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M for Moles
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Table of content
Some Common Equations
1. Basic Concept
2. Chemical Equation
3. Balancing Equation
4. Moles
5. Answers
List of Relative Atomic Mass
How to use this e-book
Finding calculations in chemistry difficult to learn? Don’t know
how/when to or why use moles in chemistry? Confuse about
molecular weights and their relationships to moles and chemical
equations? This e-book teaches you how to do mole calculations that
are often encountered in chemistry.
Readers are assumed to have only a basic science concept and with
minimum knowledge in chemistry. This e-book will show you how to
solve those tricky moles calculations with plenty of examples,
showing in clear and concise ways by following through stepwise and
simple procedures.
The first three chapters are foundations essential to mole
calculations. This include topics such as chemical formula, balancing
equations and molecular weight calculations. However, this e-book is
emphasised in solving mathematical questions in chemistry and its
concept is kept to a minimum. For instance, this book will show you
how to determine the chemical formula for a chemical substance but
will not show you why the formula is as such.
It is particularly suitable for pupils from Year 10 onwards who are
taking GCSE or A/S levels in UK or Ninth Grade onwards in the US.
Words in bold - learn it, pay particular attention and make sure you
understand them.
Equations and statements come with
Questions come with
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learn it!
practice it!
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Some common equations (best to memorise it)
This section is just a quick reminder of some
of the important equations. Please refer to the
appropriate chapters to see how they are used.
Chapter 2
(1) Percentage by mass of an element in a molecule
Total relative mass of atom
Percentage by mass =
X 100 %
Relative molecular mass
(2) Step-by-step procedures to calculate empirical formula
(a) Write down element symbols.
(b) Write down the mass or percentage of each element.
(c) Divide the mass by the relative atomic mass.
(d) Divide the mole numbers to get the smallest whole ratio.
(e) You get the empirical formula.
Chapter 4
(1) Avogadro’s constant (NA) = 6.022 x 1023 mol-1
(2) Mole equation (mol)
Mass of substance
Mole =
(3)
Relative molecular mass
Density =
mass
volume
(4) Concentration (molarity, M)
Moles of solute
concentration =
Volume of solvent in dm3
(5) 1 dm3 = 1000 cm3 = 1000 ml = 1 l (liter)
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1. Basic Concept
This Chapter gives definitions of some basic keywords that
that is relevant to chemistry. You need to familiarise yourself
before going into other parts of this book.
1. Atom - It is the smallest unit of a substance. The identity of a
substance will be destroyed if its atom is further divided. Different
substances will have different types of atoms. All atoms are made up
of a number of protons, neutrons and electrons.
Proton (+1)
Electron (-1)
Neutron (0)
Schematic representation of an atom (in this case is a helium atom)
A proton has a positive charge (+1) and an electron has a negative
charge (-1). Neutron does not have a charge. It is neutral (0).
All atoms have equal number of protons and electrons. In other
words, they are always neutral. Charged atoms, or ions, always
contain a different number of protons and electrons. For example a
+2 (positive 2 charged) ion contains two electrons less than the
number of protons.
A positively charged ion is called a cation, while a negatively charged
ion is called an anion.
Checkpoint 1:
(a) An ion with charge -1 contain _____ number of protons then electrons. By how many? ______
(b) A +2 charge ion contain _____ number of protons then electrons.
By how many? ______
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2. Element - It is a substance that cannot be further resolved into
simpler substance components by chemical means. It consists of a
single type of atom of same number of protons. For example, gold
and copper are elements which consists of only gold atoms and only
copper atoms, respectively.
3. Compound - It is a substance made of more than one type of
elements. They are usually formed by a chemical process and atoms
of different types are bound together by chemical bonds.
4. Molecules - This is the smallest unit of a compound. For example,
a water molecule is made of two hydrogen atoms and one oxygen
atom which bind together by chemical bonds called the covalent
bonds.
Oxygen atom
A chemical bond
Hydrogen atom
Schematic representation of a water molecule
Another example: butane gas, which is used as a camping stove.
Butane is a compound which consists of carbon and hydrogen
elements. Butane molecule is made of four carbon atoms and ten
hydrogen atoms.
Carbon atom
A chemical bond
Hydrogen atom
Schematic representation of a butane molecule
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9. Periodic Table of the Elements - All known elements are
arranged systematically in a table called the Periodic Table of the
Elements. They are arranged according to the number of protons and
grouped according to their chemical properties.
A modern layout of the Periodic Table of the Elements. The one or two alphabets are
element symbols with the number of protons shown above for each element.
To date, there are around 120 known types of atoms or elements. Of
these, about 90 elements can be found in nature. All matters are
made of these elements. The rest, usually those of heavier ones (from
uranium with atomic number 92 onwards) no longer exist or are
found only in traces. However, these heavier elements can be
produced in a nuclear reactor.
Checkpoint 2:
It is useful to remember the symbols of the first twenty elements beginning with hydrogen. Note also the element symbols for other common
but heavier elements such as:
Iron (Fe), Copper (Cu), Zinc (Zn), Lead (Pb) and Bromine (Br)
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2. Chemical Equation
2.1 Atomic symbol
A complete description of an isotope element can be simplified by
means of using the following notation:
A
ZE
For example:
238
92 U
where E is the atom symbol shown in the Periodic Table, Z is the
number of protons and A is the atomic mass number. Since the mass
number is the sum of protons and neutrons, the number of neutrons,
N, can be worked out as:
Number of neutron = Atomic mass number
Or
N= A Z
number of protons
The example given above is the atomic symbol for the most common
isotope of uranium with atomic number 92 and mass number 238.
There are 238 - 92 = 146 neutrons in the atomic nucleus.
Since Z is unique for each atom, it is common to write as 238U or
'uranium-238' in the text. In other words, the atomic number (Z) is
left out (since the symbol U is sufficient to identify the element as
uranium). This effectively refer to a particular isotope of a uranium
atom.
The complete notation shown above are more commonly used in
radiochemistry, the study of radiation, nuclear fission and stuff like
that. Otherwise, it is more commonly written as simply 'U', unless
there is a need to refer to a particular isotope.
Checkpoint 3:
Find the number of proton neutron and electron for the following
isotopes (Clue: If the proton number is not given, look at the periodic
table of the Elements):
212Bi
83
208Pb
82
127I
234U
4He
12C
13C
Notice that certain elements has more than one mass numbers. They are
called isotopes.
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Example 1 A methane molecule contains one carbon and four
hydrogen atoms. This can be illustrated as follows:
Or in molecular formula: CH4
The ‘ball-and-stick’ model diagram on the right is called structural
formula while CH4 is the corresponding chemical (molecular)
formula. The lines represent chemical bonds and notice that the
carbon atom is connected to four hydrogen atoms, whereas each
hydrogen atom can only have one bond, which is connected to the
carbon atom.
Example 2. The graphical representation of a water molecule can be
shown as
Or in molecular formula as: H2O
Once again, the oxygen atom can only form two bonds, each to a
hydrogen atoms, which in turn formed one bond with the oxygen atom.
As in the case of methane molecule, the little number 2 indicates
there are two hydrogen atoms. Atomic symbols without any number
on the right means there is only one of the atom.
Example 3. If one were to describe two water molecules, how is this
represented in molecular formula?
The formula symbol for one water molecule is H2O. For two water
molecules, this is 2H2O.
2H2O
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Example 4. Hydrogen peroxide has a formula of H2O2. Is this correct?
Hydrogen can only form one bond and oxygen can form two bonds.
This is illustrated below:
Example 5. Carbon dioxide has a chemical formula of CO2. How is this
possible? This is illustrated as follows:
In this case, the double-line of one side indicates two bonds are
required to connect between the carbon atom and an oxygen atom.
Notice that the total number of allowable bonds for both carbon and
oxygen is still maintained - four for carbon and two for oxygen.
Example 6. Ethyne has a chemical formula of C2H2. What is the
corresponding structural formula?
Notice that there are three bonds connected between the carbon
atom and the total number of connecting bonds for each atom is still
the same: Four bonds for carbon and one bond for hydrogen.
A single line is usually referred to as a single bond. A two-line
symbol as a double bond and a three-line symbol as a triple bond.
Checkpoint 4:
Table (a) can be used to deduce the chemical formula of some of the
simple covalent compounds. However, it is useful to remember the
chemical formula of some of the covalent substances:
Water: H2O
Hydrogen chloride: HCl
Ammonia: NH3
Methane: CH4
Hydrogen gas : H2
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Carbon dioxide: CO2
Carbon monoxide: CO
Sulfur dioxide: SO2
Ethane: C2H4
Chlorine gas: Cl2
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Example 1. Sodium hydroxide is made of one sodium cation and one
hydroxide anion:
Na+ and OHThere is one sodium atom, one oxygen atom and one hydrogen atom
for the sodium hydroxide molecule.
It is not necessary to show the chemical formula as Na+OH- . This is
because the charge is balanced and the chemical formula is simply
written as
NaOH
Example 2. Two sodium hydroxide molecules. In this case, this is
written as:
2NaOH
Note: Do not confuse this as ‘two sodium atoms, one oxygen atom
and one hydrogen atom’. The correct total number of atoms are two
sodium atoms, two oxygen atoms and two hydrogen atoms.
Example 3. What is the chemical formula for calcium hydroxide?
From Table (b) a calcium ion has positive two charge, whereas the
hydroxide ion has negative one charge. To get a net zero charge, you
need two hydroxide anion to cancel out the charge on the calcium
cation. So, the chemical formula is
Ca(OH)2
The bracket enclosed with a little two means there are two sets of
OH-. The bracket is not needed if there is only one set of the group,
as shown in the Example 1 above.
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Checkpoint 5:
Determine the chemical formula of the following compounds. It is best
to remember the symbols and charge values of these common ions, as
shown in Table (b).
Sodium oxide, Na+ and O2-
Iron(II) sulfate, Fe2+ and SO42-
Magnesium oxide, Mg2+ and O2-
Iron(III) sulfate, Fe3+ and SO42-
Potassium bromide, K+ and BrAmmonium sulfate, NH4+ and SO422.3 Relative molecular mass
It is also known as the molecular weight or formula weight. The
relative molecular mass is essential in mole calculations. For instance,
to determine the amount of products formed in a chemical reaction.
It is determined by summing up the relative atomic mass of all the
atomic constituents in a chemical formula. The atomic mass values
can be found in any standard Periodic Table of the Elements.
Example 1. What is the molecular mass for water?
The chemical formula of water is H2O, which consists of two hydrogen
and one oxygen atoms, with relative atomic mass of 1.0 and 16.0
respectively. Therefore:
Molecular mass of H2O = 2 x (H) + 1 x (O)
= 2 x 1 + 1 x 16
= 18.0
Example 2. What is the molecular mass for sodium chloride?
The chemical formula of sodium chloride is NaCl, which consists of
one sodium and one chloride respectively.
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2.4 Percentages of elements
A molecule consists of a number of atoms in different proportions.
The percentage content of an element in a compound can be
determined from the total relative atomic mass of the atom in the
molecular formula and the relative molecular mass of the
compound. Using the following relationship:
Total relative mass of atom
Percentage by mass =
X 100 %
Relative molecular mass
Note: The total percentage of all the elements contained in a
compound has to be 100 %.
Example 1. Work out the percentage by mass of sodium and chlorine
in sodium chloride.
From the Periodic Table, relative atomic masses of sodium (Na) and
Chlorine (Cl) are 23 and 35.45 respectively.
Molecular mass of NaCl = 1 x (Na) + 1 x (Cl)
= 1 x 23 + 1 x 35.45
= 58.45
Percentage of sodium in NaCl =
23
58.45
X 100 %
= 39.35 %
Percentage of chlorine in NaCl =
35.45
58.45
X 100 %
= 60.65 %
Example 2. Work out the percentage by mass of of all constituent
atoms in dry sodium carbonate.
The molecular symbol for dry (anhydrous) sodium carbonate is
Na2CO3. The relative atomic mass for Na = 23, C = 12 and O = 16.
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Therefore:
Molecular mass of Na2CO3 = 2 x (Na) + 1 x (C) + 3 x (O)
= 2 X 23.0 + 1 x 12.0 + 3 x 16.0
= 106.0
Since there are two Na atoms, the total relative atomic mass of
sodium atoms in Na2CO3 is 2 x 23 = 46. Therefore:
Percentage of sodium in Na2CO3 =
46
106
X 100 %
= 43.40 %
Percentage of carbon in Na2CO3 =
12
106
X 100 %
= 11.32 %
The total percentage mass of all the atoms in the molecule is 100%.
Therefore
Percentage of oxygen in Na2CO3 = 100 % - 43.40 % - 11.32 %
= 45.28%
Checkpoint 7:
Determine percentage by masses of the following compounds.
Potassium bromide, KBr
Glucose, C6H12O6
Potassium dichromate, K2Cr2O7
Ethanoic (acetic) acid, CH3COOH
Ammonium sulfate, (NH4)2SO4
Also, work out what is the percentage mass of ammonium anion in
ammonium sulfate?
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2.5 Empirical Formulae
The empirical formula of a substance is chemical formula in its
simplest form. It shows the atomic constituent of a chemical
substance in the simplest whole number ratio.
This is in contrast to its normal chemical formula which gives the
actual number of atoms of each type of element that made up a
molecule.
Example 1. The chemical formula for ethane is C2H6. What is its
empirical formula?
The ratio of carbon to hydrogen is 2:6. To find out the empirical
formula, the whole number ratio must be reduced to its simplest and
smallest whole number form.
Carbon
2
Hydrogen
6
Divide this by two
2
2
6
2
Simplest ratio
1
3
Note: always divide
with the same
Number for all atoms.
The empirical formula for ethane is therefore CH3.
Example 2. What is the empirical formula for butene (C4H8) ?
Write down the ratio:
Divide this by four
Simplest ratio
Carbon
4
Hydrogen
8
4
4
1
8
4
2
The empirical formula for butene is CH2.
Note: If you divide the ratio by 2, you get a ratio of carbon 2 and
hydrogen 4, which is not the simplest ratio as it can still be further
divided.
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Example 1. A sample compound was determined to contain only 24 g
of carbon and 32 g of oxygen. What is its empirical formula?
Refer to the step-by-step procedures from the previous page:
(a)
(b)
(c)
(d) divide by 2
(e) smallest ratio
C
24 g
24/12 = 2
2/2 = 1
1
O
32 g
32/16 = 2
2/2 = 1
1
The empirical formula of the substance is CO. (Do not confuse this as
carbon monoxide!)
Example 2. A compound is made of 48 g of carbon, 8 g of hydrogen
and 64 g of oxygen. What is its empirical formula?
Refer to the step-by-step procedures from the previous page:
(a)
(b)
(c)
(d) divide by 4
(e) smallest ratio
C
48 g
48/12 = 4
4/4 = 1
1
H
8g
8/1 = 8
8/4 = 2
2
O
64 g
64/16 = 4
4/4 = 1
1
The empirical formula of the substance is CH2O.
Example 3. A compound is made of 40% carbon, 6.67% hydrogen
and 53.33 % oxygen. What is its empirical formula?
The percentage composition is proportional to the relative mass of the
substance. The same step-by-step procedures can be used.
(a)
(b)
(c)
(d) Divide by 3.33
(e) smallest ratio
C
40%
40/12=3.33
3.33/3.33=1
1
H
6.67%
6.67/1=6.67
6.67/3.33=2
2
O
53.33%
53.33/16=3.33
3.33/3.33=1
1
The empirical formula is CH2O. Note that to get the smallest whole
number, always look for smallest number in step (c).
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3. Balancing Equation
A chemical equation is an important piece of information for most of
the moles calculations. A chemical equation shows a complete
summary description of a chemical reaction. However, before the
equation can be used, it must be properly balanced. Balancing a
chemical equation involves two parts:
(1) Balance the individual chemical formula of a compound.
(2) Balance the overall compounds involve in a chemical reaction.
Remember that balancing chemical equations always
involving atoms rearrangement. The total number of atoms at
the left is always the same as that on the right.
Virtually all simple gas molecules are made of two-atom pairs.
They are called diatomics. For example, H2, Cl2, N2 and F2. The
exceptions are those belong to Group 0. These are called inert
gases and they exist as a single atom entity or monoatomic.
For instance He, Ar and Xe.
Example 1
The hydrogen can react explosively with oxygen to give water. Write
the chemical equation of this reaction.
So, first of all, pick up the component compounds that involved in this
reaction:
Oxygen + hydrogen
water
O2 +
H2
H2O
Remember that gases always exist as pairs as mentioned above.
You can see that the equation is not balanced: There are one less
oxygen atom on the right hand side of the equation. The only way to
balance the equation is to change the number of molecules involved
the equation.
Start by making two water molecules and the extra hydrogen atoms
introduced on the right hand side is balanced by making two
hydrogen molecules on the left hand side:
O2 + 2H2
Graphical
illustrations:
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2H2O
+
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oxygen
hydrogen
Example 2
A displacement reaction occurs when a piece of zinc metal is added
into a solution of copper sulfate: Zinc enters into solution to become
zinc cation, while copper ions in solution are changed into the copper
metal which coats the zinc metal piece. Write the chemical equation
for the reaction.
The chemical equation for the reaction is shown below:
Zn(s) + CuSO4(aq)
Cu(s) + ZnSO4(aq)
Notice that the total number of atoms and atom types are the same
for both sides and no further balancing is required.
The brackets describe the state of a compound in a reaction, or the
state symbol. There are usually four different states: solid (s), liquid
(l), aqueous (aq) and gas (g).
Since the sulfate anion, SO42-, does not actually involve in the
reaction (remains the same from both left hand and right hand sides),
the equation can also be simplified as follows:
Zn(s) + Cu2+(aq)
Zn2+(aq) + Cu(s)
This type of equation is also called the half equation because the
charge is not balanced and usually only one type of charge is shown.
Example 3
Magnesium metal burns brilliantly with oxygen to give white
magnesium oxide. Write the chemical equation for the reaction.
First of all write down all the chemical components involve in the
reaction:
Magnesium + oxygen
Magnesium oxide
Mg(s)
+ O2(g)
MgO(s)
Next, balance the equation. Again, start looking for unbalance atoms.
Here, we have one less oxygen atom on the right hand side. Start by
writing two magnesium oxide molecules (so as to make use of the two
oxygen atoms on the left) and then write two for the magnesium in
order to balance the equation:
2Mg(s) + O2(g)
2MgO(s)
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Checkpoint 10:
Balancing an equation requires experience. The more you do it, the more
familiar it becomes to you. Obviously, it is also important to know the
chemical formula of the compounds involve in a chemical reaction.
Otherwise, the equation can never be balanced properly.
Balance the following equations.
ZnO(s) + HCl(aq)
H2(g) + Cl2(g)
Cu(s) + AgNO3(aq)
CH4(g) + O2(g)
HCl(aq) + MnO2(s)
ZnCl2(aq) + H2O(l)
HCl(g)
Cu(NO3)2(aq) + Ag(s)
CO2(g) + H2O(l)
MnCl2(aq) + H2O(l) + Cl2(g)
Copper + oxygen
Copper(II) oxide
Calcium hydroxide + carbon dioxide
Calcium carbonate +
water
Hydrochloric acid + zinc
Zinc chloride + hydrogen
Hydrochloric acid + calcium carbonate
Calcium chloride +
carbon dioxide +
water
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4. Moles
This chapter marks the beginning of showing you how moles are used
in chemistry calculations. You are assumed to have some basic
understanding of chemistry. The first three chapters highlight some
of these that are relevant to mole calculations. Please make sure
that you are familiar with them before proceed any further.
4.1 Introduction
Mole is the amount of substance, measures in terms of elementary
units, as there are number of atoms in 0.012 kg (12.0 g) of carbon12. The elementary units can be atoms, electrons, molecules, etc.
It is abbreviated as mol and is the standard SI unit for the amount of
substance. One mole of carbon-12 contains about 6.02214 x 1023
carbon-12 atoms.
1 mol of
12C
contains 6.02214 x 1023 carbon atoms
This number is also known as the Avogadro’s number (a constant)
and is in unit mol-1 (per mole).
Avogadro’s number = 6.02214 x 1023 mol-1
Because of the relationship of the atomic mass unit with the
Avogadro’s number, one mole of a chemical compound has a
mass equal to its relative molecular mass expressed in grams.
This is also true for the chemical elements. This is called molar mass.
For example, aluminium (Al) has a relative atomic mass of 26.98. This
means that there are 6.02214 x 1023 aluminium atoms in 26.98 g of
aluminium.
1 mol of Al weighs 26.98 g and contains 6.02214 x 1023 Al atoms
Water (H2O) has a mass unit of 18.0. This means that there are
6.02214 x 1023 water molecules in 18.0 g of water.
1 mol of H2O weighs 18 g and contains 6.02214 x 1023 H2O molecules
Therefore:
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Mass of substance
Mole =
Relative molecular mass
In fact, you have already come across calculation of moles in finding
empirical formula (step (c) in Page 22).
Example 1
What is the mole value for 15 g of a pure magnesium sample?
The atomic weight of magnesium is 24.305 and the molar mass is
24.305 g/mol. Using the equation shown above
15.0 g
= 0.617 mol
Mole =
24.305 g/mol
Example 2
What is the mole value for 22.3 g of a pure sodium carbonate sample?
The molecular weight of sodium carbonate is 106.0 (see Example 3
of Section 2.3) and therefore the molar mass is 106.0 g/mol. Using
the equation shown above
Mole =
22.3 g
= 0.210 mol
106.0 g/mol
Example 3
How many molecules are there in 5.6 g of pure potassium chloride?
The chemical formula of potassium chloride is KCl and the molecular
weight is 39.098 + 35.453 = 74.551 or 74.551 g/mol.
Therefore, the mole value is:
5.6 g
= 0.075 mol
74.551 g/mol
Since there are 6.02214 x 1023 molecules in 1 mol of KCl, for 0.075
mol, there are
6.02214 x 1023 mol-1 x 0.075 mol = 4.517 x 1022 molecules
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4.2 Moles in Chemical Reactions
Mole is an important feature to determine the amount of a product
being produced in a chemical reaction. Consider the following
chemical reaction (see Example 1 of Chapter 3):
2H2(g) + O2(g)
2H2O(l)
Here two hydrogen molecules react with one oxygen molecule to
produce two water molecules.
One may perhaps think that a mass unit can be added to both sides
of a reaction, just like a normal mathematical equation. For instance,
the above chemical equation wrongly reads ‘two grams of hydrogen
react with one grams of oxygen to produce two grams of water’.
Clearly this does not make sense as the total mass of the reactants
and the product must be the same from both sides of the equation.
The correct interpretation of the above reaction is ‘two moles of
hydrogen molecule react with one mole of oxygen molecule to
produce two moles of water molecules. The actual mass can then be
determined by using the mole equation in Chapter 4.
Example 1
How many moles of the oxygen is required when 0.4 mol of
magnesium ribbon burns?
The chemical reaction proceeds as follows:
2Mg(s) + O2(g)
2MgO(s)
So the reaction shows two moles of magnesium reacts with one mole
of oxygen (ignore magnesium oxide since it is not relevant to the
question). To easily work out the answer for 0.4 mol of magnesium,
arrange the number in the following way:
2 Mg
0.4 Mg
The double arrows mean ‘relate to’ and
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O2
y O2
y is the answer.
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Cross multiply as the green dotted line shows. This gives
Or
2 x y mol = 1 x 0.4 mol
2y = 0.4 mol
y = 0.4/2 = 0.2 mol
So 0.4 mol of magnesium will react with 0.2 mol of oxygen.
The above ‘cross’ method works for any chemical reaction provided it
is balanced.
Example 2
How many grams of carbon dioxide is produced when 2.5 mol of
propane is burnt in the presence of excess air?
The reaction of propane with the oxygen in air is as follows (see
Example 4 of Chapter 3)
C3H8(g) + 5O2(g)
3CO2(g) + 4H2O(g)
1 mol of propane will produce 3 mol of carbon dioxide. Using the cross
method
1 C3H8
2.5 C3H8
3 CO2
y CO2
And so
y x 1 = 3 x 2.5
y = 7.5 mol (of carbon dioxide)
Now the molar mass of CO2 is 44 g/mol. Therefore, mass for 7.5 mol
of CO2 is (see Example 4 of Chapter 4.1)
7.5 mol x 44 g/mol = 330.0 g
Example 3
How many grams of magnesium is needed to produce 50.0 g of
magnesium sulfate from the reaction with excess sulfuric acid?
The equation is as follows:
H2SO4(aq) + Mg(s)
34
MgSO4(aq) + H2O(l)
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The equation for the chemical process is as follows:
CaCO3(s)
CaO(s) + CO2(g)
From Chapter 4.1, we see that the standard mole definition of a
chemical compound bears the relationship of a mass equal to its
relative molecular mass expressed in grams (gram-mole). However,
we can also express the mole in other mass units, such as kilogram,
pound, ton and so on. Of course, any mass units other than gram will
have different numbers than the Avogadro’s constant (6.022 x 1023).
So, the molecular weight of calcium carbonate is
40.078 + 12.011 + (3 x 16.00) = 100.089
To express the molar mass in kilogram-mole, there is 100.089 kg of
calcium carbonate in one mole or 100.089 kg/mol. For 250 kg
250 kg
= 2.498 mol
100.089 kg/mol
From the equation
1 mol CaCO3
2.498 mol CaCO3
1 mol CaO
y mol CaO
y = 2.498 mol
Therefore
The molecular weight of CaO is 56.078, which is 56.078 kg/mol.
Therefore, the mass of CaO produced is
56.078 kg/mol x 2.498 mol = 140.08 kg
Alternatively, without calculating the mole (as in Example 3)
1 x 100.089 kg of CaCO3
250 kg of CaCO3
x=
1 x 56.078 kg of CaO
x kg of CaO
56.078 kg x 250 kg
= 140.07 kg
100.089 kg
You can see that only the mass unit is changed, but the calculation
method remains the same.
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Checkpoint 12:
(1) How many grams of magnesium oxide is produced when 8.6 g of
magnesium metal is burnt in air?
(2) Aluminium burns in oxygen to produce aluminium oxide:
4Al(s) + 3O2(g)
2Al2O3(s)
How much aluminium oxide can be produced if 25.8 g of aluminium is
burnt?
(3) How grams of calcium carbonate is needed in order to produce 34.9
grams of calcium chloride from a reaction with an excess of hydrochloric
acid?
CaCO3(s) + 2HCl(aq)
CaCl2(aq) + H2O(l) + CO2(g)
(4) Calcium hydroxide (slaked lime) is used in agriculture to reduce soil
acidity. It is manufactured in industry by adding water to calcium oxide
(quicklime):
CaO(s) + H2O(l)
Ca(OH)2(s)
Given that there is 350 kg of calcium oxide. How much water is needed
to convert it to calcium hydroxide?
(5) In the industry, iron is extracted by heating iron ore (haematite,
iron(III) oxide) with carbon in a blast furnace. One reaction where iron
is formed is as follows:
2Fe2O3 + 3C
4Fe + 3CO2
Given that there is 12400 kg of iron(III) oxide contained in the iron ore,
how much iron can be produced?
(6) Kroll process is an industrial process to produce pure titanium metal.
This process basically involves reaction of magnesium with titanium(IV)
chloride in a stainless steel retort at 820°C:
TiCl4(g) + 2Mg(l)
Ti(s) + 2MgCl2(l)
If a metal extraction plant can produce 1000 tonnes of titanium in a day,
how much magnesium metal is needed in a day, given that there is plenty
of supply of TiCl4?
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Examples below show how a solution of known concentration can be
determined, and the amount of chemical substance needed to give
such concentration.
Example 1
What is the mass of sodium chloride (NaCl) needed to prepare a
solution of 0.1 M concentration?
A concentration of 0.1 M of sodium chloride solution means 0.1 mol
of sodium chloride dissolves in 1 dm3 of water.
Molar mass of NaCl = 22.99 + 35.45 = 58.44 g/mol
To find out mass of 0.1 mol of sodium chloride rearrange mole equation
Mass of substance = mole x molecular weight (molar mass)
= 0.1 mol x 58.44 g/mol
= 5.844 g
Hence, 5.844 g of sodium chloride is needed to dissolve in 1 dm3 of
water to give a concentration of 0.1 M.
Example 2
Given that a 250 cm3 volumetric flask is used, what is the amount of
sodium chloride (NaCl) needed to prepare a 0.30 M solution?
Molecular weight of NaCl = 58.44 g/mol
A concentration of 0.30 M requires 0.30 mol of NaCl dissolves in 1
dm3. Therefore, the amount of sodium chloride needed
0.30 mol x 58.44 g/mol = 17.532 g
So this is the amount of NaCl needed to dissolve in 1 dm3 of water.
However, to make a 250 cm3 of 0.30 M solution, the amount of NaCl
needed is obviously less than that. This can be determined by using
the same proportionality relationship as first described in Example 1
of Chapter 4.2. Given that 1 dm3 = 1000 cm3:
17.532 g
yg
40
1000 cm3
(to make 0.25 M solution)
250 cm3
(to make 0.25 M solution)
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y = 17.532 x 250 = 4.383 g
1000
y x 1000 = 17.532 x 250
Therefore
You need 4.383 g of NaCl to dissolve in 250 cm3 of water to make 0.3
M solution.
Alternatively, you can determine the amount needed in terms of mole
and then converts the answer to grams:
0.30 mol
x mol
1000 cm3
(to make 0.25 M solution)
250 cm3
(to make 0.25 M solution)
y = 0.30 x 250 = 0.075 mol
Therefore
1000
And the amount of sodium chloride needed
0.075 mol x 58.44 g/mol = 4.383 g
Example 3
How much sodium hydroxide is needed in order to prepare a 200 cm3
6 M of solution?
First of all, work out the molar mass of sodium hydroxide:
22.99 + 16.00 + 1.00 = 39.99 g/mol
Use the relationship:
6 mol
x mol
1000 cm3
200
cm3
x = 6.0 x 200
1000
(to make 6 M solution)
(to make 6 M solution)
= 1.2 mol
Therefore, the amount of sodium hydroxide needed
1.2 mol x 39.99 g/mol = 47.99 g
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Note that this is the number of moles in 300 cm3 of water. To find
out the concentration (as at 1000 cm3), use the relationship:
300 cm3
0.054 mol
x mol
1000 cm3
x = 1000 x 0.054 = 0.18 mol
300
Therefore, 13.5 g of copper sulfate dissolved in 300 cm3 water will
give a solution concentration of 0.18 M.
Hence, to prepare 1000 cm3 of 0.18 M solution will need 0.18 mol of
copper sulfate. This is equal to
0.18 mol x 249.62 g/mol = 44.93 g of copper sulfate
This example demonstrates how concentration of solution remains
the same in different volumes but the actual amount of substance
dissolved can be different.
Example 6
Given a 250 cm3 hydrochloric acid of 2 M concentration, how much
water must be added to the acid in order to reduce the concentration
to 0.4 M?
Remember that the solution concentration is fixed for any given
volume. Hence, you can use the following relationship:
250 cm3
y
cm3
2M
0.4 M
Where y is the corresponding volume of water if the concentration is
0.4 M. However, since the more water is added, the less the
concentration becomes (solution becomes more dilute). This means
the relationship in this case is inversely proportional.
The cross relationship introduced in Example 1 of Chapter 4.2 only
applies to situation that is directly proportional. For instance, when
more substance is added the higher the concentration becomes, as
shown in the previous Examples.
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Therefore, the equation for this problem is now related as the green
lines as shown above:
y x 0.4 = 250 x 2
Therefore
y = 250 x 2 = 1250 cm3
0.4
And so the concentration is reduced from 2 M to 0.4 M as the water
is increased from 250 cm3 to 1250 cm3.
However, since the original solution has a volume of 250 cm3, the
extra volume of water that needs to be added in order to reduce the
concentration from 2 M to 0.4 M is
1250 cm3 - 250 cm3 = 1000 cm3
Checkpoint 13:
(1) Calculate the concentration of 14.5 g of sodium carbonate (Na2CO3)
dissolved in 1000 cm3 of water.
(2) What is the concentration of the following solutions:
(a) 1.5 g of potassium bromide (KBr) in 100 cm3 of water.
(b) 16.5 g of potassium iodide (KI) in 250 cm3 of water.
(c) 45.1 g of potassium manganate(VI) (KMnO4) in 2000 cm3
of water.
(d) 20.6 g of sodium hydroxide (NaOH) in 125 cm3 of water.
(e) 5.7 g of sodium chloride in 175 cm3 of water.
(3) What is the amount (in grams) of silver nitrate (AgNO3) in 75 cm3
of 0.05 M of solution?
(4) What is the amount (in grams) of sodium chloride (NaCl) in 25 cm3
of 0.15 M of solution?
(5) Calculate the mass of sodium sulfate (Na2SO4) needed to make 250
cm3 of 1.5 M solution.
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Example 1
A 20 cm3 standard sodium hydroxide solution of concentration 0.2 M
needed 18.5 cm3 of hydrochloric acid for neutralization. Determine
the concentration of the hydrochloric acid.
First of all, determine the amount of NaOH in 20 cm3 solution.
1000 cm3
0.2 mol
x mol
Therefore
20 cm3
x = 0.2 x 20 mol
1000
= 0.004 mol
Now, the neutralization reaction is as follows:
HCl(aq) + NaOH(aq)
NaCl(aq) + H2O(l)
The reaction shows that 1 mol HCl reacts with 1 mol NaOH. Therefore
0.004 mol HCl will react with 0.004 mol NaOH. There are 0.004 mol
HCl in 18.5 cm3 solution. To find out the concentration of HCl sample
Therefore
0.004 mol
18.5 cm3
y mol
1000 cm3
y=
0.004 x 1000
18.5
= 0.216 mol
mol
The concentration of HCl is 0.216 M.
Alternatively, the equation from the previous page can be used as a
quicker way to determine the concentration. Given that Ma and Va are
the concentration and volume of NaOH; and Mb and Vb are the
concentration and volume of HCl, from Equation (3), it follows that
0.2 M x 20.0 cm3
= 1
1
Mb x 18.5 cm3
And
Mb x 18.5 x 1 = 0.2 x 20.0 x 1
Therefore
Mb =
46
0.2 x 20.0
18.5
M = 0.216 M
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4.4 Moles in Gases
In previous examples, calculations involving gas compounds were
described in quantities such as grams and moles. However,
sometimes it is more practical to describe the quantity in terms of
volumes.
In questions 8 and 9 of Checkpoint 11 you will notice that different
gases apparently occupy a same volume at a given temperature. This
is in fact true for any other gases which makes volume calculations
for gases becomes much easier.
However, the volume of gas depends on both temperature and
pressure. In chemistry, two most commonly defined conditions are as
follows:
(a) At the temperature of 0°C (273.15 K) and 1 atmospheric
pressure, one mole of a gas occupies a volume of 22.414 dm3.
Such conditions are called the standard temperature and
pressure (s.t.p).
(b) At the temperature of 25°C (298.15 K) and 1 atmospheric
pressure, one mole of a gas occupies a volume of 24.466 dm3.
Such conditions are called the standard room temperature and
pressure (s.r.p).
For condition (a) the volume occupied is called the standard molar
volume. For condition (b) the volume occupied is called the molar
volume. And so
1 mole of a gas contains 6.022 x 1023 mol-1 particles and occupies
a volume of 22.4 dm3 (standard molar volume) at s.t.p.
1 mole of a gas contains 6.022 x 1023 mol-1 particles and occupies
a volume of 24.0 dm3 (molar volume) at r.t.p.
Note that the volumes are approximated to 22.4 dm3 /mol and 24.0
dm3 /mol respectively for our calculations.
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Example 1
Given a 2.5 mol carbon dioxide, what is the corresponding volume
occupied by the gas at s.t.p.?
Since more gas substance will occupy more volume, the
proportionality relationship that first introduced in Example 1 of
Chapter 4.2 can be used:
1 mol CO2
22.4 dm3 (at s.t.p.)
2.5 mol CO2
x dm3 (at s.t.p.)
x = 22.4 x 2.5 dm3
Therefore
= 56 dm3
Example 2
Given that 10.5 g of nitrogen gas is produced in a chemical reaction
at room temperature. What is the volume occupied by the gas?
The molar mass of N2 = 14.0 x 2 = 28 g/mol. Using Mole Equation:
Moles of N2 =
10.5 g
= 0.375 mol
28.0 g/mol
And
1 mol N2
0.375 mol N2
Therefore
24.0 dm3 (at r.t.p.)
x dm3 (at r.t.p.)
x = 24.0 x 0.375 dm3
= 9 dm3
Example 3
What is the volume of carbon dioxide produced when 50.0 dm3
(measured at room temperature) of propane is burnt in the presence
of excess air?
If x is the moles of propane gas that occupies 50 dm3 at room
temperature, then
50
1 mol propane
24.0 dm3 (at r.t.p.)
x mol propane
50.0 dm3 (at r.t.p.)
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Example 4
Carbon disulfide is a highly flammable liquid that burns in oxygen to
produce sulfur dioxide and carbon dioxide. What is the volume of
oxygen needed to burn in excess carbon disulfide in order to produce
125 dm3 (measured at 25 °C) of sulfur dioxide?
The reaction is shown as follows:
CS2(l) + 3O2(g)
CO2(g) + 2SO2(g)
The above equation shows ‘3 to 2’ relationship between the oxygen
and sulfur dioxide. Therefore, to determine the volume of oxygen
involves in the reaction:
And
3 unit volumes O2
2 unit volumes SO2
z unit volume O2
125 dm3 SO2
2x
z = 3 x 125
z = 3 x 125
2
= 187.5 dm3
Example 5
Potassium chlorate will decompose and release oxygen when it is
heated to above 400 °C. What is the volume of oxygen gas released
(measured at r.t.p.) when 5.4 g of potassium chlorate is completely
decomposed?
The reaction is shown as follows:
2KClO3(s)
3O2(g) + 2KCl(s)
First of all, work out the molar mass of potassium chlorate:
39.098 + 35.45 + 3 x 16.0 = 122.55 g/mol
Then work out 5.4 g of potassium chlorate in moles:
5.4 g
= 0.044 mol
122.55 g/mol
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5. Answers
If you want detailed work out for the answers, then
please email [email protected] using the same
email you have purchased this e-book, including
the name that you have registered in (see Page 2).
You will be sent the work out free of charge provided
that your email and the registered name is matched.
Checkpoint 1
(a) ...less...one
(b) ...more...two
Checkpoint 2
Just memorize it!
Checkpoint 3
Number of
neutron
Isotopes
proton
electron
222
Bi
83
83
129
83
208
Pb
82
82
126
82
127
I =
127
I
53
53
74
53
234
U =
234
U
92
92
142
92
2
2
2
4
He =
4
2 He
Checkpoint 4
Memorize it!
Checkpoint 5
Sodium oxide, Na2O
Magnesium oxide, MgO
Potassium bromide, KBr
54
Iron(II) sulfate, FeSO4
Iron(III) sulfate, Fe2(SO4)3
Ammonium sulfate, (NH4)2SO4
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Checkpoint 6
Atomic weight (note: if you use different values, your answer will be
slightly different than the model answer)
K = 39.10
Cr = 52.00
N = 14.01
H = 1.00
Br = 79.90
Na = 22.99
Cl = 35.45
C =12.00
Cu = 63.55
S = 32.07
O = 16.00
KBr = 119.00
CuSO4.5H2O = 249.62
C6H12O6 = 180.00
K2Cr2O7 = 294.20
CH3COOH = 60.00
(NH4)2SO4 = 132.09
NaClO3 = 106.44
Checkpoint 7
Potassium bromide, KBr
K = 32.86 %
Br = 67.14 %
Glucose, C6H12O6
C = 40 %
H = 6.67 %
O = 53.55 %
Potassium dichromate, K2Cr2O7
K = 26.58 %
Cr = 35.35 %
O = 38.07 %
Ethanoic acid, CH3COOH
C = 40 %
H = 6.67 %
O = 53.55 %
Ammonium sulfate, (NH4)2SO4
N = 21.21 %
H = 6.06 %
S = 24.28 %
O = 48.45 %
Percentage of ammonium = 48.48 %
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Checkpoint 8
Empirical formulae for:
Ethyne, C2H2 = CH
Benzene, C6H6 =CH
Aluminium chloride, Al2Cl6 = AlCl3
Octane, C8H18 = C4H9
Phosphorus pentoxide, P4O10 = P2O5
Ethanedioic (oxalic) acid, (COOH)2 = CO2H
Methane, CH4 (empirical formula is the same as its molecular formula)
Glucose, C6H12O6 = CH2O
Sucrose, C12H22O11 (same as its molecular formula)
Checkpoint 9
(1)
(2)
(3)
(4)
CH2
Empirical formula: CH3
Empirical formula: C2H6O
XeF2O2
Molecular formula: C2H6
Molecular formula: C4H12O2
Checkpoint 10
ZnO + 2HCl
ZnCl2 + H2O
H2 + Cl2
2HCl
Cu + 2AgNO3
Cu(NO3)2 + 2Ag
2CH4 + 4O2
2CO2 + 4H2O
2Al + 2NaOH + 3H2O
2NaAl(OH)4 + 3H2
4HCl + MnO2
MnCl2 + 2H2O + Cl2
(Words equation)
Cu + O2
CuO2
Ca(OH)2 + CO2
CaCO3 + H2O
2HCl + Zn
ZnCl2 + H2
2HCl + CaCO3
CaCl2 +CO2 + H2O
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List of relative atomic mass for some elements (in alphabetical
order)
Elements
Aluminium
Bromine
Calcium
Carbon
Chlorine
Chromium
Copper
Fluorine
Hydrogen
Iodine
Iron
Magnesium
Manganese
Nitrogen
Oxygen
Phosphorus
Potassium
Sodium
Sulfur
Xenon
Zinc
Symbols
Al
Br
Ca
C
Cl
Cr
Cu
F
H
I
Fe
Mg
Mn
N
O
P
K
Na
S
Xe
Zn
Relative atomic mass
26.98
79.90
40.08
12.00
35.45
52.00
63.55
19.00
1.00
126.9
55.85
24.31
54.94
14.00
16.00
30.97
39.10
22.99
32.07
131.3
65.39
For a complete list please consult a good Periodic Table of the
Elements.
List of other useful on-line resources
The Periodic Table of the Elements
Chemistry glossary in Etacude.com’s Science Dictionary
On-line chemistry lesson
Fun Science Projects
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