Mass Spectrometry: Identifying and Measuring Water Pollutants

Imagine a creek running past an agricultural paddock. After heavy rain, a water sample is collected downstream and sent to a laboratory. The mass spectrum contains a strong signal at \(m/z = 216.1\).

Is the pollutant atrazine? And if it is, how much atrazine is actually in the water?

These are two different questions. Mass spectrometry can help answer both, but identifying a chemical and measuring its concentration require different evidence. A strong peak might help identify an ion, but it does not automatically tell you how much of the original pollutant was present.

We’ll follow this water sample through liquid chromatography-tandem mass spectrometry (LC-MS/MS), use its fragmentation pattern to investigate chemical identity, and then use calibration data, blanks, and dilution factors to estimate concentration.

Throughout the article, atrazine is our illustrative pollutant. The instrument transitions are based on established analytical methods, while the calibration measurements and sample results are original teaching data rather than measurements from an actual creek.

01One sample, two different analytical problems

Before looking at the instrument, separate the two tasks.

Think about what would happen if another chemical produced an ion with the same mass-to-charge ratio as atrazine.

A mass spectrum might show a peak in exactly the expected position. However, that alone would not prove the chemical was atrazine.

Likewise, even if we identify atrazine correctly, we still cannot determine its concentration simply by looking at how tall its peak appears.

The two problems need to be investigated separately.

02Following the sample through an LC-MS/MS instrument

A common instrument for measuring trace environmental pollutants is a triple-quadrupole LC-MS/MS system.

The name describes two techniques working together:

• Liquid chromatography (LC) separates the chemicals in a mixture.
• Tandem mass spectrometry (MS/MS) selects particular ions, fragments them, and analyses the resulting ions.

The physical arrangement matters because each component solves a different problem.

Step 1: Separate the chemicals using liquid chromatography

The filtered creek-water sample is placed into a sample vial.

An autosampler injects a small volume into a stream of liquid called the mobile phase. This carries the sample through a chromatography column.

Inside the column, different substances interact differently with the stationary phase. Some travel through quickly, while others are retained for longer.

As a result, the substances leave the column at different times.

The time taken for a particular substance to leave the column is called its retention time.

For example, under one particular analytical method, an atrazine standard might produce a peak at 8.90 minutes.

That value is not universal. Change the column, mobile phase, temperature, or flow rate, and the retention time may change.

What matters is whether the unknown sample and a known atrazine standard behave consistently under the same conditions.

Step 2: Convert the molecules into ions

After leaving the chromatography column, the sample enters an ionisation source.

For many environmental pesticides, electrospray ionisation (ESI) is suitable.

In positive-ion mode, the instrument applies a high voltage to the liquid as it exits a narrow capillary. This produces tiny charged droplets.

As the solvent evaporates, the droplets become smaller. Eventually, gas-phase ions can enter the mass spectrometer.

Atrazine can gain a proton, forming a positively charged ion:

\[ \ce{M + H+ -> [M+H]+} \]

Here, \(M\) represents a neutral atrazine molecule.

The resulting ion is written as \(\ce{[M+H]+}\).

This is important because a mass spectrometer analyses ions, not neutral molecules.

Step 3: Select the precursor ion using Q1

The first quadrupole, Q1, acts as a mass filter.

It selects ions with a particular mass-to-charge ratio and allows them to pass through.

For our atrazine investigation, we select the ion at:

\[ m/z = 216.1 \]

This is called the precursor ion.

Other ions are excluded from the selected pathway.

Step 4: Break the precursor ion into fragments

The selected ions enter Q2, the collision cell.

The collision cell contains an inert gas, commonly nitrogen or argon. The precursor ions collide with gas particles, gaining internal energy that can cause chemical bonds to break.

This process is called collision-induced dissociation (CID).

The smaller ions produced are called product ions or fragment ions.

Different chemical structures can produce different fragmentation patterns. This gives us additional evidence about the identity of the original molecule.

Step 5: Analyse the fragments using Q3

The product ions enter Q3, which separates them according to their mass-to-charge ratios.

A detector records the ions that arrive.

The instrument can therefore tell us which fragments were produced from the original precursor ion.

This is the key advantage of tandem mass spectrometry: instead of asking only whether an ion has the expected mass, we investigate how that ion breaks apart.

The triple-quadrupole arrangement and collision-induced dissociation are explained further in Shimadzu’s mass spectrometry fundamentals.

03Identifying atrazine using mass-to-charge ratio and fragmentation

Let’s examine the original signal at \(m/z = 216.1\).

The first thing to understand is what this measurement actually represents.

What does mass-to-charge ratio mean?

The mass-to-charge ratio is:

\[ \frac{m}{|z|} \]

where:

• \(m\) is the mass of the ion,
• \(z\) is its charge number.

For a singly charged positive ion, \(z=+1\).

Therefore, its mass-to-charge ratio is numerically approximately equal to its mass in atomic mass units per elementary charge.

For a doubly charged ion, the same mass would produce half the \(m/z\) value.

This distinction matters because mass spectrometers separate ions according to both mass and charge, not mass alone.

Atrazine’s precursor ion

Atrazine has the molecular formula:

\[ \ce{C8H14ClN5} \]

Its protonated molecular ion, \(\ce{[M+H]+}\), has a characteristic \(m/z\) value of approximately 216.1.

This gives us our first identification clue.

However, other substances can produce ions with the same nominal mass. We need more evidence.

Look at the chlorine isotope pattern

Atrazine contains one chlorine atom.

Naturally occurring chlorine has two major isotopes:

• Chlorine-35, with an abundance of approximately 75.8%.
• Chlorine-37, with an abundance of approximately 24.2%.

These isotopes have different masses.

If an atrazine molecule contains chlorine-35, its protonated ion appears around \(m/z = 216.1\).

If it contains chlorine-37 instead, the corresponding ion appears around \(m/z = 218.1\).

The expected intensity ratio is approximately:

\[ \frac{I_{216}}{I_{218}} \approx \frac{75.8}{24.2} \approx 3.1 \]

In other words, the peak at 216.1 should be roughly three times as intense as the peak at 218.1.

This is a useful clue that the ion contains one chlorine atom.

It is not absolute proof of atrazine, because many other chlorine-containing compounds have similar isotope patterns.

Now examine the fragment ions

The next step is to investigate how atrazine breaks apart.

Published LC-MS/MS methods commonly monitor transitions such as:

\[ 216.1 \rightarrow 174.0 \]

and

\[ 216.1 \rightarrow 104.0 \]

The first number represents the precursor ion selected by Q1. The second represents a product ion detected by Q3.

Consider the first transition.

The difference between the precursor and product ion is:

\[ 216.1-174.0=42.1 \]

This corresponds approximately to a loss of \(\ce{C3H6}\), a neutral propene molecule.

This fragmentation is consistent with the structure of atrazine, which contains an isopropylamino substituent.

The second product ion at \(m/z=104.0\) provides another piece of fragmentation evidence.

We do not need to assume that every fragment uniquely identifies one chemical. Instead, we compare the complete pattern with a known atrazine reference standard.

Published pesticide methods use these precursor-to-product transitions for atrazine identification and monitoring. The exact transitions and collision conditions depend on the analytical method.

What makes the identification convincing?

A reliable identification combines several observations.

The more independently consistent evidence we have, the stronger the identification.

One important limitation remains: different compounds can sometimes produce similar fragments. Even an apparently convincing fragmentation pattern should be checked against the reference standard, including retention time and the relative intensities of selected product ions.

04Why relative abundance does not tell you concentration

Now suppose the atrazine peak is much taller than another pollutant’s peak.

Does this mean there is more atrazine in the water?

Not necessarily.

This is one of the most important distinctions in quantitative mass spectrometry.

Understanding a relative-abundance spectrum

A mass spectrum usually plots:

• Mass-to-charge ratio on the horizontal axis.
• Relative abundance on the vertical axis.

The most intense peak is called the base peak.

Its relative abundance is assigned a value of 100%. The intensities of the other peaks are expressed relative to it.

For example:

These are illustrative relative intensities.

They describe the distribution of detected ions in that spectrum. They do not directly describe the concentration of atrazine in the original water.

Why peak height is unreliable

The signal produced by an analyte depends on more than its concentration.

Factors include:

• How efficiently the analyte becomes ionised.
• The chemical structure of the analyte.
• The presence of other substances in the sample.
• Instrument settings and detector response.
• The width and shape of the chromatographic peak.

Even two samples containing identical atrazine concentrations can produce different signal intensities if their chemical matrices differ.

There is also a difference between peak height and peak area.

Peak height measures the maximum signal at one point.

Peak area represents the integrated signal across the entire chromatographic peak.

For quantitative LC-MS/MS, integrated peak area is commonly used because it accounts for the full signal as the analyte passes through the detector.

However, even peak area is not automatically a concentration.

We still need calibration.

05Using calibration data to calculate concentration

A calibration curve establishes the relationship between the concentration of a known substance and the instrument’s measured response.

The basic process is straightforward:

1. Prepare standards containing known atrazine concentrations.
2. Analyse each standard using the same instrument method.
3. Record the corresponding chromatographic peak areas.
4. Plot instrument response against known concentration.
5. Use the resulting relationship to calculate the concentration of an unknown sample.

Why use an internal standard?

An internal standard is a known substance added in the same amount to every calibration standard and sample.

For atrazine analysis, a useful choice is an isotopically labelled atrazine compound, such as atrazine-d5.

It behaves very similarly to atrazine during sample preparation, chromatography, and ionisation, but its different mass allows the instrument to distinguish it.

Instead of using the atrazine peak area alone, we calculate an area ratio:

\[ R=\frac{A_{\text{atrazine}}} {A_{\text{internal standard}}} \]

Here:

• \(R\) is the dimensionless area ratio.
• \(A_{\text{atrazine}}\) is the integrated atrazine peak area.
• \(A_{\text{internal standard}}\) is the integrated internal-standard peak area.

This ratio helps compensate for variations in injection volume and ionisation efficiency.

It does not guarantee that every matrix effect disappears, but it generally makes the measurement more reliable.

Constructing a calibration curve

Suppose a laboratory prepares six atrazine standards.

The following results are obtained:

The data follow a linear relationship:

\[ R=0.0200c+0.020 \]

where \(c\) is the atrazine concentration in the prepared vial, measured in ng/L.

The gradient is 0.0200 per ng/L, and the intercept is 0.020.

Rearranging gives:

\[ c=\frac{R-0.020}{0.0200} \]

This equation allows us to calculate an unknown concentration from its measured area ratio.

Why blanks matter

A laboratory needs to check whether atrazine contamination is being introduced during sampling or sample preparation.

A blank contains everything required for the analytical procedure except the pollutant being investigated.

There are several useful types.

A procedural blank is particularly useful because it passes through the same preparation steps as the actual sample.

Suppose the blank produces a small signal.

This does not automatically mean that the sample contains atrazine. The signal may come from contamination introduced during preparation.

The laboratory must compare the blank response with the method’s detection limit and its established quality-control requirements.

If a blank contains measurable contamination, a validated method may permit blank correction. If the blank is unusually high or inconsistent, the laboratory should investigate the contamination rather than blindly subtracting it.

Blanks help identify contamination, but they do not reveal whether atrazine has been lost during sample preparation. A separate recovery test is needed for that.

07Matrix effects: When the water changes the measurement

Our calibration standards were prepared in relatively clean water.

The creek sample, however, contains other dissolved substances.

These might include:

• Dissolved organic matter.
• Mineral ions and salts.
• Other pesticides.
• Naturally occurring organic compounds.

These substances form the sample’s matrix.

The matrix can change the efficiency with which atrazine becomes ionised.

This is called a matrix effect.

Ion suppression and ion enhancement

There are two main possibilities.

Ion suppression occurs when other substances reduce the signal produced by atrazine.

The actual atrazine concentration may be unchanged, but fewer atrazine ions reach the detector.

Ion enhancement occurs when the matrix increases the measured response.

Both effects can cause errors if we assume that the sample behaves exactly like a clean calibration standard.

How can we test for matrix effects?

One useful method is a matrix spike.

A matrix spike is prepared by adding a known amount of atrazine to a portion of the actual creek-water sample.

We then compare the measured increase in response with the increase we would expect from the calibration curve.

The percentage recovery is:

\[ \text{Recovery} = \frac{\text{Measured increase}} {\text{Expected increase}} \times100\% \]

If the recovery is close to 100%, the method is measuring the added atrazine reasonably well.

A recovery substantially below 100% may indicate analyte loss or ion suppression.

A recovery above 100% may indicate ion enhancement or another positive bias.

The laboratory should compare the result with its validated acceptance range.

Can we reduce matrix effects?

There are several approaches.

1. Use an isotopically labelled internal standard.

A compound such as atrazine-d5 behaves similarly to atrazine during ionisation.

If both experience similar suppression, their area ratio can compensate for some of the variation.

2. Use matrix-matched calibration.

Prepare calibration standards in a blank matrix that resembles the actual creek water.

This makes the calibration standards behave more like the samples.

3. Use standard addition.

Divide the original sample into several aliquots and add different known amounts of atrazine to each.

Measure the response for every aliquot and plot response against the added concentration.

The relationship can be extrapolated to estimate the original concentration.

This is useful when a suitable blank matrix is difficult to obtain, although it assumes a sufficiently linear response and consistent matrix effects across the additions.

4. Dilute the sample.

Dilution reduces the concentration of both atrazine and interfering substances.

If the analyte concentration remains above the LOQ, dilution can improve the reliability of the measurement.

However, dilution is not always beneficial. If the sample becomes too dilute, atrazine may fall below the method’s quantitation limit.

The goal is to reduce matrix interference without losing the ability to measure the pollutant accurately.

08Detection limits: When is a measurement reliable?

Even a sensitive mass spectrometer cannot reliably measure every concentration.

At very low concentrations, the analyte signal becomes difficult to distinguish from background noise.

Two important terms describe this limitation.

Limit of detection (LOD)

The limit of detection is the lowest concentration that can be reliably distinguished from background noise using the analytical method.

A signal below the LOD cannot be confidently distinguished from the background.

This does not mean the pollutant is definitely absent.

It means the method cannot reliably detect it at that concentration.

Limit of quantitation (LOQ)

The limit of quantitation is the lowest concentration that can be measured with acceptable accuracy and precision.

The LOQ is normally higher than the LOD.

A substance may be detected without being measured accurately enough to report a reliable numerical concentration.

For example, consider a method with:

• LOD = 0.50 ng/L.
• LOQ = 1.50 ng/L.

An important point: the LOD and LOQ belong to the analytical method, not just the instrument.

Sample filtration, extraction, dilution, contamination, and matrix effects can all influence the practical detection and quantitation limits.

If a sample is diluted fourfold, an LOD of 0.50 ng/L in the prepared vial corresponds to an original-water concentration of 2.0 ng/L, assuming no other concentration or recovery steps.

Also, the LOD is not a toxicity threshold.

A result below the LOD does not establish that the water is safe. Whether a measured concentration presents an environmental concern requires comparison with appropriate environmental guidelines and consideration of exposure conditions.

The Australian Drinking Water Guidelines’ explanation of detection limits explains the distinction between detecting and reliably quantifying a substance.

09Putting the evidence together

Let’s return to the original creek sample.

Suppose the laboratory obtains the following results:

The combined evidence supports the identification of atrazine.

The calibration curve gives a concentration of 30.0 ng/L in the diluted sample.

The dilution factor gives an estimated concentration of 120 ng/L in the original filtered creek water.

The blank result does not indicate quantifiable contamination, and the matrix spike recovery falls within the assumed acceptable range.

We can therefore report an estimated dissolved atrazine concentration of 120 ng/L, subject to the method’s measurement uncertainty and validation requirements.

There is one final distinction worth keeping in mind.

The result describes the filtered water sample at the time and location it was collected. It does not necessarily represent atrazine attached to suspended sediment, the entire creek, or conditions at a different time.

Environmental monitoring therefore requires more than a sophisticated instrument. Sampling procedures, preservation, blanks, replicate measurements, and appropriate analytical methods all influence the reliability of the final result.

The Australian and New Zealand Guidelines for Fresh and Marine Water Quality provide useful guidance on these quality-control requirements, including calibration checks, blank analysis, and recovery of known additions.

10Questions and solutions

The following questions build from straightforward identification towards more challenging problems involving calibration, matrix effects, and competing measurements.

Question 1

A laboratory analyses a filtered water sample for atrazine.

The reference standard produces the following results:

• Retention time: 7.40 minutes.
• Precursor ion: \(m/z=216.1\).
• Product ions: \(m/z=174.0\) and \(104.0\).

The unknown sample produces:

• Retention time: 7.42 minutes.
• Precursor ion: \(m/z=216.1\).
• Product ions: \(m/z=174.0\) and \(104.0\).
• An isotope peak at \(m/z=218.1\), approximately 32% as intense as the precursor peak at 216.1.

Explain why these results support the identification of atrazine. Why would the precursor ion alone be insufficient?

Solution 1

The results strongly support the identification of atrazine because the unknown matches the reference standard across several independent measurements.

First, compare the retention times.

The standard elutes at 7.40 minutes, while the sample elutes at 7.42 minutes.

This small difference is consistent with the same substance under the analytical conditions.

Next, compare the precursor ions.

Both produce an ion at \(m/z=216.1\), which is consistent with protonated atrazine.

The isotope pattern provides further evidence. The peak at 218.1 has approximately 32% of the intensity of the 216.1 peak.

This is close to the expected one-chlorine isotope pattern of approximately 3:1.

Finally, both samples produce the same two diagnostic product ions at 174.0 and 104.0.

The precursor ion alone would be insufficient because another substance could have the same nominal mass-to-charge ratio.

The combination of retention time, precursor mass, isotope pattern, and fragmentation evidence makes the identification much more convincing.

Question 2

A laboratory constructs a calibration curve for an environmental pollutant.

The relationship between area ratio \(R\) and concentration \(c\) is:

\[ R=0.0250c+0.015 \]

where \(c\) is measured in ng/L.

Three procedural blanks produce area ratios of:

\[ 0.063,\quad 0.065,\quad 0.067 \]

A creek-water sample is diluted by a factor of 3.00 before analysis and produces an area ratio of 0.615.

The laboratory’s validated procedure permits subtraction of the mean procedural blank because the blank results are stable and within the method’s acceptance limits.

Calculate the pollutant concentration in the original creek water.

Solution 2

The estimated concentration in the original creek water is 66.0 ng/L.

We first calculate the concentration represented by the sample response, then account for the blank and dilution.

Step 1: Calculate the mean blank response.

\[ \bar R_{\text{blank}} = \frac{0.063+0.065+0.067}{3} \]

\[ \bar R_{\text{blank}}=0.065 \]

Step 2: Convert the blank response into an equivalent concentration.

Rearrange the calibration equation:

\[ c=\frac{R-0.015}{0.0250} \]

Substitute the blank response:

\[ c_{\text{blank}} = \frac{0.065-0.015}{0.0250} \]

\[ c_{\text{blank}}=2.00\ \text{ng/L} \]

Step 3: Calculate the concentration measured in the diluted sample.

The sample area ratio is 0.615.

\[ c_{\text{sample}} = \frac{0.615-0.015}{0.0250} \]

\[ c_{\text{sample}}=24.0\ \text{ng/L} \]

Step 4: Correct for the procedural blank.

\[ c_{\text{corrected}} =24.0-2.00 \]

\[ c_{\text{corrected}}=22.0\ \text{ng/L} \]

Step 5: Apply the dilution factor.

\[ C_{\text{original}}=22.0\times3.00 \]

\[ \boxed{C_{\text{original}}=66.0\ \text{ng/L}} \]

The concentration is higher in the original creek water because the sample was diluted before analysis.

The important detail is that blank correction occurs before applying the dilution factor. The blank concentration is expressed on the same prepared-sample basis as the measured sample concentration.

Question 3

An environmental laboratory analyses a creek-water sample using the calibration relationship:

\[ R=0.0200c+0.020 \]

The sample is diluted fivefold and produces an area ratio of 0.420.

A procedural blank corresponds to an atrazine concentration of 1.0 ng/L in the prepared vial.

The laboratory then performs a matrix spike by adding 10.0 ng/L of atrazine to another aliquot of the diluted sample.

The spiked aliquot produces an area ratio of 0.600.

The laboratory’s acceptable matrix-spike recovery range is 80-120%.

Calculate:

(a) The blank-corrected atrazine concentration in the original creek water.

(b) The percentage recovery of the matrix spike.

(c) Whether the matrix-spike result is acceptable.

Solution 3

The original creek-water concentration is 95.0 ng/L, and the matrix-spike recovery is 90%, which falls within the accepted range.

Step 1: Calculate the concentration in the diluted sample.

Rearrange the calibration equation:

\[ c=\frac{R-0.020}{0.0200} \]

Substitute the measured response:

\[ c=\frac{0.420-0.020}{0.0200} \]

\[ c=20.0\ \text{ng/L} \]

Step 2: Correct for the procedural blank.

\[ c_{\text{corrected}}=20.0-1.0 \]

\[ c_{\text{corrected}}=19.0\ \text{ng/L} \]

Step 3: Apply the dilution factor.

\[ C_{\text{original}}=19.0\times5.00 \]

\[ \boxed{C_{\text{original}}=95.0\ \text{ng/L}} \]

Step 4: Calculate the expected increase in area ratio.

The added concentration is 10.0 ng/L.

Using the calibration gradient:

\[ \Delta R_{\text{expected}} =10.0\times0.0200 \]

\[ \Delta R_{\text{expected}}=0.200 \]

Step 5: Calculate the actual increase.

\[ \Delta R_{\text{measured}}=0.600-0.420 \]

\[ \Delta R_{\text{measured}}=0.180 \]

Step 6: Calculate percentage recovery.

\[ \text{Recovery} = \frac{0.180}{0.200}\times100\% \]

\[ \boxed{\text{Recovery}=90\%} \]

The recovery is within the accepted range of 80-120%.

This indicates that the method can detect the added atrazine reasonably well in the creek-water matrix.

The result does not prove that matrix effects are completely absent. It shows that the observed recovery is acceptable under the laboratory’s stated criteria.

Question 4

A laboratory uses LC-MS/MS to investigate two chromatographic peaks that both produce an ion near \(m/z=216.1\).

The reference atrazine standard has:

• Retention time: 8.90 minutes.
• Product-ion intensity ratio \(I_{174}/I_{104}=2.0\).
• The expected chlorine isotope pattern.

The unknown sample produces two peaks.

The method has:

• LOD = 0.50 ng/L.
• LOQ = 2.50 ng/L.

(a) Which peak is consistent with atrazine?

(b) Can the concentration of atrazine be reliably quantified?

(c) Peak B produces a much stronger signal. Does this mean that it contains more atrazine?

Solution 4

Peak A is consistent with atrazine, but its estimated concentration is below the LOQ, so it cannot be reliably quantified. Peak B is not confirmed as atrazine despite its stronger signal.

Step 1: Compare retention times.

Peak A occurs at 8.89 minutes, almost exactly matching the reference standard at 8.90 minutes.

Peak B occurs at 9.31 minutes, which is significantly different under the same analytical method.

This supports Peak A as the atrazine candidate.

Step 2: Compare fragmentation evidence.

The reference product-ion ratio is:

\[ \frac{I_{174}}{I_{104}}=2.0 \]

Peak A has a ratio of 1.9, which is consistent with the standard.

Peak B has a ratio of 0.4, which is substantially different.

Peak A also displays the expected chlorine isotope pattern, whereas Peak B does not.

The combined evidence strongly favours Peak A as atrazine.

Step 3: Compare Peak A’s estimated concentration with the detection limits.

Peak A has an estimated concentration of 2.2 ng/L.

The method has:

\[ LOD=0.50\ \text{ng/L} \]

\[ LOQ=2.50\ \text{ng/L} \]

Therefore:

\[ 0.50<2.2<2.50 \]

The atrazine is detected, but its concentration is below the method’s quantitation limit.

The appropriate reporting statement is that atrazine was detected below the LOQ, rather than presenting 2.2 ng/L as a reliably quantified result.

Step 4: Interpret Peak B’s stronger signal.

Peak B is taller, but it does not match the retention time, product-ion ratio, or chlorine isotope evidence expected for atrazine.

It should therefore be treated as an interfering signal or another compound, not as confirmed atrazine.

A stronger signal does not establish chemical identity or concentration.

The mistake would be to interpret the largest peak as the most abundant atrazine signal without checking whether it actually belongs to atrazine.

Question 5

A laboratory investigates the same filtered creek-water sample using two different dilution factors.

The calibration relationship is:

\[ R=0.0200c+0.020 \]

where \(c\) is the concentration in the prepared vial, measured in ng/L.

The calibration range is 0-60 ng/L, and the LOQ is 3.0 ng/L.

A procedural blank is below the LOD.

An isotopically labelled internal standard has been added before the samples are diluted. However, the laboratory knows that its correction for matrix effects may not be perfect.

The following results are obtained:

The laboratory’s validated protocol requires matrix-spike recoveries between 90% and 110%. It also specifies that a dilution failing this criterion must not be used for quantitative reporting. No numerical recovery correction is permitted.

(a) Calculate the concentration in the original creek water using each dilution.

(b) Calculate the matrix-spike recovery for each dilution.

(c) Decide which concentration should be reported.

(d) Explain why the more concentrated aliquot produces a larger signal but a less reliable estimate of the original concentration. Why should the laboratory not simply average the two results?

Solution 5

The fivefold dilution provides the reportable concentration of 98 ng/L. The twofold dilution produces an apparently stronger signal but fails the laboratory’s matrix-spike acceptance criterion.

This question tests an important interaction: dilution changes both the analyte concentration and the concentration of interfering substances.

A student might initially assume that the less diluted sample must give the more accurate result because it contains more atrazine per litre of prepared sample.

That assumption ignores matrix effects.

Step 1: Calculate the concentration in the twofold-diluted aliquot.

The calibration equation is:

\[ R=0.0200c+0.020 \]

Rearrange:

\[ c=\frac{R-0.020}{0.0200} \]

Substitute the measured response:

\[ c=\frac{0.820-0.020}{0.0200} \]

\[ c=40.0\ \text{ng/L} \]

Apply the dilution factor:

\[ C_{\text{original}}=40.0\times2.00 \]

\[ \boxed{C_{\text{original}}=80.0\ \text{ng/L}} \]

Step 2: Calculate the concentration in the fivefold-diluted aliquot.

\[ c=\frac{0.412-0.020}{0.0200} \]

\[ c=19.6\ \text{ng/L} \]

Apply the dilution factor:

\[ C_{\text{original}}=19.6\times5.00 \]

\[ \boxed{C_{\text{original}}=98.0\ \text{ng/L}} \]

Both measured concentrations fall within the calibration range and above the LOQ.

However, this does not mean both results are equally reliable.

Step 3: Calculate the expected matrix-spike response increase.

The laboratory adds 10.0 ng/L to each diluted aliquot.

Using the calibration gradient:

\[ \Delta R_{\text{expected}} =10.0\times0.0200 \]

\[ \Delta R_{\text{expected}}=0.200 \]

Step 4: Calculate recovery for the twofold dilution.

The measured response increase is 0.160.

\[ \text{Recovery} = \frac{0.160}{0.200}\times100\% \]

\[ \boxed{\text{Recovery}=80\%} \]

This is below the laboratory’s required minimum of 90%.

The twofold-diluted aliquot fails the quality-control requirement.

Step 5: Calculate recovery for the fivefold dilution.

The measured response increase is 0.196.

\[ \text{Recovery} = \frac{0.196}{0.200}\times100\% \]

\[ \boxed{\text{Recovery}=98\%} \]

This falls within the accepted range of 90-110%.

The fivefold-diluted aliquot passes the matrix-spike requirement.

Step 6: Select the reportable concentration.

The laboratory’s protocol requires a dilution to pass the matrix-spike test before its concentration can be reported.

The twofold dilution fails, while the fivefold dilution passes.

Therefore:

\[ \boxed{C_{\text{reported}}=98\ \text{ng/L}} \]

Step 7: Explain the difference between the results.

The twofold-diluted sample contains more atrazine per litre of prepared solution than the fivefold-diluted sample.

That is why its measured area ratio is larger:

\[ 0.820>0.412 \]

However, the twofold dilution also contains a higher concentration of interfering matrix components.

These substances suppress the ionisation response, causing the instrument to underestimate the amount of atrazine.

The fivefold dilution reduces the concentration of both atrazine and the interfering substances.

Although the atrazine signal becomes smaller, the matrix-spike recovery improves from 80% to 98%.

The fivefold dilution therefore gives a more reliable measurement.

The two results should not be averaged. The twofold result has failed the stated quality-control requirement, so it is not a valid quantitative result under this method.

Nor should we automatically divide the twofold result by 0.80 to create a corrected concentration. The laboratory has explicitly specified that spike recovery is a quality-control test, not a numerical correction factor.

The important conclusion is that a larger instrument response does not necessarily produce a more accurate concentration measurement.

The reliability of the measurement depends on the calibration, sample matrix, and quality-control evidence, not simply the size of the peak.

11Where to take this next

Mass spectrometry becomes much more useful when you connect it to the broader process of analytical chemistry.

The key ideas to carry forward are:

• Use precursor ions, fragmentation patterns, isotope evidence, and retention times to investigate chemical identity.
• Use integrated chromatographic peak areas and calibration curves to estimate concentration.
• Use blanks to investigate contamination and matrix spikes to test recovery.
• Apply dilution factors carefully, and check that the final result is within the method’s validated range.
• Treat the LOD and LOQ as measurement limitations, not environmental safety thresholds.

For further reading:

• Shimadzu – Mass spectrometry core principles: A useful foundation for understanding quadrupoles, collision-induced dissociation, and tandem mass spectrometry.
• US Geological Survey – Method TM 5-A12: An actual analytical method for determining pesticides in water and suspended sediment using LC-MS/MS.
• Australian and New Zealand Guidelines – Laboratory analysis: Explains calibration checks, blanks, spike recovery, and other quality-assurance procedures used in environmental monitoring.

The next useful step is to understand how analytical measurements are translated into environmental decisions: comparing measured pollutant concentrations with guideline values, accounting for uncertainty, and determining whether repeated measurements show a genuine change in water quality.