Calculation of atomic mass from isotopic abundance: Exercises

Intuition How to use this page

These 12 problems climb a ladder: from recognising the formula (L1) to mastering messy, real-world variants (L5). Try each one on paper before opening its solution. Every solution is inside a collapsible [!recall]- callout so this page doubles as a self-test. The one tool you need everywhere is the weighted average: $A_{r} = \sum_{i} m_{i} f_{i}, \sum_{i} f_{i} = 1,$ where $m_{i}$ is the isotopic mass in u and $f_{i}$ is the fractional abundance of isotope $i$ . When abundances are given as percentages, we write $p_{i}$ for the percentage abundance of isotope $i$ (so $p_{i} = 100 f_{i}$ , e.g. $75.77%$ means $p_{i} = 75.77$ ). In that percentage form, $A_{r} = \frac{\sum _{i} m _{i} p _{i}}{\sum _{i} p _{i}},$ where the denominator $\sum_{i} p_{i}$ is the total weight (which is $100$ when all listed abundances add up to $100%$ ). All of this is built from scratch in the parent note: see the parent topic.

Level 1 — Recognition

Can you spot the pieces and plug them in?

Exercise L1.1

Neon has two main isotopes: $^{20}$ Ne (mass $19.992$ u, abundance $90.48%$ ) and $^{22}$ Ne (mass $21.991$ u, abundance $9.52%$ ). Ignoring the tiny $^{21}$ Ne, compute $A_{r}$ .

Recall Solution

WHAT we do: multiply each mass by its percentage $p_{i}$ , add, divide by 100 (the total weight $\sum p_{i} = 90.48 + 9.52 = 100$ ). $A_{r} = \frac{19.992 ( 90.48 ) + 21.991 ( 9.52 )}{100}$ $= \frac{1808.88 + 2.0935 \times 1 0 ^{2}}{100} = \frac{1808.88 + 209.35}{100} = \frac{2018.23}{100}$ $A_{r} \approx 20.18 u$ Sanity check: the answer sits between $19.99$ and $21.99$ , hugging the light side because $^{20}$ Ne is ~90% of the sample. ✔️

Exercise L1.2

Which of these is the correct fractional abundance for an isotope reported as $37.20%$ ? (a) $37.20$ (b) $3.720$ (c) $0.3720$ (d) $0.03720$ . State why.

Recall Solution

A percentage is a "per-hundred". To get the fraction $f$ , divide the percentage $p$ by 100: $f = \frac{p}{100} = \frac{37.20}{100} = 0.3720.$ Answer: (c). Options (a),(b),(d) are off by factors of 100, 10, and 1000 respectively.

Level 2 — Application

Run the full machine on real elements.

Exercise L2.1 — Copper

$^{63}$ Cu (mass $62.930$ u, $69.17%$ ) and $^{65}$ Cu (mass $64.928$ u, $30.83%$ ). Find $A_{r}$ .

Recall Solution

$A_{r} = \frac{62.930 ( 69.17 ) + 64.928 ( 30.83 )}{100} = \frac{4352.87 + 2001.73}{100} = \frac{6354.60}{100}$ $A_{r} \approx 63.55 u$ Check: matches the periodic-table value for copper (63.55). ✔️

Exercise L2.2 — Silicon (three isotopes)

$^{28}$ Si ( $27.977$ u, $92.23%$ ), $^{29}$ Si ( $28.976$ u, $4.67%$ ), $^{30}$ Si ( $29.974$ u, $3.10%$ ). Find $A_{r}$ .

Recall Solution

WHY three terms: the formula $\sum_{i} m_{i} f_{i}$ extends to any number of isotopes — just add one term per isotope. $A_{r} = \frac{27.977 ( 92.23 ) + 28.976 ( 4.67 ) + 29.974 ( 3.10 )}{100}$ $= \frac{2580.32 + 135.32 + 92.92}{100} = \frac{2808.56}{100}$ $A_{r} \approx 28.09 u$ Sanity: dominated by $^{28}$ Si (92%), so the answer barely leaves 28. ✔️

Level 3 — Analysis

Reverse the formula, or reason about which way the average leans.

Exercise L3.1 — Solving backwards (Gallium)

Gallium has $^{69}$ Ga ( $68.926$ u) and $^{71}$ Ga ( $70.925$ u), with $A_{r} = 69.723$ u. Find the percentage abundance of each isotope.

Recall Solution

WHAT we set up: let $f$ = fraction of $^{69}$ Ga. Then $^{71}$ Ga has fraction $1 - f$ because the two must sum to 1. $69.723 = 68.926 f + 70.925 (1 - f)$ Expand the right side step by step. First distribute the $70.925$ over the bracket: $68.926 f + 70.925 (1 - f) = 68.926 f + 70.925 - 70.925 f .$ Now collect the two terms that contain $f$ (they are like terms): $68.926 f - 70.925 f = (68.926 - 70.925) f = - 1.999 f .$ So the whole right side becomes $70.925 - 1.999 f,$ and our equation reads $69.723 = 70.925 - 1.999 f .$ Move the constant across and solve: $1.999 f = 70.925 - 69.723 = 1.202$ $f = \frac{1.202}{1.999} = 0.6013$ So $^{69}$ Ga $\approx 60.1%$ and $^{71}$ Ga $\approx 39.9%$ . Check: $68.926 (0.6013) + 70.925 (0.3987) = 69.72$ ✔️

Exercise L3.2 — Which isotope is more abundant?

Bromine's $A_{r}$ is $79.90$ u. Its isotopes are $^{79}$ Br ( $78.918$ u) and $^{81}$ Br ( $80.916$ u). Without full calculation, argue whether $^{79}$ Br or $^{81}$ Br is more abundant, then confirm with numbers.

Recall Solution

Reasoning first: $79.90$ lies almost exactly halfway between $78.918$ and $80.916$ (midpoint $= 79.917$ ). It is a hair below the midpoint, so the lighter isotope $^{79}$ Br is slightly more abundant — the average is pulled toward the heavier side that has fewer votes only slightly. Confirm: let $f$ = fraction of $^{79}$ Br. Distributing and collecting like terms as in L3.1, $78.918 f + 80.916 (1 - f) = 80.916 - 1.998 f$ : $79.90 = 80.916 - 1.998 f$ $1.998 f = 1.016 \Rightarrow f = 0.5085$ So $^{79}$ Br $\approx 50.9%$ , $^{81}$ Br $\approx 49.1%$ . The lighter isotope wins by a whisker, exactly as predicted. ✔️

The figure below is a number line of this situation. The two black dots mark the isotopic masses $78.918$ u and $80.916$ u; the short black tick is their exact midpoint $79.917$ u. The red dot is the actual $A_{r} = 79.90$ u. Notice the red dot sits just to the left of the midpoint tick — that tiny leftward pull is the whole story: the average leans toward the lighter mass, which is why $^{79}$ Br must be the (slightly) more abundant isotope.

Figure — Calculation of atomic mass from isotopic abundance

Level 4 — Synthesis

Combine the formula with unit-conversion, normalisation, or the mole concept.

Exercise L4.1 — Abundances that don't sum to 100

A meteorite sample of an element reports only two isotopes: mass $23.985$ u at $79.10%$ and mass $25.983$ u at $10.99%$ (the middle isotope was not measured). Using only these two, estimate the ratio $A_{r}$ of the measured portion — i.e. divide by the true total weight, not 100.

Recall Solution

WHY divide by the actual sum: a proper average divides by the total weight present, which here is $\sum p_{i} = 79.10 + 10.99 = 90.09$ , not $100$ . $A_{r} = \frac{23.985 ( 79.10 ) + 25.983 ( 10.99 )}{79.10 + 10.99}$ $= \frac{1897.21 + 285.55}{90.09} = \frac{2182.76}{90.09}$ $A_{r} \approx 24.23 u$ Note: this is the average of only the two measured isotopes; the missing $^{25}$ Mg would shift the true element value. The point: divide by $\sum p_{i}$ , never blindly by 100.

Exercise L4.2 — From $A_{r}$ to a mole

Chlorine's $A_{r} = 35.45$ u. Using the mole concept, what is the mass of exactly one mole of chlorine atoms? What about one mole of Cl $_{2}$ molecules?

Recall Solution

Key link: the relative atomic mass in u equals the molar mass in grams per mole (this is what the u/gram bridge via Avogadro's number gives us).

One mole of Cl atoms: $35.45 g$ .
One mole of Cl $_{2}$ molecules: two atoms per molecule, so $2 \times 35.45 = 70.90 g .$ Why $\times 2$ : the average automatically accounts for the isotope mix per atom; a molecule just contains two such average atoms.

Level 5 — Mastery

Multi-step, edge cases, and conceptual synthesis.

Exercise L5.1 — Two-unknown system with a purity constraint

An element $X$ has isotopes of mass $A$ u and $B$ u with $B > A$ . A sample gives $A_{r}$ . Show that the fraction of the lighter isotope is $f_{light} = \frac{B - A _{r}}{B - A},$ then apply it to lithium: $^{6}$ Li ( $6.015$ u), $^{7}$ Li ( $7.016$ u), $A_{r} = 6.941$ u.

Recall Solution

Derive: let $f$ = fraction of the light ( $A$ ) isotope. Then $A_{r} = A f + B (1 - f) = B - (B - A) f .$ Solve for $f$ : $(B - A) f = B - A_{r} \Rightarrow f = \frac{B - A _{r}}{B - A} . ✓$ Apply to lithium ( $A = 6.015$ , $B = 7.016$ , $A_{r} = 6.941$ ): $f_{^{6} Li} = \frac{7.016 - 6.941}{7.016 - 6.015} = \frac{0.075}{1.001} = 0.0749.$ So $^{6}$ Li $\approx 7.49%$ , $^{7}$ Li $\approx 92.5%$ . Sanity: $A_{r} = 6.941$ is very close to $7.016$ , so the heavy isotope must dominate — and it does (92.5%). ✔️

Exercise L5.2 — Degenerate case (single isotope)

Fluorine exists in nature as essentially one isotope: $^{19}$ F, mass $18.998$ u, abundance $100%$ . Apply the weighted-average formula and explain what "average" means when there is only one isotope.

Recall Solution

With one isotope, $f = 1$ : $A_{r} = m \cdot f = 18.998 \times 1 = 18.998 u .$ What it means: a "weighted average" of a single value is just that value — there is nothing to pull it away from. This is the degenerate case, and it's why monoisotopic elements (F, Na, Al, P) have $A_{r}$ almost exactly equal to a single isotopic mass. It is also why their $A_{r}$ is closer to a whole number than mixed elements like Cl.

Exercise L5.3 — Sensitivity / limiting behaviour

For a two-isotope element with masses $A$ and $B$ ( $B > A$ ), what does $A_{r}$ approach as the light isotope's fraction $f \to 1$ ? As $f \to 0$ ? Describe how $A_{r}$ varies with $f$ .

Recall Solution

Use $A_{r} = B - (B - A) f$ from L5.1 — a straight line in $f$ :

As $f \to 1$ (all light): $A_{r} \to B - (B - A) = A$ . The average becomes the light mass.
As $f \to 0$ (all heavy): $A_{r} \to B$ . The average becomes the heavy mass.
In between it slides linearly from $B$ down to $A$ as $f$ goes $0 \to 1$ . WHY linear: each 1% you move from heavy to light removes a fixed chunk $(B - A) /100$ of mass — a constant rate of change, which is exactly what a straight line is.

The figure below plots exactly this. The horizontal axis is $f$ (fraction of the light isotope) running $0 \to 1$ ; the vertical axis is $A_{r}$ in u. The red line is $A_{r} = B - (B - A) f$ : read it left to right and $A_{r}$ falls in a perfectly straight line from $B = 7.016$ (at $f = 0$ , all heavy) down to $A = 6.015$ (at $f = 1$ , all light). The two black dots are those endpoints; the black square marks lithium at $f = 0.075$ , $A_{r} = 6.941$ — sitting near the top-left because lithium is mostly the heavy isotope. The key thing to see is that the red line never bends and never leaves the band between $A$ and $B$ .

Recall wrap-up

Recall One-line skills checklist

Convert % to fraction before using $\sum m_{i} f_{i}$ ::: divide by 100.
Divide by the actual total weight ::: use $\sum p_{i}$ , not always 100.
Find one abundance from $A_{r}$ ::: $f_{light} = \frac{B - A _{r}}{B - A}$ .
$A_{r}$ in u equals molar mass in ::: grams per mole.
Single isotope ::: $A_{r}$ equals that isotope's mass exactly.

Connections

Parent topic — the derivation these exercises drill.
Weighted Average (mathematics) — the linear-combination idea behind L5.3.
Isotopes and Mass Number — what the masses and abundances describe.
Mole Concept and Molar Mass — used in L4.2.
Mass Spectrometry — where these abundances are measured.