Infer isotope composition from a mass spectrum

A spectrum answers a declared particle question

Imagine two peaks from a sample containing one element. The taller peak represents more signal, but its height alone does not tell you an absolute atom count. We need to know which species produced each peak, how the instrument responds to them and how the signals were normalized.

This investigation stipulates singly charged monatomic chlorine ions and equal response for the two isotopes. The integer labels 35 and 37 are mass numbers, counting protons plus neutrons. They are not exact isotope masses. The table uses selected NIST reference neutral-atom masses, while the schematic places peaks at nominal isotope positions. At this display resolution we neglect the small electron-mass difference between atom and ion. These are different levels of numerical precision, not contradictory definitions.

Chlorine isotopes share 17 protons but have different neutron counts. A charged particle can also differ in electron count without changing its isotope. Return to atoms, isotopes and ions if those distinctions are unclear. The required spectrum stays with one element and singly charged monatomic ions; molecular fragments and mixed-element spectra require additional interpretation and are outside this opening task.

Normalize before averaging

The original synthetic signals 75 and 25 have total 100. Their fractions are 75/100 = 0.75 and 25/100 = 0.25. Both are dimensionless and add to one. Signals 750 and 250 have total 1000 and give the same fractions. A common sensitivity multiplier changes total intensity but cancels in the normalized ratio.

The weighted mean uses isotope masses, not just their integer labels:

Mean mass = 0.75 × 34.968852682 + 0.25 × 36.965902602 = 35.468115162 u.

Here u is the unified atomic mass unit. Read the calculation as “three quarters of the lighter reference mass plus one quarter of the heavier reference mass.” The mean must lie between the two masses. It is a property of this stipulated sample composition, not a third isotope whose atoms all have that mass. A typical periodic-table atomic-weight entry is also not a claim that every possible sample has exactly one isotope ratio.

Capture the original and scaled signals. Compare the total, both fractions and mean, then explain what changed. At each pure-isotope endpoint the mean equals the surviving isotope’s reference mass. With no signal, 0/0 is undefined: no fraction or mass estimate follows. Zero signal does not mean zero-mass atoms or proven absence of the element without a detection model.

Keep signal bounds separate from probabilities

For the 75:25 record, turn on an absolute bound of one signal unit per channel. The underlying lighter signal is in [74,76] and heavier signal in [24,26]. The smallest heavy fraction uses low-heavy and high-light values: 24/(76+24) = 0.24. The largest uses high-heavy and low-light: 26/(74+26) = 0.26. Shared normalization matters; changing one channel changes the denominator too.

The corresponding conditional mass interval is [35.4481446628,35.4880856612] u. This is an endpoint calculation under the declared bounds and equal-response assumption. It is not a 95% confidence interval: no repeated-sampling distribution was supplied. Applying the same absolute one-unit bound to 750:250 gives a narrower fractional range because the signal is larger relative to that bound; it does not establish that a real instrument has this error model.

A chosen zero bound means exact synthetic signals in this exercise. It does not erase the source reference-mass uncertainty or claim that real detector measurements are exact. The displayed mass digits expose the calculation; they should not be mistaken for the practical precision of a physical assay.

Source evidence and sample limits

The NIST chlorine table supplies the selected isotope masses and source uncertainty notation. The CIAAW isotopic-composition table, inspected on 28 October 2026 and labeled 2024, reports ranges for natural chlorine isotope fractions. The table in the widget states those ranges. The synthetic 75:25 record is deliberately not presented as the representative natural ratio or as a measured mass spectrum.

The source masses and simulated detector signals have distinct provenance. Replaying the same settings creates no independent measurement. If relative sensitivity were unknown or different between isotope channels, normalized signal alone would not establish isotope fractions. State that limitation without inventing corrections that were not measured.

Apply to a changed record

Before checking the result, analyze a new original pair: lighter signal 60, heavier signal 40. State the heavy fraction, weighted mean and the effect of changing both signals to 6 and 4. Sketch the mean between the reference masses and defend why it moves toward the heavier isotope.

Check the changed-record reasoning

The heavy fraction is 0.40; common rescaling leaves it unchanged. The mean is 0.60 × 34.968852682 + 0.40 × 36.965902602 = 35.767672650 u. This is a larger mean because the heavier isotope’s weight increased, not because either isotope changed its mass. Both records remain conditional on the same species and response assumptions.

In your reflection, include raw signals, normalization, units, an interval interpretation and a limitation. Continue to composition and identifiability to examine a different inference problem. These are different supplied samples; do not merge them into one unknown analysis.