How to Interpret Mass Spectrum Peaks with Charge Greater Than One

A peak at \(m/z=50\) does not necessarily represent an ion with a mass of 50 Da. If the ion has a charge of \(+2\), its actual mass is approximately 100 Da. The mass spectrometer divides the ion’s mass by its charge number, so a doubly charged ion appears at half the \(m/z\) value of an equivalent singly charged ion.

The key is to identify the charge state before calculating the mass. Fortunately, isotope patterns often give you a way to do this. For example, isotope peaks separated by approximately 0.50 \(m/z\) units can indicate a doubly charged ion, while peaks separated by approximately 0.33 units can indicate a triply charged ion.

Let’s work through how to recognise these patterns, calculate the actual mass of an ion, and avoid confusing a multiply charged fragment with a much smaller fragment.

01Why a mass spectrum does not directly show molecular mass

A mass spectrometer measures the mass-to-charge ratio of ions, written as \(m/z\).

The relationship is:

\[
\boxed{\frac{m}{z}=\frac{m_{\text{ion}}}{|z|}}
\]

where:

• \(m/z\) is the mass-to-charge ratio shown on the horizontal axis.
• \(m_{\text{ion}}\) is the mass of the ion, measured in daltons (Da) or atomic mass units (u).
• \(z\) is the charge number of the ion. Use its magnitude, so a \(+2\) or \(-2\) ion has \(|z|=2\).

Rearranging gives the equation you need when interpreting a peak:

\[
\boxed{m_{\text{ion}}=(m/z)\times |z|}
\]

Consider three ions with exactly the same mass of 120 Da.

All three ions have the same mass. They appear at different positions because they carry different amounts of charge.

This creates an important problem when interpreting a spectrum. A peak at \(m/z=60\) could represent:

• A singly charged fragment with a mass of 60 Da.
• A doubly charged fragment with a mass of 120 Da.
• A triply charged fragment with a mass of 180 Da.

The peak position alone cannot distinguish these possibilities.

You need additional evidence, particularly the isotope pattern.

02How a mass spectrometer produces an \(m/z\) measurement

It helps to understand what the instrument actually does.

Consider a simplified time-of-flight mass spectrometer. The sample enters an ionisation chamber, where molecules are converted into charged particles. Electric fields then accelerate the ions into a long flight tube.

The ions travel towards a detector at the other end. The detector records when they arrive, and the instrument uses their flight times to determine their mass-to-charge ratios.

The important prediction is that an ion with a greater charge experiences a stronger electric force for the same electric field.

For ions accelerated through the same potential difference, their arrival times depend on their mass-to-charge ratios. A lighter ion and a heavier ion can therefore reach the detector at equivalent times if their \(m/z\) values are equal.

This is why the instrument does not simply report the mass of each original molecule.

A single peak position is insufficient to identify both mass and charge. We need to look more closely at the ions’ isotope patterns.

03Use isotope spacing to determine the charge

This is the most useful diagnostic method when a spectrum contains multiply charged ions.

Remember that elements can exist as different isotopes. These isotopes have different masses but the same number of protons, so they have the same chemical properties.

Carbon provides a particularly useful example:

• Carbon-12 has a mass of approximately 12 Da.
• Carbon-13 has a mass of approximately 13 Da.

Replacing one carbon-12 atom with carbon-13 increases the mass of an ion by approximately 1 Da.

But what happens to the position of the corresponding isotope peak if the ion has a charge greater than one?

The entire mass difference is divided by the charge number.

Therefore:

\[
\Delta(m/z)=\frac{\Delta m_{\text{isotope}}}{|z|}
\]

For a carbon-13 substitution, the approximate isotope mass difference is 1 Da.

This gives us the following pattern:

The pattern is straightforward: as the charge increases, the spacing between isotope peaks decreases.

Why does this work?

Imagine an ion with a mass of 200 Da.

If it has a charge of \(+1\), its \(m/z\) value is:

\[
\frac{200}{1}=200
\]

If one carbon-12 atom is replaced by carbon-13, the new ion has a mass of approximately 201 Da:

\[
\frac{201}{1}=201
\]

The isotope peaks are separated by 1 \(m/z\) unit.

Now consider the same molecule with a charge of \(+2\).

Its original position is:

\[
\frac{200}{2}=100
\]

The heavier isotope has a position of:

\[
\frac{201}{2}=100.5
\]

The isotope peaks are now separated by only 0.5 \(m/z\) units.

The isotope mass difference has not changed. Only the charge has changed.

What about chlorine and bromine?

Chlorine and bromine provide another useful way to identify isotope patterns.

One chlorine atom has two naturally occurring isotopes that are particularly important in mass spectra:

• Chlorine-35, with a nominal mass of 35 Da.
• Chlorine-37, with a nominal mass of 37 Da.

Their approximate abundance ratio is 3:1.

The mass difference is therefore approximately 2 Da, rather than the 1 Da difference associated with carbon-13.

For a singly charged ion, the chlorine isotope peaks are separated by approximately 2 \(m/z\) units.

For a doubly charged ion, they are separated by approximately 1 \(m/z\) unit.

Bromine also has two major isotopes separated by approximately 2 Da, but their abundances are approximately equal. A compound containing one bromine atom therefore commonly produces an isotope pair with an intensity ratio close to 1:1.

This gives you two complementary clues:

• Carbon-13 isotope spacing helps determine the charge state.
• Chlorine and bromine isotope patterns can help identify which halogen is present and provide another check on the charge assignment.

Be careful with the interpretation. A chlorine isotope pair separated by 1 \(m/z\) unit is consistent with a doubly charged ion, but the spacing and intensity ratio should be considered together. One isolated pair does not automatically establish either the charge state or the molecular structure.

A useful distinction: isotope peaks versus fragment peaks

Not every pair of nearby peaks represents two isotopes of the same ion.

Isotope peaks belong to the same underlying ion, with one or more atoms replaced by heavier isotopes. Their spacing follows a consistent relationship with the charge state.

Fragment peaks represent different ions produced when chemical bonds break. Their masses depend on the fragments formed.

For example, peaks at \(m/z=50\) and \(m/z=50.5\) might be an isotope pair from a doubly charged ion. However, they could also represent two different ions.

You should only assign an isotope relationship when the surrounding pattern and chemical evidence support it.

04Calculate the actual ion mass correctly

Once you know the charge state, calculating the ion mass is usually straightforward.

However, there is another detail that matters: the way the ion was formed.

A mass spectrometer can create ions through different ionisation processes. The ion’s mass may therefore differ slightly from the mass of the original neutral molecule.

Electron ionisation: removing electrons

In electron ionisation (EI), a high-energy electron strikes a molecule and removes an electron.

The molecular ion is commonly represented as:

\[
\ce{M -> M^{+\bullet} + e^-}
\]

The dot indicates an unpaired electron.

Because the mass of an electron is extremely small compared with the mass of an atom, the molecular ion’s mass is approximately equal to the original molecule’s mass.

If two electrons are removed, a doubly charged ion can form:

\[
\ce{M -> M^2+ + 2e^-}
\]

For this ion:

\[
m/z\approx\frac{M}{2}
\]

where \(M\) is the neutral molecular mass.

This approximation is appropriate because the mass removed with the electrons is negligible.

Electrospray ionisation: adding protons

Electrospray ionisation (ESI) commonly produces ions by adding protons to molecules.

A molecule can accept one, two, or more protons, depending on its chemical structure.

For example:

\[
\ce{[M+H]^+}
\]

represents a molecule that has accepted one proton, while:

\[
\ce{[M+2H]^2+}
\]

represents a molecule that has accepted two protons.

The mass of a proton is approximately 1.0073 Da.

For a molecule with a neutral mass \(M\), a doubly protonated ion therefore has:

\[
m/z=\frac{M+2(1.0073)}{2}
\]

Simplifying:

\[
m/z=\frac{M}{2}+1.0073
\]

Notice that this is not exactly half the neutral molecular mass.

The two added protons contribute approximately 2.0146 Da to the ion’s mass. You must include them when calculating the neutral molecular mass.

For a molecule that has accepted \(n\) protons, the general relationship is:

\[
\boxed{\frac{m}{z}=\frac{M+n(1.0073)}{n}}
\]

This formula assumes positive protonation and a charge of \(+n\).

For negative ions, use the magnitude of the charge in the \(m/z\) calculation and account for any atoms or protons removed when determining the ion mass.

05What changes when you interpret fragment peaks?

The same mass-to-charge relationship applies to fragment ions.

This is where a common mistake can lead to an incorrect molecular formula.

Suppose a fragment peak appears at \(m/z=43\).

If the fragment has a charge of \(+1\), its mass is approximately 43 Da.

If it has a charge of \(+2\), its mass is approximately 86 Da.

The peak position alone does not tell you which interpretation is correct.

You must determine the fragment’s own charge state. Do not automatically assume it has retained the same charge as the original molecular ion.

During fragmentation, charge can remain on one fragment, or it can be distributed between two charged fragments. The total charge must be conserved across all products.

This means a doubly charged precursor can produce two singly charged fragment ions.

Check a fragment using more than its peak position

Once you have established the fragment’s charge, use its corrected mass to check a proposed formula.

For example, suppose:

• The observed fragment peak is at \(m/z=43\).
• Its isotope peaks are separated by approximately 0.50.
• A proposed fragment is \(\ce{[C4H8N2+2H]^2+}\).

The isotope spacing indicates a charge of \(+2\).

Calculate the ion mass:

\[
m_{\text{ion}}=43\times2=86\text{ Da}
\]

Now calculate the mass of the proposed ion:

\[
m=4(12)+10(1)+2(14)
\]

\[
m=48+10+28
\]

\[
m=86\text{ Da}
\]

The proposed ion has the correct mass.

However, matching the mass is only one test. You must also check that the parent molecule contains enough carbon, hydrogen, and nitrogen atoms to produce the fragment.

You should then consider whether the proposed fragmentation pathway is chemically plausible.

A matching mass does not uniquely establish a fragment’s structure. Different ions can have the same nominal mass, and several structures may be consistent with the available evidence.

For a more detailed method of checking fragment formulae, see How to Check Whether a Fragment Fits the Molecular Formula.

06A reliable decision process

When you encounter a peak that might have a charge greater than one, use this sequence.

Keep these common mistakes in mind:

• Treating every \(m/z\) value as a molecular mass: This is only a valid approximation for a singly charged ion.
• Ignoring added protons: A doubly protonated ion has two more proton masses than the original neutral molecule.
• Assuming every fragment retains the precursor’s charge: Product ions can have different charge states.
• Using one isotope pair as absolute proof: Check the spacing, intensity pattern, and surrounding peaks.
• Claiming a unique structure from one matching mass: Mass spectrometry provides evidence, not necessarily a complete structural identification.

If you need to combine molecular mass with elemental composition to establish a molecular formula, see Finding Molecular Formulae from Percentage Composition and Mass Spectra.

07Questions and solutions

The following questions move from straightforward charge calculations to more demanding problems involving isotope patterns, protonation, fragmentation, and charge conservation.

Question 1

A positive ion produces a mass spectrum with a peak at \(m/z=48.0\). Its isotope peaks occur at \(m/z=48.0\), \(48.5\), and \(49.0\).

Assume the spacing is caused by successive carbon-13 substitutions, each increasing the ion’s mass by approximately 1 Da.

Determine:

a) The charge state of the ion.

b) The actual mass of the ion.

Solution 1

a) The ion has a charge of \(+2\).

The isotope peaks are separated by 0.50 \(m/z\) units.

Using:

\[
\Delta(m/z)=\frac{\Delta m}{|z|}
\]

Substitute the carbon isotope mass difference:

\[
0.50=\frac{1}{|z|}
\]

Therefore:

\[
|z|=2
\]

Because the question specifies a positive ion, its charge is \(+2\).

b) The actual ion mass is 96 Da.

Use:

\[
m_{\text{ion}}=(m/z)\times |z|
\]

\[
m_{\text{ion}}=48.0\times2
\]

\[
\boxed{m_{\text{ion}}=96.0\text{ Da}}
\]

The ion has a mass of 96 Da, even though its peak appears at \(m/z=48\).

The tempting interpretation is to report 48 Da because that is the horizontal position of the peak. The isotope spacing shows why that would be incorrect.

Question 2

A compound produces a positive-ion electrospray mass spectrum containing the following precursor isotope peaks:

The strong peaks at 75.0 and 76.0 have an approximate intensity ratio of 3:1. The weaker peaks occur approximately 0.50 units above each strong peak.

Assume:

• The compound contains one chlorine atom.
• The 0.50 spacing is associated with carbon-13 substitution.
• The precursor is a doubly protonated molecule, \(\ce{[M+2H]^2+}\).
• The listed positions are rounded to one decimal place.

Determine:

a) Why the ion must have a charge of \(+2\).

b) Which chlorine isotopes correspond to the two major peaks.

c) The approximate neutral molecular mass of the compound.

Solution 2

a) The ion has a charge of \(+2\).

The weaker isotope peaks are separated from their corresponding main peaks by approximately 0.50 \(m/z\) units.

A carbon-13 substitution increases the ion’s mass by approximately 1 Da.

Therefore:

\[
0.50=\frac{1}{|z|}
\]

\[
\boxed{|z|=2}
\]

The ion is positively charged, so its charge is \(+2\).

The important clue is the half-unit spacing. A singly charged ion would show an approximately 1 \(m/z\) separation for the same carbon isotope substitution.

b) The major peaks correspond to chlorine-35 and chlorine-37.

The two major peaks are separated by:

\[
76.0-75.0=1.0
\]

Chlorine-35 and chlorine-37 differ in mass by approximately 2 Da.

For a doubly charged ion, their separation becomes:

\[
\Delta(m/z)=\frac{2}{2}=1.0
\]

The peak at \(m/z=75.0\) corresponds to the chlorine-35 isotopologue.

The peak at \(m/z=76.0\) corresponds to the chlorine-37 isotopologue.

Their approximate 3:1 intensity ratio provides further support for one chlorine atom.

Notice that the 1.0 separation does not mean the ion is singly charged. The smaller 0.50 isotope spacing establishes the charge state, while the 1.0 separation is consistent with the chlorine isotope difference.

c) The neutral molecular mass is approximately 148 Da.

The peak at \(m/z=75.0\) represents the doubly protonated chlorine-35 isotopologue.

First, calculate its total ion mass:

\[
m_{\text{ion}}=75.0\times2
\]

\[
m_{\text{ion}}=150.0\text{ Da}
\]

Next, subtract the two added protons:

\[
M=150.0-2(1.0073)
\]

\[
M=150.0-2.0146
\]

\[
\boxed{M\approx148.0\text{ Da}}
\]

The neutral compound has an approximate molecular mass of 148 Da.

The key distinction is that the chlorine isotope pattern identifies the halogen, while the carbon isotope spacing identifies the charge state. Using both clues produces a much stronger assignment than relying on either one alone.

Question 3

An unknown compound is analysed using positive-ion electrospray mass spectrometry.

Three molecular-ion isotope envelopes are observed:

Assume all three envelopes originate from the same neutral molecule, and the ions are formed by adding one, two, or three protons.

Use a proton mass of 1.0073 Da.

a) Determine the charge state of each envelope.

b) Calculate the neutral molecular mass independently from each envelope.

c) Explain why the three envelopes should not be interpreted as three different compounds.

Solution 3

a) The three envelopes represent ions with charges of \(+1\), \(+2\), and \(+3\), respectively.

For carbon-13 isotope substitutions, the approximate mass difference is 1 Da.

Envelope A has a spacing of 1.00:

\[
|z|=\frac{1}{1.00}=1
\]

Envelope B has a spacing of 0.50:

\[
|z|=\frac{1}{0.50}=2
\]

Envelope C has a spacing of 0.33:

\[
|z|\approx\frac{1}{0.33}\approx3
\]

Therefore:

• Envelope A is singly charged.
• Envelope B is doubly charged.
• Envelope C is triply charged.

b) All three envelopes give a neutral molecular mass of approximately 198 Da.

For Envelope A, the ion is \(\ce{[M+H]^+}\).

\[
M=199.0-1.0073
\]

\[
\boxed{M\approx198.0\text{ Da}}
\]

For Envelope B, the ion is \(\ce{[M+2H]^2+}\).

\[
M=2(100.0)-2(1.0073)
\]

\[
M=200.0-2.0146
\]

\[
\boxed{M\approx198.0\text{ Da}}
\]

For Envelope C, the ion is \(\ce{[M+3H]^3+}\).

\[
M=3(67.0)-3(1.0073)
\]

\[
M=201.0-3.0219
\]

\[
\boxed{M\approx198.0\text{ Da}}
\]

The small differences in the unrounded calculations are due to the peak positions being supplied to only one decimal place.

c) The three envelopes represent different charge states of the same molecule.

The neutral molecule has a mass of approximately 198 Da.

In electrospray ionisation, it can accept different numbers of protons:

\[
\ce{[M+H]^+}
\]

\[
\ce{[M+2H]^2+}
\]

\[
\ce{[M+3H]^3+}
\]

Each ion has a different total mass and charge, producing a different \(m/z\) value.

The tempting interpretation is that three different molecular masses have been detected because the spectrum contains three distinct peak groups.

However, the isotope spacings and protonation calculations show that all three groups are consistent with one neutral molecule.

This is particularly useful when interpreting electrospray spectra, where one compound can produce several charge states.

Question 4

A neutral compound has the molecular formula:

\[
\ce{C10H22N4}
\]

It is analysed using electrospray ionisation, producing a doubly protonated precursor:

\[
\ce{[M+2H]^2+}
\]

The precursor appears at \(m/z=100.0\).

During tandem mass spectrometry, a product ion appears at \(m/z=43.0\). Its isotope peaks are separated by approximately 0.50 \(m/z\) units.

A proposed product-ion formula is:

\[
\ce{[C4H8N2+2H]^2+}
\]

Assume nominal atomic masses of C = 12, H = 1, and N = 14.

a) Confirm the charge state and calculate the actual mass of the product ion.

b) Check whether the proposed formula fits the observed peak.

c) Determine the formula and mass of the neutral fragment lost during this fragmentation.

d) Explain why subtracting the product-ion mass directly from the neutral molecular mass gives an incorrect neutral-loss calculation.

Solution 4

a) The product ion has a charge of \(+2\) and a mass of 86 Da.

The isotope spacing is approximately 0.50 \(m/z\) units.

Therefore:

\[
0.50=\frac{1}{|z|}
\]

\[
|z|=2
\]

The observed ion is positively charged, so its charge is \(+2\).

Its actual mass is:

\[
m_{\text{ion}}=43.0\times2
\]

\[
\boxed{m_{\text{ion}}=86\text{ Da}}
\]

b) The proposed formula is consistent with the observed peak.

The proposed ion contains four carbon atoms, ten hydrogen atoms, and two nitrogen atoms.

Its nominal mass is:

\[
m=4(12)+10(1)+2(14)
\]

\[
m=48+10+28
\]

\[
m=86\text{ Da}
\]

Its predicted mass-to-charge ratio is:

\[
m/z=\frac{86}{2}
\]

\[
\boxed{m/z=43}
\]

This matches the observed peak.

The formula also passes the atom-inventory test. The original molecule contains ten carbon atoms, 22 hydrogen atoms, and four nitrogen atoms. The proposed neutral fragment \(\ce{C4H8N2}\) does not exceed any of these available atom counts.

c) The neutral fragment lost has the formula \(\ce{C6H14N2}\) and a mass of 114 Da.

First, calculate the neutral molecular mass:

\[
M=10(12)+22(1)+4(14)
\]

\[
M=120+22+56
\]

\[
M=198\text{ Da}
\]

The proposed neutral fragment that forms the detected doubly protonated product is:

\[
\ce{C4H8N2}
\]

Its neutral mass is:

\[
m=4(12)+8(1)+2(14)
\]

\[
m=48+8+28
\]

\[
m=84\text{ Da}
\]

The remaining neutral fragment must therefore have the formula:

\[
\ce{C10H22N4-C4H8N2}
\]

\[
\ce{C6H14N2}
\]

Its mass is:

\[
m=6(12)+14(1)+2(14)
\]

\[
m=72+14+28
\]

\[
\boxed{m_{\text{neutral loss}}=114\text{ Da}}
\]

Check the complete fragmentation equation:

\[
\ce{[C10H22N4+2H]^2+}
\]

\[
\ce{->[C4H8N2+2H]^2++C6H14N2}
\]

The precursor ion has a total mass of:

\[
198+2(1)=200\text{ Da}
\]

The products have a combined mass of:

\[
86+114=200\text{ Da}
\]

Both the atom count and total charge are conserved.

d) Subtracting 86 from 198 ignores the two protons added during ionisation.

The tempting calculation is:

\[
198-86=112\text{ Da}
\]

However, 198 Da is the mass of the neutral molecule, while 86 Da is the mass of a product ion containing the two added protons.

These quantities cannot be subtracted directly to determine the neutral loss.

The precursor ion actually has a mass of 200 Da. Therefore:

\[
200-86=114\text{ Da}
\]

The correct neutral loss is 114 Da.

This question illustrates why you must keep track of the ionisation process throughout a fragmentation calculation, rather than switching between neutral molecular masses and charged-ion masses without adjustment.

Question 5

An unknown neutral molecule has a nominal mass of 300 Da. Positive-ion electrospray mass spectrometry produces a precursor isotope envelope centred at:

\[
m/z=151.0
\]

Its carbon isotope peaks are separated by approximately 0.50 \(m/z\) units.

The precursor is therefore assigned as:

\[
\ce{[M+2H]^2+}
\]

A tandem mass spectrometry experiment then causes this precursor to split into two complementary product ions, A and B.

The experiment provides the following information:

Assume:

• The precursor’s two added protons are distributed between the two complementary product ions, with one proton retained by each.
• Both product ions are detected.
• No atoms or electrons enter or leave the system during fragmentation.
• All masses are nominal, and proton mass may be approximated as 1 Da.

Determine:

a) The charge state and actual mass of the precursor ion.

b) The charge state and actual mass of each product ion.

c) Whether the two product assignments satisfy conservation of charge.

d) Whether they satisfy conservation of mass.

e) Explain why it is incorrect to assume that both product ions retain the precursor’s \(+2\) charge.

Solution 5

a) The precursor has a charge of \(+2\) and a mass of 302 Da.

The isotope spacing is 0.50 \(m/z\) units.

Using the carbon-13 isotope mass difference:

\[
0.50=\frac{1}{|z|}
\]

\[
|z|=2
\]

The precursor is positively charged, so its charge is \(+2\).

Its actual mass is:

\[
m_{\text{precursor}}=151.0\times2
\]

\[
\boxed{m_{\text{precursor}}=302\text{ Da}}
\]

This agrees with a neutral molecule of 300 Da accepting two protons.

b) Both product ions have a charge of \(+1\). Their actual masses are 43 Da and 259 Da.

This is the first important change in the problem.

The precursor has a charge of \(+2\), but the isotope spacing of each product is 1.00 \(m/z\) unit.

A carbon-13 substitution produces an approximate 1 Da mass difference.

Therefore, for each product:

\[
1.00=\frac{1}{|z|}
\]

\[
|z|=1
\]

Both products have a charge of \(+1\).

For Product A:

\[
m_A=43.0\times1
\]

\[
\boxed{m_A=43\text{ Da}}
\]

For Product B:

\[
m_B=259.0\times1
\]

\[
\boxed{m_B=259\text{ Da}}
\]

The precursor’s charge state does not automatically determine the charge state of each product.

c) The products satisfy conservation of charge.

The precursor begins with a total charge of \(+2\).

The products have charges of \(+1\) and \(+1\).

Therefore:

\[
(+1)+(+1)=+2
\]

The total charge is conserved.

This is consistent with the two added protons being distributed between the complementary fragments.

d) The products also satisfy conservation of mass.

The combined mass of the two product ions is:

\[
m_{\text{products}}=43+259
\]

\[
\boxed{m_{\text{products}}=302\text{ Da}}
\]

The precursor mass is also 302 Da.

Therefore:

\[
m_{\text{precursor}}=m_A+m_B
\]

Both mass and charge are conserved.

The neutral fragments before protonation have masses of approximately:

\[
43-1=42\text{ Da}
\]

and:

\[
259-1=258\text{ Da}
\]

Their combined neutral mass is:

\[
42+258=300\text{ Da}
\]

This independently matches the original neutral molecular mass.

e) Assuming both products retain a \(+2\) charge contradicts the isotope evidence and conservation of charge.

A tempting approach is to assume that the precursor’s \(+2\) charge remains on each fragment.

If that were true, Product A would have a mass of:

\[
43\times2=86\text{ Da}
\]

Product B would have a mass of:

\[
259\times2=518\text{ Da}
\]

Their combined charge would be \(+4\), even though the precursor only had a total charge of \(+2\).

Their combined mass would be:

\[
86+518=604\text{ Da}
\]

This is also inconsistent with the precursor’s mass of 302 Da.

More importantly, the isotope spacing independently rejects this interpretation. Both product envelopes have a 1.00 \(m/z\) spacing, indicating singly charged ions.

The correct interpretation is that the original doubly charged molecule splits into two singly charged fragments, with one positive charge on each.

This is the conceptual point to take away: charge belongs to the complete ion, not permanently to a particular piece of the original molecule. When fragmentation occurs, the charge can be redistributed between the products, but the total charge must still be conserved.

08Where to go next

Once you can identify the charge state, you can interpret molecular-ion and fragment peaks without confusing a small fragment with a larger multiply charged ion.

The next step is to combine charge-corrected masses with molecular formulae, isotope patterns, and plausible fragmentation pathways. That is how you move from recognising individual peaks to building a consistent picture of an unknown molecule.