Chapter 1

MATTER
1.1.1 Fundamental Particles of Atoms
1. Matters are anything that have a mass and volume.

Matter can exist as atoms, molecules or ions.
2. The basic particle of a matter which is made up of smaller subatomic particles in the nucleus called protons and neutrons.

which are surrounded by a cloud of electrons which is negatively charged.
3. According to modern concepts of the atom, its electrons are portrayed as occupying a volume of space called an orbital

around the nucleus.

Simplified structure of an atom
4. An electron is a very light particle, about 5.46 x 10-4 times lighter than a proton.
5. Thus the mass of the atom is contributed by its protons and neutrons which are found in the nucleus.
6. Protons and neutrons are found in the nucleus of the atom and are collectively known as the nucleon.
7. A proton and a neutron have almost the same mass while an electron has a negligible mass.
8. The relative masses and charges of these three subatomic particles are given in the table below.

Mass and charge of sub-atomic particles [a.m.u.= atomic mass unit]
9. A big section of the atom consists of empty space .

The mass of an atom is very small and is almost 99.98 % of the mass of an atom is concentrated in the
nucleus. Its mass is usually expressed in atomic mass unit.

1 a.m.u = 1 x mass of one atom of 12C
12

= 1.66 x 10-27 kg
10. Any particle with a charge (protons and electrons) will be deflected by both electric and magnetic fields.

The manner and angle of deflection varies based on the type and magnitude of the charge carried by the
particle as well as on its momentum (combination of mass and velocity).
(a) The positively charged protons are deflected to the negative plate.
(b) The negatively charged electrons are deflected to the positive plate.
(c) The neutrons being neutral are not deflected by electric or magnetic fields.
(d) The lighter electrons are more greatly deflected than protons due to its smaller (or lighter) mass.

1

Path of proton, electron and neutron in an electric field

11. Angle of deflection  1  e where e is the relative charge of the positive ion
mm

e

and m is the relative mass of the positive ion
12. Proton is H+ ion and Deuterium is D+ ion. Both are deflected in the same direction.

Angle of deflection of H+ ion is twice of that of D+ as mass of H+ is half that of D+.

12. Behavior of protons, electrons and neutrons in a magnetic field

(a) The neutron beam, being neutral, is not affected by the magnetic field.

(b) Proton beam is deflected to the south pole of the magnetic field.

(c) Electron beam is deflected to the north pole.

Path of proton, electron and neutron in a magnetic field
13. Atoms of one element differ from the atoms of another element because they contain different numbers of

sub-atomic particles.

1.1.2 Proton Number and Nucleon Number

1. Atoms can be characterised by the number of sub-atomic particles (protons, neutrons and electrons) they contain.

2. The proton number (Z) is the number of protons present in the nucleus of an atom. Atoms are neutral particles.

Hence, the proton number is equal to its number of electrons. The chemical identity of an atom is determined solely by its proton

number. For example, any atom having 7 protons in the nucleus corresponds to the nitrogen atom.

Number of protons = number of electrons = proton number

3. Nucleon number (A) is the total number of protons and neutrons present in the nucleus of the atom of an element.

Nucleon number = number of protons + number of neutrons Note : Nucleons are nuclear particles

4. The relationship between number of neutrons with the proton number and the nucleon number is given by

A=N+Z where N = number of neutrons.

5. The symbolic form is used to represent the proton and nucleon numbers of an atom of the element X :

Examples are 12 C ; 16 O ; 35 Cl ; 15331I
6 8 17

6. In a neutral atom, the number of protons is equal to the number of electrons. In a positive ion, the number of protons is more than

the number of electrons. In a negative ion, the number of protons is less than the number of electrons.

(a) A positive ion (cation) is formed when an atom loses electron.

Atom  Cation + electrons

(b) A negative ion (anion) is formed when an atom gains electron.

Atom + electrons  Anion

7. The symbol 2 4 Mg2+ denotes a magnesium ion with a positive charge of +2. It has 12 protons, 10 electrons and 12 neutrons.
1 2

8. Polyatomic ions such as OH-, NO3- , CO32- and NH4+ are ions that contain two or more atoms in it.

EXAMPLE 1.1

Determine the number of protons, electrons and neutrons in the following species.

24 Mg ; 6 4 Cu ; 3 2 S2- ; 4 0 Ca2+ ; 16 O2+ ; Al 3+ ; Na+ ; N3- ; O2- ; F- ; [ 2 H2 16 O]+ ; [ 12 C 16 O3]2-
12 2 9 1 6 2 0 8 1 8 6 8

2

ANSWER

Particle Mg Cu S2- Ca 2+ O2+ Al 3+ Na+ N3- O2- F- [ 2 H2 16 O]+ [ 12 C 16 O3]2-
1 8 6 8

no of protons 12 29 16 20 8 13 11 7 8 9 2(1) + 8 = 10 6 + 3(8) = 30

no of electrons 12 29 18 18 6 10 10 10 10 10 9 32

no of neutrons 12 35 16 20 8 14 12 7 8 10 2(1) + 8 = 10 30

Note :
(a) Positively charged ions have more protons than electrons.
(b) Negatively charged ions have more electrons than protons.
(c) Species that have the same number of electrons are said to be isoelectronic or isoelectric .

S2- and Ca 2+ are isoelectronic / isoelectric. Similarly Na+, Mg2+, Al3+, N3-, O2- and F- are isoelectronic / isoelectric.
Nitrogen molecule (28 electrons) is isoelectronic / isoelectric with carbon monoxide.

1.1.3 Isotopes

1. (a) If two atoms possess the same number of protons, then it is of the same element.
(b) Atoms of the same element sometimes do not have the same nucleon number since the number of neutrons in the nucleus
may not always be the same.
(c) Hence, they are called isotopes of that particular element.

2. Isotopes are atoms of the same element with the same proton number but different nucleon number.
They have different number of neutrons but same number of protons.

3. Most naturally occurring elements consist of a mixture of isotopes.
For example, chlorine (17Cl) has two isotopes and tin (50Sn) has ten, while fluorine (9F) is monoisotopic.

4. The isotopes of hydrogen are as follows:

Isotope Hydrogen-1 (protium) Hydrogen-2 (deuterium) Hydrogen-3 (tritium)

Symbols 1 H 2 H or 2 D 3 H
1 1 1 1

No of protons 1 1 1

No of neutrons 0 1 2

No of electrons 1 1 1

Structure

5. The relative abundance (% composition) of the isotopes in the sample of an element is not the same for all the isotopes
present. The table below gives the relative abundance of the isotopes of iron. (The proton number of iron = 26)

Isotope 54Fe 56Fe 57Fe 58Fe

% composition 5.8 91.6 2.2 0.4

No of protons 26 26 26 26

No of electrons 26 26 26 26

No of neutrons 28 30 31 32

6. Isotopes of an element have the same (b) number of electrons in a neutral atom
(a) number of protons (proton number) (d) chemical properties
(c) electronic configuration

3

7. Isotopes of an element have different

(a) number of neutrons (nucleon number) (b) mass

(c) density (d) rates of diffusion

(e) molecular speed (f) melting points and boiling points

8. Isotopes of an element with fewer neutrons have

(a) lower masses (b) lower densities

(c) higher rates of diffusion or effusion (d) lower melting and boiling points

9. Isotopes are differentiated using the mass spectrometer based on its relative isotopic mass (nucleon number).

1.1.4 Stable Isotopes and Unstable Isotopes

1. The nuclei of some isotopes are unstable and will disintegrate (split up) spontaneously to form smaller atoms (known as

daughter nuclei), with the emission of high energy particles/radiation.

2. (a) An unstable isotope is also called a radioactive isotope and it has a nucleus which is unstable as they may contain

either too many protons or too many neutrons..

(b) Radioactive isotope is defined as an isotope in which the nucleus of the atom is not stable and will disintegrate

spontaneously, by the emission of high energy particles/radiation, into other smaller atoms.

(c) Examples of unstable isotopes are hydrogen-3, carbon-14, oxygen-18, cobalt-60, iodine-131 and lead-210.

3. When these isotopes disintegrate spontaneously, alpha (α, 4 He2+ ) particles, beta (β, 0 e) particles and
2 1

gamma (γ-rays) rays, which are electromagnetic rays are emitted.

4. The process of disintegration by which a radioisotope releases rays to achieve stability is called a radioactive decay.

5. The decay will continue until a stable isotope is formed.

6. The stability of the radioisotopes is measured by its half-life. The shorter the half-life, the less stable the radioisotope.

7. Elements with proton number greater than 83 have unstable nuclei and are radioactive.

There are certain isotopes of lighter elements such as carbon-14 and sodium-24 that are also radioactive.

Alpha particle

1. The alpha particle has a helium nuclei and consists of two

protons and two neutrons.

He4 2+

2 These are

2. However it differs from a helium atom in that it carries a charge

of +2. Since alpha (α) particles are positively charged, they attract electrons and have high ionising power.

3. The positive charge of the alpha particles causes them to be deflected towards the negative plate of an

electric field.

Beta particles

1. The beta particles are high energy electrons emitted from the nucleus of the radioactive isotope when a

neutron splits up to form proton and electron. The electron is ejected from the nucleus as β-particle.

2. They move at speeds comparable to the speed of light (3.0  108 ms-1)
Neutron  proton + high energy electron (beta particle)

3. Due to its small size and high velocity, the penetrating power of β-particles is higher than that of α particles.
4. The negative charge β-particles are deflected towards the positive plate of an electric field.

Gamma Radiation
1. Gamma radiation is a high energy electromagnetic radiation of high frequency that is with very short

wavelength (10-10 to 10-12 m).
2. It is not a particle at all but it's actually an electro-magnetic wave even more

dangerous than ultra violet, or even X-rays.
3. Gamma rays have no mass, and travel at the speed of light.
4. γ-radiation is neutral and is not affected by electric or magnetic fields.

4

Properties of  ,  and  particles/radiation

1. They are affected by an electric field as shown below.

2. The table below summarises the properties of the three radioactive particles/radiation.

Particles / radiation Alpha (α) Beta (β ) Gamma (γ )
High energy High energy electromagnetic
Nature Nucleus of helium atom
electron radiation
Symbol 4 He2 γ
2 Electrons
100 Photons
Composition 2 protons and 2 neutrons -1 10000

Relative penetrating 1 1 0
power 1834
0
Relative charge +2

Relative mass/ a.m.u. 4
1.1.5 Radioactive Decay

1. Radioactivity or radioactive decay involves only the nucleus of an atom and not the extranuclear electrons.

Hence, radioactive decay is known as a nuclear reaction.

2 (a) In an equation for a nuclear reaction, the sum of the proton numbers and the sum of the nucleon numbers are

the same on both sides of the equation.

(b) The alpha rays, beta rays, gamma rays and neutron particles are represented by the following symbols :

4 He2 (alpha) ; (beta) ; γ (gamma) and 1 n
2 0

3. If the proton number is greater than 83 and has no stable isotopes, the isotope will undergo beta decay.

Therefore alpha particles and sometimes gamma rays are also emitted.

For example the α-decay of uranium-235 produces thorium-231.

Balancing the proton number: 92 = z + 2  z = 90
Balancing the nucleon number: 235 = A + 4  A = 231

Atom with proton number 90 is Thorium,Th.

Hence, the balanced equation is:

U235  29301Th + 4 He
2
92

4. If the proton number is less than 83 and its n/p ratio is larger than the stable n/p ratio, the isotope will undergo

beta decay.

(a) β-decay produces an extra proton in the nucleus and causes the proton number to increase by one.

This results in the production of a new element.

(b) An example of β-decay is radium-225:

225Ra 0 e + 225 Ac
1 89
88

5. (a) After emitting  or  -particles, the nucleus of the new element is sometimes left in the excited

state with excess state with excess energy. This excess energy is then radiated in the form of γ

radiation.

(b) In the γ-decay, there is no change in the proton number or nucleon number from that of the original

radioisotope.

(c) An example of gamma radiation is Pulodium-240.

240P u* 240 Pu +γ
94
94

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1.1.6 Nuclear Stability

1. If the number of neutrons is plotted against the number of protons for all stable, non-radioactive
isotopes, the following graph would be obtained.

2. The shaded area of the graph shows the variation of the number of neutrons and the number of protons for all known stable,
non- radioactive isotopes. The straight line corresponds to neutron = 1.
proton

Stable isotopes do not undergo spontaneous radioactive decay. Examples are 12 C and 16 O.
3. Heavier nuclei have more neutrons than protons to increase their nuclear size and to reduce excessive repulsion

among the protons.
4. All isotopes with neutron ratio outside the stability belt are unstable and thus radioactive.

proton

5. Hence the radioactivity is caused by either too high or too low a neutron ratio of the isotopes.
proton

1.2 RELATIVE MASS

1. The mass of an atom is approximately equal to the sum of the mass of all the sub-atomic particles (protons,
electrons and neutrons) present. For example, the mass of one atom of 54Fe can be calculated as follows:

Particles Proton neutron electron

Total number 26 28 26

Total mass/kg 26 x (1.67 x 10-27) = 4.34 x 10-26 28 x (1.67 x 10-27) = 4.68 x 10-26 26 x (9.11 x 10-31) = 2.37 x 10-29
Total mass = (4.34 x 10-26) + (4.68 x 10-26) + (2.37 x 10-29) = 9.02 x 10-26 kg

The value is so small that, it is not practical to talk about the absolute mass of an atom. Relative mass is used instead.
2. In this method, the mass of an atom is compared to the mass of another atom which is used as a reference (or

standard). Relative masses of particles were initially compared with hydrogen atom because the hydrogen atom is
the lightest atom known. However, oxygen atom was later used as the reference atom.

1.2.1 Relative Isotopic Mass

1. The relative isotopic mass of an isotope is the mass of one atom of the isotope relative to 1 times the mass
12

of one atom of 12C.

Relative isotopic mass = mass of one atom of the isotope mass of one atom of the isotope12
1  mass of one atom of C -12
12 OR

mass of one atom of C -12

2. The relative isotopic mass of an isotope is approximately equal to its nucleon number. For example ,

Isotope Nucleon number Relative isotopic mass

14N 14 14.0031

23Na 23 22.990

12I 127 126.910

However, the relative isotopic mass of an isotope is almost the same as its nucleon number.

6

3. In this method, the mass of an atom is compared to the mass of another atom which is used as a reference (or

standard). Initially the hydrogen atom was used as the reference atom because it is the lightest. Subsequently

the oxygen atom was used as the reference atom.
4. Carbon, 12C, was chosen as the standard for comparing relative atomic masses because

(a) it is a solid in its pure form at room conditions.

Hence, as a solid it is also easier to store and handle compared to hydrogen which is a gas.
(b) pure 12C is easily/commonly available and abundant in nature.
(c) 12C is easily detected and identified in a mass spectrum.
(d) 12C can form many compounds.
5. This is known as the carbon, 12C scale and the mass of an atom of 12C has a mass of exactly 12.00 atomic mass

unit (a.m.u.).

EXAMPLE 1.2

The mass of one atom of 12C is 1.99 X 10-23 g. What is the mass of 1 a.m.u?

ANSWER

The relative mass of one atom of 12C = 12 a.m.u.
12 a.m.u. = 1.99 X 10-23 g

1 a.m.u. = 1 x (1.99 X 10-23) g = 1.66 X 10-24 g
12

1.2.2 R e l a t i v e A t o m i c M a s s ( R . A . M . )

1. Most elements have at least two or more isotopes with different relative abundance ( percentage composition).
The table below shows the relative abundance of the isotopes of bromine and oxygen.

Isotope Relative abundance

79Br 50.54

81Br 49.46

160 99.76
17O 0.04

18O 0.20

Hence when calculating the average relative mass of an atom of an element, the relative abundance of the isotopes has to
be taken into consideration.

2. The relative atomic mass(Ar) of an element is defined as the average mass of one atom of the element

relative to 1 times the mass of one atom of 12C.
12

Relative atomic mass = average mass of one atom of the element
1  mass of one atom of C -12
12

or = average mass of one atom of the element 12
mass of one atom of C -12

EXAMPLE 1.3

Naturally occuring silicon consists of three isotopes 2 8 S i , 2 9 S i and 3 0 S i in the ratio of 92.21 : 4.70 : 3.09.
(a) Calculate the relative atomic mass of silicon.
(b) Explain why the relative atomic mass of silicon is not rounded to the nearest whole number.

ANSWER

(a) Relative atomic mass = (28 x 92.21) + (29 X 4.70) + (30 x 3.09) = 28.1
92.21 + 4.70 + 3.09

(b) Rounding the relative atomic masses to 28 would result in a significant error.

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EXAMPLE 1.4
(a) Chlorine consists of two isotopes 3 5 C l and 37Cl in the ratio of 3 : 1. Calculate the relative atomic mass of chlorine.
(b) What is the mass of a chlorine atom relative to the mass of one oxygen atom?

ANSWER

(a) Relative atomic mass = (35 x 3) + (37 x 1) = 35.5000
(1 + 3)

(b) Mass of one atom of chlorine = 35.5000 = 2.2181

Mass of one atom of oxygen 16.0044

i.e. one atom of chlorine is 2.2181 times heavier than one atom of oxygen.

1.2.3 Relative Molecular Mass (R.M.M.) and Relative Formula Mass

1. The relative molecular mass (Mr) of a molecular substance (either element or compound) is the mass of
one molecule of the substance relative to 1 times the mass of one atom of 12C.
12

Relative molecular mass = average mass of one molecule of the substance
1  mass of one atom of C -12
12

or = average mass of one molecule of the substance12
mass of one atom of C -12

2. The relative molecular mass is equal to the sum of the relative mass of all the atoms shown in the molecular formula.
For example, the relative molecular mass of ethanoic acid, CH3COOH
= (1x12) + (3x1) + (1x12) + (2x16) + (1x1) = 60
The relative molecular mass of chlorine, Cl2 = 35.5 x 2 = 71.0

3. For ionic compounds, the term relative formula mass is used because such compounds do not exist as
discrete molecules but consist of an infinite array of ions.

4. Relative formula mass is defined as the average mass of the formula unit of x compound relative to 1
12

times the mass of an atom of 12C.

5 (a) The relative mass of an ion is taken to be the same as the relative atomic mass of the element concerned

since the mass of an ion is very close to the mass of an atom.
(b) Hence, the relative formula mass of an ionic compound is found by adding the relative mass of all the atoms

shown in one formula unit of the compound.
(c) For example, the relative formula mass of sodium nitrate, NaNO3 = mass of one Na+ ions + mass of one NO3- ion

= mass of two Na atoms + mass of one O atom = 23.0 + 14.0 + (16.0 x 3) = 101.0
6. If the relative molecular mass of a compound XY is 56, the average mass of one molecule of the compound XY is 56 times

larger than 1 of the mass of one atom 12C or about 4.7 times larger than the mass of one atom 12C.
12

1.2.4 THE MASS SPECTROMETER

1. Mass spectrometer is an instrument used to determine relative masses of atoms or molecules of elements or
compounds..

2. It is a modern technique widely used to determine the
(a) relative isotopic mass of an atom
(b) the relative abundance of an isotope in a sample of the element
(c) relative atomic mass
(d) relative molecular mass
(e) structural formula of organic compounds

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1.2.5 Determination of Relative Atomic Mass (Atomic Mass Spectrum)
A full diagram of a mass spectrometer

An outline of what happens in a mass spectrometer

1. Atoms can be deflected by magnetic fields - provided the atom is first turned into an ion. Electrically charged
particles are affected by a magnetic field although electrically neutral ones aren't.

2. (a) The mass spectrometer is maintained at low pressure by a vacuum pump. The vacuum pump attached to the mass
spectrometer ensures the inside of a mass spectrometer is vacuum where air and unionised particles are removed.
This is to enable a free path for the ions to reach the ion detector.

(b) Presence of air molecules will scatter the ions. They will hinder the movements of the ions from reaching the ion
detector and interfere with angles of deflection.

3. The sequence is :

Stage 1 : Vaporisation : (The sample is vapourised).

A sample of the element(a gas, a liquid or a volatile solid) is injected into the vapourisation chamber where it

is converted into gaseous atoms. This is to enable the atoms or molecules to move freely.

Stage 2 : Ionisation (Positive ions are produced)

(a) The gaseous atoms or molecules are bombarded with a stream of high energy electrons emitted by a hot

cathode to form unipositive ions.

M(g) + e-  M+(g) + 2e-

high energy electron

(b) The electron beam knocks out other electrons from the sample atoms(argon chlorine, etc) or sample molecules to form

positive ions.

(c) In general unipositive ions are formed, that is positive ions that are singly charged and will carry a charge of +1

because it is much more difficult to remove further electrons from an already positive ion.

Stage 3: Acceleration (The positive ions are then accelerated by an electric field)
(a) The positive ions (with different masses) are then accelerated to a high and constant velocity by using electric field

of two negatively charged plates, X and Y so that they all have the same kinetic energy.
This is to ensure that the deflection of the ions is not due to differences in velocities of the particles.
(b) All the ions are accelerated into a finely focused beam.

Stage 4: Deflection (The positive ions are then deflected by a magnetic field)
(a) Different ions are deflected by the magnetic field by different amounts by a variable magnetic field.

The amount of deflection depends on :
(i) the mass of the ion. Lighter ions are deflected more than heavier ones.
(ii) the number of positive charges on the ion. Ions with 2 (or more) positive charges are deflected more

than ones with only 1 positive charge.
(b) These two factors are combined into the mass/charge ratio. Mass/charge ratio is given the symbol m/e or (m/z).
(c) By varying the strength of the magnetic field, an ion with a particular m/e can be chosen for detection.

Ions with the smaller mass (or low mass ratio) will be deflected
ch arg e

more than the heavier ones (or those with high mass ratio).
ch arg e

9

(i) If two isotope 7 0 X2+ and 65 X+ are analysed in a mass spectrometer, isotope 70 X2+ will have a lower m
4 0 40 40 e

and hence a larger deflection than 65 X+.
40

(ii) This is because m of doubly-charged ion of 7 0 X2+ = 70 = 35 and m of singly-charged 6 5 X+ = 65 = 65
e 4 0 2 e 4 0 1

(d) For example, if an ion had a mass of 28 and a charge of 1+, its mass/charge ratio would be 28. An ion with a

mass of 56 and a charge of 2+ would also have a mass/charge ratio of 28.

Stage 5: Detection (The positive ions are detected and a record is made)

(a) The beam of ions passing through the machine is detected electrically by the ion detector. The information

( m values and their relative abundance) are fed into a computer where a graph known as a mass spectrum of the
e

element is plotted.

(b) The degree of deflection of charged particles in the magnetic field is directly proportional to the charge on the

particles, and is inversely proportional to their masses.

Example : If the angle of deflection of a proton is 2o, calculate the degree of deflection of an alpha particle,  .

Degree of deflection of an alpha particle, 4 He2+ = ch arg e  2o = 2  2o = 1o
2 mass 4

Summary

• Gaseous atoms undergo ionisation by electron bombardment
• Charged particles can be accelerated by an electric field
• Charged particles will be deflected by a magnetic or electric field
• The radius of the path (deflection) depends on mass/charge ratio (m/e or m/z)
• Heavier ions with larger m/e values are deflected less
• If an ion acquires a 2+ charge it will be deflected more.
• Ions detected by electric or photographic methods
• Mass spectra can also be used to find relative molecular mass.

What the mass spectrometer output looks like

(a) The output from the chart recorder is usually simplified into a "stick
diagram". This shows the relative current produced by ions of
varying mass/charge ratio.

(b) The stick diagram for molybdenum is as follows :

(c) Vertical axis is labeled as either "relative abundance" or "relative
intensity".

(d) The vertical scale is related to the current received by the chart
recorder - and so to the number of ions arriving at the detector: the greater the current, the more abundant the ion.

(e) Base peak is the tallest peak in the mass spectrum. The base peak of molybdenum has a mass/charge ratio of 98.
Other ions have mass/charge ratios of 92, 94, 95, 96, 97 and 100.

(f) That means that molybdenum consists of 7 different isotopes. Assuming that the ions all have a charge of 1+,
that means that the masses of the 7 isotopes on the carbon-12 scale are 92, 94, 95, 96, 97, 98 and 100.

10

Mass spectrum of an element

(a) The mass spectrum above shows that the element consists of three isotopes with relative masses ml , m2 and
m3 in the ratio of b : a : c.

(b) The relative atomic mass (R.A.M.) of the element can be calculated from the formula below:
R.A.M. = (m1 x b) + (m2 x a) + (m3 x c)
(a+b+c)

EXAMPLE 1.5

The diagram above shows the mass spectrum of lead. What information can be obtained from a study of the mass spectrum?

ANSWER

(a) Lead consists of four isotopes with relative masses of 204, 206, 207 and 208 in the ratio of 1.5 : 23.6 : 22.6 : 52.3
(b) The relative atomic mass of lead

(2041.5)  (206 23.6)  (207  22.6)  (208 52.3)

= = 207.242

1.5  23.6  22.6  52.3

1.2.6 Mass Spectra of Molecules

1. The mass spectrometer can also be used to determine the relative molecular mass of molecular substances.

2. The mass spectrum for molecular substances is more complicated because the molecules can be split into smaller

fragments by the high energy electrons emitted by the hot cathode.

3. The mass spectrum for molecular substances contains two types of lines.
 A line given by the whole molecule (where the whole molecule loses one electron to become a molecular ion, M+).

This line has the largest mass or m/e value) which corresponds to the relative molecular mass.

Other lines which are due to fragments of the molecule produced when the molecule breaks. For example:

M(g) + e-  M+(g) + 2e-

M(g) + e-  X+(g) + Y+(g) + 3e-

4. From the molecular mass spectrum, one can :

(a) determine the relative molecular mass of a molecule.

(b) predict the structural formula of an organic compound.

1.2.7(a) The Mass Spectrum of Chlorine

1. When a sample of chlorine gas is analysed in the mass spectrometer, the mass spectrum which is not drawn to scale is obtained.

2. Chlorine gas consists of a mixture of three different molecules:

Molecule 35Cl- 35Cl 35Cl- 37Cl 37Cl- 37Cl

Relative mass 70 72 74

The lines at m/e values of 70, 72 and 74 are caused by the three molecules above when each of them loses one

electron to become a positive ion.

Cl - Cl + e-  [Cl - Cl]+ + 2e-

3. The lines at m/e values of 35 and 37 are caused by the 35Cl+ ions and 37Cl+ ions which are produced when the chlorine

molecules break down in the mass spectrometer.

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Example 1.6

The relative atomic mass of an element X and chlorine, relative to the carbon-12 scale is 121.547 and 35.453 respectively. What is
the relative atomic mass of 12C on a scale where one atom of X has a relative mass of 120.000?

Answer

On the 12C scale

Massof one atom of 12C 12.000

=

Massof one atom of X 121.547

On the 120X scale ,

Massof one atom of 12C = a

Massof one atom of X 120.000

12.000 = a

121.547 120.000

a = 11.847

1.2.7(b) The Mass Spectrum of Methane, CH4
1. When methane gas is analysed in the mass spectrometer, the following mass spectrum is obtained.

2. The line at m/e = 16 corresponds to the molecular ion, CH4+

CH4(g) + e-  CH4+ (g) + e-

3. The other lines are caused by fragment ions formed when the methane molecule undergoes fragmentation when

bombarded by the high energy electrons. For example:

CH4+  CH3+ + H

4. The table below gives the species responsible for the various lines in the mass spectrum for methane.

Relative mass 1 12 13 14 15 16

Species H+ C+ CH+ CH2+ CH3+ CH4+

5. Carbon consists of two isotopes, 12C and 13C in the ratio of 0.989 : 0.011.

13C is an isotope of C. Hence the mass spectrum at m/e = 17 contains a peak which is very weak due to low % abundance of 13C

Peak at m/e = 17 is caused by the 13CH4+ ion and is due to presence of 13C. This peak is known as the (M + 1) peak.

EXAMPLE 1.7

The mass spectrum (not to scale) of ethanol, C2H50H is shown below.

(a) Give the formula for the species responsible for the lines at m values of 14, 15, 17, 29, 31, 45 and 46.
e

(b) Suggest the origin of the line at m value of 47
e

12

ANSWER

(a)

14 15 17 29 31 45 46

CH2+ CH3+ OH+ CH3CH2+ CH2OH+ CH3CH2O+ CH3CH2OH+

(b) 13C is an isotope of C.

The line at m/e = 47 (known as the M + 1) line is caused by the molecule of ethanol containing an isotope of 13C, i.e.

13 CH3CH20H+. The peak of the (M + 1) line is usually very weak/low compared to the other lines because the
% abundance of 13C is very low.

1.2.8 Structural Formula and Mass Spectrum

1. The mass spectrometer can also be used to determine the structural formula of a compound, especially organic compounds.
2. Different substances undergo different fragmentation in the mass spectrometer. The structure of the substance can be

deduced from a study of the fragments.
3. However, the mass spectrum of a compound is usually complicated, and other additional information such as ultra-violet

spectroscopy and nuclear magnetic resonance (N.M.R.) is often required far analysis.

EXAMPLE 1.8

A sample of chlorine gas was analysed in a mass spectrometer. Part of the mass spectrum is shown below. How do you account for
the relative abundance of the three lines? [Chlorine consists of two isotopes 35C1 and 37Cl].

ANSWER

The lines are all caused by the molecular ions of Cl2+

M/e 70 72 74
37Cl – 37Cl+
Ions 35Cl – 35Cl+ 35Cl – 37Cl+ or 37Cl – 35Cl+

Their relative abundance are calculated as follows:

Species Relative abundance

35Cl – 35Cl+ 3X3=9
35Cl – 37Cl+ 3X1=3
37Cl – 35Cl+ 1X3=3

37Cl – 37Cl+ 1X1=1

Thus the relative abundance of m/e 70 : 72 : 74 = 9 : (3 + 3) : 1 =9:6:1

Example 1.9

The mass spectrum of propane, C3H8, consists of a peak of m/e = 44 (caused by the molecular ion, M, C3H8+) and a peak of low intensity
of m/e = 45 (caused by the molecular ion, M, C3H8+ that contains a 13C isotope), which is called the M + 1 peak.
If the relative abundance of 12C : 13C is 99.0 : 1.0, what is the relative abundance of the M : M + 1 peaks?

Answer

The relative abundance of 12C - 12C - 12C = 99 x 99 x 99 = 993
The relative abundance of 12C - 12C - 13C = (99 x 99 x 1) x 3

= 3 x 992
[ Note : There are three types of (M + 1) molecular ion : 12C 12C 13C , 12C 13C 12C , 13C 12C 12C ]
Hence, the relative abundance of M : M + 1 = 993 : 3 x 992

= 99 : 3

= 33 : 1

13

1.3 MOLE

1.3.1 The Mole And Avogadro Constant

1. One mole of a substance is the amount of substance that contains the same number of particles (atoms,
molecules, ions or electrons) as the number of atoms in exactly 12.000 g of the C-12 isotope.

2 The number of particles in one mole of any substance (or in exactly 12.000 g of 12C) is a constant known as the
Avogadro constant (L).
The Avogadro constant = 6.02 x 1023 mol-1

3. One mole of any substance is the same as the relative atomic mass, or relative molecular/formula mass
of that substance expressed in gram.

Summary
 Mole is abbreviated to mol and given the symbol n
 1 mole contains the same number of particles as there are in 12.000 g of carbon-12 atoms by definition.
This number is called Avogadro's number or Avogadro's constant (NA) and is equal to 6.02 x 1023 particles.
 1 mole of a pure substance has a mass in grams equal to its molecular mass (MM) [also known as molecular weight
(MW) or formula mass (FM) or formula weight (FW)].
This is often referred to as the molar mass.
 1 mole of an ideal gas has a volume of
 22.4 dm3 (22.4 dm3) at S.T.P. [Standard Temperature and Pressure, 0oC (273K) and 101 kPa (1 atm)]
 24.0 dm3 (24.47 dm3) at R.T.P. [Standard Room Conditions, 25oC (298K) and 101 kPa (1 atm)]

Avogadro's Number (NA)
 1 mole of atoms or molecules contains 6.02 x 1023 atoms or molecules
1 mole of helium atoms (He) contains 6.02 x 1023 helium atoms (He)
 To find the number of atoms in a known number of moles, multiply the moles by 6.02 x 1023
2 moles of helium atoms (He) contains 2 x 6.02 x 1023 = 1.204 x 1024 helium atoms. (He)
 To find the moles of atoms, divide the number of atoms by 6.02 x 1023
If we have 4.22 x 1023 neon atoms, how many moles of neon atoms are there?
Moles of neon atoms = (4.22 x 1023) ÷ (6.02 x 1023) = 0.7 mol
 1 mole of molecules contains 6.02 x 1023 molecules.
1 mole of oxygen molecules (O2) contains 6.02 x 1023 oxygen molecules (O2).
 To find the moles of molecules, multiply the number of molecules by 6.02 x 1023
½ mole of oxygen molecules (O2) contains ½ x 6.02 x 1023 = 3.01 x 1023 oxygen molecules (O2).
 To find the number of molecules, divide the moles of molecules by 6.02 x 1023
If we have 6.02 x 1021 chlorine molecules (Cl2), how many moles of chlorine molecules are there?
Moles of chlorine molecules = (6.02 x 1021) ÷ (6.02 x 1023) = 0.01 mol
 1 mole of ionic compounds contain 6.02 x 1023 formula units
 1 mole of Ca(NO3)2 contains 6.02 x 1023 formula units
 1 mole of Ca(NO3)2 contains 3 x 6.02 x 1023 ions
[ 1 Ca(NO3)2 contains 1 Ca2+ ion and 2NO3- ions ; the total is 3 ions].
 1 mole of molecules does not necessarily contain 1 mole of atoms of each element in the formula
 1 mole of HCl WILL contain 1 mole of hydrogen atoms (H) and 1 mole of chlorine atoms (Cl)
1 mole of HCl contains 6.02 x 1023 hydrogen atoms and 6.02 x 1023 chlorine atoms
 5 moles of oxygen molecules (O2) contains 5 x 2 = 10 moles of oxygen atoms (O)
5 moles of oxygen molecules contains 10 x 6.02 x 1023 = 6.02 x 1024 oxygen atoms.
 1 mole of ammonia molecules (NH3) will contain 1 mole of nitrogen atoms (N) and 3 moles of hydrogen atoms
(H)
1 mole of ammonia molecules contains 6.02 x 1023 nitrogen atoms and 3 x 6.02 x 1023 = 1.81 x 1024 hydrogen
atoms

Molar Mass
 1 mole of a pure substance has a mass in grams equal to its molecular mass (MM).
 1 mole of Helium (a monatomic gas with the formula He) has a mass equal to its relative atomic mass, 4.003g
 1 mole of hydrogen gas (a diatomic gas with the formula H2) has a mass equal to 2 x 1.008 = 2.016g
 1 mole of ammonia gas (NH3) has a mass equal to 14.01 + (3 x 1.008) = 17.034g
 1 mole of water (H2O) has a mass equal to (2 x 1.008) + 16.00 = 18.016g

14

EXAMPLE 1.10

Determine the number of atoms in each of the following substances.
(a) 1 mol of aluminium.
(b) 0.2 mol of nitrogen gas.
(c) 5 mol of ammonia

ANSWER

(a) 1 mol of Al contains 6.02 x 1023 atoms
(b) 0.2 mol of nitrogen gas = 0 . 2 x 6.02 x 1023 molecules of N2

= 2 x ( 0 . 2 x 6.02 x 1023) atoms of nitrogen
= 2.41 x 1023 atoms
(c) 5 mol of NH3 = 5 x (6.02 x 1023) molecules.
But one molecule of NH3 contains 4 atoms.

 5 mol of NH3 = 4 x (5 x 6.02 x 1023) atoms = 1.20 x 1025 atoms

EXAMPLE 1.11

(a) How many atoms are there in 0.25 mol of zinc?
(b) How many moles of chlorine are there in 2.25 x 1024 molecules of chlorine?

ANSWER

(a) 1 mol of zinc consists of 6.02 x 1023 atoms.

 No of atoms in 0.45 mol of zinc = 0.25 x (6.02 x 1023) = 1.51 x 1023 atoms

(b) 1 mol of chlorine contains 6.02 x 1023 molecules.
No of moles of chlorine in 2.25 x 1024 molecules = 2.25 1024 = 3.74 mol
6.02 1023

1.3.2 Moles Of Gases

1. In reactions involving gases, the volume of the gases that takes part in the reaction is usually more important
than the mass of the gases involved.

2. The relationship between the amount of gas (no of moles of gas) and the volume of gas is given in Avogadro's law.
3. Avogadro's law states that under the same conditions of temperature and pressure, an equal volume of

gases contains equal number of moles (and therefore an equal number of molecules).
volume α number of moles

4. As an example, under room conditions,10 cm3 of 02 will contain the same number of moles/molecules as 10 cm3 of N2.
5. Ideal Gas Volumes

 at standard temperature and pressure, S.T.P. [0oC (273K), 101 kPa (1 atm)], 1 mole of all gases will occupy a
volume of 22.4 dm3. This volume is known as the molar volume (Vm).
 To find the volume of a certain number of moles of gas, multiply the moles by 22.4 dm3.
What is the volume of 2.5 moles of chlorine gas at S.T.P?
Volume of chlorine gas = 2.5 x 22.4 = 56.0 dm3
 To find the moles of a certain volume of gas, divide the volume by 22.4 dm3
How many moles of argon are in 3.36 dm3 of argon gas at S.T.P?
Moles of argon gas = 3.36 ÷ 22.4 = 0.15 mol

 at room conditions , R.T.P. [25oC (298K), 101.3 kPa (1 atm)], one mole of all gases will occupy 24.0 dm3.
 To find the volume of a certain number of moles of gas, multiply the moles by 24.0 dm3.
What is the volume of 0.2 moles of hydrogen sulfide gas at room conditions?
Volume of hydrogen sulfide gas = 0.2 x 24.0 = 4.80 dm3.
 To find the moles of a certain volume of gas, divide the volume by 24.0 dm3.
How many moles of carbon monoxide are in 70.5 dm3. of carbon monoxide gas at R.T.P ?
Moles of argon gas = 70.5 ÷ 24.0 = 2.94 mol

EXAMPLE 1.12

Calculate the volume occupied by the following gases at room conditions.
(a) 40 g of NH3
(b) 1.5 g of C02

15

(c) 24 x 1023 molecules of 02

ANSWER

(a) 40 g of NH3 = 40 mol of NH3
17

 Volume = 40 x 24 dm3 = 56.5 dm3

17

(b) 1.5 g of CO2 = 1.5 mol of CO2
44

 Volume = 1.5 x 24 dm3 = 0.818 dm3

44

(c) No. of moles of 02 = 24 1023 = 4 mol
6 1023

 Volume = 4 x 24 dm3 = 96.0 dm3

EXAMPLE 1.13

Calculate the mass of methane gas, CH4, which will occupy the same volume as 8.0 g of nitrogen gas under the same
conditions.

ANSWER

No. of moles of N2 = 8 = 0.286 mol
28

No. of moles of CH4 will also be = 0.286 mol

 mass of CH4 = 0.286 x 16 = 4.58 g

EXAMPLE 1.14

The equation for the complete combustion of methane is:

CH4(g) + 202(g)  C02(g) + 2H20(g)

What volume of oxygen gas is required to completely burn 60.0 cm3 of methane?

(All volumes to be measured under the same conditions).

ANSWER

From the equation:

1 mol of CH4  2 mol of 02
 1 volume of CH4 = 2 volume of 02 (under the same conditions)
 Volume of 02 required = 2 x volume of CH4

= 2 x 60.0 = 120.0 cm3

1.3.3 Moles and Solutions

1. The concentration of a solution is usually expressed as the mass of solute per 1.0 dm3 of solution (g dm-3) or expressed as
the number of moles of solute in 1.0 dm3 of solution (mol dm-3).

2. The concentration in the unit of mol dm-3 is also known as the molarity of the solution (symbol: M)
3. The relationship between molarity and concentration (in g dm-3) is given by the expression

Molarity (M) = concentration (g dm-3)
relative molecular mass of the solute

4. The number of moles of solute present in a given volume of solution (of molarity M) is
Number of moles of solute = molarity x volume (in dm3)

5. A short way to write that the concentration of a solution of hydrochloric acid is 0.01 mol/dm3 is to write [HCl] = 0.01 M

The square brackets around the substance indicate concentration.
6. M = n ÷ V (M=concentration of solution in mol/dm3, n=moles of substance, V = volume of solution in dm3

7. When a solution is diluted, more solvent is added to it.

Since M = n ÷ V, and n (the moles of solute) is the same for the original solution and the new diluted solution, it

follows that M1V1 = M2V2

where M1 = original concentration of solution

16

V1 = original volume of solution
M2 = new concentration of solution after dilution
V2 = new volume of solution after dilution
8. To calculate the new concentration (M2) of a solution given its new volume (V2) and its original concentration (M1) and
original volume (V1):
M2 = (M1 x V1) ÷ V2
9. To calculate the new volume (V2) of a solution given its new concentration (M2) and its original concentration
(M1) and original volume (V1):
V2 = (M1 x V1) ÷ M2
10. General formula for titration :

aA(aq) + bB(aq) → cC(aq) + dD(aq)

M AVA = a
M bVb b

Examples 1.15

1. M = n ÷ V
Calculate the concentration (molarity) of a sodium chloride solution containing 0.125 moles sodium chloride in 0.5 dm3

of water.
 M = n ÷ V in cubic decimeter, dm3

 n = 0.125 mol
 V in cubic decimeter = 0.5 dm3
 [NaCl(aq)] = M = 0.125 ÷ 0.5 = 0.25 mol dm-3 (or 0.25 mol/ dm3)

2. n = M x V
Calculate the moles of copper sulphate in 250 cm3 of 0.02 mol dm-3 copper sulphate solution.

 n=MxV
 M = 0.02 mol dm-3
 V = 250 cm3 = 250 ÷ 1000 = 250 x 10-3 dm3 = 0.250 dm3 (since there are 1000 cm3 in 1 dm3)
 n = 0.02 x 250 x 10-3 = 0.005 mol

3. V = n ÷ M
Calculate the volume of a 0.80 mol dm-3 potassium bromide solution containing 1.6 moles of potassium bromide.
 V=n÷M
 n = 1.6 mol
 M = 0.80 mol dm-3
 V = 1.6 ÷ 0.80 = 2.00 dm3

EXAMPLE 1.16

(M1V1) ÷ V2
Calculate the new concentration (molarity) if enough water is added to 100 cm3 of 0.25 mol dm-3 sodium chloride to make up
1.5 dm3.

 M2=(M1V1) ÷ V2
 M1 = 0.25 mol dm-3
 V1 = 100 cm3 = 100 ÷ 1000 = 0.100 dm3 (volume must be in cubic decimeter)
 V2 = 1.5 dm3
 [NaCl(aq)]new = M2 = (0.25 x 0.100) ÷ 1.5 = 0.017 mol dm-3 (or 0.0.017 mol/ dm3 or 0.0.17 mol dm-3)

2. V2=(M1V1) ÷ M2
Calculate the volume to which 500 cm3 of 0.02 mol dm-3 copper sulphate solution must be diluted to make a new concentration
of 0.001 mol dm-3.

 V2=(M1V1) ÷ M2
 M1 = 0.02 mol dm-3
 V1 = 500 cm3 = 500 ÷ 1000 = 500 x 10-3 dm3 = 0.500 dm3 (since there are 1000 cm3 in 1 dm3)
 M2 = 0.001 mol dm-3
 V(CuSO4)new V2 = (0.02 x 0.500) ÷ 0.001 = 10.00 dm3

EXAMPLE 1.17

17

(a) Calculate the molarity of a solution containing 7.3 g HCl in 1 dm3 of solution.
(b) Calculate the concentration (in g dm-3) a solution of 0.50 mol of HNO3 in 200 cm3 of solution .
(c) Calculate the molarity of sodium hydroxide which contains 23.0 g NaOH in 250 cm3 of solution.
(d) Calculate the number of moles of sodium chloride, NaCl in 25.0 cm3 of an aqueous solution of concentration 2.00 mol dm-3.
(e) Calculate the number of ions in 250.0 cm3 of 0.3 mol dm-3 aqueous aluminium chloride.

ANSWER

7.3 = 0.20 mol

(a) No. of moles of HCl=

1  35.5

.'. Concentration of HCl = 0.20 mol dm-3

(b) Molarity of HNO3 = 0.5 mol dm-3 = 2.50 mol dm-3
0.2

Concentration = molarity x relative molecular mass = 2.50 x [1 + 14 + 3(16)]

= 157.5 g dm-3

(c) Concentration of NaOH = 23.0 x 1000 g dm-3

250

= 92.0 g dm-3

Molarity of NaOH = 92 mol dm-3 = 2.30 mol dm-3

40

(d) No. of moles of NaCl in 25 cm3 = 2.00 x (25 x 10-3) = 0.05 mol
(e) No. of moles of aluminium chloride, AlCl3 = 0.3 x (250 x 10-3)

= 7.5 x 10-2 mol

No. of moles of ions = (7.5 x 10-2) x 4

No. of ions = (7.5 x 10-2) x 4 x (6.02 x 1023) = 1.81 x 1023 ions.

1.6 EMPIRICAL FORMULA AND MOLECULAR FORMULA
1.6.1 Composition By Mass

1. The empirical formula of a compound shows the simplest whole number ratio for the atoms of all the different elements

present in one molecule of the compound. The molecular formula of a compound shows the actual number of atoms of

different elements in one molecule of the compound.

2. The table below shows examples of the empirical formula and molecular formula for some compounds.

Compound Molecular formula Empirical formula

Ethene C2H4 CH2
Phosphorous(V) oxide P4O10 P205
Hydrogen peroxide H202 HO

Ethanoic acid C2H402 CH20

3. The empirical formula of a compound can be calculated if we know the composition by mass of the different elements in

that compound.

The molecular formula is a simple multiple of the empirical formula.

EXAMPLE 1.18

A hydrocarbon has the following composition by mass: C, 92.3%; H, 7.6%
a) Calculate its empirical formula.
b) Given that the Mr for the hydrocarbon is 78, determine its molecular formula.

ANSWER

Elements present Carbon Hydrogen

Mass per 100 g of compound 92.3 g 7.6 g

Number of moles 92.3 = 7.7 7.6 = 7.6
12 1

Simplest mole ratio 7.7 = 1 7.6 = 1
7.6 7.6
 The empirical formula is CH

(b) Let the molecular formula be (CH)n
Mr of compound = (12 + 1)n = 13n

18

13n = 78
n = 78 = 6
13

Molecule formula is C6H8

EXAMPLE 1.19

A compound Y has the following composition by mass: Na, 29.1%; S, 40.5%; O, 30.4% Calculate the empirical formula of Y

ANSWER

Elements present Na S O
Mass per 100 g 29.1 40.5 30.4

No. of moles 29.1 = 1.27 40.5 = 1.27 30.4 = 1.9
32 16
Simplest mole ratio 23
(Multiply by 2 to get whole numbers) 1.27 = 1 1.90 = 1.5
The empirical formula of Y is Na2S203 1.27 = 1 1.27 1.27
1.6.2 Combustion data 1.27
2 3
2

The empirical formula or molecular formula of a compound can also be deter mined by the combustion data when the compound is burnt
completely in oxygen.

EXAMPLE 1.20

4.6 gram of a compound of nitrogen and oxygen is formed when 1.4 g of nitrogen combines with oxygen. What is the empirical formula
of the compound formed ?

ANSWER

Mass of oxygen combined with 1.4 g of N = (4.6 – 1.4) g = 3.2 g

Composition by mass of hydrogen is 1.4 g N : 3.2 g O

Mole ratio of N : O = 1.4 3.2
:
14 16

= 0.1 : 0.2 = 1 : 2
Empirical formula is NO2

EXAMPLE 1.21

100 cm3 of a gaseous hydrocarbon require 450 cm3 of oxygen for complete combustion to give 300 cm3 of carbon dioxide. All volumes
are measured under the same conditions. Calculate the molecular formula of the hydrocarbon.

ANSWER

Let the formula of the hydrocarbon be CxHy.

The balanced equation for its complete combustion is:

CxH y + (x + y ) 02  x C02 + y H2O
4
2

From the equation:

100 cm3 CxHy will produce 100x cm3 CO2

19

 300 = 100x

x =3

From the equation:

100 cm3 CxHy require 100(x + y ) cm3 of O2
4

100(x + y ) = 450
4

3 + y = 4.5
4

y=6

 Molecular formula of the hydrocarbon is C3H6

EXAMPLE 1.22

When 2.50g of an organic compound B burns in excess oxygen, 7.86 g of CO2 and 3.21 g of water,H2O is produced.
0.25 g of the vapour of B occupies 80.0 cm3 at 273 K and 101 kPa.

(a) Calculate the empirical formula of B.

(b) Calculate the molar mass of B.

(c) Determine the molecular formula of B.

Solution

(a) 1 mol of CO2 (44g) contains 1 mol of carbon atoms (12 g)

 Mass of carbon atom 7.86 g CO2 = 7.86  12
44

= 2.14 g

1 mol of H2O(18 g) contains 2 mol of hydrogen atoms (2g)

 Mass of hydrogen in 3.21 g H2O = 3.21  2
18

= 0.357 g

Hence, mole ratio of C : H = 2.14 0.357
:
12 1

= 0.18 : 0.357

= 1: 2

Empirical formula of B is CH2

(b) Using : pV = nRT

(101  103) (80  10-6) = n  8.31  273

n = 0.00356 mol

 Molar mass of B = 0.25
0.00356

= 70.2 g mol-1

(c) Let the molecular formula be (CH2)n
(12 + 2)n = 70.2

n=5

Molecular formula of B is C5H10

20

Molecular Mass (Molecular Weight)

 In theory, the relative molecular mass or molecular weight of a compound is the mass of a molecule of the compound
relative to the mass of a carbon atom taken as exactly 12.

 In practice, the molecular mass, MM, (molecular weight, MW) of a compound is the sum of the atomic masses (atomic
weights) of the atomic species as given in the molecular formula.

 In theory we can only refer to the Molecular Mass or Molecular Weight of a covlaent compound since only covalent
compounds are composed of molecules.

Formula Mass (Formula Weight)

 The relative formula mass, FM, (formula weight, FW) of a compound is the sum of the atomic masses (atomic weights)
of the atomic species as given in the formula of the compound.

 Formula Mass (Formula Weight) is a more general term that can be applied to compounds that are not composed of
molecules, such as ionic compounds.

 In practice, the terms, molecular mass, molecular weight, formula mass and formula weight are used interchangeably
by Chemists.

Molecular Mass Calculations

A. Calculate the Molecular Mass (MM) of the compound carbon monoxide, CO
 The formula for carbon monoxide is composed of one atom of carbon and one atom of oxygen

21

 Atomic mass carbon = 12.01 (from the Periodic Table)
Atomic mass of oxygen = 16.00 (from the Periodic Table)

 Molecular Mass (MM) for cabon monoxide = atomic mass carbon + atomic mass oxygen
Molecular mass (MM) = 12.01 + 16.00 = 28.01 g/mol

B. Calculate the Molecular Mass (MM) of the compound carbon dioxide, CO2
 The formula for carbon dioxide is composed of one atom of carbon and two atoms of oxygen
 Atomic mass carbon = 12.01 (from the Periodic Table)

Atomic mass of oxygen = 16.00 (from the Periodic Table)
 Molecular Mass (MM) for carbon dioxide = atomic mass carbon + (2 x atomic mass oxygen)

Molecular Mass (MM) = 12.01 + (2 x 16.00) = 12.01 + 32.00 = 44.01g/mol
C. Calculate the Molecular Mass (MM) of the compound water, H2O
 The formula for water is composed of two hydrogen atoms and one oxygen atom
 Atomic mass hydrogen = 1.008 (from the Periodic Table)

Atomic mass of oxygen = 16.00 (from the Periodic Table)
 Molecular Mass (MM) for water = (2 x atomic mass hydrogen) + atomic mass oxygen

Molecular Mass (MM) = (2 x 1.008) + 16.00 = 2.016 + 16.00 = 18.016g/mol
D. Calculate the Molecular Mass (MM) of the compound calcium hydroxide, Ca(OH)2
 The formula for calcium hydroxide is composed of one calcium "atom" (actually an ion) and two hydroxide

ions. Each hydroxide ion is composed of one hydrogen "atom" (actually an ion) and one oxygen "atom" (also an
ion)
 Atomic mass calcium = 40.08 (from the Periodic Table)
Atomic mass hydrogen = 1.008 (from the Periodic Table)
Atomic mass of oxygen = 16.00 (from the Periodic Table)
 Molecular Mass (MM) for calcium hydroxide = atomic mass calcium + (2 x atomic mass oxygen) + (2 x atomic
mass hydrogen)
Molecular Mass (MM) = 40.08 + (2 x 16.00) + (2 x 1.008) = 40.08 + 32.00 + 2.016

= 74.096 g/mol

Alternatively, Moelcular Mass = atomic mass of calcium + (2 x molecular mass of hydroxide ions)
Molecular Mass (MM) = 40.08 + [2 x (16.00 + 1.008)] = 40.08 + [2 x 17.008] = 40.08 + 34.016 =
74.096g/mole

E. Calculate the Molecular Mass (MM) of the compound ammonium sulfate, (NH4)2SO4

 The formula of ammonium sulfate is composed of two atoms of nitrogen, eight atoms of hydrogen, one atom
of sulfur and four atoms of oxygen

 Atomic mass nitrogen = 14.01 (from the Periodic Table)
Atomic mass hydrogen = 1.008 (from the Periodic Table)
Atomic mass of sulfur = 32.06 (from the Periodic Table)
Atomic mass of oxygen = 16.00 (from the Periodic Table)

 Molecular Mass (MM) for ammonium sulfate = (2 x atomic mass nitrogen) + (8 x atomic mass hydrogen) +
atomic mass sulfur + (4 x atomic mass oxygen)
Molecular Mass (MM) = (2 x 14.01) + (8 x 1.008) + 32.06 + (4 x 16.00) = 28.02 + 8.064 + 32.06 +
64.00 = 132.144g/mole
Alternatively, Molecular Mass (MM) = [2 x molecular mass ammonium ions (NH4)] + atomic mass sulfur + (4 x
atomic mass oxygen)
Molecular Mass (MM) = {2 x [14.01 + (4 x 1.008)]} + 32.06 + (4 x 16.00) = {2 x [14.01 + 4.032]} +
32.06 + 64.00 = {2 x 18.042} + 32.06 + 64.00 = 36.084 + 32.06 + 64.00 = 132.144g/mol

F. Calculate the Molecular Mass (MM) of the compound ethanoic acid (acetic acid), CH3COOH

 The formula of ethanoic acid is composed of two carbon atoms, four hydrogen atoms and two oxygen atoms

 Atomic mass carbon = 12.01 (from the Periodic Table)
Atomic mass hydrogen = 1.008 (from the Periodic Table)
Atomic mass of oxygen = 16.00 (from the Periodic Table)

 Molecular Mass (MM) for ethanoic acid = (2 x atomic mass carbon) + (4 x atomic mass hydrogen) + (2 x
atomic mass oxygen)

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Molecular Mass (MM) = (2 x 12.01) + (4 x 1.008) + (2 x 16.00) = 24.02 + 4.032 + 32.00 =
60.052g/mol

Percent Composition (Percentage Composition)

The percent composition (percentage composition) of a compound is a relative measure of the mass of each different element
present in the compound.

To calculate the percent composition (percentage composition) of a compound

 Calculate the molecular mass (molecular weight, formula mass, formula weight), MM, of the compound

 Calculate the total mass of each element present in the formula of the compound

 Calculate the percent composition (percentage composition):
% by weight (mass) of element = (total mass of element present ÷ molecular mass) x 100

Example 1

Calculate the percent by weight of sodium (Na) and chlorine (Cl) in sodium chloride (NaCl)

 Calculate the molecular mass (MM):
MM = 22.99 + 35.45 = 58.44

 Calculate the total mass of Na present:
1 Na is present in the formula, mass = 22.99

 Calculate the percent by weight of Na in NaCl:
%Na = (mass Na ÷ MM) x 100 = (22.99 ÷ 58.44) x 100 = 39.34%

 Calculate the total mass of Cl present:
1 Cl is present in the formula, mass = 35.45

 Calculate the percent by weight of Cl in NaCl:
%Cl = (mass Cl ÷ MM) x 100 = (35.45 ÷ 58.44) x 100 = 60.66%

The answers above are probably correct if %Na + %Cl = 100, that is, 39.34 + 60.66 = 100.
Example 2

Calculate the percent by weight of each element present in sodium sulfate (Na2SO4).

 Calculate the molecular mass (MM):
MM = (2 x 22.99) + 32.06 + (4 x 16.00) = 142.04

 Calculate the total mass of Na present:
2 Na are present in the formula, mass = 2 x 22.99 = 45.98

 Calculate the percent by weight of Na in Na2SO4:
%Na = (mass Na ÷ MM) x 100 = (45.98 ÷ 142.04) x 100 = 32.37%

 Calculate the total mass of S present in Na2SO4:
1 S is present in the formula, mass = 32.06

 Calculate the percent by weight of S present:
%S = (mass S ÷ MM) x 100 = (32.06 ÷ 142.04) x 100 = 22.57%

 Calculate the total mass of O present in Na2SO4:
4 O are present in the formula, mass = 4 x 16.00 = 64.00

 Calculate the percent by weight of O in Na2SO4:
%O = (mass O ÷ MM) x 100 = (64.00 ÷ 142.04) x 100 = 45.06%

The answers above are probably correct if %Na + %S + %O = 100, that is, 32.37 + 22.57 + 45.06 = 100

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Example 3

Calculate the percent by weight of each element present in ammonium phosphate [(NH4)3PO4]

 Calculate the molecular mass (MM) of (NH4)3PO4:
MM = 3x[14.01 + (4 x 1.008)] + 30.97 + (4 x 16.00) = 3 x [14.01 + 4.032] + 30.97 + 64.00 = (3 x 18.042) + 30.97 + 64.00
= 54.126 + 30.97 + 64.00 = 149.096

 Calculate the total mass of N present:
3 N are present, mass = 3 x 14.01 = 42.03

 Calculate the percent by mass of N present in (NH4)3PO4:
%N = (mass N ÷ MM) x 100 = (42.03 ÷ 149.096) x 100 = 28.19%

 Calculate the total mass of H present:
12 H are present in the formula, mass = 12 x 1.008 = 12.096

 Calculate the percent by mass of H present in (NH4)3PO4:
%H = (mass H ÷ MM) x 100 = (12.096 ÷ 149.096) x 100 = 8.11%

 Calculate the total mass of P present:
1 P is present in the formula, mass = 30.97

 Calculate the percent by mass P in (NH4)3PO4:
%P = (mass P ÷ MM) x 100 = (30.97 ÷ 149.096) x 100 = 20.77%

 Calculate the total mass of O present:
4 O are present in the formula, mass = 4 x 16.00 = 64.00

 Calculate the percent by mass of O in (NH4)3PO4:
%O = (mass O ÷ MM) x 100 = (64.00 ÷ 149.096) x 100 = 42.93%

The answers above are probably correct if %N + %H + %P + %O =100, that is,
28.19 + 8.11 + 20.77 + 42.93 = 100

Summary

 1 mole of a pure substance has a mass equal to its molecular mass (MM)
So 2 moles would have a mass = 2 x MM
3 moles would have a mass = 3 x MM etc
This leads to the formula:

 mass = n x MM (mass in grams, n=moles of pure substance, MM=molecular mass of the pure substance)
This formula can be rearranged to give the following:

 n = mass ÷ MM (n=moles of pure substance, MM=molecular mass of the pure substance, mass in grams )

 MM = mass ÷ n (n=moles of pure substance, MM=molecular mass of the pure substance, mass in grams )
Examples

1. mass = n x MM
Calculate the mass of 0.25 moles of water

 mass (in grams) = n x MM

 n = 0.25mol

 MM of water (H2O) = (2 x 1.008) + 16.00 = 18.016g mol-1

 mass = 0.25 x 18.016 = 4.504g

1.1.7 Uses of radioisotopes

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1. Radioisotopes can be used as tracers. If a radioisotope isotope is incorporated into a molecule, then the path of the molecule can
be followed using suitable detecting apparatus.

2. The rate of certain biological process can be followed by adding radioactive samples to the substance, which when taken up by
the organism, allows its path to be followed.

3. Leaks in deep underground pipe-lines can be detected by injecting a small amount of radioactive substance and measuring the
activity from the surface.

4. Radioactive isotopes are also used to sterilize medical equipments and food, for measuring the thickness of metal, and in the
treatment of cancerous tumours.

EXERCISE 1.1

1. With the help of the Periodic Table, identify the following species.
No of sub-atomic particles

Species Proton Electron Neutron

W8 9 8
X1 0 0

Y 17 18 18

2. Determine the number of sub-atomic particles in the following species.

3. Identify the radioactive particles emitted in each step of the radioactive

EXERCISE 1.2

1. Lead consists of four isotopes with relative masses of 204, 206, 207 and 208 in the ratio of 1.5 : 23.6 :
22.6 : 52.3. Calculate (to 4 decimal places) the relative atomic mass of lead.

2. Bromine consists of two isotopes, 79Br and 81Br. If the relative atomic mass of bromine is 79.9, calculate the
relative abundance of each of the isotopes.

3. Calculate the relative atomic mass of lithium and silicon from the following data:

Isotope Relative abundance
Lithium-6 7.4
Lithium-7 92.6
Silicon-28 92.2
Silicon-29 4.7
Silicon-30 3.1

EXERCISE 1.3
1. The mass spectrum of an element X (proton number = 60) is shown below:

(a) Write the symbol to represent all the isotopes of X.
(b) Calculate the relative atomic mass of X.
2. The mass spectrum for hydrogen chloride is shown below:

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Give the formula of the species responsible for each of the peaks.
3. A sample of bromine vapour (which contains the isotopes 79Br and 81Br in the ratio of 1 : 1) is analysed in a mass

spectrometer. How many peaks due to molecular ions are expected to be observed in the spectrum?
What are the m/e values and the relative abundance of the molecular ion peaks?

EXERCISE 1.4

1. Between 1.10 g of hydrogen gas and 14.7 g of chromium, which contains more atoms?
2. Calculate the mass of 0.20 mol of chlorophyll, C55H72MgN405.
3. The mass of 0.372 mol of a substance is 152 g. What is its relative molecular mass?
4. The density of water at 4°C is 1.00 g cm-3. Calculate the number of water molecules in 4.5 cm3 of water at the same temperature.
5. 0.96 g of an element D (R.A.M. = 52) displaces 0.52 g of element from an aqueous solution containing the E3+ ions and itself gets

converted to D2+. What is the relative atomic mass of E?
6. Calculate the number of ions in the following solutions.

(a) 1 dm3 0.050 mol dm-3 H2SO4
(b) 2.6 dm3 1.20 mol dm-3 Ca3(PO4)2
(c) 50.0 cm3 0.30 mol dm-3 NaCl

7. When heated, potassium chlorate(V) decomposes according to the equation

2KCl103(s) 2KCl(s) + 302(s)

Calculate the mass of potassium chlorate(V) required to produce 0.55 dm3 (measured at room conditions) of oxygen

gas.

8. Haemoglobin contains 0.33% by mass of iron. Each molecule of haemoglobin has two iron atoms. Calculate the

relative molecular mass of haemoglobin.

9. The relative molecular mass of water is 18.0. Calculate the mass of one water molecule.

10. An element X (R.A.M. = 35.5) reacts with Y- aqueous ions according to the equation:

X2 + 2Y- Y2 + 2X

In an experiment, 14.2 g of X2 produced 50.8 g of Y2. Calculate the relative atomic mass of Y.
11. 5.12 g of impure sodium carbonate was dissolved in 250 cm3 of water.

25.0 cm3 of the solution required 29.50 cm 3 of 0.050 mol dm -3 of sulphuric acid for complete reaction

Na2CO3 + H2SO4 Na2SO4 + H2O + C02

(a) Calculate the number of moles of H2SO4 in 29.50 cm3 of the acid solution.

(b) Calculate the number of moles of pure sodium carbonate in the impure sample.

(c) Calculate the mass of sodium carbonate and hence determine the percentage purity of the sample of sodium

carbonate.
12. 25.0 cm3 of sodium ethanedioate 0.200 mol dm-3 requires 18.30 cm3 of potassium manganate(VII) for complete reaction.

Calculate the molarity of the potassium manganate(VII) solution.

[Given: 2 moles of potassium manganate react with 5 moles of sodium ethanedioate]

13. The mass spectrum for a sample of tetrachloromethane, CCl4 is shown below:

(a) Give the formula of the ions responsible for each of the peaks.
(b) There are two other peaks (with relative abundance of about 0.01) of values of 48 and 83. Suggest the origin

of these two peaks.
(c) A small peak is also observed at value of 41. Suggest the species responsible for this peak.
14. Chlorine has two isotopes of 35Cl and 37C1 in the ratio of 3 : 1. Phosphorous has only one isotope, 31P.
The mass spectrum of one the chloride of phosphorous contains the following peaks:

66, 68, 101, 103, 105, 136, 138, 140 and 142
Suggest the molecular formula of the chloride and determine the species responsible for the peaks.

EXERCISE 1.5

1. Calculate the empirical formulae of the following compounds with the following composition by mass.
(a) 70.0% Fe; 30.0% O

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(b) 81.8% C; 18.2% H
(c) 40.2% K; 26.9% Cr; 32.9% O
2. An organic compound (Mr = 102) has the following composition by mass:

C: 58.8%; H: 9.8% and O: 31.4%
(a) Calculate the empirical formula, and
(b) determine the molecular formula of the compound.
3 A polymer of empirical formula CH2 has a relative molecular mass of 63,000. Determine the molecular formula of the polymer.
4 24 cm3 of a mixture of methane (CH4) and ethane (C2H6) was mixed with 90 cm3 of oxygen. All volumes were measured at room
temperature and pressure. The mixture was then sparked, and the resulting mixture cooled to room conditions. When the residual
gas was passed through aqueous sodium hydroxide, a contraction of volume 32 cm3 occurs. What is the composition of the mixture
of methane and ethane?

Topic 1 Test

1. (a) Define the term isotope.
................................................................................................................................................................................ (1)
(b) Define relative atomic mass
............................................................................................................ .........................................................................
................................................................................................... ...............................................................................(2)
(c) Define relative isopotic mass.
................................................................ ................................................................................. .....................................
.......................................... ............................................................................................................. ...........................(2)
(d) The diagram shows part of the mass spectrum for chloroethane, C2H5Cl.

What ions are responsible for the peaks at m/e 66 and 64, and 29 and 28?

66: .................................................................................................................................................................................
64: .................................................................................................................................................................................
29: .................................................................................................................................................................................
28: .............................................................................................................................................................................(4)
(e) Define the term first ionisation energy.
........................................................................................... ............................................................................................
.................................................................................... .......................................................... .........................................
...................................................................................................................................................................................(2)
(f) The graph shows a plot of lg(ionisation energy) vs number of the electron removed for sodium. Explain the form of this
graph in terms of the electron structure of sodium.

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.......................................................................................................................................................................................
............................................................................................... .................................................... ....................................
.......................................................................................... ................................................... ......................................(3)
(g) Give the electronic configurations showing electrons in boxes for atoms of argon and iron then the following ions. (4 marks)
Ar........................................................................................... ........................................................................................
Fe............................................................................................................... ....................................................................

TOTAL 18 marks

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Lesson: The Mass Spectrometer

1. A Simple Mass Spectrometer

1.1 Define the following terms:
a. Nucleon number.
b. Relative Atomic Mass.
c. Isotopes.

1.2 Why must the air be pumped out of a mass spectrometer, keeping it continuously under low pressure?

1.3 Explain how electrons are emitted in the ionisation chamber.

1.4 The flow diagram below describes how a mass spectrometer works.
Use the words in the grey box to fill in the blanks labelled A to M.
(Each word may be used more than once, or not at all)

electrons protons ions magnetic singly doubly

mass spectrum electric positively detector vaporised

most lowest number mass-to-charge heated

Before entering the mass spectrometer, substances have to be (A) ________.
For liquids and solids, this means they will have to be (B) _________.
A stream of fast (C) __________ bombards the neutral atoms or molecules.
When these collide with an atom or molecule, one or more (D) _________ can be knocked off, producing (E)
_______________ charged ions.
The ions are accelerated by an (F) _________________ field.
The ions enter a variable (G) ________________ field.
The (H) __________ field deflects the ions according to their (I) _____________ ratio into a (J) ____________.
For singly charged ions, those with the (K) ________ mass will be deflected most. The heights of the peaks are proportional
to the (L) ______ of ions reaching the detector.

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Scanning through the magnetic field produces a (M) _________________________ showing the relative masses of the
ions reaching the detector.
2. Mass Spectra
2.1 The mass spectrum of a sample of lead shows 4 peaks at m/e 204, 206, 207 and 208.

a. Explain why there are 4 peaks appearing in the mass spectrum of lead.
b. The intensity of the peak at 208 is greater than that of the other peaks.

What can you deduce from this?
2.2 The mass spectrum of neon contains peaks at 20, 21 and 22
a. Assuming that only singly charged particles are produced, give the formula of the species responsible for the

peaks at 20, 21 and 22.
b. The proton number of neon is 10. Calculate the number of neutrons present in each of the isotopes of neon.

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