SCI8 Q2W3D3: Atomic mass calculations (Part 2)

🎯 Learning Goals

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

1. Solve at least 4 out of 5 problem-based exercises on atomic mass with accuracy.
2. Interpret data tables of isotopes to explain trends in atomic mass values of elements.
3. Collaborate in small groups to discuss the importance of isotopes and present findings in class within the given time.

🧩 Key Ideas & Terms

1. Average Atomic Mass - the weighted mean of isotope masses, based on their natural abundance.
2. Data Table - a structured format for presenting isotope masses and abundances for calculations.
3. Problem-Solving Strategy - the step-by-step method of converting percentages to decimals, multiplying by isotope mass, summing, and rounding.
4. Collaboration - working in groups to solve isotope problems and interpret real-world applications.

🔄 Prior Knowledge 

Activity: Review of Atomic Mass Calculations

1. Recall the formula for calculating atomic mass.
2. Why must we convert percent abundance into decimal form before using it in the formula?
3. Given the isotopes of Boron:
   • B-10: 10.0129 amu, 19.9%
   • B-11: 11.0093 amu, 80.1%
   Write the first step you would do in solving for Boron’s atomic mass.

Show Answer

1. Atomic Mass = Σ (Isotope Mass × Relative Abundance).
2. Because the formula requires a fraction (decimal), not a percent, for correct multiplication.
3. Convert the percentages to decimals: 19.9% → 0.199 and 80.1% → 0.801.

📖 Explore the Lesson – Day 3: Extended Applications of Atomic Mass

1) Introduction – Why More Practice Matters 🧠

Students often find atomic mass easy in theory but tricky in practice because of:

• Converting percent abundances to decimals correctly.
• Handling significant figures and rounding properly.
• Managing multi-isotope problems (more than two isotopes).
• Interpreting data tables in exams.

In today’s lesson, we will go deeper by solving more complex problems, working step by step, and connecting calculations to real-world uses of isotopes.

2) Formula Refresher 🧮

The formula remains the same:

Atomic Mass = Σ (Isotope Mass × Relative Abundance)

Where:

• Isotope Mass = the actual mass of a particular isotope (in amu).
• Relative Abundance = the proportion of that isotope in nature, expressed as a decimal.

Key reminders:

• Always convert percent (%) → decimal before multiplying. Example: 75% → 0.75
• Keep 4-5 significant figures during calculation, round only at the final step.
• Units are atomic mass units (amu).

3) Guided Example A – Nitrogen’s Atomic Mass ✍️

Given Data

• N-14: 14.0031 amu, 99.63%
• N-15: 15.0001 amu, 0.37%

Step 1. Convert percentages to decimals
99.63% = 0.9963
0.37% = 0.0037

Step 2. Multiply isotope mass × abundance
14.0031 × 0.9963 = 13.9517
15.0001 × 0.0037 = 0.0555

Step 3. Add products
13.9517 + 0.0555 = 14.0072 amu

Final Answer: The average atomic mass of Nitrogen ≈ 14.01 amu

Notice how the answer is very close to 14 because N-14 is overwhelmingly abundant.

4) Guided Example B – Magnesium’s Atomic Mass ⚖️

Given Data

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

Step 1. Convert percentages to decimals
0.7899, 0.1000, 0.1101

Step 2. Multiply isotope mass × abundance
23.985 × 0.7899 = 18.9456
24.986 × 0.1000 = 2.4986
25.983 × 0.1101 = 2.8607

Step 3. Add products
18.9456 + 2.4986 + 2.8607 = 24.3049 amu

Final Answer: The average atomic mass of Magnesium ≈ 24.31 amu

Why is this number closer to 24 than 25 or 26? Because Mg-24 has the highest abundance (78.99%).

5) Worked Challenge – Silicon Example 🔬

Data

• Si-28: 27.9769 amu, 92.23%
• Si-29: 28.9765 amu, 4.67%
• Si-30: 29.9738 amu, 3.10%

Step 1. Convert to decimals
0.9223, 0.0467, 0.0310

Step 2. Multiply
27.9769 × 0.9223 = 25.7939
28.9765 × 0.0467 = 1.3530
29.9738 × 0.0310 = 0.9292

Step 3. Add
25.7939 + 1.3530 + 0.9292 = 28.0761 amu

Final Answer: Silicon’s average atomic mass ≈ 28.09 amu

Why? Because Si-28 dominates at 92.23% abundance.

6) Step-by-Step Strategy for Students 📘

How to approach ANY atomic mass problem:

1. Write down data clearly - isotope mass and abundance.
2. Convert abundance to decimal - never skip this.
3. Multiply mass × abundance for each isotope.
4. Add up all products.
5. Round correctly to significant figures.
6. Check if your answer is close to the most abundant isotope’s mass (for intuition check).

Mnemonic: C-M-A-R-C
Convert - Multiply - Add - Round - Check

7) Common Errors and Fixes 🚫✅

• Using percentages directly in multiplication - Always convert to decimal (e.g., 75% → 0.75)
• Rounding each product too early - Keep full precision until the final answer
• Forgetting units - Always state the result in amu
• Averaging isotope masses without weighting - Always multiply by abundance first

8) Real-World Applications 🌍

• Medicine: PET scans use isotopes like Carbon-11 and Fluorine-18.
• Archaeology: Carbon-14 dating helps estimate the age of fossils.
• Environment: Isotopic oxygen ratios (O-16, O-18) track climate change.
• Energy: Uranium isotopes (U-235, U-238) determine nuclear fuel properties.

In each case, knowing isotopic composition and relative abundance is essential.

9) Extended Practice Problem 💡

Data for Copper (Cu):

• Cu-63: 62.9296 amu, 69.17%
• Cu-65: 64.9278 amu, 30.83%

Task: Calculate the atomic mass of Copper.

Show Answer

Step 1: Convert → 0.6917, 0.3083
Step 2: Multiply → (62.9296 × 0.6917 = 43.5363), (64.9278 × 0.3083 = 20.0131)
Step 3: Add → 43.5363 + 20.0131 = 63.5494 amu
Matches periodic table value of Cu ≈ 63.55 amu.

10) Closing Summary ✅

• Solve two-isotope problems (Chlorine, Nitrogen, Copper).
• Solve three-isotope problems (Magnesium, Silicon).
• Apply a consistent problem-solving strategy.
• Recognize the link between isotopic abundance and atomic mass.
• Understand real-world significance of isotopes in medicine, archaeology, energy, and the environment.

References

• Chang, R., & Goldsby, K. A. (2016). Chemistry (12th ed.). McGraw-Hill Education.
• Hill, J. W., & Kolb, D. K. (2001). Chemistry for Changing Times (9th ed.). Prentice Hall.
• U.S. Department of Energy – National Isotope Development Center. Isotope Basics.
• Elearnin. (2020). Uses of Radioactive Isotopes – Chemistry [Video].

💡 Example in Action – Day 3: Extended Atomic Mass Practice

Worked Example 1 – Oxygen

Data:
O-16: 15.9949 amu, 99.76%
O-17: 16.9991 amu, 0.04%
O-18: 17.9992 amu, 0.20%

Step 1: Convert to decimals
0.9976, 0.0004, 0.0020

Step 2: Multiply
15.9949 × 0.9976 = 15.9561
16.9991 × 0.0004 = 0.0068
17.9992 × 0.0020 = 0.0360

Step 3: Add
15.9561 + 0.0068 + 0.0360 = 15.9989 amu

Matches periodic table value (≈ 16.00 amu).

Worked Example 2 – Carbon

Data:
C-12: 12.0000 amu, 98.89%
C-13: 13.0034 amu, 1.11%

Convert to decimals: 0.9889, 0.0111

Multiply:
12.0000 × 0.9889 = 11.8668
13.0034 × 0.0111 = 0.1447

Add: 11.8668 + 0.1447 = 12.0115 amu

Matches periodic table value ≈ 12.01.

Worked Example 3 – Chlorine

Data:
Cl-35: 34.9689 amu, 75.78%
Cl-37: 36.9659 amu, 24.22%

Convert to decimals: 0.7578, 0.2422

Multiply:
34.9689 × 0.7578 = 26.4949
36.9659 × 0.2422 = 8.9710

Add: 26.4949 + 8.9710 = 35.4659 amu

Matches periodic table value ≈ 35.45.

Worked Example 4 – Copper

Data:
Cu-63: 62.9296 amu, 69.17%
Cu-65: 64.9278 amu, 30.83%

Convert: 0.6917, 0.3083

Multiply:
62.9296 × 0.6917 = 43.5363
64.9278 × 0.3083 = 20.0131

Add: 43.5363 + 20.0131 = 63.5494 amu

Matches periodic table value ≈ 63.55.

Worked Example 5 – Neon

Data:
Ne-20: 19.9924 amu, 90.48%
Ne-21: 20.9938 amu, 0.27%
Ne-22: 21.9914 amu, 9.25%

Convert: 0.9048, 0.0027, 0.0925

Multiply:
19.9924 × 0.9048 = 18.0962
20.9938 × 0.0027 = 0.0567
21.9914 × 0.0925 = 2.0356

Add: 18.0962 + 0.0567 + 2.0356 = 20.1885 amu

Matches periodic table value ≈ 20.18.

✍️ Now You Try (Mini-Tasks)

Task 1 – Boron

B-10: 10.0129 amu, 19.9%
B-11: 11.0093 amu, 80.1%

Show Answer

Convert → 0.199, 0.801
Multiply → 10.0129 × 0.199 = 1.9926; 11.0093 × 0.801 = 8.8184
Add → 1.9926 + 8.8184 = 10.8110 amu ≈ 10.81

Task 2 – Magnesium

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

Show Answer

Convert → 0.7899, 0.1000, 0.1101
Multiply → 18.9456, 2.4986, 2.8607
Add → 24.3049 amu ≈ 24.31

Task 3 – Silicon

Si-28: 27.9769 amu, 92.23%
Si-29: 28.9765 amu, 4.67%
Si-30: 29.9738 amu, 3.10%

Show Answer

Convert → 0.9223, 0.0467, 0.0310
Multiply → 25.7939, 1.3530, 0.9292
Add → 28.0761 amu ≈ 28.09

Task 4 – Gallium

Ga-69: 68.9256 amu, 60.11%
Ga-71: 70.9247 amu, 39.89%

Show Answer

Convert → 0.6011, 0.3989
Multiply → 41.4396, 28.2897
Add → 69.7293 amu ≈ 69.73

Task 5 – Hydrogen

H-1: 1.0078 amu, 99.98%
H-2 (Deuterium): 2.0141 amu, 0.02%

Show Answer

Convert → 0.9998, 0.0002
Multiply → 1.0076, 0.0004
Add → 1.0080 amu ≈ 1.01

📝 Try It Out – Day 3 (10 items)

Show full working for all calculations. Use: Atomic Mass = Σ (isotope mass × relative abundance as decimal).

1. Compute the average atomic mass of Bromine (Br):
   Br-79: 78.9183 amu, 50.69%
   Br-81: 80.9163 amu, 49.31%

Show Answer

Decimals: 0.5069, 0.4931
Products: 78.9183×0.5069 = 39.9890; 80.9163×0.4931 = 39.8847
Sum = 79.8737 amu ≈ 79.87

2. Compute the average atomic mass of Lithium (Li):
   Li-6: 6.0151 amu, 7.59%
   Li-7: 7.0160 amu, 92.41%

Show Answer

Decimals: 0.0759, 0.9241
Products: 6.0151×0.0759 = 0.456; 7.0160×0.9241 = 6.486
Sum = 6.942 amu ≈ 6.94

3. Reverse problem – find missing abundance (2 isotopes)
   Element Q has isotopes Q-100 (100.000 amu) and Q-102 (102.000 amu). The average atomic mass is 101.20 amu. If the abundance of Q-100 is x, what are the abundances of each isotope?

Show Answer

101.20 = 100.000×x + 102.000×(1−x)
101.20 = 100x + 102 − 102x = 102 − 2x
2x = 102 − 101.20 = 0.80 → x = 0.40
Q-100 = 40%, Q-102 = 60%

4. Three-isotope mix – Germanium (Ge)
   Ge-70: 69.9242 amu, 20.52%
   Ge-72: 71.9221 amu, 27.45%
   Ge-74: 73.9212 amu, 36.52%
   (Ignore minor isotopes for this exercise.)
   Find Ge’s average atomic mass for this subset.

Show Answer

Decimals: 0.2052, 0.2745, 0.3652
Products: 69.9242×0.2052=14.341; 71.9221×0.2745=19.747; 73.9212×0.3652=26.992
Sum = 61.080 amu (subset value; full-periodic value would include other isotopes)

5. Error check – what went wrong?
   A student computed Mg atomic mass using:
   (23.985×78.99) + (24.986×10.00) + (25.983×11.01) = 2403.49 → 2403.49 amu.
   Identify the mistake and show the correct first step.

Show Answer

Mistake: Used percent values directly instead of decimals.
Correct first step: Convert to decimals → 0.7899, 0.1000, 0.1101.

6. Reasoning – which pulls the average more?
   Element R has isotopes: R-50 (50 amu, 10%), R-52 (52 amu, 35%), R-55 (55 amu, 55%).
   Which isotope has the largest influence on the average atomic mass, and why?

Show Answer

R-55, because it has the highest relative abundance (55%), so its mass contributes the most to the weighted average.

7. Find a missing mass
   Element S has two isotopes. S-29 has mass 28.976 amu with 4.70% abundance. The other isotope S-30 has 95.30% abundance. The average atomic mass is 28.090 amu. Find the mass of S-30.

Show Answer

28.090 = (28.976×0.0470) + (m×0.9530)
28.090 = 1.3619 + 0.9530m
0.9530m = 26.7281 → m = 28.044 amu (≈ 28.04)

8. Data-table interpretation – which two to approximate?
   You must do a fast mental estimate for Cl average atomic mass. Which two numbers would you multiply first and why?
   Cl-35: 34.97 amu, 75.78%
   Cl-37: 36.97 amu, 24.22%

Show Answer

34.97 × 0.7578 first, because its abundance is largest and will dominate the weighted average; then add 36.97 × 0.2422.

9. Four-isotope practice – Tin (Sn)
   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%
   Compute the average atomic mass for this subset only.

Show Answer

Decimals: 0.1454, 0.2422, 0.0859, 0.3258
Products:
115.902×0.1454=16.855
117.901×0.2422=28.550
118.903×0.0859=10.206
119.902×0.3258=39.048
Sum = 94.659 amu (subset; full value differs with remaining isotopes)

10. Word problem – tuning composition
    A lab adjusts isotope mix of element T from:
    T-20 (20 amu, 60%) and T-22 (22 amu, 40%)
    to a new mix: T-20 (55%) and T-22 (45%).
    By how much does the average atomic mass change?

Show Answer

Old avg: (20×0.60)+(22×0.40)=12+8.8=20.8 amu
New avg: (20×0.55)+(22×0.45)=11+9.9=20.9 amu
Increase of 0.1 amu.

✅ Check Yourself – Day 3 (10 Items, Mixed)

Part A – Multiple Choice (4 items)

1. The atomic mass of an element is:
   a) The mass of its most abundant isotope only
   b) The simple average of all isotope masses
   c) The weighted average of isotope masses based on natural abundances
   d) The total number of protons and neutrons
2. Why is chlorine’s atomic mass closer to 35 than to 37?
   a) Cl-35 is lighter than Cl-37
   b) Cl-35 has higher abundance than Cl-37
   c) Both isotopes are equally abundant
   d) The periodic table rounds toward the smaller mass
3. Which isotope would influence the average atomic mass the most?
   a) An isotope with high mass but very low abundance
   b) An isotope with low mass but high abundance
   c) An isotope with equal protons and neutrons
   d) An isotope with 0% abundance
4. When solving atomic mass problems, why must percent abundances be converted to decimals?
   a) To make the numbers smaller
   b) To avoid division by zero
   c) Because percentages are fractions of 100 and the formula requires fractions
   d) To change units from amu to grams

Part B – True or False (3 items)

5. The average atomic mass can never be a decimal.
6. An isotope with higher abundance has more influence on the average atomic mass.
7. If all isotopes of an element had equal abundances, the average atomic mass would equal the simple mean of their masses.

Part C – Short Answer (2 items)

8. Define average atomic mass in one sentence.
9. Explain why isotopes of the same element have the same chemical properties even though they have different masses.

Part D – Computational (1 item)

10. Calculate the atomic mass of Gallium (Ga):
    Ga-69: 68.9256 amu, 60.11%
    Ga-71: 70.9247 amu, 39.89%

🔑 Answer Key (hidden)

Show Answer

1. c) The weighted average of isotope masses based on natural abundances.
2. b) Cl-35 has higher abundance than Cl-37.
3. b) An isotope with low mass but high abundance.
4. c) Because percentages are fractions of 100 and the formula requires fractions.
5. False – it is often a decimal.
6. True.
7. True.
8. Average atomic mass is the weighted average mass of all isotopes of an element, based on their relative abundance.
9. Because isotopes have the same number of protons and electrons, which determine chemical properties.
10. Convert: 0.6011, 0.3989 → (68.9256×0.6011=41.4396) + (70.9247×0.3989=28.2897) = 69.7293 amu ≈ 69.73.

🚀 Go Further – Day 3 (5 Activities)

1. Isotope Role-Play 🎭

Students role-play as isotopes of the same element. Each “isotope” carries a placard showing its mass and abundance. Together, the class “calculates” the element’s average atomic mass by adding up the contributions.

Show Answer

Example: 3 students represent Cl-35 (mass 35, abundance 75%) and 1 student represents Cl-37 (mass 37, abundance 25%). The class combines their contributions to approximate 35.5 amu.

2. Real-Life Applications Journal 🌍

Research and write a short journal entry on one real-life application of isotopes. Connect it with atomic mass (e.g., carbon-14 dating, uranium isotopes in energy, or technetium in medicine).

Show Answer

Examples: Carbon-14 is used in archaeology to date fossils, Technetium-99m in medical imaging, Uranium-235 for nuclear reactors.

3. Group Challenge Quiz 📝

In groups, solve a 3-isotope problem (e.g., Silicon or Magnesium). Each group explains the steps aloud to the class.

Show Answer

For Silicon (Si-28, Si-29, Si-30), the class should arrive at ~28.09 amu.

4. Error Analysis Task ❌✅

Teacher provides a sample incorrect solution where a student mistakenly used percentages instead of decimals. Learners identify the error and rewrite the correct solution.

Show Answer

Correction: Convert percent abundance to decimals (e.g., 75% → 0.75) before multiplying.

5. Debate – Accuracy in Science ⚖️

Class debate: “Small errors in isotope abundance measurement don’t matter much in science.” One group argues “for,” the other argues “against.”

Show Answer

Strong arguments show that small measurement errors can cause incorrect results in medical isotope dosages, dating artifacts, or predicting nuclear properties.

🔗 My Reflection – Day 3

✍️ Guiding Questions

Answer these in your notebook (3-5 sentences each):

1. What strategies helped me calculate atomic mass more accurately today?
2. Which example or exercise did I find most challenging, and how did I solve it?
3. How does knowing isotopes and their abundances help me understand the periodic table better?
4. In what ways can I connect isotope knowledge to real-life applications like medicine or archaeology?

☑️ Checklist Reflection

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

• I can solve atomic mass problems involving two isotopes.
• I can solve atomic mass problems involving three or more isotopes.
• I know how to spot and fix common calculation errors.
• I can explain why isotopes with higher abundance pull the average closer to their mass.
• I can give at least one real-world example where isotopes are important.