The atomic mass is the amount of matter present in an atom, which can be expressed in ordinary physical units or in atomic mass units (uma or u). An atom is empty in most of its structure; electrons that are diffused in regions called orbitals, where there is a certain probability of finding them, and their nucleus.
Protons and neutrons are found in the nucleus of the atom; the former with positive charges, while the latter with neutral charges. 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 the electrons.
The mass of an electron is approximately 9.110-31 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 often 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..
There being so much "vacuum" in the atoms, and since 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 isotopes more abundant 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 spatial region to another.
Article index
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 the element fifteenX 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, whose sum 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 9Be).
The 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, u or 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 amu represent? There must be a reference that allows mass relations to be established. For this, the atom was used as a reference 12C, 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.
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. In addition, 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 fifteenX and 9Be, 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).
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 that 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 9Be, with trace amounts of 10Be. However, this example is not very helpful in understanding the concept of average atomic mass; we need one with more isotopes.
Suppose the element exists 88J, this being the main isotope of J with an abundance of 60%. J also has two other isotopes: 86J, with an abundance of 20%, and 90J, with an abundance also of 20%. This means that of 100 J atoms that we collect on Earth, 60 of them are 88J, and the remaining 40 a mixture of 86J and 90J.
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 86J 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.
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 / am)
= 1,447956 10-22 g / atom J
This means that on average the J atoms have an absolute mass of 1.447956 · 10-22 g.
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 these concepts are sometimes 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 the 12C.
Thus, an element with a relative atomic mass of 77 means that it has a mass 77 times greater than 1/12 of the 12C.
Those who have looked at the elements in the periodic table will see that their masses are relatively expressed. They have no amu units, and it 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 of the 12C, and which can also be expressed as 55.846 amu or 55.846 g / mol.
Mathematically, an example was given of how to calculate it with the example of element J. In general terms, the weighted average formula must be applied, which would be:
P = Σ (isotope atomic mass) (abundance in decimals)
In other words, 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..
To calculate the average atomic mass of carbon we need its natural isotopes with their respective abundances. In the case of carbon these are: 12C (98.89%) and 13C (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 14C, has almost no influence on this average.
All terrestrial sodium atoms consist of the isotope 2. 3Na, 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.
The three isotopes of oxygen with their respective abundances are: 16O (99.762%), 17O (0.038%) and 18O (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.
Repeating the same steps with oxygen we have: 14N (99.634%) and fifteenN (0.366%). Then:
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, a little higher than the one we calculated.
The isotopes of chlorine with their respective abundances are: 35Cl (75.77%) and 37Cl (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).
And finally, the average mass of an element with many natural isotopes will be calculated: dysprosium. These and their respective abundances are: 156Dy (0.06%), 158Dy (0.10%), 160Dy (2.34%), 161Dy (18.91%), 162Dy (25.51%), 163Dy (24.90%) and 164Dy (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 isotopes 156Dy, 158Dy and 160Dy are few abundant; while those that predominate are 162Dy, 163Dy and 164Dy.
Yet No Comments