LC–MS ion assignment: worked solutions
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Fictional teaching model. No peptide sequence, measured sample or instrument performance is represented. The ±5 ppm criterion is invented for this exercise.
Four observations, one fictional retention zone
All four rows belong to the invented retention group R1. Their positive charge numbers are supplied. No isotope envelope is provided from which to infer charge; no peak intensities are implied.
Your starting data — use these six-decimal observations as written
Scroll sideways, or focus the table area and use the left and right arrow keys.
| Row | Synthetic observed m/z | Supplied charge z |
|---|---|---|
| A | 1235.577638 | +1 |
| B | 618.289367 | +2 |
| C | 1257.558369 | +1 |
| D | 1235.589993 | +1 |
Check your reasoning
Closest charge-compatible hypothesis in each row; D remains outside
Scroll sideways, or focus the table area and use the left and right arrow keys.
| Row | Hypothesis | Calculated m/z | Signed ppm | Fictional ±5 ppm |
|---|---|---|---|---|
| A | [M + H]⁺ | 1235.575166 | +2.000 | Within |
| B | [M + 2H]²⁺ | 618.291221 | −2.999 | Within |
| C | [M + Na]⁺ | 1257.557111 | +1.000 | Within |
| D | [M + H]⁺ | 1235.575166 | +12.000 | Outside |
The summary displays calculated m/z to six decimals and ppm to three. The calculations below and the CSV retain the full supplied mass inputs. Differences from exactly +2, −3, +1 and +12 ppm arise because the observations are fixed at six decimals.
Row A · [M + H]⁺
Calculated m/z: 1234.567890 + 1.0072764665789 = 1235.5751664665789
Signed ppm: (1235.577638 − 1235.5751664665789) / 1235.5751664665789 × 10⁶ = +2.000310 ppm
A is consistent with the singly protonated hypothesis within the fictional criterion. The sodium hypothesis is far outside; passing one comparison does not establish a unique molecular structure.
Row B · [M + 2H]²⁺
Calculated m/z: (1234.567890 + 2 × 1.0072764665789) / 2 = 618.2912214665789
Signed ppm: (618.289367 − 618.2912214665789) / 618.2912214665789 × 10⁶ = −2.999342 ppm
B is consistent with the doubly protonated hypothesis. Add two proton masses before dividing by 2. Divide its m/z difference by 618.2912214665789, not by the neutral mass 1234.567890.
Row C · [M + Na]⁺
Calculated m/z: 1234.567890 + 22.989221 = 1257.557111
Signed ppm: (1257.558369 − 1257.557111) / 1257.557111 × 10⁶ = +1.000352 ppm
C is consistent with the sodium-adduct hypothesis within this limited candidate set. Treat it as a candidate assignment; the exercise does not provide independent evidence that sodium caused a real signal.
Row D · [M + H]⁺
Calculated m/z: 1234.567890 + 1.0072764665789 = 1235.5751664665789
Signed ppm: (1235.589993 − 1235.5751664665789) / 1235.5751664665789 × 10⁶ = +11.999702 ppm
D is closest to the singly protonated hypothesis, but its approximately +12 ppm error is outside ±5 ppm. Do not force the assignment, widen the criterion after seeing the answer, or call the row an identified impurity.
Why the denominator matters
For B, using the neutral mass in the denominator gives about −1.502 ppm instead of −2.999 ppm. It still falls inside this invented limit, but it answers the wrong relative-error question. The numerator is an m/z difference, so the reference in this exercise is the corresponding calculated ion m/z.
What remains unresolved?
The calculation tests consistency with a small list of hypotheses. It cannot establish a sequence, distinguish all isomers, locate a modification, or prove purity or amount. A shared retention zone does not remove those ambiguities. There is no elemental formula or fragment spectrum in this exercise, so the model cannot be promoted to a structural identification.