B. Matter, Chemical Trends, and Chemical Bonding · Matter, Chemical Trends, and Chemical BondingIsotopic Abundance and Relative Atomic Mass
B3.2 B3.1 A1.13 A1.6About 65 minutes
Learning goals- I can explain what the atomic mass on the periodic table means.
- I can calculate an element's average atomic mass from isotope masses and abundances.
- I can work backwards from an average atomic mass to find isotopic abundances.
- I can read a simple mass spectrum and report answers with suitable significant figures.
Words to know
atomic mass unit (u)
A unit of mass equal to exactly 1/12 of the mass of a carbon-12 atom.
isotopic abundance
The percentage of an element's atoms, in a natural sample, that are a particular isotope.
relative atomic mass (average atomic mass)
The weighted average mass of the naturally occurring isotopes of an element, in u.
weighted average
An average in which each value counts in proportion to how often it occurs.
mass spectrometer
An instrument that separates ions by mass and measures how many of each there are.
mass spectrum
The graph a mass spectrometer produces: one peak for each isotope, with height showing abundance.
Why isn't it a whole number?
Look up chlorine on a periodic table. Its atomic mass is about 35.45. But a chlorine atom cannot contain 0.45 of a neutron! Protons and neutrons come in whole numbers, so every single chlorine atom has a whole-number mass number (35 or 37). Where does the decimal come from?
The answer is that the periodic table does not list the mass of one atom. It lists the average mass of the atoms in a natural sample of the element, and the average is weighted by how common each isotope is.
Remember from Lesson 1
Isotopes have the same number of protons but different numbers of neutrons, so they have different masses. Chlorine has two stable isotopes, chlorine-35 and chlorine-37.
The atomic mass unit
Atoms are far too light to weigh in grams one at a time, so chemists use the atomic mass unit (u). By international agreement, 1 u is exactly 1/12 of the mass of one carbon-12 atom. On this scale a proton and a neutron each have a mass of about 1 u, so an isotope's mass in u is close to (but not exactly) its mass number. For example, the mass of a chlorine-35 atom is 34.969 u.
Isotopic abundance
The isotopic abundance (or natural abundance) of an isotope is the percentage of an element's atoms that are that isotope in a natural sample. In nature, 75.78% of chlorine atoms are chlorine-35 and 24.22% are chlorine-37. These percentages are almost the same in a salt shaker in Thunder Bay, in sea water off Nova Scotia, or in a rock on Baffin Island, which is why one average mass works for everyone.
Measuring abundance: the mass spectrometer
Chemists measure isotopes with a mass spectrometer. It turns atoms into positive ions, speeds them up, and uses electric or magnetic fields to separate them by mass. Lighter ions are deflected more than heavier ones. A detector counts how many ions of each mass arrive. The result, a mass spectrum, has one peak for each isotope. The height of each peak shows that isotope's abundance.
Relative atomic mass: a weighted average
The relative atomic mass (average atomic mass) of an element is the weighted average of the masses of its isotopes:
average atomic mass = (mass₁ × abundance₁) + (mass₂ × abundance₂) + ...
with each abundance written as a decimal (75.78% = 0.7578). The abundances must add to 100% (1.0000).
A weighted average is like your course mark when tests count more than quizzes. The more abundant an isotope is, the more it pulls the average towards its own mass.
Worked exampleExample 1: Average atomic mass of chlorine
Chlorine-35 (mass 34.969 u) has an abundance of 75.78%. Chlorine-37 (mass 36.966 u) has an abundance of 24.22%. Calculate the average atomic mass of chlorine.
- Change the percentages to decimals: 0.7578 and 0.2422.
- Multiply each mass by its abundance: 34.969 × 0.7578 = 26.500 u and 36.966 × 0.2422 = 8.953 u.
- Add: 26.500 + 8.953 = 35.453 u.
- Check: the answer is between 35 and 37 and closer to 35, which makes sense because Cl-35 is more common.
Average atomic mass of chlorine = 35.45 u, which matches the periodic table.
Worked exampleExample 2: Three isotopes
Magnesium is 78.99% Mg-24 (23.985 u), 10.00% Mg-25 (24.986 u) and 11.01% Mg-26 (25.983 u). Find magnesium's average atomic mass.
- (23.985 × 0.7899) + (24.986 × 0.1000) + (25.983 × 0.1101)
- = 18.946 + 2.4986 + 2.8607
- = 24.305 u
24.31 u, matching the periodic table value for magnesium.
Worked exampleExample 3: Working backwards to find abundance
Copper has two isotopes: Cu-63 (62.930 u) and Cu-65 (64.928 u). Its average atomic mass is 63.546 u. Find the abundance of each isotope.
- Let x = the fraction that is Cu-63. Then the fraction that is Cu-65 is (1 − x), because the fractions add to 1.
- Write the weighted average: 62.930x + 64.928(1 − x) = 63.546
- Expand: 62.930x + 64.928 − 64.928x = 63.546
- Collect terms: −1.998x = 63.546 − 64.928 = −1.382
- Solve: x = 1.382 ÷ 1.998 = 0.6917
- So Cu-63 is 69.17% and Cu-65 is 100% − 69.17% = 30.83%.
About 69.2% copper-63 and 30.8% copper-65. (Published values are 69.15% and 30.85%; the small difference comes from rounding.)
Significant figures
Your answer can be no more precise than your data. Keep extra digits during the calculation and round only at the end. If the abundances have four significant figures (75.78%), report the average to four significant figures (35.45 u). Expectation A1.13 asks you to show this kind of precision in every calculation.
Just for funWhy did the chlorine atom refuse to round its mass to 35? Because it is only 75.78% sure of itself.
Hands-on activityThe "beanium" lab
You need: a kitchen scale that reads to 1 g (a 0.1 g scale is better), a cup, and a mixture of three kinds of dried beans or pasta shapes, for example 40 kidney beans, 30 chickpeas and 20 lentils (any amounts are fine). The made-up element "beanium" has three isotopes: the three kinds of bean.
- Sort the beans by kind and count each kind. Calculate each "isotope's" percent abundance (count ÷ total × 100).
- Weigh all the beans of one kind together and divide by the count to get the average mass of one bean of that kind. (Weighing many and dividing gives better precision than weighing one.) Repeat for each kind.
- Calculate the weighted average mass of a beanium atom: Σ (mass of one bean × abundance as a decimal).
- Check: weigh all the beans together and divide by the total number of beans. This should agree closely with step 3.
- Explain any difference (rounding, scale precision, beans that are not all the same size).
What to look for: the weighted average should be closer to the mass of the most common bean, just as chlorine's average is closer to 35.
Your turn
1Define the atomic mass unit.
1 u = 1/12 of the mass of one carbon-12 atom.
2Boron is 19.9% B-10 (10.013 u) and 80.1% B-11 (11.009 u). Calculate its average atomic mass.
(10.013 × 0.199) + (11.009 × 0.801) = 1.993 + 8.818 = 10.81 u.
3Lithium is 7.59% Li-6 (6.015 u) and 92.41% Li-7 (7.016 u). Calculate its average atomic mass.
(6.015 × 0.0759) + (7.016 × 0.9241) = 0.4565 + 6.4835 = 6.940 u.
4Without calculating, is the average atomic mass of bromine (Br-79 50.69%, Br-81 49.31%) closer to 79 or to 81? Explain.
Slightly closer to 79, but almost exactly in the middle, because the two isotopes are nearly equally abundant and Br-79 is a little more common. (The actual value is 79.90 u.)
5Why does no single chlorine atom have a mass of 35.45 u?
Each atom has a whole number of protons and neutrons, so each is Cl-35 (34.969 u) or Cl-37 (36.966 u). 35.45 u is an average of many atoms.
6What does the height of a peak in a mass spectrum tell you?
The relative abundance of the isotope with that mass.
More practice: warm-up
1Write 92.41% as a decimal.
2Two isotopes of an element have abundances 60.11% and ___%. Fill in the blank.
3What instrument is used to measure isotopic abundance?
4Which isotope's mass is exactly 12 u by definition?
5An element's isotopes have masses 20 u and 22 u, and its average atomic mass is 20.2 u. Which isotope is more abundant?
The 20 u isotope, because the average is much closer to 20.
More practice: on your own
1Gallium is 60.11% Ga-69 (68.926 u) and 39.89% Ga-71 (70.925 u). Find its average atomic mass to four significant figures.
(68.926 × 0.6011) + (70.925 × 0.3989) = 41.431 + 28.292 = 69.723 u ≈ 69.72 u.
2Silver is 51.84% Ag-107 (106.905 u) and the rest Ag-109 (108.905 u). Find its average atomic mass.
Ag-109 is 48.16%. (106.905 × 0.5184) + (108.905 × 0.4816) = 55.420 + 52.449 = 107.87 u.
3Bromine has an average atomic mass of 79.904 u. Its isotopes are Br-79 (78.918 u) and Br-81 (80.916 u). Find the abundance of each.
78.918x + 80.916(1 − x) = 79.904 → −1.998x = −1.012 → x = 0.5065. About 50.65% Br-79 and 49.35% Br-81 (published: 50.69% and 49.31%; the difference is rounding).
4A student finds the average atomic mass of chlorine by calculating (35 + 37) ÷ 2 = 36 u. Explain the mistake and why the true value is lower.
The student took a simple average, as if both isotopes were equally common. Chlorine-35 is about three times as common as chlorine-37, so a weighted average must be used; it is pulled towards 35, giving 35.45 u.
5Explain why the abundances of isotopes must add up to 100%.
Every atom of the element is one of its isotopes, so all the fractions together make up the whole sample.
Challenge
1Silicon has three isotopes: Si-28 (27.977 u, 92.23%), Si-29 (28.976 u, 4.68%) and Si-30 (29.974 u). Find the abundance of Si-30 and the average atomic mass of silicon.
Si-30: 100 − 92.23 − 4.68 = 3.09%. Average: (27.977 × 0.9223) + (28.976 × 0.0468) + (29.974 × 0.0309) = 25.803 + 1.356 + 0.926 = 28.09 u.
2Europium (average atomic mass 151.96 u) has two isotopes: Eu-151 (150.92 u) and Eu-153 (152.92 u). Find the percentage of Eu-153.
Let y = fraction of Eu-153: 150.92(1 − y) + 152.92y = 151.96 → 2.00y = 1.04 → y = 0.52. About 52% Eu-153 (and 48% Eu-151).
3The ratio of hydrogen-2 to hydrogen-1 in water is slightly different in rain from different places. Why does this not cause a problem when chemists use the periodic table value for hydrogen in calculations?
The variation is very small (deuterium is only about 0.01% of hydrogen atoms), so it changes the average atomic mass only in the fourth or fifth decimal place. That is far smaller than the precision needed in most calculations, so 1.01 u works everywhere.
Quiz
1. The relative atomic mass of an element is
2. An element has two isotopes, X-50 (20% abundant) and X-52 (80% abundant). Its average atomic mass is closest to
4. In a mass spectrum, the height of each peak shows
5. Rubidium's average atomic mass is 85.47 u. Its isotopes are Rb-85 and Rb-87. Which statement is true?
6. Which change would make a calculated average atomic mass incorrect?
Big idea: Expectation B3.2 asks students to explain the relationship between isotopic abundance and relative atomic mass. The key is the weighted average, the same idea as a course mark where some work counts more.
How to help: The "beanium" lab works well at a kitchen table. Ask your teen to predict, before calculating, whether the average will be closer to the heavier or the lighter bean, then check.
Common difficulties:
- Using percentages instead of decimals (75.78 instead of 0.7578), which gives answers 100 times too big.
- Taking a simple average ((35 + 37) ÷ 2 = 36) instead of a weighted one.
- Rounding too early, which shifts the last digit.
- In "find the abundance" problems, forgetting that the second fraction is (1 − x). This is a good use of the algebra from Grade 10 math; walk through it slowly.