Chpater 4 - Chemical Formulae and
Equations
4.1
Chemical Names and
Formulae
4.2
Chemical Equations for
Reactions
4.3
Relative Masses of Atoms
and Molecules
4.1 – Chemical Names and Formulae
Formulae of elements
 Each element has a chemical symbol:
– A single capital letter: hydrogen (H), oxygen(O), nitrogen(N)
– Two letters, capital letter + small letter: aluminium(Al), argon(Ar),
magnesium(Mg)

Formula of elements represented by its chemical symbol
– Elements of individual atoms (e.g. noble gases): He, Ne, Ar, Kr
– Elements of giant structures, either metallic or covalent bonding:
Cu, Mg, Fe, Na, K, C, Si, Ge
– Elements whose molecules contain more than three atoms: P, S
(strictly should be P4, S8)
4.1 – Chemical Names and Formulae
 Simple and compound ions:
 Simple ions:
– Na+, K+, Mg2+, Cl-, O2-,

Compound ions:
– Made up of atoms covalently bonded together.
– Negative ion: SO42-, CO32-, NO3– Postive ion: NH4+
4.1 – Chemical Names and Formulae
Name of Compounds
 Compound of two elements:
 A metal and a non-metal:
– name of metal given first
– and then name of non-metal, ending with -ide.
Examples: sodium chloride (NaCl), magnesium oxide
(MgO), iron(II) sulfide (FeS) .
 Compound of two non-metals:
– if one is hydrogen, that is named first
– otherwise the one with the lower group number
comes first
– and then the name of the other non-metal, ending
with -ide.
Examples: hydrogen chloride (HCl), carbon dioxide (CO2).
4.1 – Chemical Names and Formulae
The Names of Compounds
 Compound containing a compound ion(usually containing
oxygen):
 – Having names that end with -ate
Examples: calcium carbonate (CaCO3), potassium nitrate
(KNO3), magnesium sulfate(MgSO4), sodium
ethanoate(CH3COONa)
 Prefix to indicate the number of that particular atom:
Examples: carbon monoxide (CO), carbon dioxide (CO2),
nitrogen dioxide (NO2), dinitrogen tetroxide (N2O4), sulfur
dioxide (SO2), sulfur trioxide (SO3).
 Some compounds have ‘everyday’ names: water (H2O),
methane(CH4) , and ammonia(NH3).
4.1 – Chemical Names and Formulae
Finding formulae from the structure of compounds
 Every compound has a formula.
 The formula is made up of the symbols for the elements, and often has numbers
too.
 The formula of a compound is related to its structure.
Sodium chloride
Water is made up of
Silicon dioxide (silica)
Giant structure with one
molecules
Giant structure
sodium ion for every
Two hydrogen atoms are
There are two oxygen
chloride ion.
bonded to an oxygen atom.
atoms for every silicon
Formula is NaCl.
Formula is H2O.
atom. Formula is SiO2.
(Ionic compound)
(Covalent compound)
 Difference:
 In giant structures like sodium chloride and silicon dioxide, the formula tells you
the ratio of the ions or atoms in the compound.
 In a molecular compound, the formula tells you exactly how many atoms are
bonded together in each molecule.
4.1 – Chemical Names and Formulae
Valency（化合价）
 The valency of an element is the number of electrons its atoms
lose, gain or share, to form a compound.
 You can work out the structure of a compound if you know the
valency of the elements:
Click Here
4.1 – Chemical Names and Formulae
Elements
In forming a compound,
the atoms…
So the valency
of the element
is …
Group I
Lose 1 electron
1
sodium chloride, NaCl
Group II
Lose 2 electron
2
magnesium chloride, MgCl2
Group III
Lose 3 electron
3
aluminium chloride, AlCl3
Group IV
Share 4 electrons
4
methane, CH4
Group V
Gain or share 3 electrons
3
ammonia, NH3
Group VI
Gain or share 2 electrons
2
magnesium oxide, MgO;
water, H2O
Group VII
Gain or share 1 electrons
1
sodium chloride, NaCl;
hydrogen chloride, HCl
Group 0
(do not form compounds)
-
none
Hydrogen
Lose or share 1 electron
1
hydrogen bromide, HBr
Transition
Elements
Can lose different
numbers of electrons
Variable
Examples of compounds formed
(those in blue are covalent, with
shared electrons)
copper (I) chloride, CuCl; copper (II)
chloride, CuCl2
4.1 – Chemical Names and Formulae
Writing formulae using valencies (‘Cross-over’ method)
 To write the formula of a compound using valencies:
1. Write down the valencies of the two elements.
2. Write down their symbols, in the same order as the elements
in the name.
3. Add numbers after the symbols if you need to, to balance the
valencies.
Example 1
Formula of hydrogen sulfide?
1. Valencies: hydrogen, 1; sulfur (Group VI), 2
2. HS (valencies not balanced)
3. The formula is H2S (2 X 1 and 2, so the valencies are now
balanced)
Example 2
What is the formula of aluminium oxide?
1. Valencies: aluminium (Group III), 3; oxygen (Group VI), 2
2. AlO (valencies not balanced)
3. The formula is Al2O3 (2 X 3 and 3 X 2, so the valencies are
now balanced)
4.1 – Chemical Names and Formulae
4.1 – Chemical Names and Formulae
Writing formulae by balancing charges
 In an ionic compound, the total charge is zero. You can work out the formula of
an ionic compound by balancing the charges on its ions.
4.1 – Chemical Names and Formulae
4.1 – Chemical Names and Formulae
The Periodic Table will help you with these.
1. Write the chemical name for water (ending in -ide).
2. Name the compounds containing only these elements:
a sodium and fluorine
b fluorine and hydrogen
c sulfur and hydrogen
d bromine and beryllium
3. Why does silica have the formula SiO2?
4. Decide whether this formula is correct. If it is not
correct, write it correctly.
a HBr2
b ClNa
c Cl3Ca
d Ba2O
5. Write the correct formula for barium iodide.
6. See if you can give a name and formula for a
compound that forms when phosphorus reacts with
chlorine.
Chemical equations
Calculation of relative quantities of reactants and products in
chemical reactions.
Chemical equations
word equations , which provide elementary information
symbol equations, which provide much more information
Word
Equations
Symbol
Equations
Balance
Chemical
Equations
hydrogen + oxygen
water
2H2 (g) + O2 (g)
carbon + oxygen carbon
dioxide
2H2O (g) C (s) + O2 (g)
CO2 (g)
Numbers of atoms on each side of the arrow are equal.
Example
• Write down the missing product and balance the equation
1. CH4 + O2
CH4(g) + 2O2(g) ⟶ CO2(g) + 2H2O(l)
2. Mg(NO3)2 ⟶ MgO + NO2 + O2
Mg(NO3)2 (s) ⟶ MgO (s) + 2NO2(g) + 1 2O2(g)
Exercise
D
Exercise
B
Exercise
A
Summary
Word
Equations
Symbol
Equations
Balance
Chemical
Equations
hydrogen + oxygen
water
2H2 (g) + O2 (g)
carbon + oxygen carbon
dioxide
2H2O (g) C (s) + O2 (g)
CO2 (g)
Numbers of atoms on each side of the arrow are equal.
4.2 – Chemical Equations for Reactions
Equations for two sample reactions
1. The reaction between carbon and oxygen
When carbon is heated in oxygen, they react together to form carbon dioxide. Carbon
and oxygen are the reactants. Carbon dioxide is the product.
You could show the reaction using a diagram:
or by a word equation:
carbon + oxygen  carbon dioxide
or by a symbol equation:
C + O2  CO2
Note
Reactants are
sometimes called
reagents.
4.2 – Chemical Equations for Reactions
Symbol equations must be balanced
 Now look at the number of atoms on each side of
this equation:
2H2 + O2

2H2O
On the left:
4 hydrogen atoms
2 oxygen atoms
On the right:
4 hydrogen atoms
2 oxygen atoms
 The number of each type of atoms is the same on
both sides of the arrow.
 Atoms do not disappear during a reaction – they are
just rearranged.
 When the number of each type of atom is the same
on both sides, the symbol equation is balanced. If it
is not balanced, it is not correct.
4.2 – Chemical Equations for Reactions
Adding state symbols
 Reactants and products may be solids,
liquids, gases, or in solution.
 Adding state symbols to the equations:
(s) solid
(l) liquid
(g) gas
(aq) aqueous solution
(solution in water)
 For the two reactions above, the equations
with state symbols are:
C (s) + O2 (g)  CO2 (g)
2H2 (g) + O2 (g)  2H2 O (l)
4.2 – Chemical Equations for Reactions
How to write the equation for a reaction
 Writing an equation:
1. Write the equation in words.
2. Write it using symbols. Make sure all the formulae
are correct.
3. Check that the equation is balanced, for each type
of atom in turn.
Make sure you do not change any formulae.
4. Add the state symbols.
Example 1 Calcium burns in chlorine to form
calcium chloride, a solid. Write an equation for the
reaction, using the steps above.
1. calcium + chlorine  calcium chloride
2. Ca + Cl2
 CaCl2
3. Ca: 1 atom on the left and 1 atom on the right.
Cl: 2 atoms on the left and 2 atoms on the right.
The equation is balanced.
Ca (s) + Cl2 (g)  CaCl2 (s)
4.2 – Chemical Equations for Reactions
 Example 2
Hydrogen chloride is formed by burning hydrogen
in chlorine. Write an equation for the reaction.
1. hydrogen + chlorine
 hydrogen chloride
2.
H2
+ Cl2

HCl
3. H:
2 atoms on the left and 1 atom on the right.
Cl:
2 atoms on the left and 1 atom on the right.
The equation is not balanced. It needs another molecule of
hydrogen chloride on the right. So a 2 is put in front of the HCl.
H2 + Cl2 
2HCl
The equation is now balanced. Do you agree?
4. H2 (g) + Cl2 (g) 
2HCl (g)
Example 3
Magnesium burns in oxygen to form magnesium
oxide, a white solid. Write an equation for the reaction.
1. magnesium + oxygen

magnesium oxide
2. Mg
+ O2

MgO
3. Mg: 1 atom on the left and 1 atom on the right.
O:
2 atoms on the left and 1 atom on the right.
The equation is not balanced. Try this:
Mg + O2  2MgO (The 2 goes in front of the MgO.)
Another magnesium atom is now needed on the left:
2Mg + O2  2MgO The equation is balanced.
4. 2Mg (s) + O2 (g)  2MgO (s)
4.2 – Chemical Equations for Reactions
Ionic Equations
The simplified equation for a reacation involving ionic substances, only those ions
actually take part in the reaction are shown. Spectator ions are not shown.
4.2 – Chemical Equations for Reactions
1. What do “+” and “” mean, in an equation?
2. Balance the following equations:
a Na (s) + Cl2 (g)  NaCl (s)
b H2 (g) + I2 (g)  HI (g)
c Na (s) + H2O (l)  NaOH (aq) 1 H2 (g)
d NH3 (g)  N2 (g) + H2 (g)
e C (s) + CO2 (g)  CO (g)
f Al (s) + O2 (g)  Al2O3 (s)
3. Aluminium burns in chlorine to form
aluminium chloride, AlCl3, a solid. Write a
balanced equation for the reaction.
4.3 – The Masses of Atoms, Molecules, and Ions
Relative atomic mass
 A single atom weighs hardly anything. Can’t use
scales to weigh it. But scientists do need a way to
compare the masses of atoms.
 Chose an atom of carbon-12 to be the standard
atom. Fixed its mass as exactly 12 atomic mass
units. (It has 6 protons and 6 neutrons, as shown on
the right. Ignore electrons.)
▲ An atom of carbon-12. It is the main
isotope of carbon.
4.3 – The Masses of Atoms, Molecules, and Ions
Relative atomic mass
 Compare all the other atoms with this standard atom.
The standard
atom, 126 Cor carbon-12. Its
mass is taken as exactly
12.
Magnesium atom is
twice as heavy as the
carbon-12 atom. So its
mass is 24.
Hydrogen atom is 1/12 as
heavy as the carbon-12
atom. So its mass is 1.
 The mass of an atom compared with the carbon-12 atom is called its
relative atomic mass, or Ar.
 The small r stands for relative to the mass of a carbon-12 atom.
 So the Ar of hydrogen is 1, and the Ar of magnesium is 24.
4.3 – The Masses of Atoms, Molecules, and Ions
Ar and isotopes
 The atoms of an element are not always identical. Some may
have extra neutrons. Different atoms of the same element are
called isotopes. Chlorine has two isotopes:

Name
Protons
Neutrons Nucleon
% of chlorine atoms
Number
like this
chlorine-35
17
18
35
75%
chlorine-37
17
20
37
25%
 We need to take all the natural isotopes of an element into
account, to work out the relative atomic mass. This is the formula
to use:
 relative atomic mass (Ar) of an element =
(% × nucleon number for the first isotope)
+ (% × nucleon number for the second isotope)
… and so on, for all its natural isotopes
4.3 – The Masses of Atoms, Molecules, and Ions
Ar and isotopes
 Calculation for chlorine shows
that the relative atomic mass of
chlorine is 35.5.
 The relative atomic mass (Ar)
of an element is the average
mass of its isotopes compared
to an atom of carbon-12.
 For most elements, Ar is very
close to a whole number. It is
usually rounded off to a whole
number, to make calculations
easier.
▲ The two isotopes
of chlorine.
Ar for chlorine the
relative atomic mass of
chlorine:
= 75% X 35 + 25% X 37
= 75 X 35 = 25 X 37
100
100
(changing % to fractions)
= 26.25 + 9.25
= 35.5
4.3 – The Masses of Atoms, Molecules, and Ions
Ar values for some common elements
Element
Symbol
Hydrogen
H
Carbon
Ar
Element
Symbol
Ar
1
Chlorine
Cl
35.5
C
12
Potassium
K
39
Nitrogen
N
14
Calcium
Ca
40
Oxygen
O
16
Iron
Fe
56
Sodium
Na
23
Copper
Cu
64
Magnesium
Mg
24
Zinc
Zn
65
Sulfur
S
32
Iodine
I
127
Ar for chlorine
Finding the mass of an ion mass of sodium atom = 23, so mass of sodium ion = 23
since a sodium ion is just a sodium atom minus an electron (which has negligible
mass).
An ion has the same mass as the atom from which it is made.
4.3 – The Masses of Atoms, Molecules, and Ions
Finding the Masses of Molecules and Ions
 Using Ar values, it is easy to work out the mass of any molecule or group of
ions.
Hydrogen gas is made of
molecules. Each
molecule contains 2
hydrogen atoms, so its
mass is 2. (2 X 1 = 2)
The formula for water is
H2O. Each water
molecule contains 2
hydrogen atoms and 1
oxygen atom, so its
mass is 18. (2 X 1 + 16
= 18)
Sodium chloride (NaCl)
forms a giant structure
with 1 sodium ion for every
chloride ion. So the mass
of a ‘unit’ of sodium
chloride is 58.5. (23 + 35.5
= 58.5)
 If the substance is made of molecules, its mass found in this way is called
the relative molecular mass, or Mr. So the Mr for hydrogen is 2, and for water
is 18.
4.3 – The Masses of Atoms, Molecules, and Ions
Finding the Masses of Molecules and Ions
 But if the substance is made of ions, its mass is called the
relative formula mass, which is also Mr for short. So the Mr
for NaCl is 58.5.
 Two more examples
Substance
Formula
Atoms in
Formula
Ar in formula
Mr
ammonia
NH3
1N
3H
N = 14
H=1
1 X 14 = 14
3X1=3
Total = 17
magnesium
nitrate
Mg(NO3)2
1Mg
2N
6O
Mg = 24
N = 14
O = 16
1 X 24 = 24
2 X 14 = 28
6 X 16 = 96
Total = 148
4.3 – The Masses of Atoms, Molecules, and Ions
1. a What does relative atomic mass mean?
b Why does it have the word relative?
2. What is the Ar of the iodide ion, I–?
3. The relative molecular mass and formula mass are
both called Mr for short. What is the difference
between them?
4. Work out the Mr for each of these, and say whether it
is the relative molecular mass or the relative formula
mass:
a oxygen, O2
b iodine, I2 c methane, CH4
d chlorine, Cl2
e butane, C4H10
f ethanol, C2H5OH g ammonium sulfate (NH4)2SO4
4.3.1 – Some calculations about masses and %
Two laws of chemistry
 If you know the actual amounts of two substances
that react, you can:
 predict other amounts that will react
▲ A model of the
 say how much product will form.
carbon dioxide
 You just need to remember these two laws of
molecule. The
amounts of carbon
chemistry:
1. Elements always react in the same ratio, to and oxygen that react
to give this compound
form a given compound.
are always in the
same ratio.
E,g., when carbon burns in oxygen to form
carbon dioxide:
6 g of carbon combines with 16 g of oxygen, so
12 g of carbon will combine with 32 g of oxygen, and so on.
2. The total mass does not change, during a chemical reaction.
So total mass of reactants 5 total mass of products.
So 6 g of carbon and 16 g of oxygen give 22 g of carbon dioxide.
12 g of carbon and 32 g of oxygen give 44 g of carbon dioxide.
4.3.1 – Some calculations about masses and %
Calculating quantities
 Calculating quantities is quite easy, using the laws above.
 Example 64 g of copper reacts with 16 g of oxygen to give the
black compound copper(II) oxide.
a. What mass of copper will react with 32 g of oxygen?
64 g of copper reacts with 16 g of oxygen, so
2 X 64 g or 128 g of copper will react with 32 g of oxygen.
b. What mass of oxygen will react with 32 g of copper?
16 g of oxygen reacts with 64 g of copper, so
c.
d.
e.
16 or 8 g of oxygen will react with 32 g of copper.
2
What mass of copper(II) oxide will be formed, in b?
40 g of copper(II) oxide will be formed. (32 1 8 5 40)
How much copper and oxygen will give 8 g of copper(II)
oxide?
64 g of copper and 16 g of oxygen give 80 g of copper(II)
oxide, so
64 of copper and 16 g of oxygen will give 8 g of
10
10
copper(II) oxide, so
6.4 g of copper and 1.6 g of oxygen are needed.
4.3.1 – Some calculations about masses and %
Percentages: a reminder
 Calculations in chemistry often involve
percentages. Remember:
 The full amount of anything is
100%.
 To change a fraction to a %, just
multiply it by 100.
 Example 1 Change the fractions
and
to percentages.
1
2
X 100 = 50%
18
25
X 100 = 72%
 Example 2 Give 19% as a fraction.
19% =
19
100
4.3.1 – Some calculations about masses and %
Calculating the percentage composition of a
Mr for sulfur dioxide, SO2
Ar : S = 32, O = 16.
compound
So the Mr is:
 The percentage composition of a compound tells
1S
= 32
you how much of each element it contains, as a
2 O = 2 X 16
= 32
percentage of the total mass. This is how to work
Total
= 64
it out:
1. Write down the formula of the compound.
2. Using Ar values, work out its molecular or formula mass (Mr).
3. Write the mass of the element as a fraction of the Mr.
4. Multiply the fraction by 100, to give a percentage.
 Example
Calculate the percentage of oxygen in sulfur dioxide.
1. The formula of sulfur dioxide is SO2.
2. The Mr of the compound is 64, as shown on the right.
32
64
3.
Mass of oxygen as a fraction of the total =
4.
Mass of oxygen as a percentage of the total
32 X 100 = 50%
64
So the compound is 50% oxygen.
This means it is also 50% sulfur (100% - 50% = 50%).
4.3.1 – Some calculations about masses and %
Calculating % purity
 A pure substance has nothing else mixed with it.
 But substances often contain unwanted substances, or impurities. Purity is
usually given as a percentage. This is how to work it out:
 % purity of a substance = Mass of pure substance in it X 100
Total mass
 Example
Impure copper is refined (purified), to obtain pure copper for
use in computers. 20 tonnes of copper gave 18 tonnes
of pure copper, on refining.
a. What was the % purity of the copper before refining?
% purity of the copper = 18 tonnes X 100 = 90%
20 tonnes
So the copper was 90% pure.
b. How much pure copper will 50 tonnes of the impure
copper give? The impure copper is 90% pure.
90% is 90
100
90
So 50 tonnes of it will give 100 X 50 tonnes or 45 tonnes of pure copper.
4.3.1 – Some calculations about masses and %
1. Magnesium burns in chlorine to give magnesium chloride,
MgCl2. In an experiment, 24 g of magnesium was found to
react with 71 g of chlorine.
a How much magnesium chloride was obtained in the
experiment?
b How much chlorine will react with 12 g of magnesium?
c How much magnesium chloride will form, in b?
2. Methane has the formula CH4. Work out the % of carbon
and hydrogen in it. (Ar : C = 12, H = 1)
3. In an experiment, a sample of lead(II) bromide was made.
It weighed 15 g. But the sample was found to be impure.
In fact it contained only 13.5 g of lead(II) bromide.
a Calculate the % purity of the sample.
b What mass of impurity was present in the sample?