Physical Sciences Department

AS Level

OCR CHEMISTRY A

Study Guide

ATOMIC STRUCTURE & ISOTOPES

Name:………………………………………..………..

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Learning Outcomes:

After studying this unit you should be able to demonstrate and apply knowledge

& understanding of:

Atomic structure and isotopes

Isotopes as atoms of the same element with different

numbers of neutrons and different masses

Atomic structure in terms of the numbers of protons,

neutrons and electrons for atoms and ions, given the

atomic number, mass number and any ionic charge

Describe protons, neutrons and electrons in terms of relative charge and relative mass

Describe the distribution of mass and charge within an atom

Describe the contribution of protons and neutrons to the nucleus of an atom, in terms of atomic (proton) number and mass (nucleon) number

Deduce the numbers of protons, neutrons and electrons in:

- an atom given its atomic and mass number

- an ion given its atomic number, mass number and ionic charge

Relative mass

Explanation of the terms relative isotopic mass (mass

compared with 1/12th mass of carbon-12) and

relative atomic mass (weighted mean mass

compared with 1/12th mass of carbon-12), based on

the mass of a 12C atom, the standard for atomic

masses (definition required)

Calculation of the relative atomic mass of an element

from the relative abundances of its isotopes

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Key Terms:

nucleon a proton or neutron

cation a positively charged ion

anion a negatively charged ion

Definitions:

Isotopes

Atoms of the same element with a different number of

neutrons and different masses

Mass (Nucleon)

Number

Number of protons + neutrons in the nucleus of an atom

Atomic (Proton)

Number

Number of protons in the nucleus of an atom

Relative Atomic Mass

The weighted mean mass of an atom of an element when

compared to 1/12 of the mass of one atom of C-12

Relative Isotopic Mass

The mass of one atom of an isotope when compared with

1/12 of the mass of one atom of C-12

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THE ATOM

Around 460 BC the Greek philosopher Democritus was born. Democritus suggested that if a lump

of metal is cut into smaller and smaller pieces you eventually end up with a miniscule and invisible

piece of metal that cannot be cut any smaller. He called these smallest particles of a substance

‘atomos’, which means indivisible and led to the term ‘atom’.

Scientific knowledge is always evolving and since this first idea that all substances are made of

atoms, the model of the atom has been developed dramatically over the last 2000 years. The

current model of the atom studied is the (Niels) Bohr model from the early 1900s. In this model,

an atom is made up of protons, neutrons and electrons. The essential features of this model are

shown, below.

Compared with the total volume of the atom the nucleus is tiny and extremely dense. Most of the

atom is empty-space between the nucleus and the electron shells.

Example: One atom of carbon-12 has a relative mass of exactly 12 and an actual mass of

1.998 x 10−26 kilogram. Determine the actual mass of one electron; give your answer in standard

form in grammes to 3 significant figures.

1 atomic mass unit (relative mass 1) = 1.998 x 10−26 / 12 = 1.665 x 10-27 kg

Mass of 1 electron = 1.665 x 10-27 / 1836 = 9.0686 x 10-31 kg

Mass of 1 electron = 9.0686 x 10-31 x 1000 = 9.07 x 10-28 g

Particle Type Relative Charge Relative Mass Position

Proton 1+ 1 Nucleus

Neutron 0 1 Nucleus

Electron 1- 1/1836 (~1/2000) Shells

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An atom must have the same number of protons as electrons. This means that all atoms are

neutral. The number of protons tells us which element we are dealing with in the periodic table.

Every atom of the same element has the same number of protons. Electrons are held in place by

an electrostatic force from the nucleus, this is called nuclear attraction.

Isotopes

Most elements are made up of a mixture of isotopes.

Isotopes of the same element have:

different masses

the same number of protons as electrons (neutral overall)

different numbers of neutrons in the nucleus

Isotopes are represented using the following chemical notation:

All atoms of carbon have an atomic number of 6, because all carbon atoms contain 6 protons.

Carbon-12 has a mass number of 12 because it has 12 nucleons within its nucleus.

Example 1: 17

35Cl

Protons Neutrons Electrons

17 35 – 17 = 18 17

Q. Explain why isotopes of elements react in the same way in chemical reactions.

isotopes of an element have the same number of outer shell electrons

only outer shell electrons participate in chemical bonding

Exercise 1

Atomic number / proton number

Mass number

Symbol

p = 6 n = 12 – 6 = 6 e = p = 6

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RELATIVE MASSES

Q. Identify the substance used as the international standard for the measurement of isotopic and

atomic masses.

carbon-12

All the atoms of a single isotope are identical and they all have the same mass measured against

the atom carbon-12.

An atom of carbon-12 is made up of six protons, six neutrons and six electrons. It has a mass of

12 atomic mass units.

Relative Isotopic Mass

Each isotope of an element has the same number of protons but a different number of neutrons,

therefore each isotope has a different mass number.

The relative isotopic mass is the mass of an atom of an isotope when compared with 1/12 of the

mass of an atom of C-12

For any isotope, the relative isotopic mass is the same as its mass number.

The relative isotopic mass is shown in superscript before the atom, the atomic number can be

shown in subscript before the atom (but can be omitted).

Example: Bromine Isotopes

Example: Carbon Isotopes

79 35

81 35

13

12

Relative Isotopic Mass Br Br

C C

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Relative Atomic Mass Ar

Most elements contain a mixture of isotopes, each with a different mass.

The relative atomic mass is the weighted mean mass of an atom of an element when compare to

1/12 of the mass of an atom of C-12

The ‘weighted mean mass’ is an average which takes into account:

the percentage abundance of each isotope,

the relative mass of each isotope.

It is possible to calculate the relative atomic mass of an element given the relative abundances

of all of its isotopes by combining together the ‘weighted mean masses’. The relative atomic

mass for each element is shown on the Periodic Table of the Elements.

Example: Bromine Isotopes

Example 1:

Analysis of a sample of bromine in a mass spectrometer showed that it contained a mixture of

the 79Br and 81 Br isotopes in the proportions: 79Br, 55.0%; 81 Br, 45.0%. Calculate the relative

atomic mass of bromine in this sample, give your answer to 3 significant figures.

Relative atomic mass = 79 × 55.0 + 81 × 45.0 = 79.9

100 100

SAME AS

Relative atomic mass = 79 × 0.55 + 81 × 0.45 = 79.9 OR (79 × 55.0 + 81 × 45.0) = 79.9

100

Top tips: Answer has to be between 79 and 81 as the as you calculating the weighted mean

of these two masses.

Always check the question to see how many decimal places / significant figures the examiner

wants!

Exercise 2

35 79.9 Relative Atomic

Mass

Br bromine

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Mass Spectrometry

A mass spectrometer can be used to determine the relative abundance and relative mass of each

isotope of an element and hence the relative atomic mass.

There are four key processes which occur in a mass spectrometer.

Ionisation

A gaseous sample of the substance is bombarded with high energy electrons. This will result in

the removal of an electron to create a positive ion. Occasionally ions with a 2+ charge are

generated.

E(g) → E+(g) + e-

Acceleration

The ions are then accelerated by an electric field – they enter a magnetic field.

Deflection

The magnetic field deflected the ions. The degree of deflection depends on the mass of the ion.

Ions with greatest mass are deflected least. This separates the ions according to mass.

Detection

At the detector the positive ion accepts an electron and causes a current to flow.

The current is proportional to the relative abundance of the ion.

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Mass Spectra of Elements

Mass spectrometry can be used to find the relative atomic mass of a sample of an element.

The x-axis of a mass spectrum is labelled m/z, this is the mass / charge ratio. In most cases

the spectrometer is designed to select only ions with a 1+ charge therefore the x-axis is

effectively a measurement of the mass of the ions. This is calibrated against the mass of one

atom of carbon-12.

The y-axis of a mass spectrum is labelled relative abundance, this is proportional to the

number of ions of each mass that have been detected

Example 1: Mass spectrum of magnesium

What we can find out:

1) Magnesium has 3 isotopes – there are three lines on the spectrum, this means there are 3

ions with different masses – the isotopes

2) The relative abundance of each isotope of Magnesium is -

Isotope Mg-24 Mg-25 Mg-26

relative

abundance 100.00 12.66 13.92

fractional

abundance 100.00 / 126.58 12.66 / 126.58 13.92 / 126.58

Relative atomic mass = 24 x 100.00 / 126.58 + 25 x 12.66 / 126.58 + 26 x 13.92 / 126.58 = 24.3

SAME AS

Relative atomic mass = (mass of isotope1 x abundance) + (mass of isotope2 x abundance) + ……………

Total abundance of all isotopes

= (24 x 100.00) + (25 x 12.66) + (26 x 13.92) = 3078.42 = 24.3

(100 + 12.66 + 13.92) 126.58

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Example 2: A naturally occurring sample of the element boron has a relative atomic mass of 10.8 In this sample, boron exists as two isotopes, 10B and 11B a) Calculate the percentage abundance of B-10 in this naturally occurring sample of boron. Relative atomic mass = (mass of isotope1 x abundance) + (mass of isotope2 x abundance) + ……………

Total abundance of all isotopes

10.8 = (10 x %10B) + (11 x %11B) Equation 1 100

%10B + %11B = 100 because there are only two isotopes %11B = 100 - %10B

10.8 = (10 x %10B) + (11 x (100 -%10B) ) substitute for %11B in Equation 1 100

100 x 10.8 = 10 x %10B + 11 x (100 -%10B) crossmultiply

1080 = 10 x %10B + 1100 – 11 x%10B simplify and expand brackets 1080 - 1100 = 10 x +%10B – 11 x%10B transpose 10B = 20% simplify

b) Identify the species responsible for a small peak in the same spectrum at m/z value 5.5

11B2+(g)

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Fragmentation of molecular ions

When chlorine gas is injected into a mass spectrometer chlorine molecules are ionised to

form a molecular ion.

Cl-Cl(g) [Cl-Cl]+(g) + e-

Since chlorine exists as two isotopes 35Cl and 37Cl the molecular ion creates three peaks with

different m/z values.

m/z 70 = [35Cl- 35Cl]+(g)

m/z 72 = [35Cl- 37Cl]+(g)

m/z 74 = [37Cl- 37Cl]+(g)

The other peaks are formed because the chlorine molecule is liable to be broken into

fragments due to the bombardment of the high energy electrons during ionisation of the

sample. This is known as fragmentation. During fragmentation a covalent bond is broken

resulting in the formation of two fragments, one positively charged fragment and one neutral

fragment. Only the positive ion is shown as a peak.

[Cl-Cl]+(g) Cl(g) + [Cl]+(g) + e-

Task: State the species responsible for the peaks present at m/z values 35 and 37 in the mass

spectrum of Cl2

m/z 35 = [35Cl]+(g)

m/z 37 = [37Cl]+(g)

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ELEMENTAL IONS

Many atoms react by losing or gaining electrons to form ions. Ions are charged because they have

a different numbers of protons (+ve) to electrons (-ve).

Positive ions have more protons than electrons and are known as cations. Negative ions have more electrons than protons are known as anions.

Example 1: 12

24Mg2+

Protons Neutrons Electrons

12 12 10

Mg → Mg2+ + 2e- An Mg atom has lost two electrons to form a Mg2+ ion (oxidation)

(this is oxidation, Oxidation Is Loss of electrons, OIL)

Example 2: 17

35Cl–

Protons Neutrons Electrons

17 18 18

½ Cl2 + e- → Cl- Each Cl atom has gained one electron to form a Cl- ion

(this is reduction, Reduction Is Gain of electrons RIG)

Exercise 3