Isotopes
The chemical properties of an element depend on the number of protons and electrons inside the atom. So if a neutron or two is added or removed from the nucleus, then the chemical properties will not change. This means that such an atom would remain in the same place in the periodic table. For example, no matter how many neutrons we add or subtract from a nucleus with 6 protons, that element will always be called carbon and have the element symbol C (see the periodic table). Atoms which have the same number of protons (i.e. same atomic number Z), but a different number of neutrons (i.e. different N and therefore different mass number A), are called isotopes.
Interesting Fact:
In Greek, “same place” reads as ίσoςτόπoς (isos topos). This is why atoms which have the same number of protons, but different numbers of neutrons, are called isotopes. They are in the same place on the periodic table!
 Definition 1: Isotope
Isotopes of an element have the same number of protons (same Z), but a different number of neutrons (different N).
The chemical properties of the different isotopes of an element are the same, but they might vary in how stable their nucleus is. We can also write elements as E–A where the E is the element symbol and the A is the atomic mass of that element. For example Cl–35 has an atomic mass of 35 u (17 protons and 18 neutrons), while Cl–37 has an atomic mass of 37 u (17 protons and 20 neutrons).
In nature the different isotopes occur in different percentages. For example Cl–35 might make up 75% of all chlorine atoms on Earth, and Cl–37 makes up the remaining 25%. The following worked example will show you how to calculate the average atomic mass for these two isotopes:
Example 1: The relative atomic mass of an isotopic element
Question
The element chlorine has two isotopes, chlorine–35 and chlorine–37. The abundance of these isotopes when they occur naturally is 75% chlorine–35 and 25% chlorine–37. Calculate the average relative atomic mass for chlorine.
Answer
Calculate the mass contribution of chlorine–35 to the average relative atomic mass
75% of the chlorine atoms has a mass of of 35 u.
Contribution of Cl–35 = $(\frac{75}{100}\times 35)$ = 26,25 u.
Calculate the contribution of chlorine–37 to the average relative atomic mass
25% of the chlorine atoms has a mass of of 37 u.
Contribution of Cl–37 = $(\frac{25}{100}\times 37)$ = 9,25 u.
Add the two values to arrive at the average relative atomic mass of chlorine
Relative atomic mass of chlorine = 26,25 u + 9,25 u = 35,5 u.
If you look on the periodic table (see front of book), the average relative atomic mass for chlorine is 35,5 u.
Exercise 1: Isotopes
Atom A has 5 protons and 5 neutrons, and atom B has 6 protons and 5 neutrons. These atoms are:

allotropes

isotopes

isomers

atoms of different elements
atoms of different elements
For the sulphur isotopes, ${}_{16}^{32}\text{S}$ and ${}_{16}^{34}\text{S}$, give the number of:

protons

nucleons

electrons

neutrons
a) The number of protons is the subscript number (Z) so: i) 16 ii) 16
b) The number of nucleons is the superscript number (A) so: i) 32 ii) 34
c) The number of electrons in a neutral atom is the same as the number of protons and so for both i) and ii) the number of electrons is 16
d) The number of neutrons (N) is the number of nucleons  the number of protons or: A  Z = N.
For i) $AZ=3216=16$ and for ii) $AZ=3416=18$.
Which of the following are isotopes of ${}_{17}^{35}\text{Cl}$?

${}_{35}^{17}\text{Cl}$

${}_{17}^{35}\text{Cl}$

${}_{17}^{37}\text{Cl}$
 ${}_{17}^{37}\text{Cl}$
Which of the following are isotopes of U–235? (E represents an element symbol)

${}_{92}^{238}\text{E}$

${}_{90}^{238}\text{E}$

${}_{92}^{235}\text{E}$
To answer this question look at the definition of an isotope. An isotope must have the same number of protons. The neutrons can be different. So we can rule out b) as that will not be an isotope since the number of protons is different. c) is U235 so it is not an isotope. Therefore the isotope must be a).
Complete the table below:
Isotope 
Z 
A 
Protons 
Neutrons 
Electrons 
Carbon–12 

Carbon–14 

Iron–54 

Iron–56 

Iron–57 
Carbon12 means that A is 12. So we can easily fill in the A table. Z is the elements position on the periodic table, so for carbon it is 6 and for iron it is 26. The number of protons = the number of electrons = Z. And the number of neutrons = Z – A.
Isotope 
Z 
A 
Protons 
Neutrons 
Electrons 
Carbon12 
6 
12 
6 
6 
6 
Carbon14 
6 
14 
6 
8 
6 
Iron54 
26 
54 
26 
28 
26 
Iron56 
26 
56 
26 
30 
26 
Iron57  26  57  26  31  26 
If a sample contains 19,9% boron–10 and 80,1% boron–11, calculate the relative atomic mass of an atom of boron in that sample.
$\frac{19,9}{100}\times \left\{10\right\}=1,99\text{u}$ is the contribution to the relative atomic mass from boron10.
And the contribution from carbon14 is: $\frac{80,1}{100}\times 11=8,811\text{u}$.
Now we add these two together and get the relative atomic mass: $1,99\text{u}+8,811\text{u}=10.8\text{u}$.
If a sample contains 79% Mg–24, 10% Mg–25 and 11% Mg–26, calculate the relative atomic mass of an atom of magnesium in that sample.
$\frac{79}{100}\times 24=18,96u$ is the contribution to the relative atomic mass from Mg24.
The contribution from Mg25 is: $\frac{10}{100}\times 25=2,5u$
And the contribution from Mg26 is: $\frac{11}{100}\times 26=2,86u$.
Now we add these two together and get the relative atomic mass: $18.96u+2.5u+2.86=24.3u$.
For the element ${}_{92}^{234}\text{U}$ (uranium), use standard notation to describe:
 the isotope with 2 fewer neutrons
 the isotope with 4 more neutrons
 ${}_{92}^{232}\text{U}$
 ${}_{92}^{230}\text{U}$
Which of the following are isotopes of ${}_{20}^{40}\text{Ca}$?

${}_{19}^{40}\text{K}$

${}_{20}^{42}\text{Ca}$

${}_{18}^{40}\text{Ar}$
 ${}_{20}^{42}\text{Ca}$
For the sulphur isotope ${}_{16}^{33}\text{S}$, give the number of:

protons

nucleons

electrons

neutrons
 16
 35
 16
 19