SCI8 Q2W3D2: Calculation of atomic mass

🎯 Learning Goals

At the end of the lesson, the students should be able to:

1. Define atomic mass and explain its significance in understanding elements and their isotopes.
2. Correctly apply the formula for calculating the average atomic mass of an element using isotope mass and relative abundance, scoring at least 4 out of 5 in a worksheet.
3. Solve real-life problems involving atomic mass calculations and explain how isotopes influence the average atomic mass of elements.

🧩 Key Ideas & Terms (Essential Only)

1. Atomic Mass – the weighted average mass of all the isotopes of an element based on their relative abundance.
2. Relative Abundance – the proportion or percentage of a specific isotope occurring in nature, expressed as a decimal or percent.
3. Isotope Mass – the mass of a particular isotope of an element, usually measured in atomic mass units (amu).
4. Weighted Average – the method used in calculating atomic mass, where each isotope’s mass is multiplied by its relative abundance before summing the results.

🔄 Prior Knowledge (Day 2)

Activity: Recap on Isotopes

1. Define isotopes in your own words.
2. Give two examples of isotopes of the same element.
3. Why do isotopes of the same element have the same chemical properties?

Show Answer

1. Isotopes are atoms of the same element with the same number of protons but different numbers of neutrons.
2. Examples: Carbon-12 & Carbon-14, Hydrogen-1 (Protium) & Hydrogen-2 (Deuterium).
3. They have the same number of protons and electrons, which determine chemical behavior.

📖 Explore the Lesson - Day 2: Calculation of Atomic Mass

1) Why atomic mass matters 🧠

Atoms of the same element can have different numbers of neutrons. These variants are called isotopes, and they have different masses. Understanding how to combine isotope masses and their natural abundances lets us determine the element’s average atomic mass, which is the value listed on the periodic table.

In this lesson, we calculate average atomic mass using a weighted average that multiplies each isotope’s mass by its relative abundance, then sums across all isotopes.

2) Illustration 1 - The formula 🧮

Atomic Mass = Σ (Isotope Mass × Relative Abundance)

• Isotope mass - the mass of a particular isotope.
• Relative abundance - the fraction or percent of that isotope in nature. Use decimal form in calculations.
• Σ (sigma) - add the products for all isotopes.

Important reminders

• Convert percent to decimal before multiplying. Example: 75% becomes 0.75.
• The result is in atomic mass units, amu.
• Round at the end to match significant figures given in the data.

3) Illustration 2 - Worked example: Chlorine’s atomic mass ✅

Chlorine has two common isotopes in this simplified classroom example. We will use rounded masses and abundances to see the method clearly, then compare with more precise data later.

Given:

• Cl-35: mass = 35 amu, relative abundance = 75% → 0.75
• Cl-37: mass = 37 amu, relative abundance = 25% → 0.25

Isotope	Mass (amu)	Relative abundance (decimal)	Product = Mass × Abundance
Cl-35	35	0.75	26.25
Cl-37	37	0.25	9.25
Total			35.50 amu

Result: Average atomic mass of chlorine ≈ 35.5 amu.

Note - In data tables or exams you may be given more precise isotope masses and abundances. The calculation process stays the same.

4) Illustration 3 - Concept picture: Isotopes have same protons, different neutrons 🧩

Carbon-12: p = 6, n = 6, e = 6   → mass number = 12
Carbon-14: p = 6, n = 8, e = 6   → mass number = 14
Same element because protons = 6 in both. Different mass because neutrons differ.
        

Isotopes are atoms of the same element that differ in neutron number, so their masses differ. Chemical properties stay similar because protons and electrons are the same.

5) Illustration 4 - Guided practice table: Boron example with precise data ✍️

Given:

• B-10: mass = 10.0129 amu, abundance = 19.9% → 0.199
• B-11: mass = 11.0093 amu, abundance = 80.1% → 0.801

Isotope	Mass (amu)	Relative abundance (decimal)	Product = Mass × Abundance
B-10	10.0129	0.199	1.9926
B-11	11.0093	0.801	8.8184
Total			10.8110 amu

Rounded answer: 10.81 amu - matches the key.

What you should notice

• Converting percent to decimal is essential.
• The heavier isotope contributes more if its abundance is higher.
• Round only after summing the products.

6) Try one like this - Neon check for understanding 💡

Neon data:

• Ne-20: 19.9924 amu, 90.48% → 0.9048
• Ne-21: 20.9938 amu, 0.27% → 0.0027
• Ne-22: 21.9914 amu, 9.25% → 0.0925

Task: Compute the average atomic mass of Neon.

Target: Your result should round to 20.18 amu.

Show Answer

Convert to decimals: 0.9048, 0.0027, 0.0925.
Products:
19.9924×0.9048 = 18.0962
20.9938×0.0027 = 0.0567
21.9914×0.0925 = 2.0356
Sum = 20.1885 amu → rounds to 20.18 amu (matches key).

7) Common mistakes to avoid 🚫

• Mixing up percent and decimal in multiplications. Always convert percent to decimal.
• Averaging isotope masses arithmetically without weighting by abundance.
• Rounding too early. Keep extra digits until the final step.
• Forgetting units. Always include amu in your final answer.

8) Why this matters - real world links 🌍

• Archaeology: Carbon dating estimates the age of artifacts by measuring Carbon-14 relative to Carbon-12.
• Medicine: Medical imaging such as PET scans uses specific isotopes to trace biological processes inside the body.

9) Summary checklist ✅

• State and use the formula Atomic Mass = Σ (Isotope Mass × Relative Abundance).
• Convert percent abundance to decimal correctly and apply it in calculations.
• Reproduce the chlorine example and explain each step.
• Solve practice problems like Boron and Neon with justified working.

References

• Center for Environmental Studies. Isotopes and temperature measurement - NASA Global Climate Change Education Modules.
• Chang, R., Goldsby, K. A. Chemistry, 12th ed., McGraw-Hill Education.
• Hill, J. W., Kolb, D. K. Chemistry for Changing Times, 9th ed., Prentice Hall.
• U.S. Department of Energy - National Isotope Development Center. Isotope basics.
• Elearnin. Uses of radioactive isotopes - chemistry [Video].

💡 Example in Action

Worked Example: Average atomic mass of Chlorine

Given data
- Chlorine-35: mass = 34.9689 amu, relative abundance = 75.78%
- Chlorine-37: mass = 36.9659 amu, relative abundance = 24.22%

Goal
Compute the average atomic mass using:
Atomic Mass = Σ (isotope mass × relative abundance as a decimal)

Step-by-step solution

1. Convert percent to decimal
   - 75.78% → 0.7578
   - 24.22% → 0.2422
2. Multiply each mass by its decimal abundance
   - 34.9689 × 0.7578 = 26.4949 (keep extra digits until final rounding)
   - 36.9659 × 0.2422 = 8.9710
3. Sum the products
   - 26.4949 + 8.9710 = 35.4659 amu
4. Round suitably (match given data’s precision)
   - Average atomic mass ≈ 35.4658 to 35.466 amu
   - Report as 35.46 amu or 35.4658 amu depending on your teacher’s sig-fig rule.
   Key: The exemplar rounds to 35.4658 amu.

Why this works
The calculation is a weighted average. The more abundant isotope (Cl-35) contributes more to the final average than the less abundant isotope (Cl-37).

Now You Try (mini-task)

Task A - Neon
Use the data to compute Neon’s average atomic mass:
- Ne-20: mass = 19.9924 amu, abundance = 90.48%
- Ne-21: mass = 20.9938 amu, abundance = 0.27%
- Ne-22: mass = 21.9914 amu, abundance = 9.25%

Show Answer

Convert to decimals: 0.9048, 0.0027, 0.0925.
Products:
19.9924×0.9048 = 18.0962
20.9938×0.0027 = 0.0567
21.9914×0.0925 = 2.0356
Sum = 20.1885 amu → rounds to 20.18 amu (matches key).

Task B - Magnesium
Compute the average atomic mass:
- Mg-24: mass = 23.985 amu, abundance = 78.99%
- Mg-25: mass = 24.986 amu, abundance = 10.00%
- Mg-26: mass = 25.983 amu, abundance = 11.01%

Show Answer

Convert to decimals: 0.7899, 0.1000, 0.1101.
Products:
23.985×0.7899 = 18.9456
24.986×0.1000 = 2.4986
25.983×0.1101 = 2.8607
Sum = 24.3049 amu → 24.31 amu (matches key).

Task C - Explain your reasoning
In 2-3 sentences, explain why the calculated average atomic mass of chlorine is closer to 35 than to 37. Reference the data above in your explanation.

Show Answer

Chlorine-35 has a larger relative abundance (75.78 percent) than Chlorine-37 (24.22 percent), so it contributes more to the weighted average. Because 35 amu is multiplied by a larger decimal (0.7578) than 37 amu (0.2422), the final average is pulled closer to 35.

📝 Try It Out (10 items)

Solve and answer the following. Show solutions for calculation problems.

1. Define atomic mass in your own words.
2. State the formula used to calculate the average atomic mass of an element.
3. Why must percent abundance be converted to a decimal before applying the formula?
4. Calculate the atomic mass of Boron:
   • B-10: 10.0129 amu, 19.9%
   • B-11: 11.0093 amu, 80.1%
5. Calculate the atomic mass of Neon:
   • Ne-20: 19.9924 amu, 90.48%
   • Ne-21: 20.9938 amu, 0.27%
   • Ne-22: 21.9914 amu, 9.25%
6. Calculate the atomic mass of Magnesium:
   • Mg-24: 23.985 amu, 78.99%
   • Mg-25: 24.986 amu, 10.00%
   • Mg-26: 25.983 amu, 11.01%
7. Nitrogen has two isotopes:
   • N-14: 14.0031 amu, 99.63%
   • N-15: 15.0001 amu, 0.37%
   Find the average atomic mass of nitrogen.
8. A sample of element X has isotopes X-50 (50 amu, 60%) and X-52 (52 amu, 40%). Compute its average atomic mass.
9. Which isotope contributes more to the average atomic mass of Chlorine: Cl-35 (75.78%) or Cl-37 (24.22%)? Explain briefly.
10. Reflection: Write two real-world applications of isotopes that rely on the concept of atomic mass.

🔑 Answer Key (hidden)

Show Answer

1. Atomic mass is the weighted average mass of all isotopes of an element, based on their relative abundance.
2. Atomic Mass = Σ (Isotope Mass × Relative Abundance).
3. Because the formula requires a fraction, not a percent.
4. Boron = 10.81 amu.
5. Neon = 20.18 amu.
6. Magnesium = 24.31 amu.
7. Nitrogen = 14.01 amu.
8. X = (50×0.60) + (52×0.40) = 30 + 20.8 = 50.8 amu.
9. Cl-35 contributes more because it has a higher abundance (75.78%).
10. Examples: Carbon-14 in archaeology (dating), Technetium-99m in medical imaging, Uranium-235 in nuclear energy.

✅ Check Yourself (10 items, Mixed)

Part A – Multiple Choice (4 items)

1. What does the term atomic mass represent?
   a) The mass of protons only
   b) The average mass of all isotopes of an element, weighted by their abundance
   c) The sum of protons and neutrons only
   d) The most common isotope of an element
2. Which symbol (Σ) in the formula for atomic mass means:
   a) Subtraction
   b) Division
   c) Summation (adding products)
   d) Multiplication
3. Chlorine has two isotopes: Cl-35 and Cl-37. Which isotope contributes more to the average atomic mass?
   a) Cl-35
   b) Cl-37
   c) Both equally
   d) Neither contributes
4. Which of the following is expressed as a decimal fraction in the calculation?
   a) Atomic mass unit (amu)
   b) Isotope mass
   c) Relative abundance
   d) Mass number

Part B – True or False (3 items)

5. Relative abundance should be converted to decimal form before multiplying by isotope mass.
6. The atomic mass of an element is always the simple average of its isotopes.
7. Isotopes of the same element have identical numbers of protons.

Part C – Short Answer (2 items)

8. Define relative abundance in one sentence.
9. Why does the atomic mass of chlorine (35.46 amu) lie closer to 35 than to 37?

Part D – Computational (1 item)

10. A sample of element Z has isotopes: Z-60: 60 amu, 25% and Z-62: 62 amu, 75%. Compute the average atomic mass of element Z. Show your solution.

🔑 Answer Key (hidden)

Show Answer

1. b) The average mass of all isotopes of an element, weighted by their abundance.
2. c) Summation (adding products).
3. a) Cl-35.
4. c) Relative abundance.
5. True.
6. False – it is a weighted average, not simple.
7. True.
8. Relative abundance is the proportion or percentage of an isotope in a sample of the element.
9. Because Cl-35 has higher abundance (75.78%) than Cl-37 (24.22%), pulling the average closer to 35.
10. (60 × 0.25) + (62 × 0.75) = 15 + 46.5 = 61.5 amu.

🚀 Go Further (5 Activities)

1. Think-Pair-Share
   Question: Why is atomic mass often a decimal number on the periodic table instead of a whole number?
   
   Show Answer

   Because atomic mass is a weighted average of isotopes, it rarely equals a whole number.
2. Isotope Detective Game 🔍
   Teacher gives each group an “isotope card” (with mass and abundance). Students calculate the atomic mass of the “mystery element” and identify which element it matches on the periodic table.
   
   Show Answer

   Students’ answers will vary depending on isotope data. Correct match should correspond to the element’s known average atomic mass.
3. Connection to Real Life 🌍
   Research task: Each student finds one application of isotopes in medicine, archaeology, or energy. Share findings in 3–4 sentences in a short journal entry.
   
   Show Answer

   Examples: Carbon-14 in dating fossils, Technetium-99m in PET scans, Uranium-235 in nuclear energy.
4. Math Challenge ✍️
   An element has 3 isotopes:
   X-28: 27.9769 amu, 92.23%
   X-29: 28.9765 amu, 4.67%
   X-30: 29.9738 amu, 3.10%
   Calculate its atomic mass.
   
   Show Answer

   (27.9769×0.9223) + (28.9765×0.0467) + (29.9738×0.0310) = 28.0855 amu → Silicon.
5. Debate: Why Accuracy Matters in Isotope Data ⚖️
   Small groups discuss: “If the relative abundance of isotopes is measured inaccurately, how would it affect industries like medicine and archaeology?” Groups present a short summary of their argument.
   
   Show Answer

   Inaccurate isotope data would mislead atomic mass values, affecting medical diagnostics (wrong tracer doses) and archaeological dating (incorrect ages).

🔗 My Reflection

✍️ Write 3–5 Sentences

In your notebook, write 3–5 sentences reflecting on today’s lesson. You may use the guide below:

• What did I learn about calculating atomic mass?
• Which part of the calculation was easiest or hardest for me?
• How can I apply the idea of isotopes and atomic mass to real life (medicine, archaeology, environment, etc.)?

☑️ Checklist Reflection

Put a ✔️ if true, and ✖️ if false. Then explain briefly in your notebook.

• I can explain why atomic mass is a weighted average.
• I can solve atomic mass problems without forgetting to convert percent to decimal.
• I understand why atomic mass values are not always whole numbers.
• I can connect isotopes to real-world applications like dating fossils or medical scans.