Hydrogen has an average atomic mass of 1.00794 amu. Oxygen atoms have an average mass of 15.9994 amu. The average mass of a molecule of H 2 O equals (1.00794) (2) + 15.9994 = 18.01528 amu, equivalent to 18.01528 g/mol. Average atomic masses listed by IUPAC are based on a study of experimental results. Bromine has two isotopes, 79 Br and 81 Br, whose masses (78.9183 and 80.9163 amu) and abundances (50.69% and 49.31%) were determined in earlier experiments. Calculate the average atomic mass of Br based on these experiments. 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.

1. Calculating Average Atomic Mass Quiz
2. Calculating Average Atomic Mass From Percent Abundance
3. Calculating Average Atomic Mass
4. Calculating Average Atomic Mass Worksheet Pdf

The average atomic mass of an element is the sum of the masses of its isotopes, each multiplied by its natural abundance.

Learning Objectives

• 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.

Key Terms

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

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 +… + f_nM_nwhere f is the fraction representing the natural abundance of the isotope and M is the mass number (weight) of the isotope.

Calculating Average Atomic Mass Quiz

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 ⋅⋅ 35 amu) + (0.2423 ⋅⋅ 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⋅⋅10 amu) + (0.801⋅⋅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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Calculate the average atomic weight when given isotopic weights and abundances
Fifteen Examples

To do these problems you need some information: the exact atomic weight for each naturally-occuring stable isotope and its percent abundance. These values can be looked up in a standard reference book such as the 'Handbook of Chemistry and Physics.' The unit associated with the answer can be either amu or g/mol, depending on the context of the question. If it is not clear from the context that g/mol is the desired answer, go with amu (which means atomic mass unit).

This problem can also be reversed, as in having to calculate the isotopic abundances when given the atomic weight and isotopic weights. Study the tutorial below and then look at the problems done in the reverse direction.

Example #1: Carbon

mass number	exact weight	percent abundance
12	12.000000	98.90
13	13.003355	1.10

To calculate the average atomic weight, each exact atomic weight is multiplied by its percent abundance (expressed as a decimal). Then, add the results together and round off to an appropriate number of significant figures.

This is the solution for carbon:

(12.000000) (0.9890) + (13.003355) (0.0110) = 12.011 amu

Example #2: Nitrogen

mass number	exact weight	percent abundance
14	14.003074	99.63
15	15.000108	0.37

This is the solution for nitrogen:

(14.003074) (0.9963) + (15.000108) (0.0037) = 14.007 amu

Example #3: Chlorine			Example #4: Silicon
mass number	exact weight	percent abundance	mass number	exact weight	percent abundance
35	34.968852	75.77	28	27.976927	92.23
37	36.965903	24.23	29	28.976495	4.67
30	29.973770	3.10
The answer for chlorine: 35.453			The answer for silicon: 28.086

This type of calculation can be done in reverse, where the isotopic abundances can be calculated knowing the average atomic weight. Go to tutorial on reverse direction.

Example #5: In a sample of 400 lithium atoms, it is found that 30 atoms are lithium-6 (6.015 g/mol) and 370 atoms are lithium-7 (7.016 g/mol). Calculate the average atomic mass of lithium.

Solution:

1) Calculate the percent abundance for each isotope:

Li-6: 30/400 = 0.075
Li-7: 370/400 = 0.925

2) Calculate the average atomic weight:

x = (6.015) (0.075) + (7.016) (0.925)

x = 6.94 g/mol

Example #6: A sample of element X contains 100 atoms with a mass of 12.00 and 10 atoms with a mass of 14.00. Calculate the average atomic mass (in amu) of element X.

Solution:

1) Calculate the percent abundance for each isotope:

X-12: 100/110 = 0.909
X-14: 10/110 = 0.091

2) Calculate the average atomic weight:

x = (12.00) (0.909) + (14.00) (0.091)

x = 12.18 amu (to four sig figs)

3) Here's another way:

100 atoms with mass 12 = total atom mass of 1200

10 atoms with mass 14 = total atom mass of 140

1200 + 140 = 1340 (total mass of all atoms)

Total number of atoms = 100 + 10 = 110

1340/110 = 12.18 amu

The first way is the standard technique for solving this type of problem. That's because we do not generally know the specific number of atoms in a given sample. More commonly, we know the percent abundances, which is different from the specific number of atoms in a sample.

Example #7: Boron has an atomic mass of 10.81 amu according to the periodic table. However, no single atom of boron has a mass of 10.81 amu. How can you explain this difference?

Solution:

10.81 amu is an average, specifically a weighted average. It turns out that there are two stable isotopes of boron: boron-10 and boron-11.

Neither isotope weighs 10.81 amu, but you can arrive at 10.81 amu like this:x = (10.013) (0.199) + (11.009) (0.801)

x = 1.99 + 8.82 = 10.81

Example #8: Copper occurs naturally as Cu-63 and Cu-65. Which isotope is more abundant?

Solution:

Look up the atomic weight of copper: 63.546 amu

Since our average value is closer to 63 than to 65, we concude that Cu-63 is the more abundant isotope.

Example #9: Copper has two naturally occuring isotopes. Cu-63 has an atomic mass of 62.9296 amu and an abundance of 69.15%. What is the atomic mass of the second isotope? What is its nuclear symbol?

Calculating Average Atomic Mass From Percent Abundance

Solution:

1) Look up the atomic weight of copper:

63.546 amu

2) Set up the following and solve:

(62.9296) (0.6915) + (x) (0.3085) = 63.546

43.5158 + 0.3085x = 63.546

0.3085x = 20.0302

x = 64.9277 amu

3) The nuclear symbol is:

${29}_{65}^{}\text{Cu}$

4) You might see this

29-Cu-65

This is used in situations, such as the Internet, where the subscript/superscript notation cannot be reproduced. You might also see this:

65/29Cu

Example #10: Naturally occurring iodine has an atomic mass of 126.9045. A 12.3849 g sample of iodine is accidentally contaminated with 1.0007 g of I-129, a synthetic radioisotope of iodine used in the treatment of certain diseases of the thyroid gland. The mass of I-129 is 128.9050 amu.Find the apparent 'atomic mass' of the contaminated iodine.

Solution:

1) Calculate mass of contaminated sample:

12.3849 g + 1.0007g = 13.3856 g

2) Calculate percent abundances of (a) natural iodine and (b) I-129 in the contaminated sample:

(a) 12.3849 g / 13.3856 g = 0.92524
(b) 1.0007 g / 13.3856 g = 0.07476

3) Calculate 'atomic mass' of contaminated sample:

(126.9045) (0.92524) + (128.9050) (0.07476) = x

x = 127.0540 amu

Example #11: Neon has two major isotopes, Neon-20 and Neon-22. Out of every 250 neon atoms, 225 will be Neon-20 (19.992 g/mol), and 25 will be Neon-22 (21.991 g/mol). What is the average atomic mass of neon?

Solution:

1) Determine the percent abundances (but leave as a decimal):

Ne-20 ---> 225 / 250 = 0.90
Ne-22 ---> 25 / 250 = 0.10

The last value can also be done by subtraction, in this case 1 - 0.9 = 0.1

2) Calculate the average atomic weight:

(19.992) (0.90) + (21.991) (0.10) = 20.19

Example #12: Calculate the average atomic weight for magnesium:

mass number	exact weight	percent abundance
24	23.985042	78.99
25	24.985837	10.00
26	25.982593	11.01

The answer? Find magnesium on the periodic table:

Remember that the above is the method by which the average atomic weight for the element is computed. No one single atom of the element has the given atomic weight because the atomic weight of the element is an average, specifically called a 'weighted' average. Given that, here is a question you could be asked.

Example #13: Silver has an atomic mass of 107.868 amu. Does any atom of any isotope of silver have a mass of 107.868 amu? Explain why or why not.

Solution:

The specific question is about silver, but it could be any element. The answer, of course, is no. The atomic weight of silver is a weighted average. Silver is not composed of atoms each of which weighs 107.868.

Example #14: Given that the average atomic mass of hydrogen in nature is 1.0079, what does that tell you about the percent composition of H-1 and H-2 in nature?

Solution:

It tells you that the proportion of H-1 is much much greater than the proportion of H-2 in nature.

Example #15: The relative atomic mass of neon is 20.18 It consists of three isotopes with the masses of 20, 21and 22. It consists of 90.5% of Ne-20. Determine the percent abundances of the other two isotopes.

Solution:

1) Let y% be the relative abundance of Ne-21.

2) Then, the relative abundance of Ne-22 is:

(100 − 90.5 − y)% = (9.5 − y)%

3) Relative atomic mass of Ne (note use of decimal abundances, not percent abundances):

(20) (0.905) + (21) (y) + (22) (0.095 − y) = 20.18

Calculating Average Atomic Mass

18.10 + 21y + 2.09 - 22y = 20.18

y = 0.01

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Relative abundance of (note use of percents):

Ne-21 = 1%

Calculating Average Atomic Mass Worksheet Pdf

Ne-22 = (9.5 − 1)% = 8.5%