ATOMIC MASS: DEFINITION, TYPES, HOW TO CALCULATE IT, EXAMPLES - CHEMISTRY - 2025

The atomic mass is the quantity of material present in an atom, which can be expressed in ordinary physical units or in units of atomic mass (uma ou). An atom is empty in almost all its structure; electrons that are diffused in regions called orbitals, where there is a certain probability of finding them, and their nucleus.

In the nucleus of the atom are the protons and neutrons; the former with positive charges, while the latter with neutral charge. These two subatomic particles have a mass much greater than that of the electron; therefore, the mass of an atom is governed by its nucleus and not by the vacuum nor by electrons.

The main subatomic particles and the mass of the nucleus. Source: Gabriel Bolívar.

The mass of an electron is approximately 9.1 · 10 ^-31 kg, while that of the proton 1.67 · 10 ^-27 kg, the mass ratio being 1,800; that is, a proton “weighs” 1,800 times more than an electron. Similarly the same happens with the masses of the neutron and electron. That is why the mass contribution of the electron for ordinary purposes is considered negligible.

Because of this, it is usually assumed that the mass of the atom, or atomic mass, depends only on the mass of the nucleus; which in turn, consists of the sum of the matter of the neutrons and protons. Two concepts emerge from this reasoning: mass number and atomic mass, both closely related.

With so much "void" in atoms, and given that their mass is almost entirely a function of the nucleus, it is to be expected that the latter is extraordinarily dense.

If we remove said void from any body or object, its dimensions would contract drastically. Also, if we could build a small object based on atomic nuclei (without electrons), then it would have a mass of millions of tons.

On the other hand, atomic masses help to distinguish different atoms of the same element; These are the isotopes. As there are more abundant isotopes than others, an average of the masses of the atoms must be estimated for a given element; average that can vary from planet to planet, or from one space region to another.

Definition and concept

By definition, the atomic mass is the sum of the masses of its protons and neutrons expressed with uma or u. The resulting number (also sometimes called a mass number) is placed dimensionless in the upper left corner in the notation used for nuclides. For example, for element ¹⁵ X its atomic mass is 15uma or 15u.

The atomic mass cannot tell much about the true identity of this element X. Instead, the atomic number is used, which corresponds to the protons in the nucleus of X. If this number is 7, then the difference (15-7) will be equal to 8; that is, X has 7 protons and 8 neutrons, the sum of which is 15.

Returning to the image, the nucleus has 5 neutrons and 4 protons, so its mass number is 9; and in turn 9 amu is the mass of its atom. By having 4 protons, and consulting the periodic table, it can be seen that this nucleus corresponds to that of the element beryllium, Be (or ⁹ Be).

Atomic mass unit

Atoms are too small to be able to measure their masses by conventional methods or ordinary balances. It is for this reason that the uma, uo Da (color blind) was invented. These units devised for atoms allow you to have an idea of ​​how massive the atoms of an element are in relation to each other.

But what exactly does an uma represent? There must be a reference to establish mass relations. For this, the ¹² C atom was used as a reference, which is the most abundant and stable isotope for carbon. Having 6 protons (its atomic number Z), and 6 neutrons, its atomic mass is therefore 12.

The assumption is made that protons and neutrons have the same masses, so that each contributes 1 amu. The atomic mass unit is then defined as one twelfth (1/12) of the mass of a carbon-12 atom; this is the mass of a proton or neutron.

Equivalence in grams

And now the following question arises: how many grams does 1 amu equal? Since at first there were not sufficiently advanced techniques to measure it, chemists had to settle for expressing all masses with amu; however, this was an advantage and not a disadvantage.

Why? Because subatomic particles are so small, their mass, expressed in grams, must be just as small. In fact, 1 amu equals 1.6605 · 10 ^-24 grams. Furthermore, with the use of the concept of mole, it was not a problem to work the masses of the elements and their isotopes with amu knowing that such units could be modified to g / mol.

For example, going back to ¹⁵ X and ⁹ Be, we have that their atomic masses are 15 amu and 9 amu, respectively. As these units are so small and do not directly tell how much matter one must "weigh" to manipulate them, they are transformed into their respective molar masses: 15 g / mol and 9 g / mol (introducing the concepts of moles and Avogadro's number).

Average atomic mass

Not all atoms of the same element have the same mass. This means that they must have more subatomic particles in the nucleus. Being the same element, the atomic number or number of protons must remain constant; therefore, there is only variation in the amounts of neutrons they possess.

This is how it appears from the definition of isotopes: atoms of the same element but with different atomic masses. For example, beryllium almost entirely consists of the isotope ⁹ Be, with trace amounts of ¹⁰ Be. However, this example is not very helpful in understanding the concept of average atomic mass; we need one with more isotopes.

Example

Suppose that element ⁸⁸ J exists, this being the main isotope of J with an abundance of 60%. J also has two other isotopes: ⁸⁶ J, with an abundance of 20%, and ⁹⁰ J, with an abundance also of 20%. This means that of 100 J atoms that we collect on Earth, 60 of them are ⁸⁸ J, and the remaining 40 a mixture of ⁸⁶ J and ⁹⁰ J.

Each of the three isotopes of J has its own atomic mass; that is, their sum of neutrons and protons. However, these masses must be averaged in order to have an atomic mass for J on hand; here on Earth, as there may be other regions of the Universe where the abundance of ⁸⁶ J is 56% and not 60%.

To calculate the average atomic mass of J, the weighted average of the masses of its isotopes must be obtained; that is, taking into account the percentage of abundance for each of them. Thus we have:

Average Mass (J) = (86 amu) (0.60) + (88 amu) (0.20) + (90 amu) (0.20)

= 87.2 amu

That is, the average atomic mass (also known as the atomic weight) of J is 87.2 amu. Meanwhile, its molar mass is 87.2 g / mol. Note that 87.2 is closer to 88 than to 86, and it is also distant from 90.

Absolute atomic mass

The absolute atomic mass is the atomic mass expressed in grams. Starting from the example of the hypothetical element J, we can calculate its absolute atomic mass (that of the average) knowing that each amu is equivalent to 1.6605 · 10 ^-24 grams:

Absolute atomic mass (J) = 87.2 amu * (1.6605 · 10 ^-24 g / amu)

= 1.447956 · 10 ^-22 g / J atom

This means that on average the J atoms have an absolute mass of 1.447956 · 10 ^-22 g.

Relative atomic mass

The relative atomic mass numerically is identical to the average atomic mass for a given element; However, unlike the second, the first lacks unity. Therefore, it is dimensionless. For example, the average atomic mass of beryllium is 9.012182 u; while its relative atomic mass is simply 9.012182.

That is why sometimes these concepts are often misinterpreted as synonyms, since they are very similar and the differences between them are subtle. But what are these masses relative to? Relative to one twelfth of the mass of ¹² C.

Thus, an element with a relative atomic mass of 77 means that it has a mass 77 times greater than 1 / ^12th of ¹² C.

Those who have looked at the elements in the periodic table will see that their masses are relatively expressed. They do not have units of amu, and is interpreted as: iron has an atomic mass of 55,846, which means that it is 55,846 times more mass than the mass of 1/12 part of ¹² C, and that it can also be expressed as 55,846 amu or 55.846 g / mol.

How to calculate atomic mass

Mathematically, an example was given of how to calculate it with the example of element J. In general terms, we must apply the weighted average formula, which would be:

P = Σ (isotope atomic mass) (abundance in decimals)

That is, having the atomic masses (neutrons + protons) of each isotope (normally natural) for a given element, as well as their respective terrestrial abundances (or whatever the region considered), then said weighted average can be calculated.

And why not just the arithmetic average? For example, the average atomic mass of J is 87.2 amu. If we calculate this mass again but arithmetically we will have:

Average mass (J) = (88 amu + 86 amu + 90 amu) / 3

= 88 amu

Note that there is an important difference between 88 and 87.2. This is because the arithmetic average assumes that the abundance of all isotopes is the same; Since there are three isotopes of J, each should have an abundance of 100/3 (33.33%). But this is not the case in reality: there are much more abundant isotopes than others.

That is why the weighted average is calculated, since it does take into account how abundant one isotope is with respect to another.

Examples

Carbon

To calculate the average atomic mass of carbon we need its natural isotopes with their respective abundances. In the case of carbon, these are: ¹² C (98.89%) and ¹³ C (1.11%). Their relative atomic masses are 12 and 13, respectively, which in turn are equal to 12 amu and 13 amu. Solving:

Average atomic mass (C) = (12 amu) (0.9889) + (13 amu) (0.0111)

= 12.0111 amu

Therefore, the mass of a carbon atom is on average 12.01 amu. As there are trace amounts of ¹⁴ C, it has almost no influence on this average.

Sodium

All terrestrial sodium atoms consist of the ²³ Na isotope, so its abundance is 100%. That is why in ordinary calculations its mass can be assumed to be simply 23 amu or 23 g / mol. However, its exact mass is 22.98976928 amu.

Oxygen

The three oxygen isotopes with their respective abundances are: ¹⁶ O (99.762%), ¹⁷ O (0.038%) and ¹⁸ O (0.2%). We have everything to calculate its average atomic mass:

Average atomic mass (O) = (16 amu) (0.99762) + (17 amu) (0.00038) + (18 amu) (0.002)

= 16.00438 amu

Although its reported exact mass is actually 15.9994 amu.

Nitrogen

Repeating the same steps with oxygen we have: ¹⁴ N (99.634%) and ¹⁵ N (0.366%). So:

Average atomic mass (N) = (14 amu) (0.99634) + (15 amu) (0.00366)

= 14.00366 amu

Note that the reported mass for nitrogen is 14.0067 amu, slightly higher than what we calculated.

Chlorine

The isotopes of chlorine with their respective abundances are: ³⁵ Cl (75.77%) and ³⁷ Cl (24.23%). Calculating its average atomic mass we have:

Average atomic mass (Cl) = (35 amu) (0.7577) + (37 amu) (0.2423)

= 35.4846 amu

Very similar to the one reported (35,453 amu).

Dysprosium

And finally, the average mass of an element with many natural isotopes will be calculated: dysprosium. These and their respective abundances are: ¹⁵⁶ Dy (0.06%), ¹⁵⁸ Dy (0.10%), ¹⁶⁰ Dy (2.34%), ¹⁶¹ Dy (18.91%), ¹⁶² Dy (25.51 %), ¹⁶³ Dy (24.90%) and ¹⁶⁴ Dy (28.18%).

We proceed as in the previous examples to calculate the atomic mass of this metal:

Average atomic mass (Dy) = (156 amu) (0.0006%) + (158 amu) (0.0010) + (160 amu) (0.0234) + (161 amu) (0.1891) + (162 amu) (0.2551) + (163 amu) (0.2490) + (164 amu) (0.2818)

= 162.5691 amu

The reported mass is 162,500 amu. Note that this average is between 162 and 163, since the ¹⁵⁶ Dy, ¹⁵⁸ Dy and ¹⁶⁰ Dy isotopes are few abundant; while those that predominate are ¹⁶² Dy, ¹⁶³ Dy and ¹⁶⁴ Dy.

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