Calculation of atomic mass from isotopic abundance: Worked examples

The scenario matrix

Every problem this topic can throw at you is one of these cell classes:

Cell	Case class	What is unknown?	The twist
A	Forward, 2 isotopes, % given	$A_{r}$	none — the baseline
B	Forward, 3+ isotopes	$A_{r}$	more terms to add
C	Backward, solve for abundance	one $f_{i}$	one equation, one constraint
D	Percentages don't sum to 100	$A_{r}$	must divide by the real total
E	Degenerate: single isotope	$A_{r}$	answer is that isotope's mass
F	Limiting / bounds check	is my answer sane?	must lie between the extremes
G	Word problem (marbles / real sample)	$A_{r}$ or count	translate words → $m_{i}, f_{i}$
H	Exam twist: mass number vs true mass, or missing isotope	$A_{r}$ or a missing $f$	don't confuse $A$ with $m$ ; fill the gap

The examples below hit cells A, B, C, D, E, F, G, H in order. Symbols used everywhere:

Cell A — Forward, two isotopes, percentages

Worked example Copper (Cell A)

$^{63}$ Cu: mass $= 62.930$ u, abundance $= 69.15%$ . $^{65}$ Cu: mass $= 64.928$ u, abundance $= 30.85%$ . Find $A_{r}$ .

Forecast: Both percentages sum to $100$ . $^{63}$ Cu is roughly twice as common, so guess an answer nearer $63$ than $65$ — around $63.5$ ?

Step 1 — Multiply each mass by its percentage. Why this step? A weighted average lets the abundant isotope "vote" more; multiplying by the abundance is the vote. $62.930 \times 69.15 = 4351.6, 64.928 \times 30.85 = 2003.0$

Step 2 — Add the votes and divide by the total weight (100). Why this step? The percentages add to $100$ , so dividing by $100$ turns the total back into a per-atom average (the $/100$ is the " $\div$ total weight" from the parent derivation). $A_{r} = \frac{4351.6 + 2003.0}{100} = \frac{6354.6}{100} = 63.55 u$

Verify: $63.55$ lies between $62.930$ and $64.928$ ✔️ and leans toward the lighter, more abundant isotope ✔️. The periodic table lists Cu as $63.55$ — match.

Cell B — Forward, three isotopes

Cell C — Backward, solve for the abundance

Worked example Gallium — find the percentages (Cell C)

Gallium has $^{69}$ Ga (68.926 u) and $^{71}$ Ga (70.925 u) and a known $A_{r} = 69.723$ u. What is the percent abundance of each?

Forecast: $69.723$ is closer to $68.926$ (the light one). So the light isotope must be the majority — expect $^{69}$ Ga above 50%.

Step 1 — Use one variable for both abundances. Why this step? The two fractions must sum to 1, so if $f$ is the fraction of $^{69}$ Ga, then $^{71}$ Ga is $1 - f$ . One constraint kills one unknown. $A_{r} = 68.926 f + 70.925 (1 - f)$

Step 2 — Expand and collect. Why this step? We want $f$ alone, so gather the $f$ terms. $69.723 = 70.925 - 1.999 f$

Step 3 — Solve for $f$ . Why this step? Basic algebra: isolate $f$ . $1.999 f = 70.925 - 69.723 = 1.202 \Rightarrow f = \frac{1.202}{1.999} = 0.601$ So $^{69}$ Ga $\approx 60.1%$ and $^{71}$ Ga $\approx 39.9%$ .

Verify: plug back — $68.926 (0.601) + 70.925 (0.399) = 69.72$ ✔️, and the light isotope is indeed the majority as forecast ✔️.

Cell D — Percentages that don't sum to 100

Worked example A trace-contaminated reading (Cell D)

A mass-spectrometry readout gives isotope X-63 (62.930 u, 68.50%) and X-65 (64.928 u, 30.50%). These sum to only $99.00%$ (the missing 1% was unreadable noise). Estimate $A_{r}$ from the two clean peaks.

Forecast: If we lazily divide by 100 we'd get an answer that's slightly too small, because we'd be treating the missing 1% as if it had zero mass. Dividing by the true total (99) fixes this.

Step 1 — Compute the weighted sum. Why this step? Same numerator as always: mass × its weight, summed. $62.930 \times 68.50 = 4310.7, 64.928 \times 30.50 = 1980.3$ $sum = 6291.0$

Step 2 — Divide by the actual total weight, not 100. Why this step? A proper average divides by the sum of the weights you actually used. Here that sum is $99.00$ , not $100$ . This is the $A_{r} = \frac{\sum m _{i} p _{i}}{\sum p _{i}}$ form. $A_{r} = \frac{6291.0}{99.00} = 63.55 u$

Verify: if we'd wrongly divided by 100 we'd get $62.91$ — too low, and outside the range near the light isotope. Dividing by $99$ gives $63.55$ , safely between $62.930$ and $64.928$ ✔️.

Cell E — Degenerate case: only one isotope

Cell F — Limiting behaviour: how the average slides between extremes

The weighted average is a slider. As the light isotope's fraction $f$ runs from $0$ to $1$ , $A_{r}$ slides from the heavy mass all the way down to the light mass. The picture below shows this straight-line slide for chlorine ( $^{35}$ Cl at $34.969$ u, $^{37}$ Cl at $36.966$ u).

Figure — Calculation of atomic mass from isotopic abundance

Worked example Reading the limits off the line (Cell F)

Using $A_{r} (f) = 34.969 f + 36.966 (1 - f)$ for chlorine, evaluate the three landmark points $f = 0$ , $f = 0.5$ , $f = 0.7577$ .

Forecast: at $f = 0$ (no light atoms) the answer must be the heavy mass $36.966$ ; at $f = 1$ the light mass $34.969$ ; at $f = 0.5$ the plain midpoint.

Step 1 — $f = 0$ (all heavy). Why this step? It tests the top of the slider. $A_{r} = 34.969 (0) + 36.966 (1) = 36.966 u$

Step 2 — $f = 0.5$ (equal mix). Why this step? Equal abundances is the only case where the naive midpoint is correct. $A_{r} = 34.969 (0.5) + 36.966 (0.5) = 35.968 u$

Step 3 — $f = 0.7577$ (natural chlorine). Why this step? The real-world value — it must sit low, near $35$ . $A_{r} = 34.969 (0.7577) + 36.966 (0.2423) = 35.453 u$

Verify: all three lie in $[34.969, 36.966]$ ✔️, and they decrease as $f$ (fraction of light isotope) increases ✔️ — exactly the downward line in the figure.

Cell G — Word problem (translate the story)

Worked example The marble bag made real (Cell G)

A tray holds $5000$ atoms of element Z. Analysis shows $3800$ are $^{85}$ Z (84.912 u) and the rest are $^{87}$ Z (86.909 u). Find $A_{r}$ without first converting to percentages.

Forecast: most atoms ( $3800/5000 = 76%$ ) are the light one, so $A_{r}$ leans toward $84.9$ .

Step 1 — Turn counts into fractions. Why this step? Fraction = count ÷ total. This is the "fraction × total = count" step from the parent derivation, run backwards. $f_{85} = \frac{3800}{5000} = 0.760, f_{87} = \frac{5000 - 3800}{5000} = \frac{1200}{5000} = 0.240$

Step 2 — Weighted sum with fractions (no $/100$ needed). Why this step? Fractions already sum to $1$ , so we do not divide by anything — that's the beauty of the fraction form. $A_{r} = 84.912 (0.760) + 86.909 (0.240)$ $= 64.533 + 20.858 = 85.39 u$

Verify: answer sits between $84.912$ and $86.909$ and near the light end ✔️. (This element is rubidium; true $A_{r} \approx 85.47$ — our tray's proportions are close to nature.)

Cell H — Exam twist: a missing isotope + mass-number trap

Worked example Three isotopes, one abundance hidden (Cell H)

Element M has three isotopes with true masses $23.985$ , $24.986$ , $25.983$ u. You are told the first is $79.00%$ and the second is $10.00%$ , and that $A_{r} = 24.31$ u. Find the percentage of the third isotope — and note why you must not just use mass numbers $24, 25, 26$ .

Forecast: the three abundances must sum to $100$ , so the third is $100 - 79 - 10 = 11%$ . But the question wants us to derive it from $A_{r}$ to confirm consistency.

Step 1 — Name the unknown. Why this step? Let $p_{3}$ be the missing percentage. The constraint $\sum p_{i} = 100$ is available, but let's confirm it via $A_{r}$ instead.

Step 2 — Write the $A_{r}$ equation with the unknown. Why this step? $A_{r}$ ties all three isotopes together. $24.31 = \frac{23.985 ( 79.00 ) + 24.986 ( 10.00 ) + 25.983 p _{3}}{100}$

Step 3 — Isolate $p_{3}$ . Why this step? Standard algebra. First the known part of the numerator: $23.985 (79.00) = 1894.8, 24.986 (10.00) = 249.86$ $2431 = 1894.8 + 249.86 + 25.983 p_{3}$ $25.983 p_{3} = 2431 - 2144.66 = 286.34 \Rightarrow p_{3} = \frac{286.34}{25.983} = 11.02$ So the third isotope is $\approx 11%$ ✔️ (matches $100 - 79 - 10$ ).

The mass-number trap: if you had used $24, 25, 26$ (mass numbers) instead of the true masses, the sum $\frac{24 ( 79 ) + 25 ( 10 ) + 26 ( 11 )}{100} = 24.32$ would come out slightly wrong. Mass number $A$ is a whole-number count of nucleons; the true isotopic mass is a hair smaller because of nuclear binding-energy mass defect. Use the true masses when given.

Verify: with $p_{3} = 11.02$ , $A_{r} = \frac{1894.8 + 249.86 + 25.983 ( 11.02 )}{100} = 24.31$ ✔️, and $79 + 10 + 11.02 = 100.02 \approx 100$ ✔️.

Recall

Recall Which cell is each of these?

a) "Element has isotopes at 20% and 80%, find atomic mass." ::: Cell A (forward, 2 isotopes). b) "Given $A_{r}$ , find how much of each isotope." ::: Cell C (backward). c) "Peaks sum to 98%." ::: Cell D (divide by $\sum p$ , not 100). d) "Element has one stable isotope." ::: Cell E (degenerate; $A_{r}$ = that mass).

Recall Blitz answers

Copper $A_{r}$ ? ::: $63.55$ u Silicon $A_{r}$ ? ::: $28.09$ u Gallium $^{69}$ Ga fraction? ::: $\approx 0.601$ ( $60.1%$ ) Element Z (word problem) $A_{r}$ ? ::: $85.39$ u

Connections

← Back to the topic note
Weighted Average (mathematics) — the slider behind Cell F.
Mass Spectrometry — where the noisy Cell D percentages come from.
Isotopes and Mass Number — the mass-number trap in Cell H.
Nuclear Binding Energy and Mass Defect — why true mass < mass number.
Atomic Mass Unit (u) and the Carbon-12 standard — the unit on every answer.