How many protons, neutrons, and electrons are in an iron(III) ion with a mass number of 57 (57Fe3+)?
The correct answer and explanation is:
Correct Answer:
- Protons: 26
- Neutrons: 31
- Electrons: 23
Explanation:
An iron(III) ion with a mass number of 57 is written as ⁵⁷Fe³⁺. To determine the number of protons, neutrons, and electrons, each component must be examined separately.
Protons:
The atomic number of iron (Fe) is 26. The atomic number tells how many protons are in the nucleus of every atom of that element. Therefore, an iron atom or ion will always have 26 protons.
Neutrons:
The mass number of an atom or ion is the sum of its protons and neutrons. Given that the mass number of this isotope is 57, and we know iron has 26 protons, the number of neutrons is calculated as:
Neutrons = Mass number – Atomic number = 57 – 26 = 31
So, there are 31 neutrons in this isotope.
Electrons:
A neutral iron atom (not an ion) would have 26 electrons to balance its 26 protons. However, this is a Fe³⁺ ion, which means it has lost three electrons. Losing electrons results in a positive charge.
Electrons = 26 – 3 = 23
Thus, the Fe³⁺ ion has 23 electrons.
Understanding the composition of ions is important in chemistry because it influences the way atoms and ions interact in chemical reactions. The loss or gain of electrons affects chemical behavior, bond formation, and the physical properties of substances. In this case, Fe³⁺ is a common oxidation state of iron, especially important in biological systems (such as in hemoglobin) and in industrial chemistry. Knowing the exact number of subatomic particles helps in writing chemical formulas, balancing reactions, and predicting how ions behave in different environments.