What Does a Result Below the Detection Limit Mean?

A colourimeter gives a diluted water sample a raw absorbance of 0.0235. The reagent blank averages 0.0200. The sample reading is higher than the blank, so has the instrument detected phosphate?

Not necessarily. The difference is only 0.0035, and the blank readings themselves fluctuate. That small difference could be caused by background variation rather than phosphate.

A result below the detection limit (LOD) means that the measured signal has not crossed the method’s threshold for being distinguished from background noise with the required confidence. It does not mean the analyte is absent, and it does not mean its concentration is exactly zero.

The important distinction is between calculating a possible concentration and having enough evidence to report that concentration reliably.

01What does below the detection limit actually mean?

An analytical instrument measures a signal, not the presence or absence of a chemical directly.

For example, when testing water for phosphate using a colour-forming reagent, a colourimeter measures absorbance. A higher phosphate concentration generally produces a stronger absorbance signal, provided the method operates within its valid range.

However, the instrument also responds to background effects.

These can include:

• Absorption from the solvent and reagents.
• Light scattering from suspended particles.
• Small differences between cuvettes.
• Instrument fluctuations.
• Variations introduced during sample preparation.

Consequently, even a solution containing no phosphate can produce a measurable absorbance.

The detection limit establishes how large the analyte-related signal must be before it can be distinguished from this background.

There are three important reporting situations.

The LOQ is the limit of quantification. It represents the lowest concentration that can be measured with an acceptable level of uncertainty for the intended purpose.

Detection and quantification are therefore different tasks:

• Detection asks whether there is convincing evidence that the analyte is present.
• Quantification asks whether its concentration can be measured reliably.

A result can satisfy the first condition without satisfying the second.

For a broader explanation of these reporting categories, see HSC Chemistry: Understanding Results Below the Detection Limit.

02How is the detection limit established experimentally?

Imagine developing a method to measure small concentrations of phosphate in reservoir water.

The experimental problem is straightforward:

How can you tell whether a small increase in absorbance comes from phosphate rather than normal background variation?

Step 1: Set up the measurement

Suppose phosphate reacts with a suitable reagent to form a coloured complex.

The colourimeter measures how much light the solution absorbs.

The apparatus works in a specific order:

1. A light source produces light.
2. A wavelength selector allows the required wavelength through.
3. The selected light passes through a cuvette containing the prepared solution.
4. A detector measures the transmitted light.
5. The instrument calculates the absorbance.

The more strongly the solution absorbs the selected light, the higher its absorbance.

Step 2: Measure the background using a blank

Before analysing the unknown sample, prepare a reagent blank.

It contains:

• The same solvent.
• The same chemical reagents.
• The same preparation steps.
• No phosphate analyte.

The blank estimates the background absorbance.

What should happen if the blank is measured repeatedly?

You might initially expect exactly the same absorbance every time. In practice, small fluctuations occur.

For example, five illustrative blank measurements might produce:

Notice that the readings are not identical.

This is important because measuring a blank once tells you its approximate background signal, but does not tell you how much that signal normally varies.

That is the weakness of the simpler experimental design.

A better design repeats the blank measurement under the same conditions. The resulting variation can then be quantified using the standard deviation.

The standard deviation tells us how widely the blank readings are spread around their mean.

A larger standard deviation means greater background variation, making small analyte signals harder to distinguish.

Step 3: Establish the calibration relationship

Next, prepare calibration standards containing known phosphate concentrations.

Measure their absorbances under the same conditions as the blank and unknown sample.

The standards establish the relationship between concentration and instrument response.

For the illustrative experiment, assume the raw calibration equation is:

\[ A_{\mathrm{raw}}=0.0200+0.0800C \]

where:

• \(A_{\mathrm{raw}}\) is the measured absorbance.
• \(C\) is the phosphate concentration in \(\mathrm{mg\,L^{-1}}\).
• \(0.0200\) is the calibration intercept.
• \(0.0800\ \mathrm{L\,mg^{-1}}\) is the calibration gradient.

The intercept represents the background response at zero analyte concentration.

The gradient tells us how much the absorbance changes for each unit increase in concentration.

For this example, assume the calibration remains linear near the detection limit. The validated quantitative working range is \(0.20\) to \(2.00\ \mathrm{mg\,L^{-1}}\).

All numerical data in this example are illustrative, not measurements from an actual reservoir.

03From blank variation to the detection limit

The key idea is that the detection threshold must account for both the average blank response and its variability.

A general expression for the signal threshold is:

\[ S_{\mathrm{crit}}=\bar S_{\mathrm{blank}}+ks_{\mathrm{blank}} \]

where:

• \(S_{\mathrm{crit}}\) is the minimum signal required for detection.
• \(\bar S_{\mathrm{blank}}\) is the mean blank signal.
• \(s_{\mathrm{blank}}\) is the standard deviation of blank measurements.
• \(k\) is a factor selected according to the required confidence.

For this worked method, use \(k=3\).

The mean blank signal plus three standard deviations is a common introductory approximation. It is not a universal definition: the appropriate factor depends on the statistical decision rule and analytical method.

IUPAC expresses the detection limit using this general relationship, with the multiplier selected according to the required confidence. See the IUPAC Gold Book definition of limit of detection.

04How dilution changes the detection limit

Now suppose the original reservoir water is too concentrated for direct analysis, so the chemist dilutes it before measuring its absorbance.

A 10.0 mL aliquot of the original water is transferred into a 50.0 mL volumetric flask. Distilled water is added to the calibration mark, and the solution is mixed thoroughly.

The dilution factor is:

\[ DF=\frac{V_{\mathrm{final}}}{V_{\mathrm{aliquot}}} \]

Therefore:

\[ DF=\frac{50.0}{10.0}=5.00 \]

The colourimeter measures the concentration in the diluted solution, not the original water.

Consequently, the original-sample concentration must be calculated by multiplying by the dilution factor.

\[ C_{\mathrm{original}}=C_{\mathrm{diluted}}\times DF \]

The same adjustment applies to the detection and quantification limits.

\[ LOD_{\mathrm{original}}=LOD_{\mathrm{diluted}}\times DF \]

\[ LOQ_{\mathrm{original}}=LOQ_{\mathrm{diluted}}\times DF \]

For our phosphate method:

\[ LOD_{\mathrm{original}}=0.0593\times5 \]

\[ \boxed{LOD_{\mathrm{original}}=0.297\ \mathrm{mg\,L^{-1}}} \]

Similarly:

\[ LOQ_{\mathrm{original}}=0.20\times5 \]

\[ \boxed{LOQ_{\mathrm{original}}=1.00\ \mathrm{mg\,L^{-1}}} \]

The original water therefore has a detection limit of approximately \(0.30\ \mathrm{mg\,L^{-1}}\), not \(0.0593\ \mathrm{mg\,L^{-1}}\).

This distinction matters whenever a sample has been diluted before analysis.

05Comparing three possible results

The distinction becomes clearer when we compare three sample readings using the same phosphate method.

All three samples have been diluted fivefold.

The corresponding conclusions are:

Sample 1: Absorbance 0.0235

The calculated concentration is below the LOD of \(0.0593\ \mathrm{mg\,L^{-1}}\).

The analyte is not detected.

Its appropriate original-sample reporting limit is approximately \(0.30\ \mathrm{mg\,L^{-1}}\).

Sample 2: Absorbance 0.0270

The calculated concentration is above the LOD but below the LOQ.

The analyte is detected, but its concentration cannot be reliably quantified.

Sample 3: Absorbance 0.0450

The calculated concentration is:

\[ C_{\mathrm{diluted}}= \frac{0.0450-0.0200}{0.0800} \]

\[ C_{\mathrm{diluted}}=0.3125\ \mathrm{mg\,L^{-1}} \]

This exceeds the LOQ and lies within the validated quantitative range.

The original concentration is:

\[ C_{\mathrm{original}}=0.3125\times5 \]

\[ \boxed{C_{\mathrm{original}}=1.56\ \mathrm{mg\,L^{-1}}} \]

This result can be reported quantitatively, subject to the method’s measurement uncertainty.

The interesting comparison is between the first and second samples. Both have absorbances greater than the mean blank, but only the second has a signal that meets the detection criterion.

A reading greater than the blank is not automatically a detected result.

06Three experimental details that can change your answer

1. The correct blank matters

A reagent blank does not necessarily account for everything in the sample.

Suppose reservoir water contains naturally coloured dissolved substances or suspended particles. These may contribute additional absorbance or light scattering.

A reagent blank containing only the solvent and reagents cannot necessarily account for these effects.

A matrix-matched blank, where appropriate, is designed to reproduce the sample background without the analyte.

The choice of blank can therefore change the measured standard deviation and the calculated detection limit.

A detection limit established using clean distilled water should not automatically be applied to a complex environmental sample.

2. Do not subtract the blank twice

There are two ways to work with absorbance data.

If the instrument has not been zeroed using the blank, and the raw calibration equation is:

\[ A_{\mathrm{raw}}=mC+b \]

then calculate:

\[ C=\frac{A_{\mathrm{raw}}-b}{m} \]

If the calibration uses blank-corrected absorbance:

\[ A_{\mathrm{net}}=mC \]

then calculate:

\[ C=\frac{A_{\mathrm{net}}}{m} \]

If the instrument has already been zeroed using the blank, its displayed reading may already be corrected.

Do not subtract the blank again.

The rule is to account for the background once, using the calibration relationship that matches the measurement.

For a fuller explanation of this distinction, see How to Correct Absorbance Readings for a Blank.

3. A detection limit belongs to the method, not just the instrument

An instrument may have a very low detection limit when measuring clean standard solutions.

However, the actual analytical procedure might involve:

• Filtering the sample.
• Transferring measured volumes.
• Adding chemical reagents.
• Heating or digesting the sample.
• Diluting the solution.
• Introducing contamination during preparation.

These steps can introduce additional background variation or analyte losses.

A method detection limit should account for the complete analytical procedure, rather than just the instrument’s response to a clean blank.

The detection limit also depends on the calibration gradient.

A steeper calibration gradient means a smaller change in concentration produces a larger change in signal. This generally improves detection, provided the background noise stays the same.

The relationship is:

\[ LOD\approx\frac{ks_{\mathrm{blank}}}{m} \]

This gives a useful way to compare methods:

• Lower blank variation improves the detection limit.
• Greater calibration sensitivity improves the detection limit.
• Excessive dilution generally makes the original-sample detection limit worse.
• However, dilution can sometimes reduce matrix interference enough to improve the overall detection limit.

That final point is worth testing.

07Questions and solutions

Question 1

A colourimeter produces five blank measurements with a standard deviation of 0.00080 absorbance units.

The blank-corrected calibration gradient is:

\[ m=0.0400\ \mathrm{L\,mg^{-1}} \]

Use \(k=3\) to calculate the detection limit.

An unknown sample produces a blank-corrected absorbance of 0.0018.

Determine whether the analyte is detected.

Solution 1

The analyte is not detected, and the detection limit is \(0.0600\ \mathrm{mg\,L^{-1}}\).

Step 1: Calculate the detection limit.

Use:

\[ LOD=\frac{3s_{\mathrm{blank}}}{m} \]

Substitute:

\[ LOD=\frac{3(0.00080)}{0.0400} \]

\[ \boxed{LOD=0.0600\ \mathrm{mg\,L^{-1}}} \]

Step 2: Calculate the sample’s apparent concentration.

Using the blank-corrected calibration relationship:

\[ C=\frac{A_{\mathrm{net}}}{m} \]

\[ C=\frac{0.0018}{0.0400} \]

\[ C=0.0450\ \mathrm{mg\,L^{-1}} \]

Step 3: Compare the concentration with the detection limit.

\[ 0.0450<0.0600 \]

The calculated concentration is below the detection limit.

The signal is therefore not sufficiently distinguishable from background variation.

The appropriate report is:

\[ \boxed{\text{Not detected; }C<0.0600\ \mathrm{mg\,L^{-1}}} \]

Do not report \(0.0450\ \mathrm{mg\,L^{-1}}\) as a reliable measured concentration.

Question 2

A chemist transfers 25.0 mL of a water sample into a 100.0 mL volumetric flask and dilutes it to the calibration mark.

The analytical method has the following characteristics:

• Raw calibration equation: \(A=0.0150+0.0600C\).
• Blank standard deviation: 0.00150 absorbance units.
• Detection criterion: \(k=3\).
• Validated quantification limit: \(0.20\ \mathrm{mg\,L^{-1}}\) in the diluted solution.

The measured raw absorbance is 0.0240.

Calculate the dilution factor, the diluted concentration, and the original-sample detection and quantification limits. State the appropriate conclusion.

Solution 2

The analyte is detected but cannot be reliably quantified. The original-sample quantification limit is \(0.80\ \mathrm{mg\,L^{-1}}\).

Step 1: Calculate the dilution factor.

\[ DF=\frac{100.0}{25.0} \]

\[ \boxed{DF=4.00} \]

Step 2: Correct the sample absorbance.

The calibration equation includes the background intercept.

Therefore:

\[ A_{\mathrm{net}}=0.0240-0.0150 \]

\[ A_{\mathrm{net}}=0.0090 \]

Step 3: Calculate the diluted concentration.

\[ C_{\mathrm{diluted}}=\frac{0.0090}{0.0600} \]

\[ \boxed{C_{\mathrm{diluted}}=0.150\ \mathrm{mg\,L^{-1}}} \]

Step 4: Calculate the detection limit.

\[ LOD=\frac{3s}{m} \]

\[ LOD=\frac{3(0.00150)}{0.0600} \]

\[ \boxed{LOD=0.0750\ \mathrm{mg\,L^{-1}}} \]

The measured concentration is greater than the LOD:

\[ 0.150>0.0750 \]

However, it is below the LOQ:

\[ 0.150<0.20 \]

The analyte has been detected, but it cannot be quantified reliably.

Step 5: Calculate the original-sample limits.

\[ LOD_{\mathrm{original}}=0.0750\times4 \]

\[ \boxed{LOD_{\mathrm{original}}=0.300\ \mathrm{mg\,L^{-1}}} \]

Similarly:

\[ LOQ_{\mathrm{original}}=0.20\times4 \]

\[ \boxed{LOQ_{\mathrm{original}}=0.800\ \mathrm{mg\,L^{-1}}} \]

The apparent original concentration is:

\[ C_{\mathrm{original}}=0.150\times4 \]

\[ C_{\mathrm{original}}=0.600\ \mathrm{mg\,L^{-1}} \]

But this value is below the original-sample LOQ.

The correct conclusion is that the analyte is detected but cannot be reliably quantified. The calculated \(0.600\ \mathrm{mg\,L^{-1}}\) should not be presented as an accurate concentration.

Question 3

A chemist analyses coloured water using a spectrophotometer.

Two types of blank are available:

The calibration gradient is:

\[ m=0.0800\ \mathrm{L\,mg^{-1}} \]

The unknown sample has a raw absorbance of 0.0360.

Assume the matrix-matched blank correctly represents the background variation in the sample, and use \(k=3\).

A student uses the reagent blank and concludes that the analyte is present at \(0.200\ \mathrm{mg\,L^{-1}}\).

Evaluate the student’s conclusion. Calculate the detection limit using the appropriate blank and explain why the two approaches give different answers.

Solution 3

The student’s conclusion is not justified. Using the appropriate matrix-matched blank, the analyte is not detected. The detection limit is \(0.150\ \mathrm{mg\,L^{-1}}\).

Step 1: Calculate the detection limit using the matrix-matched blank.

\[ LOD=\frac{3s_{\mathrm{blank}}}{m} \]

\[ LOD=\frac{3(0.0040)}{0.0800} \]

\[ \boxed{LOD=0.150\ \mathrm{mg\,L^{-1}}} \]

Step 2: Correct the sample signal using the matrix background.

\[ A_{\mathrm{net}}=0.0360-0.0310 \]

\[ A_{\mathrm{net}}=0.0050 \]

Step 3: Calculate the apparent concentration.

\[ C=\frac{0.0050}{0.0800} \]

\[ C=0.0625\ \mathrm{mg\,L^{-1}} \]

Step 4: Compare with the detection limit.

\[ 0.0625<0.150 \]

The calculated concentration is below the detection limit.

Therefore, the analyte has not been detected reliably using this method.

Why did the student’s answer differ?

The student used the reagent blank.

Using that blank would produce:

\[ C=\frac{0.0360-0.0200}{0.0800} \]

\[ C=0.200\ \mathrm{mg\,L^{-1}} \]

This calculation attributes the entire difference between the sample and reagent blank to the analyte.

However, the water itself contributes background absorbance.

The reagent blank does not account for that additional matrix contribution.

The matrix-matched blank is the appropriate comparison because it better represents the background present in the unknown sample.

The important lesson is that choosing an inappropriate blank can make a weak background difference look like a genuine analyte signal.

Question 4

A chemist previously established the following raw calibration equation:

\[ A_{\mathrm{raw}}=0.0150+0.0800C \]

Before analysing a new batch of samples, the chemist measures several new blanks.

The new blank measurements have:

• Mean absorbance: 0.0190.
• Standard deviation: 0.0010.

A check standard confirms that the calibration gradient is unchanged and that the difference is caused by an additive background shift.

An unknown sample produces a raw absorbance of 0.0280.

For this question, use \(k=3\) for the LOD and \(10s/m\) for the LOQ.

Calculate the sample’s apparent concentration and determine whether it can be reliably quantified. Explain why using the original calibration intercept without checking the new blank would be misleading.

Solution 4

The analyte is detected but is below the quantification limit. Its estimated concentration is \(0.1125\ \mathrm{mg\,L^{-1}}\).

Step 1: Identify the change in background.

The original calibration intercept was 0.0150.

The new blank mean is 0.0190.

Because the gradient is confirmed to be unchanged and the shift is additive, the updated calibration relationship is:

\[ A_{\mathrm{raw}}=0.0190+0.0800C \]

The new blank must be used when interpreting the sample.

Step 2: Calculate the blank-corrected sample signal.

\[ A_{\mathrm{net}}=0.0280-0.0190 \]

\[ A_{\mathrm{net}}=0.0090 \]

Step 3: Calculate the detection limit.

\[ LOD=\frac{3(0.0010)}{0.0800} \]

\[ \boxed{LOD=0.0375\ \mathrm{mg\,L^{-1}}} \]

Step 4: Calculate the quantification limit.

\[ LOQ=\frac{10(0.0010)}{0.0800} \]

\[ \boxed{LOQ=0.125\ \mathrm{mg\,L^{-1}}} \]

Step 5: Calculate the sample concentration.

\[ C=\frac{0.0090}{0.0800} \]

\[ \boxed{C=0.1125\ \mathrm{mg\,L^{-1}}} \]

Compare the result with the two limits:

\[ 0.0375<0.1125<0.125 \]

The analyte is detected, but its concentration is below the LOQ.

Why is the original calibration intercept misleading?

If the chemist uses the old intercept without checking the new background:

\[ C=\frac{0.0280-0.0150}{0.0800} \]

\[ C=0.1625\ \mathrm{mg\,L^{-1}} \]

This incorrectly attributes the additional background absorbance to the analyte.

The apparent concentration is now above the LOQ, even though the correctly adjusted concentration is below it.

The problem is not the calibration gradient. It is the changed background signal.

This demonstrates why blanks and calibration checks must be performed under conditions that match the samples being analysed.

Question 5

A chemist is investigating a water sample that contains a small concentration of a coloured analyte.

Two sample-preparation methods are available.

Both methods use the same calibration gradient:

\[ m=0.0800\ \mathrm{L\,mg^{-1}} \]

The following results are obtained after measuring the corresponding blanks and samples.

Method A: Undiluted sample

• Dilution factor: \(DF=1\).
• Matrix-blank standard deviation: \(s_A=0.0040\).
• Sample’s blank-corrected absorbance: \(A_A=0.0080\).

Method B: Tenfold dilution

• Dilution factor: \(DF=10\).
• Matrix-blank standard deviation: \(s_B=0.00020\).
• Sample’s blank-corrected absorbance: \(A_B=0.00080\).

Both sets of blank measurements have been obtained after the complete preparation procedure.

For both methods, use:

\[ LOD=\frac{3s}{m} \]

and the provisional quantification criterion:

\[ LOQ=\frac{10s}{m} \]

Assume that:

• Both methods have a validated linear calibration relationship.
• No analyte is lost during dilution.
• The sample matrix is responsible for the difference in blank variability.
• The stated blank standard deviations accurately represent the background variation of each method.

Determine:

1. The diluted-solution LOD and LOQ for each method.
2. The original-sample LOD and LOQ for each method.
3. Whether the analyte is detected by either method.
4. Whether either method can quantify the analyte reliably.

Explain why the diluted method might produce a better detection result, even though dilution reduces the analyte concentration.

Solution 5

Method B provides stronger evidence that the analyte is present. It detects the analyte, but cannot quantify it reliably. Method A does not detect it.

This is an interesting result because Method B uses a tenfold dilution.

You might initially predict that dilution must make detection worse because it reduces the analyte concentration.

That prediction ignores the effect of background variation.

We need to calculate the detection and quantification limits for both methods, then convert them back to the original sample basis.

Step 1: Calculate the limits for Method A.

For the undiluted sample:

\[ LOD_A=\frac{3(0.0040)}{0.0800} \]

\[ \boxed{LOD_A=0.150\ \mathrm{mg\,L^{-1}}} \]

The provisional quantification limit is:

\[ LOQ_A=\frac{10(0.0040)}{0.0800} \]

\[ \boxed{LOQ_A=0.500\ \mathrm{mg\,L^{-1}}} \]

Because the dilution factor is one, these are also the original-sample limits.

Step 2: Determine whether Method A detects the analyte.

Calculate the apparent concentration:

\[ C_A=\frac{0.0080}{0.0800} \]

\[ C_A=0.100\ \mathrm{mg\,L^{-1}} \]

Compare with the detection limit:

\[ 0.100<0.150 \]

The sample signal is below the detection threshold.

Method A therefore does not detect the analyte.

Step 3: Calculate the limits for Method B.

For the tenfold diluted sample:

\[ LOD_B=\frac{3(0.00020)}{0.0800} \]

\[ \boxed{LOD_B=0.00750\ \mathrm{mg\,L^{-1}}} \]

The provisional quantification limit is:

\[ LOQ_B=\frac{10(0.00020)}{0.0800} \]

\[ \boxed{LOQ_B=0.0250\ \mathrm{mg\,L^{-1}}} \]

These limits apply to the diluted solution.

Convert them back to the original sample basis:

\[ LOD_{B,\mathrm{original}}=0.00750\times10 \]

\[ \boxed{LOD_{B,\mathrm{original}}=0.0750\ \mathrm{mg\,L^{-1}}} \]

Similarly:

\[ LOQ_{B,\mathrm{original}}=0.0250\times10 \]

\[ \boxed{LOQ_{B,\mathrm{original}}=0.250\ \mathrm{mg\,L^{-1}}} \]

Step 4: Determine whether Method B detects the analyte.

Calculate the concentration in the diluted solution:

\[ C_{B,\mathrm{diluted}}=\frac{0.00080}{0.0800} \]

\[ C_{B,\mathrm{diluted}}=0.0100\ \mathrm{mg\,L^{-1}} \]

Convert to the original concentration:

\[ C_{B,\mathrm{original}}=0.0100\times10 \]

\[ \boxed{C_{B,\mathrm{original}}=0.100\ \mathrm{mg\,L^{-1}}} \]

Now compare the result with the original-sample limits:

\[ 0.0750<0.100<0.250 \]

The analyte is above the detection limit but below the quantification limit.

Method B therefore detects the analyte, but cannot quantify its concentration reliably.

Step 5: Explain why dilution improved detection.

Compare the original-sample limits:

Method B has a lower original-sample detection limit.

The reason is that dilution reduces the background variation more than it reduces the analyte signal.

The analyte signal falls by a factor of ten:

\[ 0.0080\rightarrow0.00080 \]

However, the blank standard deviation falls by a factor of twenty:

\[ 0.0040\rightarrow0.00020 \]

The reduction in background variation more than compensates for the loss of analyte signal.

As a result, the diluted method can distinguish the analyte from background more effectively.

There is one final distinction.

Although Method B detects the analyte, its estimated original concentration of \(0.100\ \mathrm{mg\,L^{-1}}\) is still below the LOQ of \(0.250\ \mathrm{mg\,L^{-1}}\).

It therefore cannot provide a reliable quantitative result.

This is why dilution does not automatically make an analytical method less capable of detecting a substance. The outcome depends on how dilution affects both the analyte signal and the background variation.

08What to remember when interpreting a result below the detection limit

When a chemistry question gives you an instrument reading and a detection limit, follow the measurement chain rather than immediately calculating a concentration.

1. Identify the appropriate blank and determine whether the instrument has already been zeroed.
2. Establish the mean blank signal and its standard deviation.
3. Calculate the detection threshold using the stated statistical criterion.
4. Use the calibration relationship to convert the signal threshold into a concentration limit.
5. Compare the sample response with the LOD and LOQ.
6. Apply the dilution factor to both limits if you need to report results for the original sample.
7. State whether the analyte was not detected, detected but not reliably quantified, or quantified within the validated range.

Most importantly, a result below the detection limit is not evidence that the chemical is completely absent.

It tells you something more specific: the analytical method has not produced a sufficiently distinguishable signal to establish its presence with the required confidence.

The next useful step is understanding how calibration uncertainty and sample representativeness affect the reliability of a reported concentration. A precise calibration does not automatically guarantee that the sample being measured represents the entire water source.