Chemistry · Lesson

Isotope tracers, radiochemistry and radiation evidence

Use isotope-defined entity identities and separate isotope-pool balances to interpret ratios, mixtures and finite transfers.

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By the end, you can

Convert an isotope ratio to a fraction and a named-reference delta.
Conserve isotope amounts when mixing total-element inventories.
Retain isotope positions separately from integer mass-number sums.
Close both isotope pools in a stipulated finite transfer.

An isotope label is part of a chemical entity

Isotopes of an element have the same proton number and different neutron counts. Their electronic chemistry can be similar while mass-dependent behaviour or nuclear stability differs. A labelled molecular formula should specify which atoms carry the isotope.

A tracer can help follow a defined constituent or atom position. Its signal does not automatically track every transformation of the whole molecule, and isotope effects can matter for a proposed comparison.

A ratio should keep its isotope pair and amount basis

The original two-isotope model defines R as heavy amount divided by light amount, while its fraction divides by total element amount. That distinction changes mixture arithmetic. Convert each contributing material to separate heavy/light amounts, add those pools and then take the mixture ratio.

Isotope delta additionally compares a sample ratio with a reference ratio using the same isotope pair. Expressing delta in per mille multiplies its dimensionless value by 1000; it does not turn a ratio into an isotope fraction. Identify the actual reference scale in real work.

The teaching reference 0.0112 is hypothetical and uses no adopted VPDB value. A reference-material page gives a reporting convention, while a specific standard value needs its own verified identity, date and uncertainty.

Mix isotope amounts before calculating a ratio

Total-element contributions / mmol	Heavy amount / mmol	Light amount / mmol	Mixture heavy/light ratio
A 8.00 + B 2.00	0.261026103	9.738973897	0.0268022181
A 2.00 + B 8.00	0.747074707	9.252925293	0.0807392996

Original exactly-two-isotope mixtures. A/B ratios stay 0.0100/0.100; changing only their total-element contributions changes the mixture. No actual specimen or isotope standard is used.

An isotope ratio and an isotope fraction have different denominators

A ratio compares the amounts of two selected isotopes, while an isotope fraction compares one with the specified total. Convert between them under an explicit isotope inventory. Other isotopes or unresolved species may require a fuller account.

A mixture calculation conserves the selected isotope amounts. Weighting by the wrong amount basis can give an incorrect mixture ratio even when the arithmetic seems plausible. State what material and element amount each component contributes.

A tracer identity and its interpretation need more than a total

The selected labelled ethanol examples count nuclear charges and neutrons from fully specified carbon/oxygen isotope labels. Equal integer nucleon sums can arise from different labels. Without a spatial or atom-position assignment for a hydrogen label, preserve that positional uncertainty rather than inventing a unique tracer form.

A tracer signal can constrain where a constituent was retained, exchanged or transformed, subject to the response and chemical-discrimination model. Bulk isotope ratio and position-specific entity evidence answer different questions. Calibration, background and isotope-dependent effects remain relevant.

The finite-transfer case stipulates the removed isotope composition. Subtracting its two amounts closes the residual pools and may enrich the remaining heavy fraction. It neither derives a differential fractionation law nor identifies the chemical mechanism producing that composition.

A tracer measurement needs both an inventory and a response model

A measured signal can contain background, detection losses and more than one source. Calibration and an appropriate chemical separation or discrimination model determine what isotope inventory the response supports.

For a radioactive label, decay during the interval changes the expected inventory. A stable-isotope label instead needs the relevant species, mass and fraction account. Neither label alone identifies every path taken by an unmeasured constituent.

An isotope result constrains a defined question

Position-specific or species-specific observations can test a pathway differently from a bulk elemental isotope ratio. Exchange, fractionation, mixing and retention can affect interpretation. Independent chemical evidence helps distinguish alternatives.

The original tracer ledgers below use invented inventories and responses. They are not isotope handling procedures, age estimates for an actual specimen or measurements of radiation risk.

Worked examples

Ratio, fraction and isotope delta use different denominators

For an abstract element with exactly two stable isotopes, assign heavy/light amount ratio R=0.0142 and hypothetical reference ratio R_ref=0.0112 for the same numerator/denominator pair. Calculate heavy fraction and isotope delta.

Heavy fraction is f_H=R/(1+R)=0.0140011832; light fraction is 1/(1+R)=0.9859988168. The isotope ratio is not itself the heavy fraction.
Dimensionless isotope delta is R/R_ref−1=0.2678571429. Expressed per mille, it is 1000δ=267.8571429‰.
The reference is invented and explicitly not an adopted VPDB value. Real reporting must identify the scale, isotope pair and relevant reference material or normalization convention.

Heavy fraction 0.0140011832; delta 0.2678571429, or 267.8571429‰ relative to the hypothetical reference.

Mix isotope amounts before taking their ratio

Mix 8.00 mmol total element from A with R_A=0.0100 and 2.00 mmol from B with R_B=0.100. There are exactly two isotopes and no exchange or fractionation. Find the mixture ratio.

Heavy amounts are 8.00×0.0100/1.0100 and 2.00×0.100/1.100, totaling 0.261026103 mmol. Light amount is 10.0−0.261026103=9.738973897 mmol.
Mixture heavy fraction is 0.0261026103 and heavy/light ratio 0.261026103/9.738973897=0.0268022181.
A direct total-element-weighted average of the two ratios would use the wrong denominator. Convert each ratio to isotope amounts on its supplied total-element basis, then conserve both pools.

Mixture R=0.0268022181, from separately conserved heavy and light amounts.

Equal nucleon sums can retain different isotope identities

Assign neutral labelled ethanol ¹³CH₃−¹²CH₂−¹⁶OH, with all hydrogen ¹H. Compare with neutral ¹²C₂¹H₆¹⁶O.

The labelled molecule has proton/electron count 2×6+6×1+8=26. Neutrons total 7+6+8=21; integer nucleon sum is 47.
The unlabelled isotope-selected comparison has 20 neutrons and nucleon sum 46. A ¹³C label at the methyl carbon is part of the atom-position identity.
These integer counts are not exact molecular masses or guarantees of indistinguishable spectra. A label can follow a selected atom without proving that every constituent of the original molecule followed the same path.

The selected ¹³C ethanol has 26 electrons and 21 neutrons; the specified unlabelled form has 20 neutrons.

A finite selected transfer needs two closed pools

Assign initial total element 1.00 mmol with heavy/light R₀=0.0200. Remove 0.400 of total element in one stipulated finite transfer with R_removed=0.0160. There are exactly two isotopes. Find the residual ratio.

Initial heavy amount is 0.0200/1.0200=0.0196078431 mmol. The removed 0.400 mmol contains heavy amount 0.400×0.0160/1.0160=0.00629921260 mmol.
Remaining heavy is 0.0133086305 mmol; remaining total is 0.600 mmol, so remaining light is 0.5866913695 mmol. Residual R=0.0226842105.
Initial light 0.9803921569 mmol splits into removed light 0.3937007874 and remaining light 0.5866913695. Both isotope pools are positive and close separately.
The removed composition is stipulated for this finite event. It is not a differential Rayleigh law, a universally constant chemical fractionation factor or a real tracer procedure.

Residual heavy/light ratio 0.0226842105 under the assigned finite-transfer ledger.

Try it yourself

Use R=0.01176 with the same hypothetical R_ref=0.0112. Find heavy fraction and delta.
Show the answer and reasoning
Heavy fraction 0.0116233099; dimensionless delta 0.0500, or 50.0‰.
Fraction uses R/(1+R); isotope delta compares matching sample/reference ratios.
Swap the total-element amounts: A 2.00 mmol and B 8.00 mmol, retaining their ratios. Recalculate the mixture.
Show the answer and reasoning
Heavy 0.747074707 mmol; light 9.252925293 mmol; mixture R=0.0807392996.
Convert each component’s ratio to isotope amounts before adding; the total element remains 10.0 mmol.
Instead assign neutral ¹²C₂¹H₅²H₁¹⁶O ethanol. Count electrons, neutrons and nucleons. Does equality of nucleon sum establish the same isotope identity as the ¹³C-labelled form?
Show the answer and reasoning
26 electrons, 21 neutrons, nucleon sum 47; isotope identity differs.
A ²H replaces a ¹H rather than a ¹³C replacing ¹²C. Specify the labelled hydrogen position for a complete positional tracer identity.
Remove 0.800 of the initial 1.00 mmol in the same finite model, retaining R_removed=0.0160. Find remaining heavy/light amounts and ratio.
Show the answer and reasoning
Heavy 0.00700941794 mmol, light 0.1929905821 mmol; residual R=0.0363200.
Removed heavy is 0.0125984252 mmol; subtract both isotope pools and retain the 0.200 mmol residual total.

What to carry forward

Ratio, total fraction and isotope delta use different denominators.
Mixing conserves separate isotope amounts.
A label position remains part of chemical identity.
A finite isotope ledger does not establish a universal fractionation law.

References

Teaching resources and reference material supporting this lesson’s explanations and examples.