
➋
Molar mass of CuO
–1
= (63.5 + 16.0) g mol
–1
= 79.5 g mol
Number of moles of CuO
15.9
=
mol = 0.2 mol
79.5
➌
From the equation, mole ratio of
CuO to Cu is 1 : 1.
 number of moles of Cu = 0.2 mol
–1
Molar mass of Mg = 24.3 g mol
Number of moles of Mg
mass (g)
=
–1
molar mass (g mol )
2.43 g
=
–1
24.3 g mol
= 0.100 mol
3
Use the balanced chemical equation to calculate the number
of moles of the substance asked in the question.
From the equation, mole ratio of
Mg : MgO = 2 : 2 (or simply 1 : 1)
 number of moles of MgO
= 0.100 mol
cont’d
12
20
stoichiometric coefficient 計量系數
quantitative relationship 定量關係
stoichiometry 化學計量學
12
4
Convert the number of moles of that substance into mass
Molar mass of MgO
–1
= (24.3 + 16.0) g mol
–1
= 40.3 g mol
 mass of MgO produced
–1
= 0.100 mol × 40.3 g mol
= 4.03 g
➍
Reacting masses
Mass of Cu produced = 0.2 × 63.5 g = 12.7 g
The flow chart below illustrates the steps for determining the reacting
masses from a chemical equation.
Known
mass of A
divided by
molar
mass of A
Number of
moles of A
Number of
moles of the
substance asked
in the question
by mole ratio
(shown in the
equation)
multiplied by
molar mass of
that substance
Mass of the
substance asked
in the question
(Note: A represents the chemical formula of a particular substance.)
H20
Example 12.11
CE2005(II)37
Determining the mass of a reactant from the mass of another reactant
Magnesium reacts with copper(II) oxide according to the following equation:
Mg(s) + CuO(s)
MgO(s) + Cu(s)
Calculate the mass of magnesium required to react completely with 7.95 g of copper(II) oxide.
(Relative atomic masses: O = 16.0, Mg = 24.3, Cu = 63.5)
Solution
Step 1: Mg(s) + CuO(s)
MgO(s) + Cu(s)
–1
Step 2: Molar mass of CuO = (63.5 + 16.0) g mol
–1
= 79.5 g mol
mass of CuO (g)
Number of moles of CuO =
–1
molar mass of CuO (g mol )
7.95 g
=
–1
79.5 g mol
= 0.100 mol
Step 3: From the equation, mole ratio of Mg : CuO = 1 : 1.
 number of moles of Mg = 0.100 mol
Step 4: Molar mass of Mg = 24.3 g mol
 mass of Mg required
–1
= 0.100 mol × 24.3 g mol
= 2.43 g
–1
cont’d
21
12
III
Metals
Self-test 12.11
Self-test 12.11
–1
–1
Molar mass of PbO = (207.2 + 16.0) g mol = 223.2 g mol
10.55
Number of moles of PbO =
mol = 0.0473 mol
223.2
From the equation, mole ratio of Mg to PbO is 1 : 1.
 number of moles of Mg required = 0.0473 mol
Mass of Mg required = 0.0473 × 24.3 g = 1.15 g
Magnesium reacts with lead(II) oxide according to the following equation:
Mg(s) + PbO(s)
MgO(s) + Pb(s)
Calculate the mass of magnesium required to react completely with 10.55 g of lead(II) oxide.
(Relative atomic masses: O = 16.0, Mg = 24.3, Pb = 207.2)
H20
Example 12.12
CE2002(II)14
CE2005(I)2(a)(iii)
CE2006(II)12
CE2007(II)34
CE2011(I)5(a)(iii)
DSE2019(IA)6
Determining the mass of a product from the mass of a reactant
Sodium hydrogencarbonate decomposes on heating to give sodium carbonate, carbon dioxide and
water. What is the mass of sodium carbonate produced when 6.30 g of sodium hydrogencarbonate
undergoes complete decomposition upon heating?
(Relative atomic masses: H = 1.0, C =12.0, O = 16.0, Na = 23.0)
Solution
Step 1: 2NaHCO3(s)
Na2CO3(s) + CO2(g) + H2O(,)
–1
Step 2: Molar mass of NaHCO3 = (23.0 + 1.0 + 12.0 + 16.0 × 3) g mol
–1
= 84.0 g mol
Number of moles of NaHCO3 used
mass (g)
=
–1
molar mass (g mol )
6.30 g
=
–1
84.0 g mol
= 0.0750 mol
Step 3: From the equation, mole ratio of NaHCO3 : Na2CO3 = 2 : 1.
0.0750 mol
 number of moles of Na2CO3 =
2
= 0.0375 mol
–1
Step 4: Molar mass of Na2CO3 = (23.0 × 2 + 12.0 + 16.0 × 3) g mol
–1
= 106.0 g mol
Mass of Na2CO3 produced
Self-test 12.12
–1
8.51
= 0.0375 mol × 106.0 g mol
Number of moles of Na reacted =
mol = 0.37 mol
23.0
= 3.98 g
From the equation, mole ratio of Na to H2 is 2 : 1.
Self-test 12.12
 number of moles of H2 formed = 0.37 mol = 0.185 mol
2
Mass of H2 produced = 0.185 × 1.0 × 2 g = 0.37 g
Sodium reacts with water to give sodium hydroxide and hydrogen according to the following
equation:
2Na(s) + 2H2O(,)
2NaOH(aq) + H2(g)
Calculate the mass of hydrogen formed when 8.51 g of sodium reacts completely with water.
(Relative atomic masses: H = 1.0, O = 16.0, Na = 23.0)
12
22
12
Note 8
—
Internet
resource
(video)
Decomposition of Baking Soda
This is an online learning and teaching
resource suggested by the EDB.
https://e-aristo.hk/r/cm2bktech12i01.e
e-Learning
Flipped classroom
Calculating reacting
masses involving
limiting reactant
Experiment 12.3
PowerPoint
Reacting masses
Experiment Workbook 1
Studying the thermal decomposition of sodium hydrogencarbonate N8
In this experiment, you are going to study the thermal decomposition of
sodium hydrogencarbonate and solve the related stoichiometric problems.
Limiting reactant
DSE2014(IA)4
Consider the reaction between hydrogen and oxygen to form water:
2H2(g) + O2(g)
2H2O(,)
From the equation, only 1 molecule of oxygen is required to react with 2
molecules of hydrogen for complete reaction. See Figure 12.5.
O2 molecule
H2 molecule
H2O molecule
https://e-aristo.hk/r/
cm2fc12i02.e
Figure 12.5
Two hydrogen
molecules require one oxygen
molecule for complete reaction.
Therefore, oxygen is in excess.
2 H2 molecules + 2 O2 molecules
2 H2O molecules + 1 O2 molecule
In this case, oxygen is in excess. All hydrogen has reacted. The amount
of water produced is limited by the amount of hydrogen used. Therefore,
hydrogen is called the limiting reactant. The limiting reactant limits the
amount of the product formed in a reaction.
H20
Example 12.13
DSE2014(IA)4
DSE2014(IA)19
Calculating reacting masses involving limiting reactant
Calculate the mass of zinc formed when 8.14 g of zinc oxide are heated with 2.20 g of carbon powder.
(Relative atomic masses: C = 12.0, O = 16.0, Zn = 65.4)
Solution
Step 1: 2ZnO(s) + C(s)
2Zn(s) + CO2(g)
–1
–1
Step 2: Molar mass of ZnO = (65.4 + 16.0) g mol = 81.4 g mol
8.14 g
Number of moles of ZnO =
–1 = 0.100 mol
81.4 g mol
–1
Molar mass of C = 12.0 g mol
2.20 g
Number of moles of C =
–1 = 0.183 mol
12.0 g mol
limiting reactant 限量反應物
cont’d
23
12
III
Metals
Step 3: From the equation, mole ratio of ZnO : C = 2 : 1.
 0.100 mol of ZnO would react with 0.100 = 0.0500 mol of C
2
Since 0.183 mol of C is heated, C is in excess.
ZnO is the limiting reactant in this case, as it is all used up.
From the equation, mole ratio of ZnO : Zn = 2 : 2 = 1 : 1.
 number of moles of Zn formed = 0.100 mol
Step 4: Molar mass of Zn = 65.4 g mol
–1
 mass of Zn formed
Self-test 12.13
–1
–1
Molar mass of NO = (14.0 + 16.0) g mol = 30.0 g mol
26.58
Number of moles of NO =
mol = 0.886 mol
30.0
–1
–1
Molar mass of O2 = 16.0 × 2 g mol = 32.0 g mol
8.06
Number of moles of O2 =
mol = 0.252 mol
32.0
From the equation, mole ratio of NO to O2 = 2 : 1.
 O2 is the limiting reactant.
–1
= 0.100 mol × 65.4 g mol
= 6.54 g
Self-test 12.13
Calculate the mass of nitrogen dioxide formed when 26.58 g of nitrogen monoxide reacts with 8.06 g
–1
–1
Molar mass of NO2 = (14.0 + 16.0 × 2) g mol = 46.0 g mol
of oxygen according to the following equation:
2NO(g) + O2(g)
(Relative atomic masses: N = 14.0, O = 16.0)
From the equation, mole ratio of O2 to NO2 = 1 : 2.
2NO2(g)
 number of moles of NO2 formed = 0.252 × 2 mol = 0.504 mol
Mass of NO2 formed = 0.504 × 46.0 g = 23.2 g
Theoretical yield, actual yield and percentage
yield
Theoretical yield is the amount of product expected if the reaction
proceeds exactly as shown in the chemical equation. Actual yield is the
amount of product actually obtained from a reaction. The actual yield of
a reaction is often less than the theoretical yield because:
•
the reaction is incomplete.
•
impurities are present in the reactants.
•
side reactions occur in which unwanted side products are produced.
•
some product is lost during different experimental processes, such as
purification.
The efficiency of a chemical reaction can be expressed by the
percentage yield.
Key point
Percentage yield =
12
24
actual yield 實際產量
percentage yield 百分產率
side product 副產品
side reaction 副反應
theoretical yield 理論產量
actual yield
× 100%
theoretical yield
12
H20
Example 12.14
Reacting masses
CE2006(I)5(d)
Calculating the theoretical yield and actual yield of a product
In an experiment, 15.9 g of copper(II) oxide was heated with 0.60 g of hydrogen according to the
following equation:
CuO(s) + H2(g)
Cu(s) + H2O(,)
(a) Calculate the theoretical yield of copper.
(b) Given the percentage yield of copper is 82%, calculate the actual yield of copper.
(Relative atomic masses: H = 1.0, O = 16.0, Cu = 63.5)
Solution
(a) Step 1: CuO(s) + H2(g)
Cu(s) + H2O(,)
–1
Step 2: Molar mass of CuO = (63.5 + 16.0) g mol
–1
= 79.5 g mol
15.9 g
Number of moles of CuO =
–1
79.5 g mol
= 0.20 mol
–1
–1
Molar mass of H2 = (1.0 × 2) g mol = 2.0 g mol
0.60 g
Number of moles of H2 =
–1
2.0 g mol
= 0.30 mol
Step 3: From the equation, mole ratio of CuO : H2 = 1 : 1.
 0.20 mol of CuO would react with 0.20 mol of H2.
Since 0.30 mol of H2 is heated, H2 is in excess.
CuO is the limiting reactant in this case, as it is all used up.
From the equation, mole ratio of CuO : Cu = 1 : 1.
 number of moles of Cu formed = 0.20 mol
–1
 theoretical yield of Cu = 0.20 mol × 63.5 g mol
= 12.7 g Self-test 12.14
430
(a) Number of moles of H2 =
mol = 215 mol
1.0 × 2
(b) Actual yield of Cu = theoretical yield (g) × percentage yield (%)
–1
–1
Molar mass of CH3OH = (12.0 + 1.0 × 4 + 16.0) g mol = 32.0 g mol
= 12.7 g × 82%
From the equation, mole ratio of H2 to CH3OH = 2 : 1.
= 10.4 g
 number of moles of CH3OH produced = 215 mol = 107.5 mol
Self-test 12.14
2
Theoretical yield of CH3OH = 107.5 × 32.0 g = 3440 g
(b) Actual yield of CH3OH = 3440 g × 45% = 1548 g
Methanol (CH3OH) can be produced from carbon monoxide and hydrogen according to the following
equation:
CO(g) + 2H2(g)
CH3OH(g)
(a) Calculate the theoretical yield of methanol when 430 g hydrogen reacts with excess carbon
monoxide.
(b) Given the percentage yield of methanol is 45%, calculate the actual yield of methanol.
(Relative atomic masses: H = 1.0, C = 12.0, O = 16.0)
Try Chapter Exercise Q27
25
12
III
Metals
Class practice 12.8
1.
Upon strong heating, silver oxide decomposes to silver and oxygen.
Calculate the mass of silver obtained when 6.96 g of silver oxide is
strongly heated in air.
(Relative atomic masses: O = 16.0, Ag = 107.9)
2.
Titanium can be prepared by the reaction of titanium(IV) chloride with
molten magnesium.
TiCl4(g) + 2Mg(,)
Ti(s) + 2MgCl2(,)
6
Calculate the mass of titanium obtained when 5.42 × 10 g of magnesium
7
were allowed to react with 1.77 × 10 g of titanium(IV) chloride.
(Relative atomic masses: Mg = 24.3, Cl = 35.5, Ti = 47.9)
3.
A student performed the following experiment to prepare calcium
hydroxide. 1.50 g of calcium granules was dissolved in a large amount of
water. The calcium hydroxide precipitate was then filtered off, washed and
dried.
(a) Write an equation for the reaction of calcium with water.
(b) Calculate the theoretical mass of calcium hydroxide obtained.
(c) The mass of calcium hydroxide obtained from the experiment was
much less than the theoretical value. Explain why there was such a
difference.
(Relative atomic masses: H = 1.0, O = 16.0, Ca = 40.1)
A12.8
1. 2Ag2O(s)
4Ag(s) + O2(g)
Number of moles of Ag2O used =
6.96
mol = 0.0300 mol
107.9 × 2 + 16.0
From the equation, mole ratio of Ag2O to Ag is 1 : 2.
 number of moles of Ag produced = 0.0300 × 2 mol = 0.0600 mol
Mass of Ag produced = 0.0600 × 107.9 g = 6.47 g
6
5.42 × 10
mol = 223 045 mol
24.3
7
1.77 × 10
Number of moles of TiCl4 used =
mol = 93 207 mol
47.9 + 35.5 × 4
From the equation, mole ratio of TiCl4 to Mg is 1 : 2.
 TiCl4 is the limiting reactant.
From the equation, mole ratio of TiCl4 to Ti is 1 : 1.
 number of moles of Ti formed = 93 207 mol
Mass of Ti formed = 93 207 × 47.9 g = 4 464 615 g
2. Number of moles of Mg used =
3. (a) Ca(s) + 2H2O()
Ca(OH)2(s) + H2(g)
1.50
mol = 0.0374 mol
40.1
From the equation, mole ratio of Ca to Ca(OH)2 = 1 : 1.
 number of moles of Ca(OH)2 formed = 0.0374 mol
Theoretical mass of Ca(OH)2 formed
= 0.0374 × [40.1 + (16.0 + 1.0) × 2] g = 2.77 g
(c) Possible reasons:
The calcium used was impure.
Some calcium hydroxide was lost during filtration.
(b) Number of moles of Ca used =
12
26
12
Reacting masses
Key terms
PowerPoint
English term
Chinese translation
Page
1.
actual yield
實際產量
24
2.
Avogadro constant
亞佛加德羅常數
3
3.
composition by mass
質量組成
12
4.
empirical formula
實驗式
10
5.
limiting reactant
限量反應物
23
6.
molar mass
摩爾質量
4
7.
mole
摩爾
3
8.
molecular formula
分子式
10
9.
percentage by mass
質量百分比
7
10. percentage yield
百分產率
24
11. structural formula
結構式
10
12. theoretical yield
理論產量
24
A12.6
1.
Number of
moles of
atoms (mol)
Mg
O
28.698 – 28.092 = 0.606
29.103 – 28.698 = 0.405
0.606
= 0.0249
24.3
0.405
= 0.0253
16.0
0.0249
=1
0.0249
0.0253
= 1.02  1
0.0249
Simplest
whole
number mole
ratio of atoms
12.0
g = 0.320 g
12.0 + 16.0 × 2
1.0 × 2
Mass of H in the compound = 0.240 ×
g = 0.0267 g
1.0 × 2 + 16.0
Mass of O in the compound = (1.200 – 0.320 – 0.0267) g = 0.853 g
2. Mass of C in the compound = 1.173 ×
Number of
moles of
atoms (mol)
Simplest
whole
number mole
ratio of atoms
Mass (g)
Number of
moles of
atoms (mol)
Simplest
whole
number mole
ratio of atoms
C
H
0.857
1.000 – 0.857 = 0.143
0.857
= 0.0714
12.0
0.143
= 0.143
1.0
0.0714
=1
0.0714
0.143
=2
0.0714
 the empirical formula of the compound is CH2.
 the empirical formula of the oxide of magnesium is MgO.
Mass (g)
Extension
Mass (g)
3.
C
H
O
0.320
0.0267
0.853
0.320
= 0.0267
12.0
0.0267
= 0.0267
1.0
0.853
= 0.0533
16.0
0.0267
=1
0.0267
0.0267
=1
0.0267
0.0533
=2
0.0267
4. Assume that there are 100 g of X. Then, there are 26.95 g of
sulphur, 13.44 g of oxygen and 59.61 g of chlorine.
Mass (g)
Number of
moles of
atoms (mol)
Simplest
whole
number mole
ratio of atoms
S
O
Cl
26.95
13.44
59.61
26.95
= 0.840
32.1
13.44
= 0.84
16.0
59.61
= 1.68
35.5
0.840
=1
0.840
0.84
=1
0.840
1.68
=2
0.840
 the empirical formula of the compound is SOCl2.
 the empirical formula of the compound is CHO2.
27
12
III
Metals
Progress check
PowerPoint
Can you answer the following questions? Put a ‘✓’ in the box if you can. Otherwise, review the relevant
part on the page as shown.
Page
1.
What is the meaning of mole?
3
2.
What is the Avogadro constant?
3
3.
What is the meaning of molar mass?
4
4.
How is the mole of a substance related to its mass and number of formula units?
5
5.
How can we calculate the percentage by mass of an element in a compound?
7
6.
What are empirical formula, molecular formula and structural formula?
10
7.
How can we determine the empirical formula of a compound?
12–14
8.
How can we determine the molecular formula of a compound?
15
9.
What are the interrelationship between masses of reactants and products in a reaction?
20
10. How can we calculate masses of reactants and products in a reaction from the relevant
equation?
21
11. What is the meaning of a limiting reactant?
23
12. What are the meanings of actual yield and theoretical yield?
24
13. How can we calculate the percentage yield of a chemical reaction?
24
A12.7
1. Mass of C in the compound = 1.32 ×
12.0
g = 0.36 g
12.0 + 16.0 × 2
1.0 × 2
Mass of H in the compound = 0.45 ×
g = 0.05 g
1.0 × 2 + 16.0
Mass of O in the compound = (0.81 – 0.36 – 0.05) g = 0.40 g
Mass (g)
Number of
moles of
atoms (mol)
Simplest
whole
number mole
ratio of atoms
C
H
O
0.36
0.05
0.40
0.36
= 0.03
12.0
0.05
= 0.05
1.0
0.40
= 0.025
16.0
0.03
= 1.2
0.025
0.05
=2
0.025
0.025
=1
0.025
1.2 × 5 = 6
2 × 5 = 10
1×5=5
 the empirical formula of the compound is C6H10O5.
Let the molecular formula of the compound be (C6H10O5)n.
320.0 = n × (12.0 × 6 + 1.0 × 10 + 16.0 × 5)
n = 1.98  2
 the molecular formula of the compound is C12H20O10.
2. Assume that there are 100 g of the compound. Then, there are
40.00 g of carbon, 6.67 g of hydrogen and 53.33 g of oxygen.
Number of moles
of atoms (mol)
28
H
O
6.67
53.33
40.00
= 3.33
12.0
6.67
= 6.67
1.0
53.33
= 3.33
16.0
3.33
=1
3.33
6.67
=2
3.33
3.33
=1
3.33
Simplest whole
number mole
ratio of atoms
 the empirical formula of the compound is CH2O.
Let the molecular formula of the compound be (CH2O)n.
60.0 = n × (12.0 + 1.0 × 2 + 16.0)
n=2
 the molecular formula of the compound is C2H4O2.
3. Assume that there are 100 g of Epsom salt. Then, there are 51.22 g
of water of crystallization and (100 – 51.22) g = 48.78 g of MgSO4.
MgSO4
Mass (g)
H2O
48.78
51.22
24.3 + 32.1 + 16.0 × 4 = 120.4
1.0 × 2 + 16.0 = 18.0
Number of moles of
formula units (mol)
48.78
= 0.4051
120.4
51.22
= 2.85
18.0
Simplest whole
number mole ratio of
formula units
0.4051
=1
0.4051
2.85
= 7.04  7
0.4051
Formula mass
12
C
40.00
Mass (g)
 the value of n is 7.
12
Reacting masses
Summary
PowerPoint
12.1 The mole, Avogadro constant and molar mass
1.
Chemists use mole (abbreviation: mol) to describe the quantity of particles in a substance.
2.
The Avogadro constant (L) is the number of atoms in exactly 12.0 g of carbon-12. It is equal
23
–1
to 6.02 × 10 mol .
3.
The molar mass of a substance is its formula mass expressed in gram unit. The unit of molar
–1
mass is g mol .
4.
Important relationships involving moles:
�
•
Number of moles (mol) =
�
•
Number of moles (mol) =
mass (g)
–1
molar mass (g mol )
number of formula units
–1
Avogadro constant (mol )
12.2 Percentage by mass of an element in a compound
5.
The percentage by mass of an element in a compound can be found by the equation:
Percentage by mass of element A in a compound
relative atomic mass of A × number of atoms of A in the formula
× 100%
=
formula mass of the compound
12.3 Chemical formulae of compounds
6.
Chemical formulae are part of the language of chemistry. Some common chemical formulae
include empirical formula, molecular formula and structural formula.
12.4 Empirical formulae and molecular formulae derived from experimental data
7.
Empirical formula of a compound is the formula which shows the simplest whole number ratio
of the atoms or ions present.
8.
The empirical formula of a compound can be calculated from its composition by mass. The
composition of a compound has to be determined by experiment.
9.
Molecular formula may be determined from empirical formula and relative molecular mass. This
is because molecular formula is a whole number multiple of empirical formula.
29
12
III
Metals
12.5 Reacting masses from chemical equations
10.
The theoretical amounts of substances used up or produced in a reaction can be calculated from
its balanced equation.
11.
Limiting reactant is the reactant that is completely used up in a reaction. It limits the amount of
product(s) formed in the reaction.
12.
The theoretical amounts of product predicted by calculation from its balanced equation is called
theoretical yield. The actual yield of a reaction is often less than the theoretical yield.
13.
Percentage yield is the ratio of actual yield and theoretical yield. It is a measure of the efficiency
of a chemical reaction.
Percentage yield =
12
30
actual yield
× 100%
theoretical yield
12
Reacting masses
Concept map
PowerPoint
Complete the following concept map.
Mass
equals
Number
of moles
Molar mass
equals
Number of
formula units
equals
Avogadro constant
6.02 × 1023
formula units
without unit,
equals
Formula mass
Relative molecular mass
equals
Sum of
relative
atomic masses
of all atoms/ions in
a formula unit of a
substance
equals
Sum of relative
atomic masses of
all atoms in a
Empirical formula
determine
Molecular formula
Chemical
formulae
molecule
Structural formula
(Hints: Avogadro constant, molar mass, molecular formula, molecule, relative atomic masses, relative
molecular mass, structural formula)
31
12
III
Metals
Chapter exercise
Fill in the blanks
6.02 × 10
One mole contains
particles —
Avogadro constant
this number is called the
.
3.
molar mass
The
of a substance is the
mass in grams of one mole of the substance.
Section 12.1
1.
The relative molecular mass of a molecular
relative atomic masses
compound is the sum of
of all atoms present in a
molecule of that compound.
23
2.
Section 12.2
4.
Percentage by mass of element A in a compound =
Relative atomic mass
number
of A ×
of atoms of A in the formula
× 100%
Formula mass of the compound
Section 12.3
5.
6.
Section 12.5
empirical
The
formula of a compound is
the formula which shows the simplest whole
number ratio of atoms or ions present. It can be
composition
calculated from its
by mass, which
can be determined by experiment.
Molecular formula may be determined from
relative molecular mass
empirical formula and
.
Practice questions
Section 12.1
9.
Limiting reactant
7.
is the substance which is
all used up in a chemical reaction.
8.
Percentage yield =
actual yield
× 100%
theoretical yield
9. (a) Number of sodium atoms
23
24
= 2 × 6.02 × 10 = 1.204 × 10
(b) Number of moles of oxygen molecules
2
=
mol = 0.0625 mol
16.0 × 2
Number of oxygen atoms
23
22
= 0.0625 × 2 × 6.02 × 10 = 7.525 × 10
(c) Number of atoms in 1.5 moles of nitrogen dioxide gas
23
24
= 1.5 × 3 × 6.02 × 10 = 2.709 × 10
Calculate the total number of atoms in the following substances.
(a) 2 moles of sodium metal
(b) 2 g of oxygen gas
(c) 1.5 moles of nitrogen dioxide gas
(d) 0.5 mole of sodium carbonate-10-water
(e) 22 g of aluminium sulphate
(d) Number of atoms in 0.5 mole of sodium carbonate-10-water
23
25
= 0.5 × 36 × 6.02 × 10 = 1.084 × 10
(e) Number of moles of aluminium sulphate
22
=
mol = 0.0643 mol
27.0 × 2 + (32.1 + 16.0 × 4) × 3
Number of atoms in 0.0643 mol of aluminium sulphate
23
23
= 0.0643 × 17 × 6.02 × 10 = 6.58 × 10
10. (a) Formula mass of CH4 = 12.0 + 1.0 × 4 = 16.0
(Relative atomic masses: H = 1.0, C = 12.0, N = 14.0, O = 16.0, Na = 23.0, Al = 27.0, S = 32.1)
12.0
× 100% = 75%
16.0
(b)
Formula
mass
of
anhydrous
Na
SO
Section 12.2
2
4
= 23.0 × 2 + 32.1 + 16.0 × 4 = 142.1
10. Calculate the percentage by mass of
Percentage by mass of S in anhydrous Na2SO4
32.1
(a) carbon in methane, CH4
× 100% = 22.6%
=
142.1
(b) sulphur in anhydrous sodium sulphate, Na2SO4
(c) Formula mass of Na2CO3․10H2O
= 23.0 × 2 + 12.0 + 16.0 × 3 + 10 × (1.0 × 2 + 16.0) = 286.0
(c) water in sodium carbonate-10-water, Na2CO3․10H2O
Percentage by mass of H2O in Na2CO3․10H2O
10 × 18.0
(d) oxygen in iron(II) sulphate-7-water, FeSO4․7H2O
× 100% = 62.9%
=
286.0
Percentage by mass of C in CH4 =
(Relative atomic masses: H = 1.0, C = 12.0, O = 16.0, Na = 23.0, S = 32.1, Fe = 55.8)
12
32
(d) Formula mass of FeSO4․7H2O
= 55.8 + 32.1 + 16.0 × 4 + 7 × (1.0 × 2 + 16.0) = 277.9
Percentage by mass of O in FeSO4․7H2O
16.0 × 11
× 100% = 63.3%
=
277.9
11.
12
11. (a) Formula mass of CH4 = 12.0 + 1.0 × 4 = 16.0
1.0 × 4
× 100% = 25%
Percentage by mass of H in CH4 =
16.0
mass of Mass of H in 10 g of CH4 = 10 g × 25% = 2.5 g
Reacting masses
(b) Formula mass of anhydrous Na2SO4
= 23.0 × 2 + 32.1 + 16.0 × 4 = 142.1
Find the
Percentage by mass of Na in anhydrous Na2SO4
23.0 × 2
× 100% = 32.4%
=
(a) hydrogen in 10 g of methane, CH4
142.1
Mass of Na in 50 g of anhydrous Na2SO4
(b) sodium in 50 g of anhydrous sodium sulphate, Na2SO4
= 50 g × 32.4% = 16.2 g
(c) chlorine in 2 moles of iron(III) chloride-6-water, FeCl3․6H2O (c) Formula mass of FeCl3․6H2O
(d) water in 1.25 moles of calcium chloride-6-water, CaCl2․6H2O = 55.8 + 35.5 × 3 + 6 × (1.0 × 2 + 16.0) = 270.3
Percentage by mass of Cl in FeCl3․6H2O
(Relative atomic masses: H = 1.0, C = 12.0, O = 16.0, Na = 23.0, S = 32.1, Cl = 35.5, Ca = 40.1,
35.5 × 3
× 100% = 39.4%
=
Fe = 55.8)
270.3
Mass of 2 moles of FeCl3․6H2O = 2 × 270.3 g = 540.6 g
Mass of Cl in 2 moles of FeCl3․6H2O
= 540.6 × 39.4% = 213 g
Section 12.4
12. Find the empirical formulae of compounds having the following composition by mass:
(a) 75% carbon, 25% hydrogen
(b) 86.6% lead, 13.4% oxygen
(c) 36.5% sodium, 25.4% sulphur, 38.1%
(d) Formula mass of CaCl2․6H2O
= 40.1 + 35.5 × 2 + 6 × (1.0 × 2 + 16.0) = 219.1
6 × 18.0
× 100% = 49.3%
Percentage by mass of H2O in CaCl2․6H2O =
219.1
×
Mass
of
1.25
moles
of
CaCl
․6H
O
=
1.25
219.1
g
=
273.9
g
2
2
oxygen
Mass of H2O in 1.25 moles of CaCl2․6H2O = 273.9 × 49.3% = 135.0 g
(d) 40.67% carbon, 23.73% nitrogen, 27.13% oxygen, 8.47% hydrogen
(e) A hydrated salt containing 37.11% copper, 41.68% chlorine (the rest being water of crystallization)
(Relative atomic masses: H = 1.0, C = 12.0, N = 14.0, O = 16.0, Na = 23.0, S = 32.1, Cl = 35.5, Cu = 63.5,
Pb = 207.2) (Answers on the back page of p.12-35.)
Multiple-choice questions
Section 12.1
13. (1): The molar mass of a substance is
the mass in gram of one mole of it.
13. Which of the following statements is/are correct?
(1) The molar mass of a substance is the mass in
kilograms of one mole of it.
(2) One mole of oxygen gas has the same number
of atoms as one mole of nitrogen gas.
(3) One mole of oxygen gas has the same mass as
one mole of nitrogen gas.
A.
B.
C.
D.
(1) only
(2) only
(1) and (3) only
(2) and (3) only
(2): Both molecules of oxygen and
nitrogen are diatomic.
(3): Molar mass of oxygen is (16.0 × 2)
–1
–1
g mol = 32.0 g mol ;
B
molar mass of nitrogen is
–1
–1
(14.0 × 2) g mol = 28.0 g mol
14. Which of the following substances contains the
greatest number of atoms?
(Relative atomic masses: H = 1.0, Ne = 20.2,
Mg = 24.3, Cu = 63.5)
A.
B.
C.
D.
2.0 g of hydrogen
24.3 g of magnesium
30.3 g of neon
31.8 g of copper
2.0
mol = 1 mol
1.0 × 2
Number of moles of H atoms = 1 × 2 mol = 2 mol
24.3
(B): Number of moles of Mg atoms =
mol = 1 mol
24.3
30.3
(C): Number of moles of Ne atoms =
mol = 1.5 mol
20.2
31.8
(D): Number of moles of Cu atoms =
mol = 0.501 mol
63.5
14. (A): Number of moles of H2 molecules =
A
15. If 2 g of helium gas contains y molecules, how
many molecules are present in 38 g of fluorine
gas?
(Relative atomic masses: He = 4.0, F = 19.0)
15. Number of helium molecules present in 2 g of
2
× Avogadro constant
helium gas = y =
4.0
 Avogadro constant = 2y
Number of moles of fluorine molecules present in
38 g of fluorine gas
C
38
× Avogadro constant
=
19.0 × 2
12.2 = 38 × 2y = 2y
19.0 × 2
1y
2
B. y
C. 2y
D. 4y
A.
Section
16. The percentage by mass of water of crystallization
in Na2CO3․H2O is
(Relative atomic masses: H = 1.0, C = 12.0,
O = 16.0, Na = 23.0)
A.
B.
C.
D.
16. Percentage by mass of water of crystallization in
Na2CO3․H2O
1.0 × 2 + 16.0
× 100%
=
23.0 × 2 + 12.0 + 16.0 × 3 + 1.0 × 2 + 16.0
= 14.5%
D
13.0%.
13.5%.
14.0%.
14.5%.
17. How many grams of iron can be extracted from
100 g of iron ore, which contains 70% by mass of
Fe3O4?
(Relative atomic masses: O = 16.0, Fe = 55.8)
A.
B.
C.
D.
16.9 g 17.
24.1 g
50.6 g
72.3 g
Mass of Fe3O4 in 100 g of iron ore = 100 g × 70%
= 70 g
Mass of Fe in 70 g of Fe3O4
55.8 × 3
= 70 ×
g = 50.6 g
C
55.8 × 3 + 16.0 × 4
33
12
29
III
Metals
Section
18. Mass of metal X in the oxide = (8.42 – 2.40) g = 6.02 g
Mass of O in the oxide = 2.40 g
Number of moles of X : number of moles of O
12.4 = 6.02 : 2.40 = 0.150 : 0.150 = 1 : 1
Questions
40.1 16.0
18. The relative atomic mass of metal X is 40.1. When
metal X burns in air, an oxide forms. It is found
that 8.42 g of the oxide contains 2.40 g of oxygen.
Calculate the mole ratio of X to oxygen in the
oxide.
(Relative atomic mass of O = 16.0)
A. 1 : 1
C. 2 : 1
B. 1 : 2
D. 1 : 3
A
19. If 20.0 g of water is completely decomposed into
hydrogen and oxygen, the total mass of gases
formed is
(Relative atomic masses: H = 1.0, O = 16.0)
B. 2.22 g
D. 20.0 g
D
20. 4.76 g of metal R combine with 5.33 g of chlorine
to form a chloride in which the charge of the ion
of R is +2. What is the relative atomic mass of R?
(Relative atomic mass of Cl = 35.5)
A.
B.
C.
D.
63.4
63.6
65.3
65.5
21. Calcium carbonate decomposes on heating to give
quicklime (calcium oxide) and carbon dioxide.
What is the mass of calcium oxide produced if the
decomposition of 10.01 g of calcium carbonate is
complete? 21. CaCO (s)
CaO(s) + CO (g)
3
Section 12.5
A. 1.11 g
C. 17.8 g
21 and 22 are about the thermal
decomposition of calcium carbonate.
A
Structured questions
Section 12.2
2
(Relative atomic masses: C = 12.0, O = 16.0,
Number of moles of CaCO3 used
Ca = 40.1)
A.
B.
C.
D.
4.84 g
5.61 g
7.21 g
9.62 g
10.01
mol = 0.1 mol
40.1 + 12.0 + 16.0 × 3
From the equation, mole ratio of
CaCO3 to CaO = 1 : 1.
 mass of CaO produced = 0.1 ×
B
(40.1 + 16.0) g = 5.61 g
=
22. If 4.31 g of calcium oxide is finally obtained, what
is the percentage yield of the reaction?
A.
B.
C.
D.
44.8%
59.8%
76.8%
89.0%
4.31 g
× 100%
5.61 g
= 76.8%
22. Percentage yield of the reaction =
C
19. 2H2O()
2H2(g) + O2(g)
Atoms cannot be created or destroyed in a reaction. Hence, the mass
of products (H2(g) and O2(g)) is the same as that of the reactant, i.e.
20.0 g.
20. R(s) + Cl2(g)
RCl2(s)
Number of moles of Cl2 used =
5.33
mol
35.5 × 2
= 0.0751 mol
From the equation, mole ratio of R to Cl2 is 1 : 1.
 number of moles of R used = 0.0751 mol
–1
Let the molar mass of R be y g mol .
4.76
0.0751 =
y
y = 63.4
23. Fluoride is usually present in toothpastes. Some toothpastes contain tin(II) fluoride.
(a) Write the chemical formula of tin(II) fluoride.
(b) Calculate the formula mass of tin(II) fluoride.
(c) Calculate the percentage by mass of fluorine in tin(II) fluoride.
(d) A tube of toothpaste contains 1.50 g of tin(II) fluoride. Calculate the mass of fluorine in this tube of
toothpaste.
23. (a) SnF2
(Relative atomic masses: F = 19.0, Sn = 118.7)
12
34
(b) Formula mass of SnF2 = (118.7 + 19.0 × 2) = 156.7
19.0 × 2
× 100% = 24.3%
(c) Percentage by mass of F in SnF2 =
156.7
(d) Mass of F in 1.50 g of SnF2
= 1.50 g × 24.3% = 0.365 g
12
Reacting masses
Section 12.4
24. In an experiment, when sodium was heated in air, it caught fire and formed an oxide. It was found that the
mass of the sample increased by 35% after heating. In another experiment, when sodium was heated in pure
oxygen, another oxide formed. The increase in mass was found to be 70%. Determine the empirical
formulae of the two oxides of sodium. (Answers on the back page of p.12-35.)
(Relative atomic masses: O = 16.0, Na = 23.0)
25. Paracetamol is a common ingredient in some drugs that relieve pain and reduce fever.
Paracetamol has a relative molecular mass of 151.0 and it contains 63.58% by mass of carbon, 5.96% by
mass of hydrogen, 9.27% by mass of nitrogen and 21.19% by mass of oxygen. Calculate its molecular
formula. (Answers on the back page of p.12-35.)
(Relative atomic masses: H = 1.0, C = 12.0, N = 14.0, O = 16.0)
26. 13.07 g of hydrated sodium carbonate, Na2CO3․nH2O, on strong heating, gave 8.23 g of water.
(a) Calculate the mass of anhydrous sodium carbonate (Na2CO3).
(Answers on the back page of p.12-35.)
(b) Calculate the formula mass of anhydrous sodium carbonate.
(c) Calculate the number of moles of anhydrous sodium carbonate in the sample.
(d) Calculate the number of moles of water in the sample.
(e) Calculate the value of n.
27. (a) Number of moles of C2H2 =
(Relative atomic masses: H = 1.0, C = 12.0, O = 16.0, Na = 23.0)
2.00
mol = 0.0769 mol
(12.0 × 2 + 1.0 × 2)
5.20
mol = 0.0325 mol
79.9 × 2
From the equation, mole ratio of C2H2 to Br2 is 1 : 2.
Number of moles of Br2 =
Section 12.5
27. Ethyne (C2H2) reacts with bromine (Br2) to form tetrabromoethane (C2H2Br4).
C2H2(g) + 2Br2(,)
C2H2Br4(,)
In an experiment, 2.00 g of ethyne reacted with 5.20 g of bromine. 5.02 g of tetrabromoethane was produced
 Br2 is the limiting reactant.
in the reaction.
0.0325
mol = 0.0163 mol
2
Theoretical yield of C2H2Br4 = 0.0163 × (12.0 × 2 + 1.0 × 2 + 79.9 × 4) g
= 5.63 g
(b) Calculate the percentage yield of the reaction.
5.02 g
× 100% = 89.2%
(b)
Percentage
yield
of
the
reaction =
(Relative atomic masses: H = 1.0, C = 12.0, Br = 79.9)
5.63 g
(a) Calculate the theoretical yield of C2H2Br4(,).
Number of moles of C2H2Br4 formed =
28. To prepare iron(III) oxide, 5.91 g of iron was dissolved in excess dilute hydrochloric acid to give a solution
2+
2+
containing Fe ions. The solution was then boiled with concentrated nitric acid to convert all Fe ions to
3+
3+
Fe ions. Excess sodium hydroxide solution was added to precipitate all Fe ions as iron(III) hydroxide,
Fe(OH)3. The precipitate was filtered, washed, dried and finally heated to convert the precipitate to iron(III)
oxide, Fe2O3.
28. (a) The whole process may be represented by a sequence of steps:
(a) Calculate the theoretical yield of iron(III) oxide.
2+
3+
Fe
Fe
Fe
Fe(OH)3
Fe2O3
And the whole process can be represented by the overall equation:
(b) The mass of iron(III) oxide actually obtained from the experiment was 7.95 g.
(i)
Calculate the percentage yield of iron(III) oxide.
2Fe
Fe2O3 (the ‘2’ is added to balance the number of Fe atoms)
Thus, mole ratio of Fe to Fe2O3 = 2 : 1.
(ii) Compare the actual yield with the theoretical yield and give TWO possible reasons for the
5.91
difference.
Number of moles of Fe =
mol = 0.106 mol
(Relative atomic masses: O = 16.0, Fe = 55.8)
55.8
0.106
Number of moles of Fe2O3 formed =
mol = 0.053 mol
2
–1
–1
Molar mass of Fe2O3 = (55.8 × 2 + 16.0 × 3) g mol = 159.6 g mol
Theoretical yield of Fe2O3 = 0.053 × 159.6 g = 8.46 g
7.95 g
× 100% = 94.0%
8.46 g
(ii) The actual yield is smaller than the theoretical yield. The possible reasons for the difference (Any TWO):
– The reaction was incomplete.
– The iron used might be impure.
– There was a loss of materials during various experimental processes, e.g. filtration.
28. (b) (i) Percentage yield of iron(III) oxide =
35
12
29
(This is a blank page.)
12. (a) Assume that there are 100 g of the compound.
C
H
75
25
75
= 6.25
12.0
25
= 25
1.0
6.25
=1
6.25
25
=4
6.25
Mass (g)
24. Assume that the mass of the sodium to be 1 g.
Oxide of sodium formed in air:
Simplest
whole
number mole
ratio of atoms
Number of
moles of
atoms (mol)
Simplest
whole
number mole
ratio of atoms
 the empirical formula of the compound is CH4.
(b) Assume that there are 100 g of the compound.
Pb
O
86.6
13.4
86.6
= 0.418
207.2
13.4
= 0.838
16.0
0.418
=1
0.418
0.838
=2
0.418
Mass (g)
Simplest
whole
number mole
ratio of atoms
Number of
moles of
atoms (mol)
Simplest
whole
number mole
ratio of atoms
S
O
36.5
25.4
38.1
36.5
= 1.59
23.0
25.4
= 0.79
32.1
38.1
= 2.38
16.0
1.59
= 2.01  2
0.79
0.79
=1
0.79
2.38
= 3.01  3
0.79
Simplest
whole
number mole
ratio of atoms
Mass (g)
Number of
moles of
atoms (mol)
Simplest
whole
number mole
ratio of atoms
C
N
O
H
40.67
23.73
27.13
8.47
40.67
= 3.39
12.0
23.73
= 1.70
14.0
27.13
= 1.70
16.0
8.47
= 8.47
1.0
3.39
= 1.99  2
1.70
1.70
=1
1.70
1.70
=1
1.70
8.47
= 4.98  5
1.70
 the empirical formula of the compound is C2NOH5.
(e) Assume that there are 100 g of the compound.
Cu
Cl
H2O
37.11
41.68
(100 – 37.11 –
41.68) = 21.21
Number of moles
of formula units
(mol)
37.11
= 0.584
63.5
41.68
= 1.174
35.5
21.21
= 1.178
18.0
Simplest whole
number mole ratio
of formula units
0.584
=1
0.584
1.174
= 2.01  2
0.584
1.178
= 2.02  2
0.584
Mass (g)
0.35
= 0.0219
16.0
0.0435
= 1.99  2
0.0219
0.0219
=1
0.0219
Na
O
1
0.70
1
= 0.0435
23.0
0.70
= 0.0438
16.0
0.0435
=1
0.0435
0.0438
=1
0.0435
25. Assume that there are 100 g of paracetamol.
 the empirical formula of the compound is Na2SO3.
(d) Assume that there are 100 g of the compound.
Number of
moles of
atoms (mol)
1
= 0.0435
23.0
 the empirical formula of the oxide of sodium formed in pure oxygen
is NaO.
Na
Mass (g)
0.35
 the empirical formula of the oxide of sodium formed in air is Na2O.
Oxide of sodium formed in pure oxygen:
 the empirical formula of the compound is PbO2.
(c) Assume that there are 100 g of the compound.
Simplest
whole
number mole
ratio of atoms
1
Mass (g)
Number of
moles of
atoms (mol)
Number of
moles of
atoms (mol)
O
Mass (g)
Number of
moles of
atoms (mol)
Mass (g)
Na
 the empirical formula of the compound is CuCl2․2H2O.
C
H
N
O
63.58
5.96
9.27
21.19
63.58
= 5.30
12.0
5.96
= 5.96
1.0
9.27
= 0.662
14.0
21.19
= 1.32
16.0
5.30
=8
0.662
5.96
=9
0.662
0.662
=1
0.662
1.32
=2
0.662
 the empirical formula of paracetamol is C8H9NO2.
Let the molecular formula of paracetamol be (C8H9NO2)n.
n × (12.0 × 8 + 1.0 × 9 + 14.0 + 16.0 × 2) = 151.0
n=1
 the molecular formula of paracetamol is C8H9NO2.
26. (a) Mass of anhydrous Na2CO3 = (13.07 – 8.23) g = 4.84 g
(b) Formula mass of anhydrous Na2CO3
= (23.0 × 2 + 12.0 + 16.0 × 3) = 106.0
(c) Number of moles of anhydrous Na2CO3 in the sample
4.84
=
mol = 0.0457 mol
106.0
(d) Number of moles of H2O in the sample
8.23
=
mol = 0.457 mol
18.0
(e)
Na2CO3
H2O
Number of moles of
formula units (mol)
Simplest whole
number mole ratio
of formula units
 the value of n is 10.
0.0457
0.457
0.0457
=1
0.0457
0.457
= 10
0.0457