3.E: Atoms and the Periodic Table (Exercises)
- Page ID
- 85149
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Concept Review Exercises
- Why is the atomic number so important to the identity of an atom?
- What is the relationship between the number of protons and the number of electrons in an atom?
- How do isotopes of an element differ from each other?
- What is the mass number of an element?
- Calculate the subatomic particles for
- Zn-67,
- radium,
- the aluminum cation,
- Mg-23, and
- the phosphorus anion.
- Which element(s) has/have 32 neutrons: germanium, cobalt, or Mn-57?
- Which natural isotope of platinum is most abundant?
- Which element has 19 electrons?
- Write A/Z format for the species in question #5.
- Calculate the weighted average of nitrogen. N-14 has an exact mass of 14.003074u and is 99.63% abundant. N-15 has a mass of 15.000108u.
Solutions
- The atomic number defines the identity of an element.
- In an electrically neutral atom, the number of protons equals the number of electrons.
- Isotopes have different numbers of neutrons in their nuclei.
- The mass number is the sum of the numbers of protons and neutrons in the nucleus of an atom.
- For ion charges, refer to the table for assistance:

| Species | Protons | Electrons | Neutrons |
|---|---|---|---|
| Zn-67 | 30 | 30 | 37 (is isotope) |
| Radium | 88 | 88 | 138 |
| Aluminum cation | 13 | 10 (look up charge, 3+) | 14 |
| Mg-23 | 12 | 12 | 11 (is isotope) |
| Phosphorus anion | 15 | 18 (look up charge, 3-) | 16 |
- Both cobalt and Mn-57 have 32 neutrons. Germanium has 41 neutrons.
- Pt-195 is closest to the weighted average (found on periodic table) and would be the most abundant.
- If an ion charge is not given, locate the electrons of the element by looking to the atomic number. Potassium is the only atom of these three that would have 19 electrons.
- \(\ce{^{67}_{30}Zn}\), \(\ce{^{226}_{88}Ra}\), \(\ce{^{27}_{13}Al^{3+}}\), \(\ce{^{23}_{12}Mg}\) and \(\ce{^{31}_{15}P^{3-}}\)
- For this problem, you need to assume that all percentages add up to 100%. Since the problem provided 99.63%, you will need to subtract this given percentage from 100.00%. \(\begin{align} \text{Atomic mass} &= \left( 0..9963 \right) \left( 14.003074 \right) + \left( 0.0037 \right) \left( 15.000108 \right) = 14.01 \: \text{amu} \end{align}\)


