Did you know that chlorine gas, Cl₂, can produce three molecular-ion peaks at m/z 70, 72 and 74 in a mass spectrum, even though natural chlorine has only two main stable isotopes? The familiar isotopes are chlorine-35 and chlorine-37. When two chlorine atoms join to form a molecule, there are three distinct isotope-mass combinations: 35+35, 35+37 and 37+37. With approximate individual isotope abundances of 75% chlorine-35 and 25% chlorine-37, the molecular-ion peaks appear in the characteristic ratio 9:6:1.
For learners searching for an H2 Chemistry tutor in Hougang, chlorine mass spectrum 9 6 1, Cl2 isotopes 70 72 74, mass spectrometry molecular ions, chlorine-35 chlorine-37 probability, M+2 isotope peaks or A-Level Chemistry calculations, this is an excellent way to learn that the probability of a molecular formula depends on every isotope combination, not simply the relative abundance of one atom.
I begin with a small puzzle: if chlorine has two common isotopes, why isn’t its Cl₂ molecular-ion pattern just two peaks in the ratio 3:1? The answer is that one molecule contains two chlorine positions. Both can be chlorine-35, both can be chlorine-37, or one can be each. The mixed combination occurs in two distinguishable sampling orders, so its probability is larger than a student may initially expect.
The result brings together isotopes, weighted averages, the mole concept, probability and mass-spectral interpretation. The revised Singapore H2 Chemistry 9476 syllabus requires sound understanding of isotopes and relative atomic masses. Detailed mass-spectrum isotope-cluster interpretation is presented here as an enrichment application, not an assertion that all molecular-ion peak analysis is mandatory under 9476; follow the exact examination-year syllabus and question data.
At eduKate, three-student small-group tutorials emphasise first-principles explanations and changed practice questions rather than remembering one pattern without knowing where it comes from. Parents looking for support from Hougang should confirm actual venue, timetable and JC class availability directly. A Hougang educational article is not evidence of a physical classroom operating in Hougang.
Hougang Chemistry learning library · Official SEAB 2027 H2 Chemistry syllabus · Chemistry LibreTexts: chlorine mass spectrum
Begin with the meaning of an isotope
Isotopes of one element have the same number of protons but different numbers of neutrons.
Chlorine has atomic number 17, so each neutral chlorine atom contains 17 protons.
The two abundant stable isotopes are:
³⁵Cl, with mass number 35.
³⁷Cl, with mass number 37.
The ³⁵Cl isotope contains 18 neutrons because 35 − 17 = 18.
The ³⁷Cl isotope contains 20 neutrons because 37 − 17 = 20.
Their different neutron counts change their masses while their shared proton count makes them the same chemical element.
For most introductory chemical calculations, isotopes of chlorine have very similar chemical behaviour, so chlorine-containing molecules form combinations according to the natural isotope abundances.
The essential question for our calculation is how likely is each combination of two atoms?
The natural isotope abundances
Natural chlorine has approximately:
³⁵Cl: 75–76%.
³⁷Cl: 24–25%.
For the simple textbook probability model, these are commonly rounded to:
p(³⁵Cl) = 0.750 = 3/4.
p(³⁷Cl) = 0.250 = 1/4.
The isotope ratio is therefore:
³⁵Cl : ³⁷Cl = 3 : 1.
More precise reference abundances are approximately 75.77% and 24.23%, so 9:6:1 is an excellent approximate ratio rather than an exact universal measurement for every natural sample.
A Chemistry LibreTexts mass-spectrometry reference gives representative natural abundances close to those values.
For school arithmetic, the 3:1 approximation makes the reasoning especially transparent.
Why chlorine normally exists as Cl₂
Under ordinary elemental gaseous conditions, chlorine consists of diatomic Cl₂ molecules, with two chlorine atoms held together by a covalent bond.
The formula Cl₂ means two chlorine atoms per molecule.
The atoms may have different isotopic mass numbers even though both are chemically chlorine.
Therefore a sample of natural chlorine molecules can contain:
³⁵Cl–³⁵Cl.
³⁵Cl–³⁷Cl.
³⁷Cl–³⁷Cl.
There is no separate chemical element called “chlorine-72”. The number 72 will arise from the combined mass numbers of one 35 and one 37 isotope.
The same molecular formula Cl₂ covers all three isotope compositions.
This is a powerful distinction between chemical formula and isotopic composition.
What does a mass spectrometer measure?
A mass spectrometer detects charged particles according to their mass-to-charge ratio, written m/z.
The instrument must produce ions because neutral molecules are not detected by ordinary mass analysis in the same way as charged species.
When a Cl₂ molecule is ionised without losing its two chlorine atoms, a molecular ion can form:
Cl₂ → Cl₂⁺ + e⁻.
That is a simplified electron-impact ionisation representation; actual charge notation may include an unpaired electron, such as Cl₂⁺•.
The newly formed positive molecular ion still contains two chlorine atoms.
If its charge is +1, its m/z value is numerically close to the relative mass of its isotope composition in the common introductory approximation.
Therefore we predict possible molecular-ion peaks at:
35 + 35 = 70.
35 + 37 = 72.
37 + 37 = 74.
The peak positions follow isotopic masses and charge state, not a guess about colour or chemical reactivity.
Why the mixed molecule is a single peak at 72
A molecule containing one ³⁵Cl and one ³⁷Cl can be described as:
³⁵Cl–³⁷Cl,
or with the isotopes written in the opposite order:
³⁷Cl–³⁵Cl.
Those are not two different chemical compounds with different m/z values. Both have combined mass number 72.
In the mass spectrum, both contribute to the same molecular-ion isotope peak at approximately m/z 72 for charge +1.
However, when calculating the probability of drawing two independent atoms from a large isotope mixture, the two sampling orders are distinct events.
That is why the probability of getting one of each isotope is:
P(35 then 37) + P(37 then 35).
The factor of two arises from the two possible orders in the sampling calculation, even though the resulting isotope composition has one observed peak.
This is the heart of the 9:6:1 pattern.
Probability of two chlorine-35 atoms
Suppose isotope choices are effectively independent and each atomic position has probability:
P(³⁵Cl) = 3/4.
For a molecule containing two chlorine-35 atoms:
P(³⁵Cl–³⁵Cl) = (3/4)(3/4).
Therefore:
P = 9/16.
As a percentage:
9/16 × 100% = 56.25%.
This isotope composition gives the molecular ion at m/z 70 in the approximate singly charged model.
The 56.25% is the expected fraction of Cl₂ molecules with that isotope composition under the simplified random-association model.
It is not necessarily 56.25% of the entire mass spectrometer’s detected signal when atomic fragments, other ions or different ionisation efficiencies also contribute.
The isotope-cluster ratio is a conditional comparison among the corresponding molecular-ion isotope compositions.
Probability of one chlorine-35 and one chlorine-37
There are two orders:
³⁵Cl then ³⁷Cl.
³⁷Cl then ³⁵Cl.
Each order has probability:
(3/4)(1/4) = 3/16.
Add the two:
P(mixed) = 3/16 + 3/16 = 6/16.
As a percentage:
6/16 × 100% = 37.50%.
The mixed isotope composition gives the molecular-ion peak at m/z 72.
A common student error is to calculate only:
(3/4)(1/4) = 3/16.
That misses the reverse ordering and underestimates the middle peak by a factor of two.
The correct expression is:
P(one of each) = 2pq,
where p and q are the probabilities of the two isotope types.
The number 2 is a combinatorial factor, not the charge of the molecular ion.
Probability of two chlorine-37 atoms
For both positions to contain chlorine-37:
P(³⁷Cl–³⁷Cl) = (1/4)(1/4).
Therefore:
P = 1/16.
As a percentage:
1/16 × 100% = 6.25%.
This isotope composition has approximate molecular-ion m/z:
37 + 37 = 74.
Now we have three expected fractions:
m/z 70: 9/16.
m/z 72: 6/16.
m/z 74: 1/16.
They add to:
(9 + 6 + 1)/16 = 16/16 = 1.
That is a useful probability check. If your fractions do not sum to one, at least one isotope combination is missing or double-counted incorrectly.
The 9:6:1 pattern in one table
| Molecular isotope composition | Approximate m/z for Cl₂⁺ | Probability with p = 3/4, q = 1/4 | Relative peak value |
|---|---|---|---|
| ³⁵Cl–³⁵Cl | 70 | p² = 9/16 | 9 |
| ³⁵Cl–³⁷Cl or ³⁷Cl–³⁵Cl | 72 | 2pq = 6/16 | 6 |
| ³⁷Cl–³⁷Cl | 74 | q² = 1/16 | 1 |
Thus the approximate relative molecular-ion peak intensities are:
70 : 72 : 74 = 9 : 6 : 1.
When ratios are scaled relative to the most intense peak:
100 : 66.7 : 11.1.
Both sets express the same ideal isotope pattern.
The exact relative measured peak intensities can deviate slightly from this approximation because actual abundances are not exactly 75/25 and instruments have their own behaviour.
The standard teaching prediction remains 9:6:1.
Why the binomial expansion appears
The probability calculation can be compressed using:
(p + q)² = p² + 2pq + q².
Here:
p = 3/4 for ³⁵Cl.
q = 1/4 for ³⁷Cl.
Expanding:
(3/4 + 1/4)² = (3/4)² + 2(3/4)(1/4) + (1/4)².
That gives:
9/16 + 6/16 + 1/16.
The coefficients 1, 2, 1 in the binomial expansion count how many ways the different isotope arrangements occur.
This is not magic numerology. It is the same probability rule used in two independent binary choices.
A molecule with three chlorine atoms would require the cubic expansion (p+q)³ and four isotope-combination mass groups, under the same simplified assumptions.
The method can transfer to molecular ions containing multiple halogen atoms without requiring students to memorise every peak ratio.
Why chlorine atoms themselves give only two main isotope peaks
Chlorine molecules can fragment during ionisation.
A simplified fragmentation is:
Cl₂⁺ → Cl⁺ + Cl.
The charged fragment Cl⁺ contains one chlorine atom.
There are only two common isotope choices for that single charged atom:
³⁵Cl⁺ at m/z 35.
³⁷Cl⁺ at m/z 37.
Their expected isotope abundance ratio is approximately:
35 : 37 peak intensity = 3 : 1.
This differs from the molecular-ion pattern because the fragment contains only one chlorine atom, not two.
Therefore a chlorine mass spectrum can contain two distinct groups:
Atomic fragment peaks near 35 and 37.
Molecular-ion peaks near 70, 72 and 74.
Recognising the different numbers of atoms in the detected ion is the first step toward interpreting the groups correctly.
Why the atomic and molecular peak groups cannot simply be compared using 3:1 versus 9:6:1
The within-group isotope ratios are predictable from isotope abundances under the simplified model.
But the total intensity of the atomic fragment group compared with the total molecular-ion group depends on how much fragmentation and ionisation occurred.
Some Cl₂⁺ ions survive intact, contributing to the molecular cluster.
Others fragment, producing Cl⁺ ions.
The relative amounts depend on the instrument’s ionisation and fragmentation conditions and the stability of the ions.
Therefore the 3:1 and 9:6:1 patterns do not tell us an automatic overall ratio between:
Total peaks 35+37, and
Total peaks 70+72+74.
This is a major exam-reading trap. The isotope abundance ratio predicts the relative distribution within a given class of ions; it does not automatically fix how much of each fragmentation class is detected.
Chemistry LibreTexts makes this distinction in its explanation of elemental chlorine spectra.
How this connects to relative atomic mass
Chlorine’s relative atomic mass is a weighted mean of the isotope masses.
Under our rounded 75/25 distribution:
Ar(Cl) ≈ 35(0.75) + 37(0.25).
Calculate:
Ar(Cl) ≈ 26.25 + 9.25 = 35.50.
That explains why the familiar Periodic Table value for chlorine is near 35.5, even though an individual stable chlorine atom is usually isotope 35 or 37.
The non-integer average is not evidence that each chlorine atom somehow contains “half a neutron”.
It is the weighted average across many atoms of naturally occurring isotope compositions.
This is one of the basic isotope calculations that transfers to many elements.
The molecular ion cluster provides another route to recover the same average molecular mass.
Calculate chlorine’s average molecular mass from 9:6:1
Take the three Cl₂ isotope compositions and their simplified ratios:
m/z 70 with weight 9.
m/z 72 with weight 6.
m/z 74 with weight 1.
The weighted mean molecular mass is:
Mr(Cl₂) ≈ [9(70) + 6(72) + 1(74)]/(9 + 6 + 1).
Calculate the numerator:
9(70) = 630.
6(72) = 432.
1(74) = 74.
Total:
630 + 432 + 74 = 1136.
Divide by 16:
1136/16 = 71.0.
This agrees with:
Mr(Cl₂) = 2 × Ar(Cl) ≈ 2 × 35.5 = 71.0.
That agreement is a powerful consistency check showing that the isotope probabilities and mass values have been combined correctly.
A student who obtains an average Cl₂ mass of 72.0 has probably overcounted the middle isotope combination or used unweighted averaging.
Why the middle peak at 72 does not imply a chlorine-36 isotope
One of the most common mistakes is to look at the peak labels:
70, 72, 74,
divide by two and conclude that chlorine contains isotopes with mass numbers:
35, 36, 37.
That conclusion is incorrect.
The peak at 72 comes from a molecule containing one chlorine-35 and one chlorine-37 atom, not two chlorine-36 atoms.
The underlying stable isotope masses in the familiar natural chlorine model remain 35 and 37.
A molecular-ion spectrum records the combined masses of atoms within each charged molecule.
To infer isotopes correctly, students must know whether the detected ion contains one or multiple atoms of the element.
That is why the formula Cl₂ is essential to the calculation.
A mass spectrum should not be interpreted as a list of isotope masses without first identifying the ions responsible for the peaks.
Why the pattern is not always 9:6:1 with different abundances
The 9:6:1 ratio uses the approximation:
p = 0.75, q = 0.25.
If the isotope abundances change, the probabilities become:
p² : 2pq : q².
For instance, suppose a hypothetical element X has isotopes of masses 10 and 12, with abundances 60% and 40%.
Then:
p = 0.60, q = 0.40.
For X₂ molecular ions, the three probabilities are:
p² = 0.36.
2pq = 0.48.
q² = 0.16.
Therefore the relative pattern is:
36 : 48 : 16 = 9 : 12 : 4.
The middle peak is now the tallest.
The method remains the same; the numerical ratio changes with the isotope abundances.
This is why memorising “two isotopes always give 9:6:1” would be wrong.
The correct general relationship for two independent isotope choices is p² : 2pq : q².
A comparison with bromine molecules
Bromine has two main stable isotopes:
⁷⁹Br and ⁸¹Br.
Their natural abundances are approximately equal, close to 50% each.
For a Br₂ molecule, the possible nominal molecular-ion masses are:
79 + 79 = 158.
79 + 81 = 160.
81 + 81 = 162.
With p = 0.50 and q = 0.50:
p² = 0.25.
2pq = 0.50.
q² = 0.25.
Thus the molecular isotope cluster is approximately:
1 : 2 : 1.
That differs from chlorine’s 9 : 6 : 1 only because the isotope abundances are different.
The number of distinct isotope compositions is three in each case, but the relative probabilities are controlled by the isotope fractions.
This is a useful changed example to test transfer rather than memory.
Why the exact naturally occurring chlorine ratio is slightly different
More precise representative abundances are about:
³⁵Cl = 75.77%, and
³⁷Cl = 24.23%.
Then:
p = 0.7577, q = 0.2423.
The probabilities are approximately:
p² ≈ 0.5741.
2pq ≈ 0.3672.
q² ≈ 0.0587.
Those values sum to approximately one after rounding.
The resulting relative ratio is close to, but not exactly, 9:6:1.
For most school isotope-cluster questions, the 3:1 atomic abundance assumption is an intentional simplification.
If the paper provides different more precise abundances, use those values.
A student should not force a measured dataset to match exact 9:6:1 values when the supplied proportions clearly differ.
The lesson is to distinguish an approximate teaching ratio from precisely measured natural isotopic abundance.
What about the ion charge?
The m/z axis reports mass divided by charge.
The peaks at 70, 72 and 74 are explained here by singly charged Cl₂⁺ molecular ions, whose charge magnitude is one elementary positive charge.
If a hypothetical ion had charge magnitude two, its m/z values would be different, roughly half the corresponding values for comparable masses.
Therefore an exam question that explicitly supplies a 2+ charge state must be read differently.
The conventional chlorine molecular-ion teaching example assumes +1 ions unless otherwise stated.
This is why the notation Cl₂⁺ matters: it identifies both the molecular composition and the charge state.
Do not interpret an m/z value as the mass number of one atom without checking what kind of ion the instrument detected.
Why fragment ions and molecular ions are different evidence
A molecular ion can preserve the complete atomic composition of the original molecule after ionisation.
A fragment ion is a charged part produced when a larger ion or molecule breaks apart during the measurement process.
For chlorine, Cl₂⁺ is a simple molecular ion. Cl⁺ may be a fragment ion.
For larger organic molecules, fragment ions can help identify pieces of the structure.
But a fragment’s mass is not necessarily equal to the mass of the intact parent compound.
A mass spectrum can therefore contain several groups of peaks from one original molecular substance.
The correct workflow is:
Identify the ion formula and charge → calculate possible isotopic masses → calculate their probabilities → interpret the relative peaks within that ion type.
That is a more reliable method than trying to assign every peak directly to a separate chemical compound.
An organic example with one chlorine atom
Suppose a hypothetical organic molecular ion contains one chlorine atom and all other relevant atoms have one dominant isotope under the question’s approximation.
Then the chlorine can be either ³⁵Cl or ³⁷Cl.
There are two main isotope alternatives, separated by two nominal mass units.
If the lower-mass peak is labelled M and the heavier one M+2, their relative intensity is approximately:
M : M+2 = 3 : 1.
This is a useful characteristic pattern of one chlorine atom in the ion.
But it is not the same as a Cl₂ molecular ion, which contains two chlorine atoms.
The number of chlorine positions is what changes the isotope combination calculation.
An ion with two chlorines gives three main chlorine isotope combinations rather than two.
That is a useful way to distinguish the patterns when interpreting an unknown halogen-containing organic molecule.
An organic example with two chlorine atoms
A molecule containing two chlorine atoms can have isotope combinations:
35+35.
35+37.
37+37.
So an intact singly charged molecular ion under the simplified isotope model gives three mass groups, separated by 2 mass units each.
Their approximate intensities are 9:6:1.
For example, dichloromethane, CH₂Cl₂, has a lightest nominal molecular-ion mass:
12 + 2(1) + 2(35) = 84.
The chlorine isotope cluster therefore appears near:
m/z 84, 86 and 88,
with approximate relative intensities 9:6:1.
This is the same probability pattern as Cl₂ but with a constant mass contribution from CH₂ added to every isotopic combination.
A student should recognise that adding a fixed set of other atoms shifts all peak positions without changing the chlorine isotope probability coefficients in the simple model.
A Chemistry LibreTexts halogen mass-spectrometry guide discusses the M/M+2/M+4 cluster for two chlorine atoms.
Why other isotopes can complicate a real molecular spectrum
Organic molecules can also contain carbon-13, hydrogen isotopes, oxygen isotopes and other naturally occurring variants.
Their small additional isotope contributions can create further peaks or subtle intensity changes around the main halogen isotope cluster.
In a simple examination problem, the question may explicitly say to ignore minor isotopes of the other elements.
That permits the clean 3:1 or 9:6:1 result.
A real high-resolution spectrum can be more complicated, including accurate isotope masses and instrument resolution effects.
The ideal nominal mass pattern remains useful, but it is not a claim that every real molecule with two chlorine atoms produces exactly three and only three detected peaks.
The student should obey the scope of the supplied model and avoid inventing additional minor isotope peaks when the question clearly asks for a simple ratio.
A complete worked problem from first principles
A hypothetical question states:
Natural chlorine consists of 75.0% ³⁵Cl and 25.0% ³⁷Cl. A mass spectrometer detects singly positively charged molecular ions of chlorine gas. Calculate the possible m/z values and their relative intensities.
Step 1: identify the original molecule.
Cl₂ contains two chlorine atoms.
Step 2: calculate masses.
35+35=70, 35+37=72, 37+37=74.
Step 3: convert abundances to probabilities.
p = 0.750, q = 0.250.
Step 4: calculate isotope combinations.
p² = 0.5625.
2pq = 0.3750.
q² = 0.0625.
Step 5: express as a ratio.
0.5625 : 0.3750 : 0.0625.
Divide by 0.0625:
9 : 6 : 1.
The final answer is:
Cl₂⁺ peaks near m/z 70, 72 and 74 in an approximate relative intensity ratio 9:6:1.
That is the complete logic, not just a memorised spectral fingerprint.
A changed exam question: three chlorine atoms
Suppose an unfamiliar molecular ion contains three chlorine atoms, with the same p = 3/4 and q = 1/4 and ignoring all other minor isotopes.
The probabilities follow:
(p + q)³ = p³ + 3p²q + 3pq² + q³.
Substitute:
p³ = 27/64.
3p²q = 27/64.
3pq² = 9/64.
q³ = 1/64.
Therefore the relative isotope groups are:
27 : 27 : 9 : 1.
There are four main combinations, corresponding to zero, one, two or three chlorine-37 atoms.
This is an enrichment extension showing how binomial coefficients scale with more isotope-bearing positions.
Do not confuse it with Cl₂’s 9:6:1 pattern. The number of isotope-bearing atoms is different, so the number of peaks and probability coefficients change.
Why isotope patterns do not determine the full molecular structure
A 9:6:1 cluster suggests two chlorine atoms in the relevant molecular or fragment ion under the simple isotope model.
But it does not uniquely tell us the rest of the molecular formula or the positions of chlorine on an organic skeleton.
For example, different compounds containing two chlorines could have similar chlorine-related isotope ratios but different molecular-ion masses and fragment patterns.
Structural isomers with the same full formula can also share the same molecular-ion isotope masses.
Thus the isotope pattern is useful evidence about elemental composition, but not automatically proof of a complete chemical structure.
A careful analyst combines the isotope distribution with the molecular mass, relevant fragments and other analytical evidence.
That is the difference between a diagnostic feature and a complete molecular identification.
A frequent error: treating m/z 72 as the most abundant peak
The middle isotopic combination has two sampling orders, making it more common than if only one order were counted.
But with ³⁵Cl abundance at 75%, the all-35 combination still has probability:
9/16 = 56.25%.
The mixed combination has:
6/16 = 37.5%.
Therefore m/z 70 is the tallest of the three molecular-ion isotope peaks in this ideal model.
The middle peak is the second tallest, not the largest.
Compare bromine, where p and q are each near 0.5. There, the mixed combination has 50% probability, greater than either all-identical combination at 25%, so the middle Br₂ peak is tallest.
This is a useful reminder that the factor of two must be considered together with the individual abundances.
The first-error diagnostic table
| Student statement | First misconception | Repair question |
|---|---|---|
| Cl₂ has two isotope peaks because chlorine has two isotopes | Ignored the two atomic positions | How many combinations can two isotope choices make? |
| The mixed Cl₂ peak has probability 3/16 | Forgot reverse sampling order | Can 37 be first and 35 second? |
| m/z 72 proves chlorine-36 exists | Confused combined molecular mass with atomic isotope mass | Could 35+37 make 72? |
| All 70/72/74 peaks must have equal height | Ignored different isotope abundances | What are p and q? |
| Atomic fragment peaks 35/37 have ratio 9:6:1 | Applied a two-atom pattern to one-atom fragments | How many chlorine atoms are in Cl⁺? |
| The atomic and molecular peak groups have a fixed overall intensity ratio | Ignored fragmentation fraction | What determines how many Cl₂⁺ survive? |
| The mean mass of Cl₂ is 72 because there are three masses 70,72,74 | Used an unweighted average | What are the probabilities of each mass? |
| Any 9:6:1 pattern proves the exact whole compound is Cl₂ | Confused isotope count with complete molecular identity | Could CH₂Cl₂ show a shifted 9:6:1 pattern? |
| The ratio is exact for every chlorine sample | Ignored 75/25 rounding and minor isotope effects | What abundances are supplied? |
Eight independent questions
- What is the difference between ³⁵Cl and ³⁷Cl atoms?
- What are the approximate natural abundance fractions of the two isotopes used for a 3:1 model?
- Calculate all possible nominal masses of singly charged Cl₂ molecular ions.
- Why does the mixed ³⁵Cl–³⁷Cl molecular ion have probability 2pq rather than pq?
- Derive the 9:6:1 isotope ratio using p = 0.75 and q = 0.25.
- Calculate the average relative molecular mass of Cl₂ from peaks 70,72,74 and weights 9,6,1.
- Explain why a chlorine spectrum may also have atomic fragment peaks at m/z 35 and 37.
- What approximate three-peak isotope ratio would Br₂ show if its two isotope abundances were each 50%?
Answers
1. Both have 17 protons, but ³⁵Cl has 18 neutrons and ³⁷Cl has 20. 2. 0.75 and 0.25. 3. 70, 72 and 74. 4. The two orders 35–37 and 37–35 both contribute to the same mixed composition. 5. p²:2pq:q² = 0.5625:0.375:0.0625 = 9:6:1. 6. (9×70+6×72+74)/16 = 71. 7. Some Cl₂⁺ molecular ions fragment to give charged Cl⁺ ions. 8. 1:2:1.
If the learner gets Question 3 right but Question 5 wrong, the issue is probability, not isotope identification. If Question 6 is wrong, focus on weighted averages.
How a 90-minute Chemistry tutorial can teach this
Begin with a short diagnostic: ask the student to calculate the weighted relative atomic mass of chlorine from a stated 3:1 abundance.
Next, draw a Cl₂ molecule with two blank atomic positions and ask the learner to fill them using ³⁵Cl and ³⁷Cl.
Then work through the three mass combinations and the two possible mixed orders. Write p², 2pq and q² beside the combinations.
Follow this with a calculation of the 9:6:1 relative intensities and a weighted check of Mr(Cl₂) = 71.
The independent section should change the abundances to 60%/40%, producing the new pattern 9:12:4 rather than the memorised 9:6:1.
Finish by distinguishing atomic Cl⁺ fragments from Cl₂⁺ molecular ions and asking which conclusions an isotope cluster supports about an unknown compound.
A three-student group should still require each learner to complete the changed arithmetic independently.
What Hougang parents should ask a Chemistry tutor
Bring a marked isotope or mass-spectrum question and ask whether the student struggled with the meaning of isotopes, probabilities, the factor of two for mixed combinations, m/z charge or weighted molecular mass.
These are different weaknesses. A good session should locate the first one and give a changed problem with a new abundance ratio.
At eduKate, small-group lessons of up to three students are designed for individual explanations and correction. Confirm the actual teaching venue, lesson schedule, JC programme and travel requirements rather than assuming this Hougang-targeted page represents a physical local branch.
Why isotope patterns matter beyond examinations
Mass spectrometry helps chemists study molecular masses, isotope composition and fragments of unknown substances.
Distinctive isotope clusters can reveal the presence and number of certain atoms in a detected ion, particularly elements such as chlorine and bromine with conspicuous multiple stable isotopes.
The transferable habit is to identify the charged species, count the isotope-bearing positions and calculate the probability of each combination.
That is useful preparation for later analytical Chemistry and laboratory research, without promising any examination score or university admission result.
Frequently asked questions
Why does Cl₂ have three molecular-ion isotope peaks?
Because two chlorine atoms can combine as 35–35, 35–37 or 37–37, producing three different nominal masses.
Why are the peaks at 70, 72 and 74?
These are the sums of the chlorine isotope mass numbers for singly charged Cl₂⁺ ions.
Where does 9:6:1 come from?
From the probabilities (3/4)², 2(3/4)(1/4) and (1/4)², giving 9/16, 6/16 and 1/16.
Why is the mixed peak multiplied by two?
The two independent sampling orders 35–37 and 37–35 have the same total mass and both contribute to the peak.
Are there three stable chlorine isotopes?
The two principal stable isotopes are ³⁵Cl and ³⁷Cl. The mass 72 peak comes from a mixed two-atom molecule, not a chlorine-36 atom.
Why are there also m/z 35 and 37 peaks?
Fragmentation of molecular chlorine ions can produce singly charged atomic chlorine ions.
Does every molecule containing chlorine show 9:6:1?
No. The pattern corresponds to an ion with two chlorine positions under the approximate 3:1 isotope-abundance model. One chlorine gives about 3:1; two bromine atoms give about 1:2:1.
Is this molecular-ion interpretation required in H2 9476?
The fundamental isotope and relative-mass ideas support the explanation. Detailed mass-spectral pattern recognition is presented as enrichment, and examination requirements should be checked against the actual 9476 syllabus.
Continue learning and references
Why iodine is solid but chlorine is gaseous · Why fluorine is a stronger oxidising agent than chlorine · Mixed chlorine and bromine isotope clusters · Isotopic abundance and relative atomic mass · Chemistry LibreTexts: mass spectra of elements · Chemistry LibreTexts: halogen isotope patterns in organic compounds · Hougang Chemistry Tutor Library · eduKate Singapore lesson enquiries.
Data and safety note: The 9:6:1 ratio assumes rounded 75%/25% chlorine isotope abundances, independent isotope selection and the corresponding singly charged molecular ions. More precise abundances produce slightly different ratios. Chlorine gas is toxic; this is a mathematical and analytical Chemistry explanation, not a laboratory procedure.