SC8 Q2W3D4: Atomic mass calculation

🎯 Learning Goals

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

1. Accurately solve at least 8 out of 10 advanced isotope problems involving missing masses or abundances within the allotted class period.
2. Analyze isotope data tables and explain how isotopic composition affects the periodic table values of atomic mass.
3. Work collaboratively in pairs or groups to present real-world applications of isotopes in fields such as medicine, archaeology, and energy.

🧩 Key Ideas & Terms (Essential Only)

1. Weighted Average – the calculation method that considers both mass and relative abundance of isotopes.
2. Relative Abundance – the fraction of each isotope present in nature, expressed as a decimal.
3. Significant Figures – the rule for keeping precision and accuracy in atomic mass calculations.
4. Isotopic Composition – the natural distribution of isotopes in an element that determines its atomic mass.
5. Periodic Table Value – the published atomic mass that reflects weighted averages of isotopes in nature.

🔄 Prior Knowledge (Day 4)

Activity: Building on Atomic Mass Basics

1. Recall: What is the formula used to calculate the average atomic mass of an element?
2. Why do we convert percent abundances into decimals before multiplying with isotope masses?
3. If an element has 2 isotopes, one very abundant and one very rare, which isotope will pull the average atomic mass closer to its value? Explain briefly.

Show Answer

1. Formula: Atomic Mass = Σ (Isotope Mass × Relative Abundance).
2. Because the formula requires fractions, not percentages, for correct multiplication.
3. The more abundant isotope pulls the average closer to its value, since its contribution is larger in the weighted average.

📖 Explore the Lesson – Day 4: Mastery and Real-World Applications of Atomic Mass

1) Introduction – Mastering the Concept 🎯

By Day 4, you already know how to:

• Use the formula for average atomic mass
• Convert percent abundances into decimals
• Multiply and sum correctly for 2- or 3-isotope elements

But mastery goes further:

• Applying the formula to reverse problems (find missing abundance or isotope mass)
• Handling real-world datasets with multiple isotopes
• Explaining why atomic mass values are decimals on the periodic table
• Linking isotopes to practical applications in medicine, archaeology, and industry

Today’s focus is critical problem-solving and interpretation, not just calculation.

2) The Formula – Full Explanation 🧮

The key relationship remains:

Atomic Mass = Σ (Isotope Mass × Relative Abundance)

Where:

• Isotope Mass (amu) – the measured mass of an isotope
• Relative Abundance – expressed as a decimal fraction (not percent)

Example Conversion:
75.0% → 0.750
10.0% → 0.100

Why decimals?
Because the formula is essentially a weighted average. Percentages must be divided by 100 to properly weight contributions.

3) Advanced Example A – Missing Abundance ✍️

Problem:
Element X has 2 isotopes:

• X-10: 10.012 amu, abundance = ?
• X-11: 11.009 amu, abundance = 65%

The average atomic mass is 10.81 amu. Find the abundance of X-10.

Solution:
Step 1. Let abundance of X-10 = x, so abundance of X-11 = 0.65. Since total = 1, then x = 0.35.

Step 2. Apply formula:
10.012(x) + 11.009(0.65) = 10.81

Step 3. Compute:
10.012x + 7.1559 = 10.81
10.012x = 3.6541
x ≈ 0.365 → 36.5%

So, X-10 = 36.5% and X-11 = 63.5%.

4) Advanced Example B – Missing Isotope Mass 🧩

Problem:
Element Y has 2 isotopes:

• Y-20: mass = 19.998 amu, abundance = 60%
• Y-? : mass = ?, abundance = 40%

Average atomic mass = 20.40 amu.

Solution:
20.40 = (19.998 × 0.60) + (m × 0.40)
20.40 = 11.9988 + 0.40m
0.40m = 20.40 − 11.9988 = 8.4012
m = 8.4012 ÷ 0.40 = 21.00 amu

5) Advanced Example C – Multi-Isotope Challenge ⚖️

Problem:
Element Z has 3 isotopes:

• Z-30: 29.978 amu, 50%
• Z-31: 30.983 amu, 30%
• Z-32: 31.974 amu, 20%

Solution:
Atomic Mass = (29.978 × 0.50) + (30.983 × 0.30) + (31.974 × 0.20)
= 14.989 + 9.295 + 6.395 = 30.679 amu

6) Common Pitfalls – Advanced Cases 🚫✅

1. Forgetting total abundance = 1.00 – Always check: x + y + z = 1.
2. Using percentages directly – Wrong: 29.978 × 50; Right: 29.978 × 0.50.
3. Rounding too early – Keep 4–5 sig figs until the final sum.
4. Mixing mass number with isotope mass – Use measured isotope mass (decimal), not the whole-number mass number.

7) Real-World Connections 🌍

• Medicine: Iodine-131 for thyroid treatment; Carbon-11 & Fluorine-18 for PET scans.
• Archaeology: Carbon-14 dating for fossils.
• Energy: Uranium-235 vs Uranium-238 in nuclear power.
• Environment: Oxygen isotope ratios in ice cores to reveal climate history.

In all cases, accuracy in isotope measurement is vital.

8) Extended Practice Problem 💡

Problem:
Chlorine has isotopes:

• Cl-35: 34.969 amu, 75.78%
• Cl-37: 36.966 amu, 24.22%

Calculate its average atomic mass.

Solution:
(34.969 × 0.7578) + (36.966 × 0.2422)
= 26.494 + 8.971 = 35.465 amu
Matches periodic table value ≈ 35.45.

9) Closing Mastery Summary ✅

• Solve direct problems (given masses + abundances → atomic mass).
• Solve reverse problems (find missing abundance or isotope mass).
• Handle 3+ isotope datasets confidently.
• Avoid common mistakes (percent → decimal, rounding errors).
• Explain why isotopes make atomic masses decimals, not whole numbers.
• Connect isotope knowledge to real-world science and technology.

References

• Zumdahl, S. & Zumdahl, S. (2014). Chemistry: An Atoms First Approach. Cengage Learning.
• Chang, R. (2010). General Chemistry: The Essential Concepts. McGraw-Hill.
• U.S. Department of Energy – National Isotope Development Center. Isotope Basics.
• International Atomic Energy Agency (IAEA) – Isotopes in Medicine and Environment.

💡 Example in Action – Day 4: Mastery Problems

Worked Example 1 – Average Atomic Mass (Direct)

Data:
Li-6: 6.015 amu, 7.59%
Li-7: 7.016 amu, 92.41%

Solution:
(6.015 × 0.0759) + (7.016 × 0.9241) = 0.456 + 6.486 = 6.942 amu
Matches periodic table (≈ 6.94).

Worked Example 2 – Missing Abundance

Data:
Element Q has 2 isotopes:
Q-10: 10.012 amu, abundance = ?
Q-11: 11.009 amu, abundance = ?
Average atomic mass = 10.81 amu.

Solution:
Let abundance of Q-10 = x, Q-11 = 1 − x.
10.012x + 11.009(1−x) = 10.81
10.012x + 11.009 − 11.009x = 10.81
−0.997x = −0.199 → x = 0.199 (19.9%)
So Q-10 = 19.9%, Q-11 = 80.1%.

Worked Example 3 – Missing Isotope Mass

Data:
Element R has 2 isotopes:
R-20: 19.998 amu, 60%
R-?: ? amu, 40%
Average atomic mass = 20.40 amu.

Solution:
20.40 = (19.998 × 0.60) + (m × 0.40)
20.40 = 11.999 + 0.40m
0.40m = 8.401 → m = 21.00 amu.

Worked Example 4 – Multi-Isotope Element

Data:
Silicon:
Si-28: 27.977 amu, 92.23%
Si-29: 28.976 amu, 4.67%
Si-30: 29.974 amu, 3.10%

Solution:
(27.977 × 0.9223) + (28.976 × 0.0467) + (29.974 × 0.0310)
= 25.793 + 1.353 + 0.929 = 28.075 amu
Matches periodic table (≈ 28.09).

Worked Example 5 – Reverse Multi-Isotope Problem

Data:
Element X has 3 isotopes:
X-30: 29.978 amu, 50%
X-31: 30.983 amu, 30%
X-32: mass = ? , 20%
Average atomic mass = 30.68 amu.

Solution:
30.68 = (29.978 × 0.50) + (30.983 × 0.30) + (m × 0.20)
30.68 = 14.989 + 9.295 + 0.20m
30.68 = 24.284 + 0.20m
0.20m = 6.396 → m = 31.98 amu.

✍️ Now You Try (Mini-Tasks)

Task 1 – Magnesium (Direct Calculation)

Mg-24: 23.985 amu, 78.99%
Mg-25: 24.986 amu, 10.00%
Mg-26: 25.983 amu, 11.01%

Show Answer

(23.985×0.7899) + (24.986×0.1000) + (25.983×0.1101) = 18.946 + 2.499 + 2.861 = 24.31 amu

Task 2 – Missing Abundance (2 Isotopes)

Element Y has isotopes:
Y-50: 49.995 amu, abundance = x
Y-52: 51.995 amu, abundance = 1 − x
Average atomic mass = 51.20 amu.

Show Answer

49.995x + 51.995(1−x) = 51.20 → x = 0.40 → Y-50 = 40%, Y-52 = 60%

Task 3 – Missing Mass (2 Isotopes)

Element Z has isotopes:
Z-60: 59.998 amu, 70%
Z-?: ? amu, 30%
Average atomic mass = 61.20 amu.

Show Answer

61.20 = (59.998×0.70) + (m×0.30) → 61.20 = 41.999 + 0.30m → m = 64.00 amu

Task 4 – Chlorine Check

Cl-35: 34.969 amu, 75.78%
Cl-37: 36.966 amu, 24.22%

Show Answer

(34.969×0.7578) + (36.966×0.2422) = 26.495 + 8.971 = 35.466 amu

Task 5 – Reverse 3-Isotope

Element W has isotopes:
W-100: 99.995 amu, 50%
W-101: 100.995 amu, 20%
W-102: ? amu, 30%
Average atomic mass = 101.40 amu.

Show Answer

101.40 = 49.998 + 20.199 + 0.30m → 0.30m = 31.203 → m = 104.01 amu

📝 Try It Out – Day 4 (10 Items)

Show full working for all calculations. Formula reminder: Atomic Mass = Σ (Isotope Mass × Relative Abundance)

1. Krypton (Direct Calculation)

Kr-78: 77.920 amu, 0.35%
Kr-80: 79.916 amu, 2.28%
Kr-82: 81.913 amu, 11.58%
Kr-83: 82.914 amu, 11.49%
Kr-84: 83.911 amu, 57.00%
Kr-86: 85.910 amu, 17.30%
Compute the average atomic mass of Krypton.

Show Answer

Weighted average ≈ 83.80 amu (matches periodic table).

2. Two-Isotope Problem

X-10: 10.012 amu, 19.9%
X-11: 11.009 amu, 80.1%
Find the atomic mass.

Show Answer

(10.012×0.199) + (11.009×0.801) = 1.993 + 8.818 = 10.811 amu

3. Missing Abundance (2 isotopes)

Y-50: 49.995 amu, x%
Y-52: 51.995 amu, (100−x)%
Average atomic mass = 51.20 amu. Find x.

Show Answer

x = 40% for Y-50, 60% for Y-52.

4. Missing Mass (2 isotopes)

Z-60: 59.998 amu, 70%
Z-?: ? amu, 30%
Average atomic mass = 61.20 amu.

Show Answer

64.00 amu

5. Multi-Isotope Calculation – Tin (Subset)

Sn-116: 115.902 amu, 14.54%
Sn-118: 117.901 amu, 24.22%
Sn-119: 118.903 amu, 8.59%
Sn-120: 119.902 amu, 32.58%
Calculate the subset atomic mass.

Show Answer

94.66 amu (subset only; full atomic mass ≈ 118.71).

6. Error Identification

A student solved Chlorine’s atomic mass as: (34.969×75.78) + (36.966×24.22) = 4299.5 amu. What mistake did they make?

Show Answer

They used percent values directly. Should convert to decimals (0.7578, 0.2422).

7. Reverse Problem – Missing Mass (3 isotopes)

Q-30: 29.978 amu, 50%
Q-31: 30.983 amu, 30%
Q-?: ? amu, 20%
Average atomic mass = 30.68 amu. Find the missing mass.

Show Answer

31.98 amu

8. Boron Problem

B-10: 10.0129 amu, 19.9%
B-11: 11.0093 amu, 80.1%
Compute the atomic mass of Boron.

Show Answer

(10.0129×0.199) + (11.0093×0.801) = 10.81 amu

9. Reasoning Question

Why is Chlorine’s atomic mass (35.45) closer to 35 than 37?

Show Answer

Because Cl-35 has higher natural abundance (75.78%) compared to Cl-37 (24.22%).

10. Real-World Application

Carbon has isotopes C-12, C-13, and C-14. Which isotope is used for dating fossils and why?

Show Answer

C-14, because it is radioactive and decays at a known rate, allowing scientists to estimate the age of ancient materials.

✅ Check Yourself – Day 4 (10 Items, Mixed)

Part A – Multiple Choice (4 items)

1. The atomic mass of chlorine is 35.45 amu. This value indicates that:
   a) All chlorine atoms weigh exactly 35.45 amu
   b) Chlorine atoms are 50% Cl-35 and 50% Cl-37
   c) It is the weighted average of Cl-35 and Cl-37 isotopes
   d) Chlorine has only one isotope
2. Which step is always first in solving average atomic mass problems?
   a) Add isotope masses directly
   b) Multiply isotope mass × abundance
   c) Convert percent abundances into decimals
   d) Round to significant figures
3. If an element has two isotopes, one with 90% abundance and the other with 10% abundance, the average atomic mass will be:
   a) Exactly halfway between both isotope masses
   b) Closer to the isotope with 90% abundance
   c) Closer to the isotope with 10% abundance
   d) The larger of the two isotope masses
4. Which real-world field uses carbon-14 to determine the age of fossils?
   a) Medicine
   b) Archaeology
   c) Nuclear engineering
   d) Astronomy

Part B – True or False (3 items)

5. Average atomic mass values on the periodic table are often decimals because they account for isotopes.
6. If isotope abundances are not converted to decimals, the calculated atomic mass will still be correct.
7. An isotope with higher abundance will have greater influence on the average atomic mass.

Part C – Short Answer (2 items)

8. Write the general formula for calculating average atomic mass.
9. Explain why isotopes of the same element have identical chemical properties but different masses.

Part D – Computational (1 item)

10. Gallium has two isotopes:
    Ga-69: 68.926 amu, 60.11%
    Ga-71: 70.925 amu, 39.89%
    Calculate its average atomic mass.

🔑 Answer Key (hidden)

Show Answer

1. c) It is the weighted average of Cl-35 and Cl-37 isotopes.
2. c) Convert percent abundances into decimals.
3. b) Closer to the isotope with 90% abundance.
4. b) Archaeology.
5. True.
6. False – it will be incorrect.
7. True.
8. Formula: Atomic Mass = Σ (Isotope Mass × Relative Abundance).
9. They have the same number of protons and electrons, which control chemical behavior; only neutrons differ, affecting mass.
10. (68.926×0.6011) + (70.925×0.3989) = 41.44 + 28.29 = 69.73 amu.

🚀 Go Further – Day 4 (3 Activities)

1. Real-World Research 🌍

Research how isotopes are used in nuclear medicine or archaeology. Write a one-page report showing how accurate atomic mass calculations are important in that field.

Show Answer

Example: Carbon-14 dating relies on the isotope’s half-life; nuclear medicine uses precise isotope doses (e.g., Iodine-131, Technetium-99m).

2. Group Debate ⚖️

Debate statement: “Errors in atomic mass calculation are minor and do not affect science significantly.”
Divide the class into For and Against groups. Each must provide evidence-based reasoning.

Show Answer

Strong rebuttal: Even small errors in isotope abundance measurement affect medicine (dosages), archaeology (dating), and nuclear energy (fuel composition).

3. Design a Quiz Game 🎲

In small groups, create 5 original atomic mass questions (direct calculation, missing mass, reasoning). Exchange with another group and compete in solving.

Show Answer

Students should include at least:
- 2 direct calculation problems
- 1 missing abundance
- 1 missing mass
- 1 reasoning question

🔗 My Reflection – Day 4

✍️ Guiding Questions

Answer in your notebook (3–5 sentences each):

1. Which type of atomic mass problem (direct, missing abundance, missing mass, multi-isotope) did I find most challenging, and how did I solve it?
2. How does practicing more complex problems help me understand the periodic table values better?
3. In what way can accurate isotope calculations impact real-world fields like medicine, energy, or archaeology?

☑️ Checklist Reflection

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

• I can correctly calculate average atomic mass for 2-isotope elements.
• I can solve problems involving missing abundance or missing isotope mass.
• I can handle 3 or more isotopes in calculations with confidence.
• I understand why atomic mass values are decimals on the periodic table.
• I can explain at least one real-world application of isotopes.