Atomic mass is the sum of the masses of all protons, neutrons and electrons that make up an atom or molecule. Compared to protons and neutrons, the mass of electrons is very small, so it is not taken into account in calculations. Although this is incorrect from a formal point of view, it is often this term used to indicate the average atomic mass of all isotopes of an element. It's actually relative atomic mass, also called atomic weight element. Atomic weight is the average of the atomic masses of all isotopes of an element found in nature. Chemists must differentiate between these two types of atomic mass when doing their work—an incorrect atomic mass may, for example, result in an incorrect result for the yield of a reaction.
Steps
Finding atomic mass from the periodic table of elements
- The atomic mass unit characterizes the mass one mole of this element in grams. This value is very useful in practical calculations, since it can be used to easily convert the mass given quantity atoms or molecules of a given substance in moles, and vice versa.
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Find the atomic mass in periodic table Mendeleev. In the majority standard tables Mendeleev contains the atomic masses (atomic weights) of each element. Typically, they are listed as a number at the bottom of the element cell, below the letters representing the chemical element. Usually this is not a whole number, but a decimal fraction.
Remember that the periodic table gives the average atomic masses of elements. As noted earlier, the relative atomic masses given for each element in the periodic table are the average of the masses of all isotopes of the atom. This average value is valuable for many practical purposes: for example, it is used in calculating the molar mass of molecules consisting of several atoms. However, when you are dealing with individual atoms, this value is usually not enough.
- Since the average atomic mass is an average of several isotopes, the value shown in the periodic table is not accurate the value of the atomic mass of any single atom.
- The atomic masses of individual atoms must be calculated taking into account the exact number of protons and neutrons in a single atom.
Calculation of the atomic mass of an individual atom
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Find the atomic number of a given element or its isotope. The atomic number is the number of protons in the atoms of an element and never changes. For example, all hydrogen atoms, and only they have one proton. The atomic number of sodium is 11 because it has eleven protons in its nucleus, while the atomic number of oxygen is eight because it has eight protons in its nucleus. You can find the atomic number of any element in the periodic table - in almost all its standard versions, this number is indicated above letter designation chemical element. The atomic number is always a positive integer.
- Suppose we are interested in the carbon atom. Carbon atoms always have six protons, so we know that its atomic number is 6. In addition, we see that in the periodic table, at the top of the cell with carbon (C) is the number "6", indicating that the atomic carbon number is six.
- Note that the atomic number of an element is not uniquely related to its relative atomic mass in the periodic table. Although, especially for the elements at the top of the table, it may appear that the element's atomic mass is twice its atomic number, it is never calculated by multiplying the atomic number by two.
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Find the number of neutrons in the nucleus. The number of neutrons may vary for different atoms the same element. When two atoms of the same element with the same number of protons have different quantities neutrons, they are different isotopes of this element. Unlike the number of protons, which never changes, the number of neutrons in the atoms of a given element can often change, so the average atomic mass of an element is written as a decimal fraction with a value lying between two adjacent whole numbers.
Add up the number of protons and neutrons. This will be the atomic mass of this atom. Ignore the number of electrons that surround the nucleus - their total mass is extremely small, so they have virtually no effect on your calculations.
Calculating the relative atomic mass (atomic weight) of an element
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Determine which isotopes are contained in the sample. Chemists often determine the ratio of isotopes in specific sample using a special instrument called a mass spectrometer. However, in training, this data will be provided to you in assignments, tests, and so on in the form of values taken from the scientific literature.
- In our case, let's say that we are dealing with two isotopes: carbon-12 and carbon-13.
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Determine the relative abundance of each isotope in the sample. For each element, different isotopes occur in different ratios. These ratios are almost always expressed as percentages. Some isotopes are very common, while others are very rare—sometimes so rare that they are difficult to detect. These values can be determined using mass spectrometry or found in a reference book.
- Let's assume that the concentration of carbon-12 is 99% and carbon-13 is 1%. Other carbon isotopes really exist, but in quantities so small that in this case they can be neglected.
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Multiply the atomic mass of each isotope by its concentration in the sample. Multiply the atomic mass of each isotope by its percentage abundance (expressed as a decimal). To convert interest to decimal, simply divide them by 100. The resulting concentrations should always add up to 1.
- Our sample contains carbon-12 and carbon-13. If carbon-12 makes up 99% of the sample and carbon-13 makes up 1%, then multiply 12 (the atomic mass of carbon-12) by 0.99 and 13 (the atomic mass of carbon-13) by 0.01.
- The reference books give percentages based on the known quantities of all isotopes of a particular element. Most chemistry textbooks contain this information in a table at the end of the book. For the sample being studied, the relative concentrations of isotopes can also be determined using a mass spectrometer.
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Add up the results. Sum up the multiplication results you got in the previous step. As a result of this operation, you will find the relative atomic mass of your element - the average value of the atomic masses of the isotopes of the element in question. When considering an element as a whole, rather than a specific isotope of a given element, this is the value used.
- In our example, 12 x 0.99 = 11.88 for carbon-12, and 13 x 0.01 = 0.13 for carbon-13. The relative atomic mass in our case is 11.88 + 0.13 = 12,01 .
- Some isotopes are less stable than others: they decay into atoms of elements with fewer protons and neutrons in the nucleus, releasing particles that make up atomic nucleus. Such isotopes are called radioactive.
Learn how atomic mass is written. Atomic mass, that is, the mass of a given atom or molecule, can be expressed in standard SI units - grams, kilograms, and so on. However, because atomic masses expressed in these units are extremely small, they are often written in unified atomic mass units, or amu for short. – atomic mass units. One atomic mass unit is equal to 1/12 the mass of the standard isotope carbon-12.
From the lesson materials you will learn that atoms of some chemical elements differ from atoms of other chemical elements in mass. The teacher will tell you how chemists measured the mass of atoms that are so small that you cannot see them even with the help of electron microscope.
Topic: Initial chemical ideas
Lesson: Relative Atomic Mass of Chemical Elements
At the beginning of the 19th century. (150 years after the work of Robert Boyle), the English scientist John Dalton proposed a method for determining the mass of atoms of chemical elements. Let's consider the essence of this method.
Dalton proposed a model according to which a molecule complex substance includes only one atom of different chemical elements. For example, he believed that a water molecule consists of 1 hydrogen atom and 1 oxygen atom. According to Dalton, simple substances also contain only one atom of a chemical element. Those. an oxygen molecule must consist of one oxygen atom.
And then, knowing mass fractions elements in a substance, it is easy to determine how many times the mass of an atom of one element differs from the mass of an atom of another element. Thus, Dalton believed that the mass fraction of an element in a substance is determined by the mass of its atom.
It is known that the mass fraction of magnesium in magnesium oxide is 60%, and the mass fraction of oxygen is 40%. Following the path of Dalton's reasoning, we can say that the mass of a magnesium atom more mass oxygen atom by 1.5 times (60/40=1.5):
The scientist noticed that the mass of the hydrogen atom is the smallest, because There is no complex substance in which the mass fraction of hydrogen would be greater than the mass fraction of another element. Therefore, he proposed to compare the masses of atoms of elements with the mass of a hydrogen atom. And in this way he calculated the first values of the relative (relative to the hydrogen atom) atomic masses of chemical elements.
The atomic mass of hydrogen was taken as unity. And the meaning relative mass sulfur turned out to be equal to 17. But all the values obtained were either approximate or incorrect, because the experimental technique of that time was far from perfect and Dalton’s assumption about the composition of the substance was incorrect.
In 1807 - 1817 Swedish chemist Jons Jakob Berzelius conducted extensive research to clarify the relative atomic masses of elements. He managed to obtain results close to modern ones.
Significantly later work Berzelius began to compare the masses of atoms of chemical elements with 1/12 of the mass of a carbon atom (Fig. 2).
Rice. 1. Model for calculating the relative atomic mass of a chemical element
The relative atomic mass of a chemical element shows how many times the mass of an atom of a chemical element is greater than 1/12 the mass of a carbon atom.
Relative atomic mass is denoted by A r; it has no units of measurement, since it shows the ratio of the masses of atoms.
For example: A r (S) = 32, i.e. a sulfur atom is 32 times heavier than 1/12 the mass of a carbon atom.
Absolute mass 1/12 of a carbon atom is a reference unit, the value of which is calculated with high accuracy and is 1.66 * 10 -24 g or 1.66 * 10 -27 kg. This reference mass is called atomic unit masses (a.e.m.).
There is no need to memorize the values of the relative atomic masses of chemical elements; they are given in any textbook or reference book on chemistry, as well as in the periodic table of D.I. Mendeleev.
When calculating, the values of relative atomic masses are usually rounded to whole numbers.
The exception is the relative atomic mass of chlorine - for chlorine a value of 35.5 is used.
1. Collection of problems and exercises in chemistry: 8th grade: to the textbook by P.A. Orzhekovsky and others. “Chemistry, 8th grade” / P.A. Orzhekovsky, N.A. Titov, F.F. Hegel. – M.: AST: Astrel, 2006.
2. Ushakova O.V. Chemistry workbook: 8th grade: to the textbook by P.A. Orzhekovsky and others. “Chemistry. 8th grade” / O.V. Ushakova, P.I. Bespalov, P.A. Orzhekovsky; under. ed. prof. P.A. Orzhekovsky - M.: AST: Astrel: Profizdat, 2006. (p. 24-25)
3. Chemistry: 8th grade: textbook. for general education institutions / P.A. Orzhekovsky, L.M. Meshcheryakova, L.S. Pontak. M.: AST: Astrel, 2005.(§10)
4. Chemistry: inorg. chemistry: textbook. for 8th grade. general education institutions / G.E. Rudzitis, Fyu Feldman. – M.: Education, OJSC “Moscow Textbooks”, 2009. (§§8,9)
5. Encyclopedia for children. Volume 17. Chemistry / Chapter. ed.V.A. Volodin, Ved. scientific ed. I. Leenson. – M.: Avanta+, 2003.
Additional web resources
1. Unified collection of digital educational resources ().
2. Electronic version of the journal “Chemistry and Life” ().
Homework
p.24-25 No. 1-7 from Workbook in chemistry: 8th grade: to the textbook by P.A. Orzhekovsky and others. “Chemistry. 8th grade” / O.V. Ushakova, P.I. Bespalov, P.A. Orzhekovsky; under. ed. prof. P.A. Orzhekovsky - M.: AST: Astrel: Profizdat, 2006.
Determination of the number of elementary particles in atoms of isotopes and isobars
Example 1. Determine the number of protons, neutrons and electrons for the isotopes 82 207 X and 82 212 X; the isobars have 81,210 Y and 84,210 Z. Name these elements.
Solution. The 82nd element of the periodic table is lead (X = Pb), the 81st element is thallium (Y = Tl), the 84th element is polonium (Z = Po). the number of electrons and protons corresponds to the atomic number of the element. The number of neutrons in the nucleus is calculated by subtracting the number of protons in the nucleus (element number) from the mass number of elements. As a result we get:
Element symbol |
Number of electrons |
Number of protons in the nucleus |
Number of neutrons in the nucleus |
Calculation of the relative atomic mass of elements from their natural isotopic composition
Example 2. Mole fractions isotopes 24 Mg, 25 Mg and 26 Mg are 79.7, respectively; 9.8 and 10.5%. Calculate the average relative atomic mass of magnesium.
Solution. The average relative atomic mass of magnesium is calculated by summing the products of the mass fractions of each isotope by its mass number:
M = 0.797 · 24 + 0,098· 25 + 0,105· 26 = 19,128 + 2,450 + 2,730 = 24,308.
The resulting value is close to the value of the atomic mass of magnesium given in the periodic table of elements (24.305).
Drawing up nuclear reaction equations
Example 3. Identify the radioactive decay products X, Y and Z:
88 226 Ra -(α-decay) X -(α-decay) Y -(β-decay) Z.
Solution. During the α-decay of 88 226 Ra, its mass number A decreases by four units and becomes equal to A X = 226-4 = 222. In this case, the charge of the nucleus decreases by two units and turns out to be equal to Z X = 88-2 = 86. Thus, the first decay leads to the formation of the radon isotope 86 222 Rn. The α-decay product of radon is determined in a similar way: A Y = 222-4 = 218, Z Y = 86-2 = 84. As a result of the second decay, we obtain the polonium isotope 84 218 Po; The β-decay of polonium does not change the mass number of the element, but increases the charge of its nucleus by one: Z Z = 84+1 = 85. The end product of this chain of decays will be the element with number 85, i.e. astatine (85,218 At). The final scheme of nuclear transformations will look like:
88 226 Ra -(α-decay) 86 222 Rn -(α-decay) 84 218 Po -(β-decay) 85 218 At.
Determination of the maximum number of electrons on electronic layers and electron shells
Example 4: Calculate the maximum number of electrons in the fifth electronic layer and on the f-shell.
Solution. The maximum possible number of electrons in the electron layer with number n is N n = 2n 2 . For the fifth electron layer we get:
Nn=5=2 · 5 2 = 50.
The maximum possible number of electrons per electron shell with given value l equals N l = 2(2l+ 1). For f-shell l= 3. As a result we get:
N l=3 = 2(2· 3 + 1) = 14.
Determining the values of quantum numbers for electrons in different states
Example 5. Determine the values of the main and side quantum numbers for the following states of electrons: 3d, 4s and 5p.
Solution. The value of the principal quantum number for various conditions electrons in atoms are denoted Arabic numeral, and the value of the side quantum number– corresponding lowercase Latin letter. As a result, we obtain for the electron states under consideration.
Problems on isotopes
Level A
1. Calculate the isotopic composition (in%) of hydrogen (average relative atomic massA r = 1.008) and lithium (A r = 6.9), assuming that each element consists of only two isotopes whose relative atomic masses differ by one.
Answer. Hydrogen: 1 H – 99.2% and 2 H – 0.8%; lithium: 6 Li – 10% and 7 Li – 90%.
2. The relative atomic mass of natural hydrogen is 1.00797. This hydrogen is a mixture of protium isotopes ( A r = 1.00782) and deuterium (A r = 2.0141). What is the percentage of deuterium in natural hydrogen?
Answer. 0,015%.
3. Among the given symbols of elements, indicate isotopes and isobars:
Answer. Isotopes have the same chemical symbols, and isobars have the same atomic masses.
4. Natural lithium (A r = 6.9) consists of isotopes with mass numbers 6 and 7. What percentage of the first isotopedoes it contain?
Answer. 10%.
5. The mass of an atom of the magnesium isotope is 4.15 10 –23 d. Determine the number of neutrons that the nucleus of this atom contains.
Answer. 13.
6. Copper has two isotopes with mass numbers 63 and 65. Mass fraction their content in natural copper is 73% and 27%, respectively. Based on these data, calculate the average relative atomic mass of natural copper.
Answer. 63,54.
7. The average relative atomic mass of natural chlorine is 35.45. Calculate the mass fractions of two of its isotopes having mass numbers 35 and 37.
Answer. 77.5% and 22.5%.
8. Determine the relative atomic mass of boron if the mass fractions of its isotopes are known ( 10 B) = 19.6% and( 11 B) = 80.4%.
Answer. 10,804.
9. Lithium consists of two natural isotopes with mass numbers 6 ( 1 = 7.52%) and 7 ( 2 = 92.48%). Calculate the relative atomic mass of lithium.
Answer. 6,9248.
10. Calculate the relative atomic mass of cobalt if it is known that two of its isotopes exist in nature: with mass numbers 57 ( 1 = 0.17%) and 59 ( 2 = 99,83%).
Answer. 58,9966.
11. The relative atomic mass of boron is 10.811. Determine the percentage of isotopes with mass numbers 10 and 11 in natural boron.
Answer. 18.9% and 81.1%.
12. Gallium has two natural isotope with mass numbers 69 and 71. What is the quantitative relationship between the numbers of atoms of these isotopes if the relative atomic mass of the element is 69.72.
Answer. 1,78:1.
13. Natural bromine has two isotopes with mass numbers 79 and 81. The relative atomic mass of bromine is 79.904. Determine the mass fraction of each isotope in natural bromine.
Answer. 54.8% and 45.2%.
Level B
1. Silicon has three stable isotope – 30 Si (3.05%(mol.)), 29 Si and 28 Si. Calculate the content (in % (mol.)) of the most common isotope of silicon. How will they differ? molar masses silicon dioxide, which has a different isotopic composition, considering that oxygen has three stable isotopes with mass numbers 16, 17 and 18?
Answer. 94.55%; 18 types of silicon dioxide molecules.
2. The sample consists of a mixture of two isotopes of one element; 30% is an isotope, the nucleus of which has 18 neutrons; 70% is an isotope, the nucleus of which has 20 neutrons. Determine the atomic number of an element if the average relative atomic mass of the element in a mixture of isotopes is 36.4.
Answer. 17.
3. A chemical element consists of two isotopes. The nucleus of an atom of the first isotope contains 10 protons and 10 neutrons. There are 2 more neutrons in the nucleus of an atom of the second isotope. For every 9 atoms of a lighter isotope there is one atom of a heavier isotope. Calculate the average relative atomic mass of the element.
Answer. 20,2.
4. Isotope 137 Cs has a half-life of 29.7 years. 1 g of this isotope reacted explosively with excess water. What is the half-life of cesium in the resulting compound? Justify your answer.
Answer. T 1/2 = 29.7 years.
5. After how many years does the amount of radioactive strontium-90 (half-life 27 years) fall out as a result of radioactive fallout? nuclear explosion, will become less than 1.5% of the amount that was discovered at the moment after the nuclear explosion?
Answer. 163.35 years.
6. In the tagged atom method, radioactive isotopes are used to “trace the route” of an element in the body. Thus, a patient with a diseased pancreas is injected with a preparation of the radioactive isotope iodine-131 (undergoes – -decay), which allows the doctor to monitor the passage of iodine through the patient’s body. Write an equation for radioactive decay and calculate how long it takes for the amount of radioactive iodine introduced into the body to decrease by 10 times (half-life 8 days).
Answer.
7. How long will it take for three quarters of the nickel to turn into copper as a result of – -decay, if the half-life of the isotope 63 28 Ni is 120 years old?
Answer. 240 years.
8. Find the mass of the isotope 81 Sr (half-life 8.5 hours) remaining after 25.5 hours of storage if the original mass was 200 mg.
Answer. 25 mg.
9. Calculate the percentage of isotope atoms 128 I (half-life 25 minutes), remaining undisintegrated after storage for 2.5 hours.
Answer. 1,5625%.
10. Half life – -radioactive isotope 24 Na is equal to 14.8 hours. Write the equation for the decay reaction and calculate how many grams of the daughter product are formed from 24 g of this isotope in 29.6 hours.
Answer.
11. Isotope 210 Ro, radiating-particles, used in a mixture with beryllium in neutron sources. After what time will the intensity of such sources decrease by 32 times? The half-life of the isotope is 138 days.
Answer. 690 days
Exercises on nuclear reactions
1. How many- And – -particles had to lose their nucleus 226 Ra to obtain a daughter element with mass number 206, belonging to group IV periodic table elements? Name this element.
Answer. 5, 4 – , 206 82 Pb.
2. Nucleus of an isotope atom 238 92 U turned into a nucleus as a result of radioactive decay 226 88 Ra. How many- And – -particles were emitted by the original nucleus?
1. Which element has more pronounced non-metallic properties: a) at oxygen or carbon; b) phosphorus or arsenic? Give a reasoned answer based on the position of the elements in the periodic table.
2. Give a description of element No. 11 according to the plan:
Position on the periodic table
Metal or non-metal
Atomic structure
Electronic formula
Number of electrons on the outer energy level whether it is complete
Superior Oxide Formula
Does the element form volatile compound with hydrogen, if it forms, what is its chemical formula
3. How and why do the properties of chemical elements change over periods? Show this using the example of elements of the 3rd period.
4. Calculate the relative atomic mass of boron if it is known that the proportion of the 10 B isotope is 19.6%, and the 11 B isotope is 80.4%. (Answer: 10.8.)
Solutions and answers:
1. Non-metallic properties are more pronounced in a) oxygen (since from left to right in periods non-metallic properties increase),
b) phosphorus (since in groups from bottom to top, non-metallic properties are enhanced due to a decrease in the radius of the atom).
3. In periods from left to right, non-metallic properties increase and metallic properties weaken, because Due to an increase in the number of electrons in the valence shell, electrons begin to be more strongly attracted to the nucleus, and the radius of the atom decreases.
