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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.
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!
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 \(\text{35}\) \(\text{u}\) (17 protons and 18 neutrons), while Cl–37 has an atomic mass of \(\text{37}\) \(\text{u}\) (17 protons and 20 neutrons).
In nature the different isotopes occur in different percentages. For example Cl–35 might make up \(\text{75}\%\) of all chlorine atoms on Earth, and Cl–37 makes up the remaining \(\text{25}\%\). The following worked example will show you how to calculate the average atomic mass for these two isotopes:
The element chlorine has two isotopes, chlorine–35 and chlorine–37. The abundance of these isotopes when they occur naturally is \(\text{75}\%\) chlorine–35 and \(\text{25}\%\) chlorine–37. Calculate the average relative atomic mass for chlorine.
\(\text{75}\%\) of the chlorine atoms has a mass of of \(\text{35}\) \(\text{u}\).
Contribution of Cl–35 \(= (\frac{75}{100} \times 35 = \text{26,25}\text{ u})\).
\(\text{25}\%\) of the chlorine atoms has a mass of of \(\text{37}\) \(\text{u}\).
Contribution of Cl–37 \(= (\frac{25}{100} \times 37 = \text{9,25}\text{ u})\).
Relative atomic mass of chlorine = \(\text{26,25}\) \(\text{u}\) + \(\text{9,25}\) \(\text{u}\) = \(\text{35,5}\) \(\text{u}\).
If you look on the periodic table (see front of book), the average relative atomic mass for chlorine is \(\text{35,5}\) \(\text{u}\).
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
For the sulfur isotopes, \(_{16}^{32}\text{S}\) and \(_{16}^{34}\text{S}\), give the number of:
protons
nucleons
electrons
neutrons
Which of the following are isotopes of \(_{17}^{35}\text{Cl}\)?
\(_{35}^{17}\text{Cl}\)
\(_{17}^{35}\text{Cl}\)
\(_{17}^{37}\text{Cl}\)
Which of the following are isotopes of U–235? (E represents an element symbol)
\(_{92}^{238}\text{E}\)
\(_{90}^{238}\text{E}\)
\(_{92}^{235}\text{E}\)
Complete the table below:
Isotope | Z | A | Protons | Neutrons | Electrons |
Carbon–12 | |||||
Carbon–14 | |||||
Iron–54 | |||||
Iron–56 | |||||
Iron–57 |
If a sample contains \(\text{19,9}\%\) boron–10 and \(\text{80,1}\%\) boron–11, calculate the relative atomic mass of an atom of boron in that sample.
If a sample contains \(\text{79}\%\) Mg–24, \(\text{10}\%\) Mg–25 and \(\text{11}\%\) Mg–26, calculate the relative atomic mass of an atom of magnesium in that sample.
For the element \(_{92}^{234}\text{U}\) (uranium), use standard notation to describe:
Which of the following are isotopes of \(_{20}^{40}\text{Ca}\)?
\(_{19}^{40}\text{K}\)
\(_{20}^{42}\text{Ca}\)
\(_{18}^{40}\text{Ar}\)
For the sulfur isotope \(_{16}^{33}\text{S}\), give the number of:
protons
nucleons
electrons
neutrons
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