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Average Atomic Mass

The atomic mass values given in the periodic table are average atomic masses. They are calculated using the exact masses and natural abundances of the stable isotopes of an element.

As you know, atoms of an element come in different varieties called
isotopes. For example, the common isotopes of nitrogen include^{15}_{7}N
and^{14}_{7}N. Isotopes are
atoms of the same element that have different mass numbers due to the fact
that they have different numbers of neutrons. The isotopes of an element are
not all equally common. Of all the nitrogen atoms on Earth 99.63% of them
are^{14}_{7}N. Only 0.37% of
all nitrogen atoms are^{15}_{7}N. That is why
the average atomic mass of N as given in the
Periodic Table is closer to 14 amu than to 15 amu.

The percentages given above for the isotopes of nitrogen are called
**percent abundances**. The percent abundance of an isotope
tells you the fraction of all atoms of an element that are a particular
isotope of that element. Think of it this way: if you have 10,000 nitrogen
atoms then 9,963 of them are^{14}_{7}N and 37 of them
are^{15}_{7}N. Percentages
are just a way to write fractions so that they all have the same denominator:
100. So the percentage 99.63% means 99.63/100 and 0.37% means 0.37/100.
Percentages need to be written as decimals in order to use them in
calculations. Therefore 99.63% (99.63/100) is 0.9963 and 0.37% (0.37/100) is
0.0037. Just move the decimal point two places to the left.

Average atomic masses are computed using a method called **weighted
averages**. Weighted averages are used when the importance of the
numbers to be averaged are different. For example, at Scarborough High School
a student’s semester average is computed based on three grades. The
First Quarter grade (Q_{1}), the Second Quarter grade
(Q_{2}), and the Semester Exam grade (R_{1}). Both quarter
grades are weighted at 40% (W_{Q1} = W_{Q2} = 0.40) of the
semester average and the exam is weighted at 20% (W_{R1} = 0.20).

Say a Chemistry student named Lyle has a Q_{1} grade of 85, a
Q_{2} grade of 80 and a R_{1} grade of 87. What is
Lyle’s semester 1 grade? Set up the calculation this way:

Q_{1}·W_{Q1}+ Q_{2}·W_{Q2}+ R_{1}·W_{R1}= S_{1}(85)·0.40 + (80)·0.40 + (87)·0.20 = 83.4

Now, what if Lyle scored poorly during the first quarter (he earned a 75) but improved during the second (he earned an 89) and he needs to know the minimum exam grade to earn an 85 for the semester? Just use a little algebra to solve the problem:

Q_{1}·W_{Q1}+ Q_{2}·W_{Q2}+ R_{1}·W_{R1}= S_{1}(75)·0.40 + (89)·0.40 + x·0.20 = 85 0.20x = 85 - (75)·0.40 - (89)·0.40 0.20x = 19.4 x = 97

In short, Lyle will have to work very hard to get the grade he needs on the semester exam.

Average atomic masses are calculated in just the same way. Each isotope’s exact mass (determined using a mass spectrometer) is multiplied by the decimal equivalent of its percent abundance and all the results are added together.

Symbol | Mass number | Exact mass | Percent abundance |

^{39}_{19}K |
39 | 38.963707 | 93.2581 |

^{40}_{19}K |
40 | 39.963999 | 0.0117 |

^{41}_{19}K |
41 | 40.961826 | 6.7302 |

Here is how to calculate the average atomic mass of potassium:

38.963707 · 0.932581 39.963999 · 0.000117 + 40.961826 · 0.067302 ------------------------------- 39.098301 amu

What if you know the percent abundance of the isotopes of an element but
not all of the exact masses? Use algebra to find the missing exact mass. Take
a look at the potassium data and imagine that the exact mass of ^{41}_{19}K is unknown. Use the
average mass of the element, the known exact masses, and the percent
abundances to find the missing exact mass data.

38.963707 · 0.932581 + 39.963999 · 0.000117 + x · 0.067302 = 39.098301 x · 0.067302 = 39.09830144 - 36.34148863 x · 0.067302 = 2.756812809 x = 2.756812809/0.067302 = 40.961826

Since we made up this problem we can check the answer against the exact
mass given in the table: it matches perfectly. The exact mass of^{41}_{19}K is 40.961826
amu.

One thing that is important to getting the calculations of this kind exactly correct is to use your calculator to hold all the digits for you while you work. Set up your math so that you can type the numbers and operations into your calculator in one long calculation without having to stop, write down an answer and re-enter it. This prevents rounding errors which will cause your answers to be wrong.

Show your work for all the following calculations. The masses are atomic masses expressed in atomic mass units (amu). Write the symbol for each isotope. Pay attention to significant figures: do not report any more numbers after the decimal point than are reported in the data you are given.

- Ryan’s grades for the first semester are
Q
_{1}: 76, Q_{2}: 84, R_{1}: 86. What is his semester grade? - Marcus would like to earn a 93 for the semester.
If his Q
_{1}grade is 90 and his Q_{2}grade is 93 what grade does he need on the semester exam? - Leslie’s first quarter grade was 82. She
wants to know what grades she needs for Q
_{2}and R_{1}to earn a 90 for the semester. Assume that she will earn the same grade for Q_{2}and R_{1}. - Find the average atomic mass of carbon
(C ).

Symbol Mass number Exact mass Percent abundance 12 12.000000 98.93 13 13.003355 1.07

- Find the average atomic mass of chlorine
(Cl).

Symbol Mass number Exact mass Percent abundance 35 34.968853 75.78 37 36.965903 24.22 - Find the average atomic mass of nitrogen
(N).
Symbol Mass number Exact mass Percent abundance 14 14.003074 99.63 15 15.000108 0.37 - Find the average atomic mass of neon
(Ne).

Symbol Mass number Exact mass Percent abundance 20 19.992440 90.48 21 20.993847 0.27 22 21.991386 9.25 - Find the average atomic mass of magnesium
(Mg).
Symbol Mass number Exact mass Percent abundance 24 23.985042 78.99 25 24.985837 10.00 26 25.982593 11.01 - Find the average atomic mass of chromium
(Cr).
Symbol Mass number Exact mass Percent abundance 50 49.946049 4.345 52 51.940511 83.789 53 52.940653 9.501 54 53.938884 2.365 - Find the average atomic mass of molybdenum
(Mo).
Symbol Mass number Exact mass Percent abundance 92 91.906808 14.84 94 93.905085 9.25 95 94.905840 15.92 96 95.904678 16.68 97 96.906020 9.55 98 97.905406 24.13 100 99.907477 9.63

- There is an easy way to check whether your answers are correct when you calculate the average atomic mass for the naturally occurring isotopes of an element. Explain this easy way to check your answers.
- An element “X” has five major
isotopes, which are listed below along with their abundances. Calculate the
average atomic mass. What is the element?
Symbol Exact mass Percent abundance ^{46}X45.952630 8.25 ^{47}X46.951764 7.44 ^{48}X47.947947 73.72 ^{49}X48.947871 5.41 ^{50}X49.944792 5.18 - An element consists of 1.4% of an isotope with mass 203.973 amu, 24.1% of an isotope with mass 205.9744 amu, 22.1% of an isotope with mass 206.9759 amu, and 52.4% of an isotope with mass 207.9766 amu. Calculate the average atomic mass and identify the element.
- The element rhenium (Re) has two naturally occurring isotopes,
^{185}Re and^{187}Re, with an average atomic mass of 186.207 amu. Rhenium is 62.60%^{187}Re, and the atomic mass of^{187}Re is 186.956 amu. Calculate the mass of^{185}Re. - The isotopes of silicon are
^{28}_{14}Si,^{29}_{14}Si, and^{30}_{14}Si. Find the percent abudance and exact mass of silicon-28 using the following information:Symbol Mass number Exact mass Percent abundance ^{28}_{14}Si28 ^{29}_{14}Si29 28.9765 4.69 ^{30}_{14}Si30 29.9737 3.10