# Mass Of Boron

Boron has two isotopes, B−10 B − 10 and B−11 B − 11. The average atomic mass of boron is found to be A =10.80 u A = 10.80 u. Isotopes of Boron (click to see decay chain): 6 B 7 B 8 B 9 B 10 B 11 B 12 B 13 B 14 B 15 B 16 B 17 B 18 B 19 B 12 B.

Learning Objectives

• to know the meaning of isotopes and atomic masses.

There are 21 elements with only one isotope, so all their atoms have identical masses. All other elements have two or more isotopes, so their atoms have at least two different masses. However, all elements obey the law of definite proportions when they combine with other elements, so they behave as if they had just one kind of atom with a definite mass. To solve this dilemma, we define the atomic mass as the weighted average mass of all naturally occurring isotopes of the element.

A atomic mass is defined as

[text{Atomic mass} = left(dfrac{%text{ abundance isotope 1}}{100}right)times left(text{mass of isotope 1}right) + left(dfrac{%text{ abundance isotope 2}}{100}right)times left(text{mass of isotope 2}right)~ ~ ~ + ~ ~ ... label{amass}]

Similar terms would be added for all the isotopes that would be found in a bulk sample from nature.

GPAs

The weighted average is analogous to the method used to calculate grade point averages in most colleges:

[text{GPA} = left(dfrac{text{Credit Hours Course 1}}{text{total credit hours}}right)times left(text{Grade in Course 1}right) + left(dfrac{text{Credit Hours Course 2}}{text{total credit hours}}right)times left(text{Grade in Course 2}right)~ + ~... nonumber]

The periodic table lists the atomic masses of all the elements. Comparing these values with those given for some of the isotopes reveals that the atomic masses given in the periodic table never correspond exactly to those of any of the isotopes Figure (PageIndex{1}). Because most elements exist as mixtures of several stable isotopes, the atomic mass of an element is defined as the weighted average of the masses of the isotopes. For example, naturally occurring carbon is largely a mixture of two isotopes: 98.89% 12C (mass = 12 amu by definition) and 1.11% 13C (mass = 13.003355 amu). The percent abundance of 14C is so low that it can be ignored in this calculation. The average atomic mass of carbon is then calculated as follows:

[ rm(0.9889 times 12 ;amu) + (0.0111 times 13.003355 ;amu) = 12.01 ;amu label{Eq5} ]

Carbon is predominantly 12C, so its average atomic mass should be close to 12 amu, which is in agreement with this calculation.

The value of 12.01 is shown under the symbol for C in the periodic table, although without the abbreviation amu, which is customarily omitted. Thus the tabulated atomic mass of carbon or any other element is the weighted average of the masses of the naturally occurring isotopes.

Naturally occurring lead is found to consist of four isotopes:

• 1.40% ({}_{text{82}}^{text{204}}text{Pb}) whose isotopic mass is 203.973.
• 24.10% ({}_{text{82}}^{text{206}}text{Pb}) whose isotopic mass is 205.974.
• 22.10% ({}_{text{82}}^{text{207}}text{Pb}) whose isotopic mass is 206.976.
• 52.40% ({}_{text{82}}^{text{208}}text{Pb}) whose isotopic mass is 207.977.

### Atomic Mass Of Boron 11

Calculate the atomic mass of an average naturally occurring sample of lead.

Solution

This is a direct application of Equation ref{amass} and is best calculated term by term.

Suppose that you had 1 mol lead. This would contain 1.40% ((dfrac{1.40}{100}) × 1 mol) ({}_{text{82}}^{text{204}}text{Pb}) whose molar mass is 203.973 g mol–1. The mass of 20482Pb would be

[begin{align*} text{m}_{text{204}} &=n_{text{204}}times text{ }M_{text{204}} [4pt] &=left( frac{text{1}text{.40}}{text{100}}times text{ 1 mol} right)text{ (203}text{.973 g mol}^{text{-1}}text{)} [4pt] &=text{2}text{0.86 g} end{align*}]

Similarly for the other isotopes

[begin{align*}text{m}_{text{206}}&=n_{text{206}}times text{ }M_{text{206}}[4pt] &=left( frac{text{24}text{.10}}{text{100}}times text{ 1 mol} right)text{ (205}text{.974 g mol}^{text{-1}}text{)}[4pt] &=text{49}text{0.64 g} [6pt]text{m}_{text{207}}&=n_{text{207}}times text{ }M_{text{207}}[4pt] &=left( frac{text{22}text{.10}}{text{100}}times text{ 1 mol} right)text{ (206}text{.976 g mol}^{text{-1}}text{)}[4pt] &=text{45}text{0.74 g} [6pt] text{m}_{text{208}}&=n_{text{208}}times text{ }M_{text{208}}[4pt] &=left( frac{text{52}text{.40}}{text{100}}times text{ 1 mol} right)text{ (207}text{.977 g mol}^{text{-1}}text{)}[4pt] &=text{108}text{0.98 g} end{align*}]

Upon summing all four results, the mass of 1 mol of the mixture of isotopes is to be found

[2.86, g + 49.64, g + 45.74, g + 108.98, g = 207.22, gnonumber]

The mass of an average lead atom, and thus lead's atomic mass, is 207.2 g/mol. This should be confirmed by consulting the Periodic Table of the Elements.

Exercise (PageIndex{1}): Boron

Boron has two naturally occurring isotopes. In a sample of boron, (20%) of the atoms are (ce{B}-10), which is an isotope of boron with 5 neutrons and mass of (10 : text{amu}). The other (80%) of the atoms are (ce{B}-11), which is an isotope of boron with 6 neutrons and a mass of (11 : text{amu}). What is the atomic mass of boron?

The mass of an average boron atom, and thus boron's atomic mass, is (10.8 : text{amu}). This should be confirmed by consulting the Periodic Table of the Elements.

But which Natural Abundance should be used?

An important corollary to the existence of isotopes should be emphasized at this point. When highly accurate results are obtained, atomic weights may vary slightly depending on where a sample of an element was obtained. For this reason, the Commission on Isotopic Abundance and Atomic Weights of IUPAC (IUPAC/CIAAWhas redefined the atomic masses of 10 elements having two or more isotopes. The percentages of different isotopes often depends on the source of the element.

For example, oxygen in Antarctic precipitation has an atomic weight of 15.99903, but oxygen in marine (ce{N2O}) has an atomic mass of 15.9997. 'Fractionation' of the isotopes results from slightly different rates of chemical and physical processes caused by small differences in their masses. The difference can be more dramatic when an isotope is derived from nuclear reactors.

## Mass Spectrometry: Measuring the Mass of Atoms and Molecules

Although the masses of the electron, the proton, and the neutron are known to a high degree of precision, the mass of any given atom is not simply the sum of the masses of its electrons, protons, and neutrons. For example, the ratio of the masses of 1H (hydrogen) and 2H (deuterium) is actually 0.500384, rather than 0.49979 as predicted from the numbers of neutrons and protons present. Although the difference in mass is small, it is extremely important because it is the source of the huge amounts of energy released in nuclear reactions.

Because atoms are much too small to measure individually and do not have charges, there is no convenient way to accurately measure absolute atomic masses. Scientists can measure relative atomic masses very accurately, however, using an instrument called a mass spectrometer. The technique is conceptually similar to the one Thomson used to determine the mass-to-charge ratio of the electron. First, electrons are removed from or added to atoms or molecules, thus producing charged particles called ions. When an electric field is applied, the ions are accelerated into a separate chamber where they are deflected from their initial trajectory by a magnetic field, like the electrons in Thomson’s experiment. The extent of the deflection depends on the mass-to-charge ratio of the ion. By measuring the relative deflection of ions that have the same charge, scientists can determine their relative masses (Figure (PageIndex{2})). Thus it is not possible to calculate absolute atomic masses accurately by simply adding together the masses of the electrons, the protons, and the neutrons, and absolute atomic masses cannot be measured, but relative masses can be measured very accurately. It is actually rather common in chemistry to encounter a quantity whose magnitude can be measured only relative to some other quantity, rather than absolutely. We will encounter many other examples later in this text. In such cases, chemists usually define a standard by arbitrarily assigning a numerical value to one of the quantities, which allows them to calculate numerical values for the rest.

The arbitrary standard that has been established for describing atomic mass is the atomic mass unit (amu or u), defined as one-twelfth of the mass of one atom of 12C. Because the masses of all other atoms are calculated relative to the 12C standard, 12C is the only atom whose exact atomic mass is equal to the mass number. Experiments have shown that 1 amu = 1.66 × 10−24 g.

Mass spectrometric experiments give a value of 0.167842 for the ratio of the mass of 2H to the mass of 12C, so the absolute mass of 2H is

[rm{text{mass of }^2H over text{mass of }^{12}C} times text{mass of }^{12}C = 0.167842 times 12 ;amu = 2.104104; amu label{Eq4}]

The masses of the other elements are determined in a similar way.

### Mass Of Boron Trifluoride

Example (PageIndex{2}): Bromine

Naturally occurring bromine consists of the two isotopes listed in the following table:

IsotopeExact Mass (amu)Percent Abundance (%)
79Br78.918350.69
81Br80.916349.31

Calculate the atomic mass of bromine.

Given: exact mass and percent abundance

Strategy:

1. Convert the percent abundances to decimal form to obtain the mass fraction of each isotope.
2. Multiply the exact mass of each isotope by its corresponding mass fraction (percent abundance ÷ 100) to obtain its weighted mass.
3. Add together the weighted masses to obtain the atomic mass of the element.

Solution:

A The atomic mass is the weighted average of the masses of the isotopes (Equation ref{amass}. In general, we can write

Bromine has only two isotopes. Converting the percent abundances to mass fractions gives

[ce{^{79}Br}: {50.69 over 100} = 0.5069 nonumber]

[ce{^{81}Br}: {49.31 over 100} = 0.4931 nonumber]

B Multiplying the exact mass of each isotope by the corresponding mass fraction gives the isotope’s weighted mass:

(ce{^{79}Br}: 79.9183 ;amu times 0.5069 = 40.00; amu)

(ce{^{81}Br}: 80.9163 ;amu times 0.4931 = 39.90 ;amu)

C The sum of the weighted masses is the atomic mass of bromine is

40.00 amu + 39.90 amu = 79.90 amu

D This value is about halfway between the masses of the two isotopes, which is expected because the percent abundance of each is approximately 50%.

Exercise (PageIndex{2})

Magnesium has the three isotopes listed in the following table:

IsotopeExact Mass (amu)Percent Abundance (%)
24Mg23.9850478.70
25Mg24.9858410.13
26Mg25.9825911.17

Use these data to calculate the atomic mass of magnesium.

24.31 amu

## Summary

The mass of an atom is a weighted average that is largely determined by the number of its protons and neutrons, whereas the number of protons and electrons determines its charge. Each atom of an element contains the same number of protons, known as the atomic number (Z). Neutral atoms have the same number of electrons and protons. Atoms of an element that contain different numbers of neutrons are called isotopes. Each isotope of a given element has the same atomic number but a different mass number (A), which is the sum of the numbers of protons and neutrons. The relative masses of atoms are reported using the atomic mass unit (amu), which is defined as one-twelfth of the mass of one atom of carbon-12, with 6 protons, 6 neutrons, and 6 electrons. The atomic mass of an element is the weighted average of the masses of the naturally occurring isotopes. When one or more electrons are added to or removed from an atom or molecule, a charged particle called an ion is produced, whose charge is indicated by a superscript after the symbol.

• Ed Vitz (Kutztown University), John W. Moore (UW-Madison), Justin Shorb (Hope College), Xavier Prat-Resina (University of Minnesota Rochester), Tim Wendorff, and Adam Hahn.

#### Learning Objective

• Calculate the average atomic mass of an element given its isotopes and their natural abundance

#### Key Points

• An element can have differing numbers of neutrons in its nucleus, but it always has the same number of protons. The versions of an element with different neutrons have different masses and are called isotopes.
• The average atomic mass for an element is calculated by summing the masses of the element’s isotopes, each multiplied by its natural abundance on Earth.
• When doing any mass calculations involving elements or compounds, always use average atomic mass, which can be found on the periodic table.

#### Terms

• natural abundanceThe abundance of a particular isotope naturally found on the planet.
• average atomic massThe mass calculated by summing the masses of an element’s isotopes, each multiplied by its natural abundance on Earth.
• mass numberThe total number of protons and neutrons in an atomic nucleus.

The atomic number of an element defines the element’s identity and signifies the number of protons in the nucleus of one atom. For example, the element hydrogen (the lightest element) will always have one proton in its nucleus. The element helium will always have two protons in its nucleus.

## Isotopes

Atoms of the same element can, however, have differing numbers of neutrons in their nucleus. For example, stable helium atoms exist that contain either one or two neutrons, but both atoms have two protons. These different types of helium atoms have different masses (3 or 4 atomic mass units), and they are called isotopes. For any given isotope, the sum of the numbers of protons and neutrons in the nucleus is called the mass number. This is because each proton and each neutron weigh one atomic mass unit (amu). By adding together the number of protons and neutrons and multiplying by 1 amu, you can calculate the mass of the atom. All elements exist as a collection of isotopes. The word ‘isotope’ comes from the Greek ‘isos’ (meaning ‘same’) and ‘topes’ (meaning ‘place’) because the elements can occupy the same place on the periodic table while being different in subatomic construction.

## Calculating Average Atomic Mass

The average atomic mass of an element is the sum of the masses of its isotopes, each multiplied by its natural abundance (the decimal associated with percent of atoms of that element that are of a given isotope).

Average atomic mass = f1M1 + f2M2 + … + fnMn where f is the fraction representing the natural abundance of the isotope and M is the mass number (weight) of the isotope.

The average atomic mass of an element can be found on the periodic table, typically under the elemental symbol. When data are available regarding the natural abundance of various isotopes of an element, it is simple to calculate the average atomic mass.

• For helium, there is approximately one isotope of Helium-3 for every million isotopes of Helium-4; therefore, the average atomic mass is very close to 4 amu (4.002602 amu).
• Chlorine consists of two major isotopes, one with 18 neutrons (75.77 percent of natural chlorine atoms) and one with 20 neutrons (24.23 percent of natural chlorine atoms). The atomic number of chlorine is 17 (it has 17 protons in its nucleus).

To calculate the average mass, first convert the percentages into fractions (divide them by 100). Then, calculate the mass numbers. The chlorine isotope with 18 neutrons has an abundance of 0.7577 and a mass number of 35 amu. To calculate the average atomic mass, multiply the fraction by the mass number for each isotope, then add them together.

Average atomic mass of chlorine = (0.7577 [latex]cdot[/latex] 35 amu) + (0.2423 [latex]cdot[/latex] 37 amu) = 35.48 amu

Another example is to calculate the atomic mass of boron (B), which has two isotopes: B-10 with 19.9% natural abundance, and B-11 with 80.1% abundance. Therefore,

Average atomic mass of boron = (0.199
[latex]cdot[/latex]

10 amu) + (0.801
[latex]cdot[/latex]

11 amu) = 10.80 amu

Whenever we do mass calculations involving elements or compounds (combinations of elements), we always use average atomic masses.

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### Mass Of Boron

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### Mass Of Boron In Grams

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