How to Compare Duplicate Sample Concentrations in HSC Chemistry

Two portions of the same water sample are analysed using atomic absorption spectroscopy (AAS). One produces an absorbance of 0.092, while the other produces 0.088. Have the two samples got different concentrations, or is the difference caused by the way they were prepared?

You cannot tell from the raw signals alone.

To compare duplicate samples properly, you must process each result independently:

1. Correct the measured signal for the method blank.
2. Use the calibration equation to calculate the concentration in the measured solution.
3. Multiply by that sample’s total dilution factor to recover the original concentration.
4. Check that both measurements are within the method’s quantifiable range.
5. Calculate the relative percentage difference (RPD) and compare it with the method’s acceptance criterion.

The important point is that you compare the original sample concentrations, not the raw instrument readings or the concentrations in the diluted flasks.

01What are you actually comparing?

A duplicate analysis involves taking two separate portions of the same well-mixed sample and processing them independently.

For example, imagine dividing a water sample into two portions:

• Duplicate A is transferred into a volumetric flask and diluted.
• Duplicate B is transferred into a separate volumetric flask and diluted.
• A method blank is prepared using the same reagents and preparation steps, but without the sample.
• Each prepared solution is introduced into the instrument, which produces a measurable signal.

For AAS, the instrument measures how much light is absorbed by atoms of the element being analysed. The absorbance is related to the element’s concentration.

The two duplicate preparations should give similar concentrations, but their signals will not necessarily be identical.

There is also an important distinction between two types of repeat measurement.

If you measure the same flask twice, you have not tested whether the sample preparation was consistent.

For this comparison, we are interested in independently prepared duplicates.

02Step 1: Correct each signal using the blank

An instrument may produce a signal even when no analyte is present.

This background response can come from reagents, solvents, the preparation procedure, or the instrument itself.

A method blank helps measure this background. It contains the same reagents as the sample preparation but does not contain the sample being analysed.

The blank correction is:

\[ S_{\text{net}}=S_{\text{sample}}-S_{\text{blank}} \]

Where:

• \(S_{\text{net}}\) is the corrected signal.
• \(S_{\text{sample}}\) is the raw sample signal.
• \(S_{\text{blank}}\) is the blank signal.

For example, suppose the method blank produces an absorbance of 0.012.

Two duplicate samples produce the following raw signals:

The corrected signals are:

\[ A_{\text{net, A}}=0.092-0.012=0.080 \]

\[ A_{\text{net, B}}=0.088-0.012=0.076 \]

Notice that the difference between the duplicates remains, but the background response has been removed.

An important warning: don’t subtract the blank twice.

There are two valid ways to handle blank correction:

• If the calibration equation uses blank-corrected signals, subtract the blank from each sample before applying that equation.
• If the calibration equation was constructed using raw signals, use the raw sample signal directly in that equation.

Mixing these approaches produces an incorrect concentration. We will examine this mistake in a later example.

03Step 2: Convert the corrected signal into concentration

A calibration curve relates instrument response to the concentration of solutions with known concentrations.

Once you have corrected the sample signal, you can use the calibration equation to determine the concentration of the diluted sample.

For a linear calibration:

\[ S_{\text{net}}=mc+b \]

Rearranging:

\[ c=\frac{S_{\text{net}}-b}{m} \]

Where:

• \(c\) is the concentration in the solution introduced into the instrument, in mg/L.
• \(m\) is the gradient of the calibration line.
• \(b\) is the intercept of the blank-corrected calibration line.
• \(S_{\text{net}}\) is the corrected instrument signal.

If the corrected calibration line passes through the origin, \(b=0\).

For example, suppose the calibration equation is:

\[ A_{\text{net}}=0.0200c \]

If a duplicate produces a corrected absorbance of 0.080, its measured-solution concentration is:

\[ c=\frac{0.080}{0.0200} \]

\[ c=4.00\text{ mg/L} \]

This is the concentration in the diluted solution, not necessarily the original sample.

Before moving on, check that the sample signal falls within the calibration’s validated working range.

A mathematical equation might produce a concentration outside that range, but this does not make the result reliable. You may need to dilute the sample further and measure it again.

04Step 3: Apply the dilution factor

This is where many duplicate comparisons go wrong.

The instrument measures the concentration in the prepared solution. However, you usually want the concentration in the original sample.

You therefore need to reverse the dilution.

For a single dilution:

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

Where:

• \(DF\) is the dilution factor.
• \(V_{\text{final}}\) is the final volume after dilution.
• \(V_{\text{aliquot}}\) is the volume of original sample transferred.

The original concentration is:

\[ C_{\text{original}}=c_{\text{measured}}\times DF \]

For example, suppose 10.00 mL of sample is transferred into a 50.00 mL volumetric flask and diluted to the mark.

\[ DF=\frac{50.00}{10.00}=5.00 \]

If the instrument measures a concentration of 4.00 mg/L:

\[ C_{\text{original}}=4.00\times5.00 \]

\[ \boxed{C_{\text{original}}=20.0\text{ mg/L}} \]

What if the sample was diluted more than once?

Multiply the individual dilution factors together.

\[ DF_{\text{total}}=DF_1\times DF_2 \]

For example:

• Transfer 5.00 mL into a 100.0 mL volumetric flask.
• Transfer 10.00 mL of that solution into a 50.00 mL volumetric flask.

The first dilution factor is:

\[ DF_1=\frac{100.0}{5.00}=20.0 \]

The second dilution factor is:

\[ DF_2=\frac{50.00}{10.00}=5.00 \]

Therefore:

\[ DF_{\text{total}}=20.0\times5.00=100 \]

If the final measured solution has a concentration of 4.00 mg/L, the original sample concentration is:

\[ C_{\text{original}}=4.00\times100 \]

\[ \boxed{C_{\text{original}}=400\text{ mg/L}} \]

Notice that the dilution factors are multiplied, not added.

For a more detailed explanation of this calculation, see How to Calculate Original Concentration After Dilution.

05Step 4: Compare the original concentrations

Once both duplicates have been corrected and their dilution factors applied, you can compare their original concentrations.

A useful measure is the relative percentage difference (RPD).

\[ RPD= \frac{|C_1-C_2|} {\frac{C_1+C_2}{2}} \times100\% \]

Where:

• \(C_1\) is the original concentration calculated from duplicate A.
• \(C_2\) is the original concentration calculated from duplicate B.
• The denominator is the mean of the two concentrations.

The absolute value ensures that the result is positive, regardless of which duplicate has the higher concentration.

A smaller RPD indicates better agreement between the duplicate results.

However, there is no universal RPD acceptance limit that applies to every analytical method. You must use the acceptance criterion specified by the experiment or method.

For example, if a method specifies an RPD limit of 10%:

• An RPD of 4% meets the criterion.
• An RPD of 8% also meets the criterion.
• An RPD of 14% does not meet the criterion.

If the duplicates meet the acceptance criterion, the method may allow you to report their mean concentration.

If they fail, investigate the difference rather than simply averaging the results.

Possible causes include:

• Inconsistent pipetting or volumetric preparation.
• Incomplete sample digestion or extraction.
• Poor mixing of the original sample.
• Contamination during preparation.
• Instrument drift or inconsistent instrument readings.

06Step 5: Check the detection and quantification limits

There is one more complication.

Even if you calculate two concentrations and obtain a small RPD, the comparison may not be valid.

You need to check the measurement limits.

Two important terms are:

Limit of detection (LOD): The lowest concentration that can be distinguished from the background signal using the method.

Limit of quantification (LOQ): The lowest concentration that can be measured quantitatively with acceptable precision and accuracy using the method.

The distinction matters.

The LOD and LOQ are method-specific. They should be supplied by the experiment or analytical procedure, rather than guessed.

Dilution changes the effective measurement limits

Suppose an instrument has an LOQ of 0.50 mg/L in the solution being measured.

If your sample has been diluted by a factor of 10, the corresponding LOQ for the original sample is:

\[ LOQ_{\text{original}}=0.50\times10 \]

\[ LOQ_{\text{original}}=5.0\text{ mg/L} \]

If another duplicate has been diluted by a factor of 50:

\[ LOQ_{\text{original}}=0.50\times50 \]

\[ LOQ_{\text{original}}=25.0\text{ mg/L} \]

The second preparation has a higher effective LOQ for the original sample.

This is why two duplicates can have different measurement limits even when they are analysed using the same instrument.

If one duplicate is below the LOQ, you should not treat its calculated concentration as an equally reliable numerical result when calculating RPD.

If a result is above the upper calibration limit, dilute it further and remeasure it.

If it is below the quantifiable range, consider using a smaller dilution factor, provided the sample concentration will remain within the instrument’s validated range.

Never assume that a mathematically calculated concentration is automatically a valid analytical result.

09A common mistake: Subtracting the blank twice

Consider a different calibration equation:

\[ A_{\text{raw}}=0.0180c+0.0120 \]

This equation was constructed using raw absorbance readings, so the intercept already accounts for the background response.

The method blank produces an absorbance of 0.0120, and the sample produces a raw absorbance of 0.0840.

The correct calculation is:

\[ c=\frac{0.0840-0.0120}{0.0180} \]

\[ \boxed{c=4.00\text{ mg/L}} \]

What happens if you subtract the blank first and then use the same raw calibration equation?

\[ c=\frac{(0.0840-0.0120)-0.0120}{0.0180} \]

\[ c=3.33\text{ mg/L} \]

This is incorrect.

You have removed the background response once from the sample reading and then effectively removed it again through the calibration intercept.

There is an equivalent way to do the calculation correctly.

If you subtract the blank from both the standards and the unknown sample, the corrected calibration equation becomes:

\[ A_{\text{net}}=0.0180c \]

Using the corrected sample signal:

\[ c=\frac{0.0720}{0.0180} \]

\[ c=4.00\text{ mg/L} \]

Both approaches give the same answer when applied consistently.

The rule is simple: match the sample’s signal treatment to the calibration equation’s signal treatment.

10A practical decision process

When two duplicate samples produce different concentrations, work through these checks in order.

One final distinction is worth remembering: a low RPD demonstrates good agreement, not necessarily accuracy.

For example, both duplicates could have been prepared using an incorrectly calculated dilution factor. They might agree almost perfectly while both reporting the wrong original concentration.

Duplicate analysis tests precision. Establishing accuracy requires additional evidence, such as suitable calibration standards, appropriate blanks, and validated analytical procedures.

11Questions and solutions

Question 1

A student analyses two duplicate portions of a water sample using AAS.

The method blank produces an absorbance of 0.0100. The blank-corrected calibration equation is:

\[ A_{\text{net}}=0.0200c \]

The raw absorbances are 0.0900 and 0.0860.

Both portions were diluted by transferring 10.00 mL into a 50.00 mL volumetric flask.

The method LOQ is 0.40 mg/L in the measured solution, and the acceptable RPD is 8.0%.

Calculate:

a) The original concentration of each duplicate.

b) The RPD.

c) Whether the duplicates meet the acceptance criterion.

Solution 1

The original concentrations are 20.0 mg/L and 19.0 mg/L. The RPD is 5.13%, so the duplicates meet the acceptance criterion.

First, subtract the blank absorbance.

For duplicate A:

\[ A_{\text{net,A}}=0.0900-0.0100=0.0800 \]

For duplicate B:

\[ A_{\text{net,B}}=0.0860-0.0100=0.0760 \]

Next, use the calibration equation to calculate the diluted-solution concentrations.

\[ c_A=\frac{0.0800}{0.0200}=4.00\text{ mg/L} \]

\[ c_B=\frac{0.0760}{0.0200}=3.80\text{ mg/L} \]

The dilution factor for both samples is:

\[ DF=\frac{50.00}{10.00}=5.00 \]

Therefore:

\[ C_A=4.00\times5.00=20.0\text{ mg/L} \]

\[ C_B=3.80\times5.00=19.0\text{ mg/L} \]

Calculate the RPD:

\[ RPD=\frac{|20.0-19.0|}{19.5}\times100 \]

\[ \boxed{RPD=5.13\%} \]

The original-sample LOQ is:

\[ LOQ=0.40\times5.00=2.00\text{ mg/L} \]

Both results are above the LOQ.

Since 5.13% is less than the specified 8.0% limit, the duplicates pass.

The mean concentration is:

\[ C_{\text{mean}}=\frac{20.0+19.0}{2} \]

\[ \boxed{C_{\text{mean}}=19.5\text{ mg/L}} \]

Question 2

Two portions of a well-mixed water sample are analysed independently.

The method blank produces an absorbance of 0.0080, and the blank-corrected calibration equation is:

\[ A_{\text{net}}=0.0200c \]

The calibration range is 0.50-8.00 mg/L, and the LOQ is 0.50 mg/L.

The following results are obtained:

A student claims that duplicate B must have a higher original concentration because it produces a higher absorbance.

Calculate the original concentrations and RPD. Assess the student’s claim using a method acceptance limit of 5.0%.

Solution 2

The original concentrations are 60.0 mg/L and 58.0 mg/L. The RPD is 3.39%, so the duplicates pass. The student’s claim is incorrect because the two portions have different dilution factors.

First, calculate the corrected absorbances.

\[ A_{\text{net,A}}=0.0680-0.0080=0.0600 \]

\[ A_{\text{net,B}}=0.1240-0.0080=0.1160 \]

Use the calibration equation:

\[ c_A=\frac{0.0600}{0.0200} =3.00\text{ mg/L} \]

\[ c_B=\frac{0.1160}{0.0200} =5.80\text{ mg/L} \]

Now calculate the dilution factors.

For duplicate A:

\[ DF_A=\frac{50.00}{5.00}\times \frac{20.00}{10.00} \]

\[ DF_A=10.0\times2.00=20.0 \]

For duplicate B:

\[ DF_B=\frac{50.00}{10.00}\times \frac{20.00}{10.00} \]

\[ DF_B=5.00\times2.00=10.0 \]

Calculate the original concentrations:

\[ C_A=3.00\times20.0=60.0\text{ mg/L} \]

\[ C_B=5.80\times10.0=58.0\text{ mg/L} \]

The RPD is:

\[ RPD=\frac{|60.0-58.0|}{(60.0+58.0)/2} \times100 \]

\[ \boxed{RPD=3.39\%} \]

Both measured-solution concentrations are above the LOQ and within the calibration range.

The RPD is below the 5.0% acceptance limit.

The student’s mistake was comparing the absorbances without considering dilution.

Duplicate B produces a higher absorbance because its final measured solution is less diluted. This does not mean that its original sample concentration is higher.

Always compare concentrations after applying the appropriate dilution factor.

Question 3

An instrument produces the following raw calibration equation:

\[ A_{\text{raw}}=0.0180c+0.0120 \]

The method blank produces an absorbance of 0.0120.

Two duplicate sample readings are 0.0840 and 0.0810. Both samples were diluted by a factor of 5.00.

Calculate the original concentration of each duplicate and their RPD.

Explain why subtracting the blank before applying the supplied calibration equation would produce an incorrect result.

Solution 3

The original concentrations are 20.0 mg/L and 19.2 mg/L, with an RPD of approximately 4.26%. The blank must not be subtracted separately because the supplied calibration equation already uses raw signals.

For duplicate A, apply the raw calibration equation directly:

\[ c_A=\frac{0.0840-0.0120}{0.0180} \]

\[ c_A=4.00\text{ mg/L} \]

For duplicate B:

\[ c_B=\frac{0.0810-0.0120}{0.0180} \]

\[ c_B=3.83\text{ mg/L} \]

Now apply the dilution factor.

\[ C_A=4.00\times5.00 \]

\[ \boxed{C_A=20.0\text{ mg/L}} \]

\[ C_B=3.833\ldots\times5.00 \]

\[ \boxed{C_B=19.2\text{ mg/L}} \]

Using the unrounded concentrations:

\[ RPD= \frac{|20.0-19.1667|} {(20.0+19.1667)/2} \times100 \]

\[ \boxed{RPD\approx4.26\%} \]

The important issue is how the calibration equation was constructed.

The equation:

\[ A_{\text{raw}}=0.0180c+0.0120 \]

already includes the background response through its intercept.

Subtracting 0.0120 from the sample readings and then using this same equation would remove the background twice.

Alternatively, the blank could be subtracted from both the calibration standards and the sample readings. The resulting corrected equation would be:

\[ A_{\text{net}}=0.0180c \]

Both approaches produce the same concentrations when used consistently.

The mistake is mixing a blank-corrected sample signal with an uncorrected calibration equation.

Question 4

Two duplicate samples are analysed using a method with the following characteristics:

• Method blank absorbance: 0.0100.
• Blank-corrected calibration equation: \(A_{\text{net}}=0.0200c\).
• Validated calibration range: 0.50-8.00 mg/L.
• LOQ: 0.50 mg/L.

Duplicate A is initially diluted by transferring 10.00 mL into a 100.0 mL volumetric flask. Its raw absorbance is 0.2500.

This signal is above the upper calibration limit, so 10.00 mL of the first diluted solution is transferred into a 20.00 mL volumetric flask and measured again. The new raw absorbance is 0.1300.

Duplicate B is diluted in two stages:

• 10.00 mL to 100.0 mL.
• 5.00 mL of that solution to 50.00 mL.

Its final raw absorbance is 0.0340.

Calculate the original concentration of each duplicate and determine whether they meet a 10.0% RPD acceptance limit.

Solution 4

Both duplicates have an original concentration of 120 mg/L. Their RPD is 0%, so they meet the acceptance criterion. However, duplicate A’s initial measurement must be rejected because it is outside the validated calibration range.

First, consider duplicate A’s initial measurement.

The corrected absorbance is:

\[ A_{\text{net}}=0.2500-0.0100=0.2400 \]

The calibration equation gives:

\[ c=\frac{0.2400}{0.0200} \]

\[ c=12.0\text{ mg/L} \]

This exceeds the upper calibration limit of 8.00 mg/L.

Although the equation produces a numerical result, it is outside the validated range. It should not be used as the final quantitative measurement.

Instead, use the second, further-diluted measurement.

For duplicate A:

\[ A_{\text{net,A}}=0.1300-0.0100=0.1200 \]

\[ c_A=\frac{0.1200}{0.0200} \]

\[ c_A=6.00\text{ mg/L} \]

This is within the calibration range.

The total dilution factor is:

\[ DF_A=\frac{100.0}{10.00}\times \frac{20.00}{10.00} \]

\[ DF_A=10.0\times2.00=20.0 \]

Therefore:

\[ C_A=6.00\times20.0 \]

\[ \boxed{C_A=120\text{ mg/L}} \]

Now calculate duplicate B.

Its corrected absorbance is:

\[ A_{\text{net,B}}=0.0340-0.0100=0.0240 \]

Its measured-solution concentration is:

\[ c_B=\frac{0.0240}{0.0200} \]

\[ c_B=1.20\text{ mg/L} \]

Its total dilution factor is:

\[ DF_B=\frac{100.0}{10.00}\times \frac{50.00}{5.00} \]

\[ DF_B=10.0\times10.0=100 \]

Therefore:

\[ C_B=1.20\times100 \]

\[ \boxed{C_B=120\text{ mg/L}} \]

Finally, calculate the RPD:

\[ RPD=\frac{|120-120|}{(120+120)/2}\times100 \]

\[ \boxed{RPD=0\%} \]

Both final measured concentrations are above the LOQ and within the validated calibration range.

The duplicates therefore meet the acceptance criterion.

The key lesson is that when a sample is diluted again, the additional dilution factor must be multiplied by the original dilution factor. You must use the complete dilution history, not just the final transfer.

Question 5

Two independently prepared portions of the same well-mixed sample are analysed.

The method has the following specifications:

• Method blank absorbance: 0.0100.
• Blank-corrected calibration equation: \(A_{\text{net}}=0.0200c\).
• Validated calibration range: 0.50-8.00 mg/L.
• LOD: 0.10 mg/L in the measured solution.
• LOQ: 0.50 mg/L in the measured solution.

The two duplicate preparations give:

A student applies the calibration equation to both samples, multiplies by their respective dilution factors, and obtains an original concentration of 20.0 mg/L for each.

The student concludes that the duplicates agree perfectly because the RPD is 0%.

Is this conclusion scientifically justified?

Determine what can legitimately be reported for each duplicate. Then calculate a suitable total dilution factor for a new preparation of duplicate A, assuming the original concentration is approximately 20 mg/L.

Solution 5

No. The apparent 0% RPD is not scientifically justified because duplicate A is below the quantification limit and outside the validated calibration range. Its calculated concentration cannot be treated as a reliable quantitative result.

Start by correcting both raw absorbances.

For duplicate A:

\[ A_{\text{net,A}}=0.0180-0.0100 \]

\[ A_{\text{net,A}}=0.0080 \]

For duplicate B:

\[ A_{\text{net,B}}=0.0500-0.0100 \]

\[ A_{\text{net,B}}=0.0400 \]

Calculate the concentrations using the calibration equation.

For duplicate A:

\[ c_A=\frac{0.0080}{0.0200} \]

\[ c_A=0.40\text{ mg/L} \]

For duplicate B:

\[ c_B=\frac{0.0400}{0.0200} \]

\[ c_B=2.00\text{ mg/L} \]

Now multiply by the respective dilution factors.

\[ C_A=0.40\times50=20.0\text{ mg/L} \]

\[ C_B=2.00\times10=20.0\text{ mg/L} \]

Mathematically, both results appear identical.

However, there is a problem.

The concentration measured for duplicate A is 0.40 mg/L, which is below the LOQ of 0.50 mg/L and below the lowest validated calibration concentration.

The calibration equation can produce a number at 0.40 mg/L, but the method has not been validated to quantify samples at that concentration.

Duplicate B, on the other hand, produces a measured concentration of 2.00 mg/L. This is within the validated range and above the LOQ.

Now consider the original-sample LOQs.

For duplicate A:

\[ LOQ_A=0.50\times50 \]

\[ LOQ_A=25.0\text{ mg/L} \]

For duplicate B:

\[ LOQ_B=0.50\times10 \]

\[ LOQ_B=5.0\text{ mg/L} \]

The estimated original concentration of 20.0 mg/L is below duplicate A’s original-sample LOQ of 25.0 mg/L.

It is above duplicate B’s original-sample LOQ of 5.0 mg/L.

Duplicate A is above its LOD, so the analyte is detected, but it cannot be reliably quantified.

A suitable report would therefore identify duplicate A as detected but below the original-sample LOQ of 25.0 mg/L, following the method’s reporting convention. Duplicate B can be reported quantitatively as 20.0 mg/L.

The calculated RPD of 0% is mathematically correct for the two provisional estimates, but it is not a valid measure of duplicate precision because one result is not reliably quantifiable.

How should duplicate A be reanalysed?

We want to reduce its dilution factor so that its measured concentration falls inside the validated range.

Assuming the original concentration is approximately 20 mg/L:

\[ c_{\text{measured}}= \frac{C_{\text{original}}}{DF} \]

We need the measured concentration to be between 0.50 and 8.00 mg/L:

\[ 0.50\leq\frac{20}{DF}\leq8.00 \]

Rearranging gives:

\[ 2.5\leq DF\leq40 \]

A suitable choice is a total dilution factor of 20.

The expected measured concentration would then be:

\[ c_{\text{measured}}=\frac{20}{20} \]

\[ \boxed{c_{\text{measured}}=1.00\text{ mg/L}} \]

This is above the LOQ and within the validated calibration range.

Duplicate A should be prepared again using an appropriate dilution factor, measured, and then compared with duplicate B using the original-sample concentrations.

The deeper lesson is that a small RPD does not automatically mean two results are reliable. Both values must first be valid quantitative measurements.

Once duplicate precision has been established, the next question is whether the analytical method is accurate. Two duplicates can agree closely and still share the same systematic error.

If you are comparing results from different analytical techniques rather than two duplicate preparations, see Why AAS and Gravimetry Give Different Water Concentrations. That comparison involves a different issue: whether two methods are measuring the same chemical quantity, not simply whether repeated measurements agree.