Calculate the average atomic weight when given isotopic weights and abundances
Fifteen Examples

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Calculate the isotopic abundances, given the atomic weight and isotopic weights


To do these problems you need some information. To wit:

(a) the exact atomic weight for each naturally-occuring stable isotope
(b) the percent abundance for each isotope

These values can be looked up in a standard reference book such as the "Handbook of Chemistry and Physics." The values can also be looked up via many online sources. The ChemTeam prefers to use Wikipedia to look up values.

The unit associated with the answers to the problems below 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).

By the way, the most correct symbol for the atomic mass unit is u. The older symbol (which the ChemTeam grew up with) is amu (sometimes seen as a.m.u.) The unit amu is still in use, but you will see u used more often.

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 tutorial linked just above.


Example #1: Calculate the average atomic weight for carbon.

mass number   isotopic weight   percent abundance
1212.00000098.93
1313.0033551.07

Solution:

To calculate the average atomic weight, each isotopic 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.
(12.000000) (0.9893) + (13.003355) (0.0107) = 12.0107 amu

This is commonly rounded to 12.011 or sometimes 12.01.

The answers to problems like this tend to not follow strict significant figure rules. Consult a periodic table to see what manner of answers are considered acceptable.


Example #2: Nitrogen

mass number   isotopic weight   percent abundance
1414.00307499.636
1515.0001080.364

Solution:

(14.003074) (0.9963) + (15.000108) (0.0037) = 14.007 amu (or 14.007 u)
(isotopic weight) (abundance) + (isotopic weight) (abundance) = average atomic weight

A point about the term 'atomic weight:' When discussing the atomic weight of an element, this value is an average. When discussing the atomic weight of an isotope, this value is a value that has been measured experimentally, not an average.


Example #3: Silicon

mass number   isotopic weight   percent abundance
2827.97692792.23
2928.9764954.67
3029.9737703.10

Solution:

(27.976927) (92.23) + (28.976495) (4.67) + (29.973770) (3.10) = 2808.55 u

There is a problem with the answer!!

The true value is 28.086 u. Our answer is too large by a factor of 100.

This is because I used percentages (92.23, 4.67, 3.10) and not the decimal equivalents (0.9223, 0.0467, 0.0310).

To obtain the correct answer, we must divide by 100.


Example #4: How to Calculate an Average Atomic Weight
 
Solution:
 
 
Two points: (1) notice I wrote the same number of decimal places in the answer as were in the isotopic weights (the 184.953 and the 186.956). This is common. (2) I forgot to put a unit on the answer, so 186.207 u would be the most correct answer.

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

I put g/mol for the unit because that what was used in the problem statement.


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

4) 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 u according to the periodic table. However, no single atom of boron has a mass of 10.81 u. 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 u, but you can arrive at 10.81 u like this:

x = (10.013) (0.199) + (11.009) (0.801)

x = 1.99 + 8.82 = 10.81 u

It's like the old joke: consider a centipede and a snake. What's the average number of legs? Answer: 50. Of course, neither one has 50.


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?

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:

2965 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 numberexact weightpercent abundance
2423.98504278.99
2524.98583710.00
2625.98259311.01

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.

See Example #7 and the example just below to see how this "no individual atom has the average weight" can be exploited.


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

18.10 + 21y + 2.09 - 22y = 20.18

y = 0.010

Relative abundance of (note use of percents):

Ne-21 = 1.0%

Ne-22 = (9.5 − 1)% = 8.5%


Calculating isotopic abundances, given the atomic weight and isotopic weights

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