O calculation of the number of atomic particles is used to indicate the amount of protons (in the nucleus), electrons (in the electrosphere) and neutrons (in the nucleus) present in an atom or any ion. To do this, it is essential to know some characteristics of atoms:
1- Atomic number (Z)
It is a mathematical code, represented by the capital letter Z, positioned on the lower left side of the initials of an atom:
ZX
It indicates the number of protons (p) in the nucleus and the number of electrons (e) in the electrosphere of an atom. So, in general terms:
Z = p = e
2- Mass number (A)
It is a mathematical code that corresponds to the sum of the number of protons (p) and neutrons (n), both present in the nucleus of any atom. The equation representing the mass number is given by:
A = p + n
Since the number of protons equals the atomic number, we can write the equation to calculate the mass number as follows:
A = Z + n
If we know the mass number and atomic number of an atom, we can determine the number of neutrons as follows:
n = A – Z
3- Ions
Atoms that lose or gain electrons. They present a positive or negative sign positioned in the upper right part of their representation, as in the following model:
X+ or X-
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4- Atomic Similarities
a) Isotopes
Atoms that have the same atomic number but different mass numbers. Example:
7X14 and 7Y16
Atoms X and Y have the same atomic number (to the left of the initials), that is, equal to 7. Atom X has a mass number (to the right of the initials) equal to 14, and atom Y has the same mass number to 16.
b) isobars
Atoms that have the same mass number but different atomic numbers. Example:
15X31 and 13Y31
Atoms X and Y have a mass number (to the right of the acronym) equal to 31. Atom X has an atomic number equal to 15, and atom Y has an atomic number equal to 13.
c) isotones
Atoms that have different mass numbers and atomic numbers, but the same number of neutrons.
d) Isoelectronic
Atoms that have the same number of electrons. Example:
12X+2 and 7Y-3
The atom X has an atomic number equal to 12 and is a cation (with a positive charge of +2), so it loses two electrons, thus having 10 electrons in its electrosphere. Atom Y, on the other hand, has an atomic number equal to 7 and is an anion (with a negative charge of -3), so it gains three electrons, thus having 10 electrons in its electrosphere.
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Mind Map: Atomic Particles
* To download the mind map in PDF, click here!
Examples of calculating the number of atomic particles
Example 1: Determine the number of protons, neutrons, and electrons in the 14X29 atom.
The following values were given for the X atom:
The number of protons is always equal to the atomic number, so atom X has 14 protons.
As the atom X is not an ion, therefore, the number of electrons is equal to the number of protons, that is, 14.
The number of neutrons is determined using the mass number and proton number in the following formula:
A = p + n
29 = 14 + n
29 – 14 =n
n = 15
Example 2: Determine the number of protons, neutrons, and electrons in the X+3 ion, knowing that its mass number and atomic number are 51 and 23, respectively.
The following values were given for the X ion:
The number of protons is always equal to the atomic number, so atom X has 23 protons.
The X ion is positive (+3), so it is a cation that has lost three electrons. So its number of electrons is 20.
NOTE: The reduction or increase in the number of electrons always occurs in relation to the atomic number.
The number of neutrons is determined using the mass number and proton number in the following formula:
A = p + n
51 = 23 + n
51 – 23 =n
n = 28
Example 3: An atom W has an atomic number and a mass equal to, respectively, 29 and 57, being isobarous with an atom Y, which has an atomic number equal to 30, which is isobarous with an atom B, whose mass number is 65. With Using this information, determine the number of protons, neutrons, and electrons in atom B.
Data provided by the exercise:
atomic number (bottom left) = 29
mass number (upper right) = 57
Isobar of Y, that is, the mass of Y is also 57.
atomic number = 30
mass number = 57
With these two values, we must determine its number of neutrons because it is isotone of element B:
A = Z + n
57 = 30 + n
57 – 30 = n
n = 27
mass number = 65
number of neutrons = 27
With these data, we must determine its atomic number, because with this we will be determining its number of protons and its number of electrons (since it is not an ion):
A = Z + n
65 = Z+27
65 – 27 = Z
Z = 38
Therefore, atom B has 38 protons, 38 electrons and 27 neutrons.
* Mind Map by Victor Ricardo Ferreira
Chemistry teacher
By Me. Diogo Lopes Dias