ACCUTEST CORPORATE BASIC TRAINING PROTOCOL MODULE 1
Area of Applicability:
Scope:
ACCUTEST ENVIRONMENTAL LABORATORIES
CORPORATE LEVEL TRAINING - STANDARDIZATION OF REQUIREMENTS FOR BASIC SKILLS TRAINING IN MATHEMATICAL SKILLS.
Written by:
David Sommers
Approved by:
_______________________________________
____________
Phil Worby, Corporate QA/QC Director
Date
_______________________________________
____________
Harry Behzadi, Vice President of Operations
Date
_______________________________________
____________
Reza Tand, Vice President of Operations
Date
Accutest
LABORATORY
ENVIRONMENTAL
TRAINING SERIES
DEFINING AND UNDERSTANDING
DATA QUALITY
BASIC TRAINING - MODULE 1
MATHEMATICAL SKILLS REQUISITE FOR
ANALYTICAL QUALITY
Table of Contents
1.0
Introduction ................................................................................................................................ 1
1.1
Purpose/Objective ........................................................................................................... 1
1.2
Calculations ..................................................................................................................... 1
1.3
Detection Limits ............................................................................................................... 1
1.4
Reportable Numbers ....................................................................................................... 2
1.5
Miscellaneous Terminology ............................................................................................. 2
1.5.1 Non-Exact Solutions ............................................................................................. 2
1.5.2 Percent Recovery (Analyte and Surrogate Spikes) ............................................... 2
N
Analyte Standard........................................................................................ 2
N
Internal Standard ....................................................................................... 2
N
Laboratory Control Standard ...................................................................... 2
N
Surrogate Standard .................................................................................... 3
<
LCS, ICV/CCV and Surrogate Standard .......................................... 3
<
Percent Recovery of Spiked Sample ............................................... 3
1.5.3 Relative Percent Difference (RPD) ....................................................................... 5
1.5.4 Molarity and Normality .......................................................................................... 7
N
Molarity ...................................................................................................... 7
N
Normality .................................................................................................... 7
1.5.5 Dimensional Analysis and Measurements ............................................................ 9
N
Weight Units .............................................................................................. 9
N
Volumetric Units ........................................................................................ 9
N
Measurement of Weight by Volume ......................................................... 10
N
Reportable Units ...................................................................................... 10
1.5.6 Units Conversions ............................................................................................... 11
1.5.7 Correcting for Moisture Content .......................................................................... 13
1.5.8 Dilution/Concentration Factors and Final Results ............................................... 14
1.5.9 Solution Preparation and Dilution ....................................................................... 16
1.5.10 Standardization ................................................................................................... 18
1.5.11 Significant Figures and Rounding ....................................................................... 19
a.
All nonzero integers are significant figures. .............................................. 19
b.
Always begin counting significant figures with the first nonzero integer ... 19
c.
Zeros may or may not be significant figures ............................................. 19
d.
Pure Numbers .......................................................................................... 20
e.
Limiting Figures ........................................................................................ 20
f.
Rounding Rules........................................................................................ 20
g.
Reportable Significant Figures ................................................................. 21
1.5.12 Calibration Calculations (Calibration Curve) ...................................................... 22
1.5.13 Reading Concentration From A Calibration Curve .............................................. 23
1.
Calibration Range and Sample Dilution ................................................... 27
2.
Response Factors .................................................................................... 27
1.5.14 Calculation of Compound or Element From Formula .......................................... 29
Table of Contents
List Of Figures
Figure 1 - Scatter Plot With r $ 0.995 .................................................................................................. 23
Figure 2 -Second Order Curve With r $ 0.995..................................................................................... 23
Figure 3 - Calibration Curve From NO3-N Table Data......................................................................... 24
List Of Tables and Data
Calculation Equivalences Table .......................................................................................................... 10
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Accutest CORPORATE TRAINING - BASIC TRAINING
MODULE 1 - MATHEMATICAL SKILLS
1.0
Introduction
1.1
Purpose/Objective
The purpose of this module is to familiarize the analyst with some basic mathematical tools that
are essential for transforming raw data collected during analyses into final or reportable results.
The analyst must be familiar with all the types of calculations that are used while reducing raw
analytical data to final data. These include such calculations as dimensional or units conversions,
equivalences, concentration units, dilution factors, and the determination of compounds or
elements from formula weights. Besides the basic calculations that will be covered in this section,
each instrumental or methodological training method will contain one or more sections dealing with
calculations appropriate to that instrument or technique. These calculations will be amplifications
of the basic calculations that all analysts working for Accutest Environmental Laboratories are
expected to be able to perform. All analysts must be familiar with these principles. Correct data
handling is essential to all work groups. The mathematical problems and setups covered in this
module are common to all work groups.
A set of mathematical problems will be presented to test the analyst's understanding and
application of the material presented in this syllabus. The analyst will be tested after he/she has
completed reading the syllabus, has been trained on the calculations, and understands the model
calculations. After completion of the test, the analyst will have his/her answers and problem setups
checked against the actual problems. It is as important to have demonstrated that the problem has
been correctly set up (in order to obtain a correct solution) as it is to simply get the correct answer
without showing the process of arriving at the answer. This is because having a correct set-up will
generally eliminate all potential errors except incorrect entries in a calculator or computer. Therefore, the problem set-up will count for at least half of the score on the test. After the test has been
completed and the answers discussed, the analyst will receive the answers to the test. The test
answer section has been designed for use by the analyst as a reference source for calculation
setups in addition to testing the analyst's skills in performing calculations.
1.2
Calculations
Since the raw data obtained by the analyst during the course of analysis are usually not in a final
form, the analyst must perform calculations based on the aliquot analyzed and the response of the
analytical system to determine the actual concentration of an analyte in the original sample. Final
results must be properly calculated, expressed in the proper units, and have the required number
of significant figures (normally two or three).
1.3
Detection Limits
Most tests have a limit on the minimum analyte concentration that can be detected under standard
conditions for the methodology employed. Thus an analyte may be present, yet undetectable by
the protocol which was used to perform the test. Unless specifically instructed otherwise, it is
not acceptable to report a zero as a result in cases where an analyte was not detected.
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Results are normally reported as less than a specified concentration (using the "<" sign preceding
the calculated number). The detection limit usually reported for each parameter (based on a
maximum sample aliquot) is maintained on a list available to each laboratory. This list is available
through the QA Department or the unit supervisor.
1.4
Reportable Numbers
Reportable numbers are the numbers that are reported to the client as the result of an analysis.
These numbers are produced by performing correct calculations using the raw data obtained
during the course of the test. Reportable numbers must be rounded to the correct number of
significant figures. They also are based on any transformation calculations that are performed.
Reportable numbers MUST be reduced to the correct or specified component (either compound(s)
or one or more elements) and MUST be expressed in the CORRECT concentration units.
1.5
Miscellaneous Terminology
1.5.1 Non-Exact Solutions
A non-exact solution is a solution whose concentration does not need to be known accurately, such as a 1:1 HCl to deionized water mixture. A non-exact solution is prepared by
dissolving solids weighed on top-loading balance in a premeasured volume of water or by
measuring liquids with graduated cylinders and mixing them together. All solutions are
expressed either in a weight/volume (e.g.: 10 grams/liter) or in a volume/volume (e.g., 30%
v/v; 1:1 mixture) ratio.
1.5.2 Percent Recovery (Analyte and Surrogate Spikes)
Measurement of percent recovery requires recording the volume and concentration of an
analyte standard, surrogate standard, laboratory control standard, or internal standard
(similar to surrogate) added to an aliquot of sample or blank. These terms are defined
below (and will be amplified in Basic Training Module 2 - Good Laboratory Practices
(GLP) and Analytical Quality):
●
Analyte Standard - A specific concentration of the chemical substance or physical
parameter that a test is designed to measure (the target parameter) that is added to
a “blank” matrix and not processed through any preparative steps. An analyte
standard is used to measure the instrument or method response to that
concentration of the reference standard without contributing effects from potential
interferents.
●
Internal Standard - An internal standard is either an organic or an inorganic element
or compound which is not a target analyte for the test being performed. It is added at
the same concentration to all standards, blanks, QC, and analytical samples. The
internal standard characterizes the effects of physical and/or chemical
characteristics of the sample on instrument response and may be used to
compensate for instrument bias resulting from these effects.
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●
Laboratory Control Standard - Addition of a measured amount of the chemical
substance or physical parameter which a test is designed to measure (the target
parameter) to deionized water or other solvent and taken through all of the sample
preparation steps to characterize the effects both of the instrument response to the
substance and the effects of any analytical processing such as sample preparation
by distillation, digestion, extraction or other preparative methods.
●
Surrogate Standard - A surrogate standard is a substance that behaves analytically
similar to one or more target analytes but which is not a target analyte and is
normally not a naturally occurring material. It is added prior to any sample
preparation or analysis and used to characterize the effects both of the instrument
response to the compound and of any analytical processing such as sample
preparation.
●
Recovery Determination - The volume and concentration of the standard allows the
analyst to determine the amount of the spiked component added to the sample
aliquot. By initially quantifying any analyte present in the sample and mathematically
subtracting this value, the amount of analyte in the spiked aliquot actually measured
by a test can be compared to the amount that theoretically should be present. In
many cases the analyte concentration in the sample is below detection limits, so that
the result may be treated mathematically as zero. Percent recovery is used as a
quality control measure in both organics and inorganics work to measure the
accuracy of a test. Percent recovery calculation formulas and examples follow:
◦
Percent Recovery of Laboratory Control Standard (LCS), Initial or Continuing
Calibration Verification Standard (ICV/CCV), or Surrogate Standard:
SR
 100
SA
◦
Percent Recovery of Spiked Sample:
(SSR  SR)
 100
SA
◦
Where:
SA
= Spike Added
SR
= Sample Result
SSR = Spiked Sample Result
With:
SA, SR, and SSR and having the same units. If SA, SR, or
SSR have different units, the analyst must either convert
all values to the same units or add conversion factors to
the above equations.
An analyst prepared a laboratory control standard (LCS) by spiking 0.5 mL of
a 10 mg/L As standard into 100 mL of acidified deionized water. This solution
was digested and then injected into a graphite furnace AA
spectrophotometer, and a reading of 0.037 mg/L was obtained. Assuming the
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control limits for the LCS to be 80-120% recovery, did the solution pass QC
acceptance?
Answer: Since the units are the same for the standard and the result, only a
dilution factor is involved in calculating the amount of spike added (SA).
Calculate SA as follows:
SA =
10 mg 0.5 mL

= 0.050 mg/L As
L
100 mL
The analyst can now determine the spike recovery since the results of the
analysis are also listed. To calculate the recovery:
Percent Recovery =
0.037 mg/L
 100% = 74.0%
0.050 mg/L
This analysis failed the QC acceptance criteria.
◦
The quality control limits for matrix spike recoveries for metals analysis by
ICP are normally 75-125% recovery. An analyst prepares an unspiked 100
mL sample aliquot and also prepares a 100 mL aliquot which is spiked with
1.0 mL of a solution containing the following target analytes at the listed
concentrations:
Analyte
Concentration (mg/L)
Copper (Cu)
25
Chromium (Cr)
20
Cobalt (Co)
50
Iron (Fe)
100
The analysis tests the digested and diluted aliquots of the sample and
obtains the following results:
Analyte
Original
Spike
Copper (Cu)
< 25 ug/L
260 ug/L
Chromium (Cr)
230 ug/L
463 ug/L
Cobalt (Co)
< 50 ug/L
399 ug/L
33,300 ug/L
33,200 ug/L
Iron (Fe)
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To calculate the spike added, the analyst must convert the units of mg/L to
ug/L and must also account for the sample dilution. (NOTE: The dilution
factors and units conversions are presented separately). The analyst
proceeds as follows:
SA Cu =
25 mg 1.0 mL 1000 ug


= 250 ug /L Cu
L
100 mL
mg
Similar calculations yield the SA amounts for Cr = 200 ug/L, Co = 500 ug/L,
and Fe = 1000 ug/L. Continuing with the Cu percent recovery calculation and
substituting 0 for the Cu sample result (SR) since it was below the detection
limit:
Recovery =
(260 ug /L)  (0 ug /L)
250 ug /L
100% = 104.0%
Performing the same calculations for the remaining elements gives the
following table of recoveries. Values that failed the QC criteria are flagged
with an asterisk (*).
Unspiked
Results
Spike Results
Percent
Recovery
Copper (Cu)
< 25 ug/L
260 ug/L
104.0
Chromium
(Cr)
230 ug/L
463 ug/L
116.5
Cobalt (Co
< 50 ug/L
399 ug/L
79.8
33,300 ug/L
33,500 ug/L
20.0*
Analyte
Iron (Fe)
Note that Fe (which had a very high concentration in the sample relative to
the spike added) failed to pass the QC criteria. This is a common occurrence
when the original (indigenous) sample concentration of an analyte is very
high relative to the spike added.
1.5.3 Relative Percent Difference (RPD)
Relative percent difference is used to measure the precision (repeatability) of a test.
The precision of a test can be calculated by analyzing 2 or more replicate aliquots of
a sample and determining the difference between the analytical runs. If the analyte
concentration in both the initial sample and any replicate(s) is below detection limits,
RPD cannot be calculated. If either the initial or the replicate sample is above the
detection limit while the other replicate is below the method detection limit, RPD will
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always be 200%; however, Accutest Environmental Laboratories do not normally
report an RPD value for this situation except when performing analyses under the
Contract Laboratory Program (CLP) protocols. Relative percent difference is
calculated by the following formula:
RPD =
SR1  SR 2
100%
½(SR1 + SR 2)
In the formula, the subscripts 1 and 2 for SR represent the first and second analyses
of the replicate aliquots of the sample. An example calculation is given below:
◦
In the analysis listed for the 4 metals under the percent recovery section, the
analyst also performed a method duplicate and obtained the following results:
Analyte
Copper (Cu)
Chromium
(Cr)
Cobalt (Co)
Iron (Fe)
Original
Results
< 25 ug/L
Duplicate
Results
< 25 ug/L
230 ug/L
285 ug/L
< 50 ug/L
33,300 ug/L
53 ug/L
33,100 ug/L
The acceptance criterion for method duplicates for ICP metals is 20% RPD.
Did any of the analytes fail for precision (RPD)?
Answer: Apply the calculation formula for determining RPD each of the data
pairs. Chromium was selected as an example for the RPD calculation. The
remaining data were calculated and the results entered in the table below.
RPD =
Analyte
Copper (Cu)
Chromium
(Cr)
Cobalt (Co)
Iron (Fe)
1
230 - 285
½(230 + 285)
 100% = 21.4%
Original
Results
< 25 ug/L
Duplicate
Results
< 25 ug/L
230 ug/L
285 ug/L
21.4%*
< 50 ug/L
33,300 ug/L
53 ug/L
33,100 ug/L
200.0%
0.602%
RPD
-----1
The RPD cannot be calculated since the calculation involves two non-detects which
are mathematically treated as zeros. Division by zero is mathematically undefined.
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In this analytical run, only chromium failed RPD QC limits. No value
can be calculated for Cu, and the RPD for the replicate pairs of Co
had one value above the detection limit and the other value below the
detection limit. This RPD is always 200% by definition and is not a
failure.
Note 1:
When one or both duplicate results are 10 times the detection limit,
the RPD control limit is normally 2 times the detection limit.
Note 2:
Accutest normally calculates RPD for the differences between matrix
spike (MS) and matrix spike duplicate (MSD) results. The rationale for
this is that RPD values measured when one or both results are close
to or less than the detection limit, the RPD becomes an arbitrary
number that may be artificially high (as noted above). For example,
assume that the metals group detected Cu in a sample at 33 ug/L and
at 36 ug/L for the duplicate result. The RPD for these values is
8.696%, which Accutest normally reports as 8.70%. However, if the
values were expressed in mg/L units with a DL of 0.03 mg/L, then
Accutest would report results of 0.3 mg/L and 0.4 mg/L, respectively.
These values yield an RPD of 28.57%, which Accutest would report as
28.6%. Obviously, since the RPD is dependent on both the DL and on
the number of significant figures, it is not a reliable QC criterion for
analyses with results 10 times the detection limit.
1.5.4 Molarity and Normality
●
Molarity - The molarity (M) of a solution expresses how many moles of a substance
are dissolved in one liter of a solvent. A mole is one gram molecular weight (also
known as the molecular weight) of a substance. The gram molecular weight is the
sum of all the atomic weights of the atoms in a molecular formula expressed in
grams. This type of calculation requires the analyst to be familiar with the periodic
table. For example, the gram molecular weight of sodium hydroxide (NaOH) is 40.0.
This is determined by adding the individual atomic mass units (amu) of the
components as follows:
Sodium (Na)
Oxygen (O)
Hydrogen (H)
=
=
Sodium Hydroxide (NaOH) =
22.9898 amu*
15.9994 amu*
=
1.0080 amu*
──────
39.9972 amu*
*It is customary to round the atomic weights to no more than 2 decimal places, so that Na
would be 22.99 amu, O would be 16.00 amu, and H would be 1.01 amu for a total of 40.0
amu NaOH. Round the amu values to one decimal place when 4 decimal precision is not
required (most cases).
The sum of these weights is 39.9972 amu (as shown above), which the analyst
normally rounds to 40.0 amu. Based on this, prepare a one molar (1M) solution of
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NaOH by dissolving 40.0 grams of NaOH in deionized water and diluting the solution
to exactly 1 liter.
●
Normality - Normality is another concentration term that is related to molarity. Normality is very useful in analytical chemistry since once it is determined, it can be
used to quantify an analyte based only on the concentration of the reagents used to
perform the test. Using molarity alone, an analyst would have to determine the
reaction products to calculate an analyte concentration or to calculate a reaction
such as a titration. For example, consider the reaction of NaOH and HCl which is as
follows:
NaOH + HCl → NaCl + H2O
In this reaction, exactly one molecule of NaOH will react with exactly one molecule
of HCl to produce exactly one molecule each of NaCl and H2O. This means that one
mole each of NaOH and HCl will produce one mole each of NaCl and H2O. This is a
simple calculation if the concentrations of NaOH and HCl are the same. However,
assume that the analyst reacts H2SO4 with NaOH to produce Na2SO4 and H2O as
follows:
(?)NaOH + (?)H2SO4 → (?)Na2SO4 + (?)H2O
Here the analyst must determine how many molecules of each reactant are required
to produce how many molecules of each product. This process is known as
balancing the chemical equation. In the case of the reaction above, the balanced
equation is:
2 NaOH + H2SO4 → Na2SO4 + 2 H2O
One way to look at this is that one molecule of H2SO4 is equivalent to two
molecules of NaOH in this reaction. Of course, chemical reactions and products are
not always as easy to balance as the above examples. This means that the analyst
needs a manner of determining how many what the molecular equivalents are
without needing to constantly balance equations. This is the basic idea behind the
concept of normality (denoted N).
In the case of the H2SO4 reaction with NaOH, there are 2 equivalents of H2SO4 for
each equivalent of NaOH. This means that 1 mole of NaOH will react completely
with 0.5 moles of H2SO4, and the analyst will require 2 moles of NaOH to produce 1
mole of each end product. Another way of stating this is that a 1M solution of NaOH
is 1N, and a 1M solution of H2SO4 is 2N. This means that in order to determine the
normality of a solution, the analyst needs to know the molarity of the solution and
the number of equivalents of the dissolved substance. Equivalents are defined as
the number of acidic hydrogen atoms transferred or accepted during a reaction (or
more technically, the number of electrons transferred or accepted during an
oxidation-reduction reaction). Normality is calculated by multiplying molarity by
the number of equivalents. This is why a 1M solution of NaOH (with the ability to
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react with one hydrogen atom per molecule) would also be 1N; while a 1M solution
of H2SO4 would be 2N (releases 2 hydrogen atoms per molecule).
The usefulness of working with normality is that it incorporates molarity but allows
for calculations based on equivalents rather than on formula reactions. In many
instances, the correct dilutions for producing standard solutions of a specific
normality are already listed in method references, method files, or in such sources
as Standard Methods. This means that the analyst will only rarely have to calculate
normality from formula weight or molarity. Most normality calculations involve
standardization of one solution against another solution with a known
normality or calculation of the concentration of an analyte based on the
reaction of a solution to a standardized titrant. Normality is frequently used in
titration calculations.
NOTE:
Molarity and normality can be calculated from the concentration
(weight/volume) of a solution. The analyst needs to know the formula weight in amu
of the component and the number of equivalents. Once this is ascertained, the
calculation becomes a units conversion problem. For example, a 1000 mg/L chloride
standard is also 0.0282N Cl . The calculation is:
1000 mg Cl1 gm Cl1M Cl1N Cl= 0.0282 N Cl


L
1000 mg Cl 35.45 gm Cl 1M Cl
-
Since Cl (with a valence of -1) can only transfer 1 electron, the number of
equivalents is 1. Thus 0.0282M Cl- = 0.0282N Cl . Another example is calculating
+2
the normality of a calcium standard where Ca exists as the Ca ion in solution. With
a valence of +2, calcium can accept 2 electrons. Assume an analyst has a 1000
mg/L solution of calcium standard. What is the molarity and normality of this
solution? First calculate the molarity of the solution:
1000 mg Ca +2
1 gm Ca +2
1M Ca +2
= 0.02495 M Ca + 2


+2
+2
L
1000 mg Ca
40.08 gm Ca
The normality of the solution is:
2 N Ca +2
 0.02495M Ca + 2 = 0.0499 N Ca + 2  0.05 N Ca + 2
+2
1M Ca
Using the normal rounding rules (discussed below in more detail), the analyst would
+2
+2
say the calcium solution is 0.025M Ca and 0.05N Ca .
1.5.5 Dimensional Analysis and Measurements
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The concept of dimension enters into almost all laboratory measurements. Only a
few laboratory measurements are dimensionless (such as pH and specific gravity).
Most analytes must be measured by weight or volume. The laboratory normally uses
standard metric system units for expressing the results of measurements.
●
Weight Units - Laboratory weights are expressed in kilograms (kg), grams (gm),
milligrams (mg), and micrograms (ug). On occasion, weight units such as
nanograms (ng) and picograms (pg) may be used, although weights of less than 1
ug normally cannot be measured using an analytical balance. These latter units are
usually measured by volume in solutions of known concentrations.
●
Volumetric Units - Laboratory volumes are usually expressed in liters (L), milliliters
(mL), and microliters (uL). On occasion the units of nanoliters (nL) and cubic
3
centimeters (cm ) may also be used. Use of the term "volumetric units" suggests
that, although these types of measurements are usually made on liquids, the
concept of volume also applies to solids and gases.
●
Measurement of Weight by Volume - By knowing the concentration of a dissolved
substance, an accurate measurement of weight can be performed using volumetric
measurement. To measure the weight of a dissolved substance, first apply one of
the concentration equivalences (also see Figure 1):
mg/L = ug/mL = ng/uL
ug/L = ng/mL = pg/uL
Grams per liter (gm/L) units should be converted into mg/L by multiplying by 1000.
Using these equivalences, the analyst can determine how many mg, ug, or ng of an
analyte are present in a standard and in calculating theoretical spike recoveries.
●
Reportable Units - In the laboratory most final results are reported in one of the
following units:
mg/L (milligrams per liter)
ug/L (micrograms per liter)
mg/kg (milligrams per kilogram)
ug/kg (micrograms per kilogram)
ppm, ppb (parts per million, parts per billion)
% by weight or weight/volume
% by volume
Also, there are some other less common units used for reporting final results (such
as mg/filter, total mg, etc.). Raw data analytical results may be in units different from
the units desired for final results. This means that a units conversion step may be
necessary in calculating final results from raw data. Unit conversions use equivalences (not to be confused with equivalents used for normalities) to convert one set
of units to another.
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Calculation Equivalences Table
Volume Conversions
Basic Unit Equivalence
Conversion Factors
1L
1 liter (L) = 1000 mL
or
1000 mL
1L
1 liter (L) = 1,000,000 uL
1,000,000  L
1 mL
1 milliliter (mL) = 1000 uL
1000  L
or
1000 mL
1L
1,000,000  L
1L
or
1000  L
1 mL
Weight Conversions
1 kg
1 kilogram (kg) = 1000 gm
or
1000 gm
1 gm
1 gram (gm) = 1,000,000 ug
1,000,000  g
1 gm
1 gram = 1000 milligrams (mg)
1000  g
Miscellaneous Conversions
1
10,000 ppm = 1% by weight or volume
1% =
or
1000 gm
1 kg
1,000,000  g
1 gm
or
ppm
1000  g
1 gm
or
10,000
10,000
1
F = 5/9
C or 1
C = 9/5
F
Weight/Volume Conversions
mg/L = ug/mL = ng/uL
ug/L = ng/mL = pg/uL
ppm =
ppm
9
5
 F =   C + 32 or  C =  ( F- 32)
5
9
Weight/Weight Conversions
mg/kg = ug/gm = ng/mg
ug/kg = ng/gm = pg/gm
mg/L
ppm = mg/kg
Specific Gravity
1
Percent by volume assumes a specific gravity of 1.000 0.001. If the specific gravity is not equal to 1.000, correct by dividing
the volumetric measure of mg/L by the specific gravity. Percent by weight does not require a specific gravity determination.
1.5.6 Units Conversions - To convert from one set of concentration units to another, one
or more equivalences may be needed. For example, if an analyst is requested to
analyze a solid sample for a specific analyte. The normal reportable units for this
analysis are either mg/kg or ug/kg (which will be based on either an extracted or a
digested aliquot of the solid). When the sample is digested, the digestate will be
diluted to a specific volume, while an extract may be either concentrated or diluted.
When an analyst analyzes an aliquot of the digested or extracted sample, he/she will
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either find the target analyte(s) present or undetected. However, the analyte
concentration units in the digestate or extract will be in weight/volume units such as
mg/L or ug/L. The normal reportable units for solids are either mg/kg or ug/kg.
These types of conversions are performed as follows:
●
Set up the problem based on the units that are known. In this case the analyst
should know the concentration of the target analyte(s) in the digestate in mg/L or in
ug/L, the weight of the solid sample that was prepared in grams, and the final
volume of the sample preparation in milliliters or in liters.
●
Determine the appropriate equivalences from the Calculation Equivalencies Table or
any other applicable units equivalences (see above). The analyst may not find a
direct conversion equivalence for the problem. The objective then becomes one of
finding appropriate equivalences that can be arranged so that all units cancel except
for the desired final units. Two examples of problems typical of units conversions
performed by analysts in the Accutest Environmental Laboratories are listed below
to demonstrate this technique:
a.
Assume that 30.0 grams of sample were extracted and the final extract was
concentrated to 1 mL. The analyst injected 1 uL of the extract and found 0.36
ng of Aldrin to be present. How much Aldrin was in the solid sample in ug/kg?
Answer: Since 0.36 ng of Aldrin were present in 1 uL of sample extract, the
Aldrin concentration in the extract was 0.36 ng/uL. The most direct
conversion is that ng/uL = ug/mL. Therefore, the Aldrin concentration in the
extract was 0.36 ug/mL. The 1 mL extract represents the entire amount
extracted from 30.0 grams of sample; thus there were 0.36 ug of Aldrin in the
30.0 gram aliquot of sample. To convert to ug/kg:
0.36 u g Aldrin 1000 gm

= 12.0 u g/kg Aldrin
30.0 gm sample
kg
b.
1.2343 grams of sample were digested and diluted to 100 mL final volume.
The digestate showed a concentration of 14.5 mg/L of copper (Cu). What
was the Cu concentration in the solid sample in mg/kg?
Answer: The Cu concentration of 14.5 mg/L may also be expressed as 14.5
ug/mL since mg/L = ug/mL. Also, the analyst should use the units conversion
factor mg/kg = ug/gm (see the Calculation Equivalencies Table). The problem
solution may be expressed either in a long or short form calculation as
illustrated below:
Long:
14.5 mg Cu
1L
100 mL 1000 gm
= 1170 mg/kg Cu



kg
1L
1000 mL 1.2343 gm
Short:
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14.5  g Cu 100 mL
= 1170 mg/kg Cu

mL
1.2343 gm
●
A similar type of problem is adding a specific weight (usually in ug or ng) of analyte
to a specific volume of solute for a total final volume. This can also be understood as
representing a total weight of the analyte contained in the specified volume of
solvent. For example, a procedure calls for adding 2.0 ug of formaldehyde to a test
tube and diluting the formaldehyde to 4.0 mL. The analyst has a working standard of
1.0 mg/L formaldehyde. How can this task be accomplished?
Answer: Since mg/L = ug/mL, add (2.0 mL) (1.0 ug/mL) = 2.0 ug formaldehyde. Since there are now 2 mL of solvent in the tube, add another 2.0 mL of
blank solvent to bring the total volume to 4.0 mL.
●
Another example of this type problem arises when determining the total mercury
(Hg) content of a sample. The AA spectrophotometer may be calibrated using either
peak height or peak area based on the total amount of mercury vapor given off by
the sample. This type of calibration is expressed in total ug of Hg, not in ug/L or in
mg/L of Hg. There are other procedures (such as the formaldehyde example above)
which also use this technique. Calculations to reduce the data to weight/volume ratio
must be performed. For example, an AA spectrophotometer is calibrated using total
Hg in ug. An aqueous sample aliquot of 50 mL, diluted to the standard test volume
of 100 mL, reads 0.05 ug of Hg, and a solid sample aliquot of 0.2035 gm reads 0.08
ug of Hg. What is the Hg content of each sample? Express the aqueous units in
mg/L and the solid units in mg/kg.
Answer: For both samples, the analyst must realize that only the sample
portion contains Hg and that the measurement represents the total amount
of Hg given off by the sample under the test conditions. This means that 0.05
ug of Hg from the water sample were contained in 50 mL of sample, while
0.08 ug of Hg from the solid sample were contained in the 0.2035 gm aliquot.
Recalling that mg/L = ug/mL and that ug/gm = mg/kg, the problems simplify
Soil =
(0.08 ug Hg)
= 0.39 mg/kg Hg
(0.2035 gm sample)
as follows:
1.5.7 Correcting for Moisture Content - Frequently, our clients want data for non-aqueous
samples to be expressed in dry weight ratio. In fact, some programs such as CLP
Water =
(0.05 ug Hg)
= 0.001 mg/L Hg (1 ug/L Hg)
(50 mL sample)
mandate that all units be reported in dry weight units. To perform the calculation to
convert wet weight units to dry weight units, the moisture content of the sample must
be determined and the analytical result corrected for the moisture content. Typically,
moisture content will be expressed in percent units (e.g., 13.4% H2O). To calculate
the dry weight ratio, the analyst must know the decimal equivalent of the solids
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content. The solids content may be expressed as percent solids or percent moisture.
Always perform the percent moisture or percent solids correction as the last
calculation step prior to rounding to correct significant figures. Proceed as
follows (analysts who understand the concept of units expressions can perform
some of the conversion steps listed by moving the decimal place in the percent
units):
●
If required, convert percent moisture to percent solids using the following formula:
(100%  H2O%)  Solids%
●
Next, convert percent solids to its decimal equivalent using the following formula:
Solids %
= Decimal Fraction Solids (DS)
100%

Finally, calculate the dry weight ratio:
mg 1

= Dry Weight Ratio
kg DS
a.
Example Calculation: An analyst finds a total dichlorophenol concentration of
1600 ug/kg in a solid sample. The percent moisture content of the soil is
20.0% What is the dichlorophenol concentration on a dry weight basis?
Answer: First, convert the percent moisture to percent solids and then
convert this to the decimal equivalent as follows:
Solids % = (100.0-20.0) = 80.0% => 0.800
Next, divide the mg/kg value by the decimal equivalent (Ds) of the percent
solids (same as multiplication by 1/Ds) as follows:
●
(1600 ug/kg Dichlorophenol)
= 2000 ug/kg Dichlorophenol (Dry Weig ht)
0.800
IMPORTANT:
Always determine if the moisture is reported as percent solids
or percent moisture. Either incorrectly converting moisture to
solids when the units are already solids or failing to convert
moisture to solids when the units are moisture are very
common errors.
1.5.8 Dilution/Concentration Factors and Final Results - Whenever a sample dilution or a
sample concentration is done during the analytical process, the raw results normally
must be multiplied by the inverse of the dilution or concentration that was made. For
example, if an analyst diluted 5 mL of a sample to 100 mL using a pipette and a
volumetric flask and then performed the analysis on the diluted sample, the final
result must be calculated back to the original, undiluted solution by multiplying the
raw data result by the inverse dilution factor. The original dilution was 5/100, so the
raw data result must be multiplied by 100/5 to calculate the final result. Further
examples are listed below:
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●
A cyanide test calls for an initial distillation aliquot of 250 mL and a color
development aliquot of 20 mL. Due to limited sample, an analyst only takes 100 mL
sample for distillation and dilutes it to 250 mL with deionized water. The sample
does contain cyanide, and to make the color development step work, the analyst
has to dilute 1.0 mL of the distillate in the color development step to 20 mL. The
concentration of cyanide is determined to be 0.150 mg/L in the final step. What is
the cyanide concentration in the sample?
Answer: Multiply the raw data result by the inverse of all dilutions and round to
correct significant figures. Be certain that all units cancel properly (leaving only the
desired final units) and that all inverse multiplications are set up correctly. The
calculation for the problems expressed above is:
0.150 mg CN

20 mL

250 mL
= 7.5 mg/L CN
L
1 mL 100 mL
In this calculation, the desired final units are mg/L, and for each dilution ratio the inverse was used to calculate the final answer. Note that the mL units all cancel,
leaving only mg/L units.
●
An extraction was performed for PCBs and pesticides in which the analyst used 745
mL of aqueous sample that was concentrated to a final volume of 5 mL. The
concentration of PCB Aroclor 1242 in the sample was very high, so the chromatographer diluted the extract by taking 1 uL and diluting it to 1 mL with hexane. The
analysis of Aroclor 1242 in the diluted sample extract showed a result of 0.53 ug/L.
What was the concentration in the original sample?
Answer: Both a concentration and a dilution were performed in this analysis.
Multiply by the inverse of all concentration and dilution steps:
NOTE 1:
NOTE 2:
0.53 u g PCB- 1242 1000 u L 5 mL
= 3.6 u g/L PCB- 1242


745 mL
1u L
L
Detection Limits and Sample Dilution - If there is limited sample and a full
aliquot cannot be taken, and if the analysis fails to detect an analyte, then
multiply the detection limit by all relevant dilution factor(s). For example, if the
analyst can only use 10 mL of sample for a sulfate test normally requiring a
100 mL standard aliquot, the aliquot must still be diluted to 100 mL. If no
sulfate is detected (with a normal detection limit of 1 mg/l), then the reportable detection limit is (100/10) X (<1 mg/L) = <10 mg/L, so the analyst would
record a result of <10 mg/L rather than the usual <1 mg/L.
Detection Limits and Sample Concentration - On occasion, the analyst
may use a larger than normal aliquot of sample and concentrate this aliquot
to allow for a lower than normal detection limit. In this case, multiply the
detection limit by the inverse of the concentration performed. For example,
assume the laboratory standard detection limit for thallium is 60 ug/L but the
client requires a detection limit of 10 ug/L or better. It may be acceptable in
this type of case to take a standard 100 mL aliquot of the sample and
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concentrate it to 10 mL. This gives a concentration factor of 100/10. If no
thallium were to be detected in the concentrate, the reportable detection limit
is determined by multiplying the inverse of the concentration factor times
the detection limit → (10/100) X (<60 ug/L) = <6 ug/L, so the analyst could
report < 6 ug/L thallium.
1.5.9 Solution Preparation and Dilution - There are two useful formulas for preparing
dilutions, determining the concentration of a diluted sample, and determining the
volume of a concentrated primary standard to use to make a secondary standard of
a different, more dilute concentration. For general solution concentration problems,
the formula V1C1= V2C2 (where V1= initial volume, V2= final volume, C1= initial
concentration, and C2= final concentration) can be used. For problems involving
normality, the formula V1N1= V2N2 (where V1 and V2 are the same and N1 and N2
are the initial and final normalities respectively) can be used. These formulas can be
rearranged algebraically to solve problems requiring the determination of working
solution concentrations and the amount of concentrate required for working standard
dilutions. Typical problems involving these relationships would be as follows:
●
How many milliliters of a nitrate primary standard with a concentration of 500 mg/L
should be used to prepare 250 mL of a 10.0 mg/L secondary nitrate standard?
Answer: Use the relationship V1C1= V2C2. The initial concentration of the nitrate
standard (C1) is 500 mg/L. The desired volume of secondary standard (final volume
or V2) is 250 mL and the desired concentration of the secondary standard (C2 is
10.0 mg/L). This means that V1 (required number of mL of primary nitrate standard)
is the unknown. The relationship can therefore be expressed:
V1 
500 mg
10 mg
= 250 mL 
L
L
Rearranging algebraically and solving for V1:
  10 mg  
 L 

   (250 mL)  10 mg  L
V1 = (250 mL)  
L
500 mg
  500 mg  


  L  
V1=
5.0 mL
Therefore, 5.0 mL of the 500 mg/L nitrate standard will be required to make 250 mL
of a 10.0 mg/L working standard. To achieve this dilution, dilute 5.0 mL to a final volume of 250 mL for the working standard.
●
A method calls for a 0.02 N solution of NaOH as a titrant to determine the acidity of
a sample. The analyst must prepare 1 liter of this solution. Describe how the solution
would be made if a stock solution of 1 N NaOH were available.
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Answer: Use the relationship V1N1 = V2N2 where the known values are N1= 1.0 N
NaOH, V2= 1 liter, and N2= 0.02 N NaOH. The setup is the same as above. The algebraic solution yields a value for N1 of 0.02 liters of 1.0 N NaOH diluted to 1 liter
are required to make a 0.02 N solution. Using the fact that 1 L = 1000 mL, the
analyst should add 20.0 mL of 1.0 N NaOH to a volumetric flask and then dilute it to
exactly 1 liter with deionized water.
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Problems - Volume/Concentration Ratios in QC Calculations:
●
An analyst spiked 1.0 mL of a 100 mg/L standard solution of cobalt (Co) into a 200
mL volume of sample. What is the theoretical concentration of this matrix spike
(spike added or SA)?
Answer: Use the relationship C1V1=C2V2. The initial concentration of the cobalt
standard (C1) is 100 mg/L and the initial volume of the cobalt standard (C2) is 1.0
mL. The final volume of the spiked sample (V2) is 200 mL. This means that C2
(concentration of Co in the spiked sample) is the unknown. The relationship can
therefore be expressed:
1.0 mL X 100 mg/L = C2 X 200 mL
Rearranging algebraically and solving for C2:
C2 =
C2=
(100 mg) (1.0 mL)

L
(200 mL)
0.50 mg/L Co as the theoretical concentration for the spike added (SA).
NOTE: One alternate solution could have been based on the equivalence mg/l =
ug/mL and therefore 1.0 mL of a 100 ug/mL Co standard contains a total of 100 ug
Co. Thus the analyst added 100 ug of Co to 200 mL or created a final weight/volume
ratio of 0.50 ug/mL. Another alternate solution would be to realize that there was the
equivalent of a 1:200 dilution made, and that 100 mg/L X (1/200) = 0.50 mg/L. Either
method of calculation is acceptable as long as the proper dilution factors or
weight/volume ratios are preserved. These are examples of alternate calculations
used as a check.
●
An analyst spiked 1.0 mL of a 50 mg/L solution of dibutylchlorendate (PCB surrogate
spike) into 1 liter of sample, and the extract of the solution was concentrated to 10.0
mL. The dibutylchlorendate (DBC) had a peak on the gas chromatograph that was
calculated to be 3.5 nanograms of DBC based on a 1 uL injection. What was the
surrogate spike recovery of the DBC, and assuming the QC DBC recovery limits to
be 24% - 154%, did the sample pass the surrogate recovery?
Answer: Use the relationship V1C1 = V2C2 to determine the theoretical spike
concentration. The first step is to determine the concentration, in appropriate units,
in the 10.0 mL of extract. The initial concentration of the dibutylchlorendate standard
(C1) is 50 mg/L. V1 is also known (1.0 mL). The final volume of the extract (V2) is
10.0 mL and the actual final concentration of the DBC (C2) was 3.5 ng in 1 uL.
(However, this is not relevant to calculation of the theoretical SA but instead is the
SSR). Recalling that mg/L = ug/mL = ng/uL and that 1 mL = 1000 uL, the
relationship can be expressed:
1000 uL X 50 ng/uL = 10,000 uL X C2
Rearranging algebraically and solving for C2:
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50 ng 1000 u L

= 5.0 ng/ u L
u L 10000 u L
C2= 5.0 ng of DBC since 1.0 uL was injected. Since C2 is the SA, 3.5 ng DBC was
the SSR and since no DBC was present in the sample, the percent recovery is:
C2 =
(SSR - SR)
(3.5 ng - 0 ng)
 100% 
 100% = 70.0% Recovery
SA
5.0 ng
The DBC recovery is within the QC limits of 24% - 154%.
NOTE: As with the previous example, alternate solutions to this problem exist. One
alternate solution would have been to realize that, even though the DBC was spiked
into 1 liter of sample, the actual extract was concentrated to 10 mL. Thus the 1 mL
of spike was diluted by a 1:10 ratio in the extract. Therefore, the SA would have
been 5 mg/L → 5 ng/uL (using the equivalencies above) with a recovery of 3.5
ng/uL. This can be the method of calculation for organics spike recoveries as long
as the analyst properly handles dilution/concentration factors. This example is
included to show that alternate calculations may be used to check the calculation
work.
1.5.10
●
Standardization - Standardization is the process of determining the concentration
or normality of a solution by comparing it against either a weighed solid primary
standard or against a solution whose concentration is known. Standardization differs
from calibration which involves analyzing a known standard, determining the response
of an analytical system to the standard, and setting the system to run based on the
standard analyzed. Standardization calculations involve the normality and concentration calculation formulas defined above. An example calculation is shown below:
-
An analyst performs triplicate titrations and finds that 50.0 mL of 0.0282 N Cl
standard is titrated each time by 14.1 mL of AgNO3. What is the normality of the
AgNO3?
Answer: Use the formula V1N1 = V2N2. In this case the volume (V1) of the AgNO3 is
known, and the volume (V2) and normality (N2) of the Cl standard are also known.
The relationship can be expressed:
14.1 mL X N1 = 0.0282 N2 X 50.0 mL
Rearranging algebraically and solving for N1:
N 1 = 0.0282 
50 mL
14.1 mL
N1 = 0.100 N AgNO3.
NOTE: When performing standardization as part of any analysis, the analyst must
always use a minimum of 3 replicate titrations and use the average (mean) value
of the titrations in the formula.
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1.5.11
Significant Figures and Rounding - For any measurement performed in the analytical
laboratory, both the precision and the accuracy of the measurement are limited by
the measuring instrument. The last digit of every measurement is always uncertain
to some degree. Significant figures are the figures in any measurement which are
known with acceptable accuracy and precision, with the last figure in the
measurement being uncertain. The rules for handling and reporting significant
figures are very important, and one of the most common calculation or reporting
errors made by analysts is failing to properly report significant figures. The rules that
follow are designed to familiarize the analyst with the concept of what constitutes
significant figures and how they are handled when performing calculations.
a.
All nonzero integers are significant figures.
b.
Always begin counting significant figures with the first nonzero integer. For
example, the numbers 0.002, 0.0019, and 0.00187 have one, two, and three
significant figures, respectively.
c.
Zeros may or may not be significant figures. Numbers can have three types
of zeros - leading, internal, and trailing. Zeros must be handled differently in
accordance with their position in the number. The basic rules for zeros are:
d.
◦
Leading zeros (0.0019, 005137) are never significant.
◦
Internal zeros (zeros between nonzero integers) are always significant
(506, 0.0101 both have three significant figures).
◦
Trailing zeros are significant only if they represent part of the
expressed accuracy of the measurement. Trailing zeros are never
significant if they are used for expressing the magnitude of a number.
Another way to state this is that trailing zeros are significant only if the
number requires a decimal point followed by one or more zeros to
express the precision and accuracy of the measurement. For
example, the number 50,000 has one significant figure since the zeros
just define the magnitude of the number, while the numbers 5000.0
and 0.50000 both have five significant figures since the trailing zeros
express the accuracy of the number. Another example is that the
numbers 5000, 5100, 5130, and 5127 have one, two, three, and four
significant figures respectively since trailing zeros preceding the
decimal point are always nonsignificant.
Pure Numbers - Some things can be counted exactly, such as 10 test tubes.
In the case of things that can be counted exactly, the resulting number is a
pure number and does not impose any limitations on the number of significant figures in a calculation. A defined number, such as a conversion
number, is another example of a pure number. Thus, one liter is defined as
being equivalent to exactly 1000 milliliters. Although the rules for trailing
zeros above state that 1000 mL would have only 1 significant figure, the
defined value implies a degree of precision and accuracy greater than any
actual measurement performed. If a conversion factor uses a defined
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number or a pure number as a part of a calculation, it has NO effect on
the number of significant figures that can be justified by the calculation.
e.
Limiting Figures - When a calculation is performed, the number with the
fewest significant figures always controls the number of significant figures
that may be reported in the final answer. For example, in the following
calculation:
1345.8 X 50.04 X 2.1 = 141,422.0472 → 140,000
The controlling number of significant figures is 2 based on the value 2.1 that
only has two significant digits. This means that the number must be rounded
to two significant figures in its final form for reporting (140,000). Another
example may be observed in the following calculation:
1569.26 30.1 10.01


= 1.34971192  1.3
5.5 30.33
2100
Again, the controlling number of significant figures is two based on the fact
that 5.5 and 2100 only have two significant digits each. This means that the
number must be rounded to two significant figures in its final form for
reporting (1.3). Note that a common error by analysts might occur with this
number due to incorrect rounding practices. A commonly encountered error
is to perform the rounding operation based on rounding the third significant
figure before the final rounding. The number rounded to 3 significant figures
is 1.35, so analysts who commit this type of error will erroneously round the
number to 1.4.
f.
Rounding Rules - In addition to the rule for limiting figures, round off in accordance with the following rules:
◦
Always round all calculation results to the correct number of significant
figures as the final calculation step. NEVER round any results
before reaching the final answer, even in a lengthy calculation.
◦
To round a reportable number, follow the steps below:
1.
Determine the number of significant figures justified by the
calculation to determine the maximum number of significant
figures that may be reported.
2.
Determine if there is any digit to the right of the final significant
figure. If so, the digit will fall into one of five possible cases. It
will be greater than, equal to, or less than 5 and either will or
will not have numbers to its right. Proceed as follows:
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g.
◦
If the number to the right of the last significant figure is
greater than 5, round to the next highest number and
drop any trailing integers. For example, 1.44791→ 1.45
◦
If the number to the right of the last significant figure is
less than 5, do not round to the next highest number.
Drop that number (and any other trailing digits). For
example, 1.44491 → 1.44
◦
If the number to the right of the last significant figure is
equal to 5, then round to the next highest number when
the preceding figure is odd, drop the 5 if the preceding
figure is even, and round upward if there are nonzero
digits to the right of the 5 regardless of whether the
preceding number is odd or even. Three possible
examples include the following:
1.435 → 1.44
1.445 → 1.44
1.4451 → 1.45
Reportable Significant Figures - Regardless of the number of significant
figures that can be justified when using the rules for significant figures, Accutest rounds inorganics results to 3 significant figures and all organics
results to 2 significant figures. If the analytical results are <10 times the
detection limit, round to one significant figure. If the analytical results are 10
times but <100 times the detection limit, report two significant figures. If the results are 100 times the detection limit, report up to three significant
figures. Be sure to follow all project-specific requirements for reporting
significant figures. Compare the result to the detection limit prior to rounding
and report any value below the detection limit as undetected ("ND").
Detection limits should be rounded to 1 significant figure. Exceptions to the
rounding rules occur when the analyte concentration is close to the detection
limit. The result cannot have more numbers to the right of the last significant
figure of the detection limit. For instance, if the detection limit is 0.005 mg/L,
proceed as follows to express the results:
Analytical Result
(mg/L)
Reported Result (mg/L)
0.0056
ND
0.0061
0.006
0.015
0.02 (1 significant figure)
0.123
0.123 (3 significant figures for inorganics)
0.12 (2 significant figures for organics)
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As an additional example, if the detection limit = 50 mg/L (the zero is significant), proceed as follows:
Analytical Result
( 49.9
/L)
Reported Result (mg/L)
51.2
51
103.2
1.5.12
1.5.13
ND
103 (3 significant figures for inorganics)
100 (2 significant figures for organics)
Calibration Calculations (Calibration Curve) - Analytical laboratories use calibration curves to establish and quantify a functional relationship between the
concentration of an analyte and the response of an instrument. The analyst
may measure the response as an absorbance reading on a spectrophotometer, as peak height or peak area (spectrophotometer, GC, GC/MS), or as a
millivolt response (ion analyzers, IC). The standard format for constructing a
manual curve is to use standard graph paper to plot the instrument reading
as the Y-axis and the analyte concentration in appropriate concentration units
as the X-axis. The analyst then measures the instrument response at several
increasing levels of analyte concentration (from a blank or "zero" concentration to the highest concentration either within the linear response range for
the instrument or within the defined method acceptance limits). Many instruments in the laboratory allow for direct readout in concentration units based
on the calibration curve established internally by a microprocessor. However,
the analyst still must ensure that the instrument response is correct and that
the calibration was performed correctly. For this reason, all analysts must
understand the basic concepts of calibration. New (initial) calibration curves
must be prepared in the following circumstances:
◦
Whenever a new test or procedure is being performed.
◦
Whenever a test does not follow the linearity laws (such as nitrate) or
has a curvilinear relationship rather than a linear relationship to
concentration.
◦
Whenever the recovery of a continuing calibration verification
standard is outside the QC range established for that technique.
◦
Whenever prescribed by the method or whenever changing reagent
lots.
◦
Quarterly (at a minimum) regardless of whether the continuing
calibration verification standard is within the QC range established for
that technique.
Reading Concentration From A Calibration Curve - In the simplest method,
an analyst draws and labels the X and Y axes on graph paper, marking units
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appropriate to each axis. The analyst then plots the (x,y) pairs on the graph
and manually draws a "best fit" line through the points from the origin to the
highest concentration/response pair. The analyst can then determine the yvalue (instrument response) for any unknown concentration(s) of an analyte
in samples, determine where the y value intersects the calibration line (x-axis
or calibration amount), and from that point determine the x value (analyte
concentration) which is equivalent to the response. However, this manual
method is cumbersome and is both imprecise and inaccurate compared to
using a mathematical calculation to determine the concentration. For a
calculation, the line should be established as linear (of the form y = ax + b)
either by having a correlation coefficient (r) 0.995 or by meeting the relative
response factor criteria specific to the analytical method. Then the analyst
can use a response factor to calculate concentration.
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Effective Date: 01/31/2012
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Note 1: The analyst should plot the curve for the calibration where possible
as a correlation of 0.995 r is not sufficient in itself to establish that a line is of
the form y = ax + b. Both curves exhibiting considerable scatter and curves
2
that are second or higher order functions (e.g., y = ax + bx + c) can yield a
correlation coefficient of 0.995 or better. Figure 1 illustrates a curve with
considerable scatter that still exhibits an r of 0.995. Figure 2 illustrates a
curve fitted to a set of points that should be expressed as a curve of the form
2
y = ax + bx + c that exhibits an r of 0.995 when expressed as a line of the
form y = ax + b.
Note 2: If a calibration curve is nonlinear or cannot be represented by a
single response factor, then the analyst will have to use either an actual
calibration curve or a calculated value using the response of the nearest
calibration standard. The calculated value, for any instrument, is an equation
of the form:
CX =
Where:
CX
IX
IS
CS
DF
CF
=
=
=
=
=
=
IX   
CS DF CF
IS
Concentration of the analyte in the sample.
Instrument reading for the sample aliquot.
Instrument reading of the nearest calibration standard.
Concentration of the calibration standard used for calculation.
Dilution factor correction (if applicable).
Conversion factor to change units (if applicable).
Two example calculations are shown below:
◦
An analyst analyzes a series of standards for nitrate (NO3) and obtains the following
instrument readout in absorbance units:
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Effective Date: 01/31/2012
Page Number: 26 of 35
Nitrate
Concentration
Absorbance
0.00 mg/L
0.000
0.10 mg/L
0.021
0.50 mg/L
0.105
1.00 mg/L
0.210
1.50 mg/L
0.245
2.00 mg/L
0.275
3.00 mg/L
0.277
The resulting curve is nonlinear (illustrated in Figure 3). The analyst reads a sample
that has been diluted 1/20 and records an absorbance of 0.095. What is the
concentration of nitrate in the sample?
Answer: The nearest standard reading is a nitrate standard of 0.50 mg/L with an
absorbance of 0.105. The sample absorbance reading was 0.097 and the sample
was diluted 1/20. Using the equation above and substituting variables:
CX =
0.097 0.50 mg 20


0.105
L
1
CX
=
9.238 mg/L NO3-N
CX
=
9.2 mg/L NO3-N
Document No: Accutest BT1 Rev 1.0
Effective Date: 01/31/2012
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◦
An analyst analyzes a series of standards for pentachlorophenol (PCP) and obtains
the following instrument readout in peak area on the GC:
PCP Concentrations
Peak Areas
0 ug/L
0
5.0 ug/L
1256
50.0 ug/L
13995
80.0 ug/L
30128
120.0 ug/L
33256
The resulting calibration curve is nonlinear. The analyst injects an aliquot of sample
extract that has been diluted 1/100 and records a peak area of 15268 (which is
identified as PCP by retention time). The sample extract was taken from 30.5 grams
of soil and was concentrated to 5.0 mL. The percent moisture is 15.3%. What is the
concentration of pentachlorophenol in the dry weight sample in ug/kg?
Answer: The nearest PCP standard reading is a standard of 50 ug/L (or 0.050 mg/L
by equivalencies) with a peak area of 13995. The sample area reading was 15268
and the sample was diluted 1/100. Using the equation above and both substituting
variables and using appropriate equivalences:
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Effective Date: 01/31/2012
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CX =
15268 50  g 100 0.005 L
1




13995
L
1 30.5 gm (1 - 0.153)
Cx = 1.05576 ug/gm → 1,100 ug/kg PCP (dry weight).
1.
Calibration Range and Sample Dilution - Whenever a sample is diluted for analysis,
the instrument reading for the dilution must fall in the mid-range of the standard
curve for the analyte(s) of interest. NEVER DILUTE A SAMPLE AND ATTEMPT TO READ IT
AT EITHER THE LOW OR THE HIGH ENDS OF THE CURVE. Using the low end of the
calibration curve to quantitate a sample is only acceptable if a full aliquot of sample
was used (no dilutions). Sometimes a full aliquot has so much analyte that the
concentration greatly exceeds the calibration range. These samples must be rechecked by dilution and reanalysis. On other occasions, one or more components of
the sample may severely interfere with the determination of the target analyte(s). In
the latter case, there may be no option but to dilute the sample until the effects of
the interfering component(s) can be controlled. On other occasions, the client may
not provide sufficient sample for a full aliquot. If a dilution or smaller than normal
aliquot and low end reading are necessary due to limited sample, report a higher
detection limit. THERE IS NEVER A JUSTIFICATION FOR QUANTITATING SAMPLES AT THE
UPPER END OF A CURVE. In particular, "extending" a curve beyond the highest standard analyzed can NEVER be justified. If a sample is analyzed and the instrument
reading is at or above the upper end of the curve, dilute and reanalyze the sample.
2.
Response Factors - Linear calibration curves may be defined in terms of the
equation:
y = ax + b
y=
Instrument response in appropriate units (either direct response such as
absorbance, peak area, etc. or calculated response such as concentration
units).
a=
Line slope.
x=
Concentration of the analyte being measured.
b=
Y-intercept - normally this should be zero since the instrument should not
have any response to a "zero" concentration of the analyte.
Since b= 0, x = the concentration of the analyte, y = the instrument response, and a
is the linear slope (or instrument response factor), the linear equation simplifies to:
y = ax  x =
y
Instrument Reading
 Concentration =
a
RF
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Where R̄F̄ is the calculated average response factor and the Instrument Reading is
the instrument readout in the units for which the instrument was set. This means that
the slope "a" for the line (the average slope of all the calibration points) becomes the
mean response factor (R̄F̄). The R̄F̄, once determined for a specific calibration
curve, may be divided by any instrument reading to yield the associated analyte
concentration as long as the value for the instrument reading is equal to or less than
the reading of the highest standard used for the curve. An example problem is
presented below:
◦
An analyst calibrates an AA for sodium in absorbance units and finds these responses:
Na
Concentration
Absorbance
RF
0.00 mg/L
0.10 mg/L
0.50 mg/L
1.00 mg/L
2.00 mg/L
3.00 mg/L
0.000
0.021
0.103
0.209
0.415
0.626
N/A
0.210
0.206
0.209
0.208
0.209
Average RF (R̄F̄)
0.208
The analyst records the following data in the laboratory notebook for the sample
analyses:
Sample Number
Dilution
Absorbance
99070100
99070101
99070102
99070103
99070115
99070117
10/10
5/10
10/10
1/10
1/10,000
10/10
0.235
0.438
0.112
0.566
0.497
0.059
Calculate the final results in mg/L that should be reported for these samples:
Sample Number
95070100
95070101
95070102
95070103
95070115
Dilution
10/10
5/10
10/10
1/10
1/10,000
Absor0.235
0.438
0.112
0.566
0.497
Na (mg/L) = (R̄F̄)/A X 1/DF
0.11 mg/L
0.42 mg/L
0.54 mg/L
2.7 mg/L
2,400 mg/L
95070117
10/10
0.059
0.03 mg/L
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Effective Date: 01/31/2012
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Where:
A
R̄F̄
DF
= Absorbance
= Mean Response Factor
= Dilution (Factor)
NOTE: Many instruments can be calibrated to read out in concentration units after
calibration. Where this is the situation, the analyst can use the direct reading
of concentration to calculate the final result since the instrument
microprocessor is performing a calculation similar to the one above. Many
other instruments (such as inorganic spectrophotometers and GCs) will
require calculations similar to the one above to reduce the raw data to
reportable numbers.
1.5.14
Calculation of Compound or Element From Formula - A calculation that is frequently
required in the inorganic groups is to change raw data results from a formula weight
concentration to a concentration as a specific element, such as reporting phosphate
(PO4) as phosphorus (P). Conversely, sometimes the analyst is required to calculate
the amount of a specified compound based on the concentration of an element,
such as reporting silicon (Si) as silica (SiO2). This is part of the calculation process
and it is subject to the same rounding rules as any other calculation. Always
complete this calculation before rounding the final answer. Complete ALL other
calculation steps before performing this calculation. Proceed as follows:
1.
The first step requires using the periodic table to determine the atomic
weights of all the components. As an example, determine the nitrogen
content of nitrate (NO3). The atomic weight of nitrogen is 14 amu, and the
atomic weight of oxygen is 16 amu.
2.
Determine exactly what computation is required. In this case, the nitrate ion
has one nitrogen atom and three oxygen atoms from the formula.
3.
Add the sum of the atomic masses of all of the components. In this case:
+
4.
5.
(1 N) X (14 amu/N) =
14 amu
(3 O) X (16 amu/O) = 48 amu
Total = 62 amu per NO3 molecule
Set up a conversion factor based on this information. There are 14 amu of
nitrogen per every 62 amu of nitrate. Therefore, the nitrogen fraction is 14/62.
This is the conversion factor for multiplying raw data results to convert NO3 to
N.
For cases where an element or a compound must be calculated as another
compound, the same general procedure must be followed. For example, to
convert carbon (C) to carbon dioxide (CO2), carbon has 12 amu, while
oxygen has 16. The amus of carbon dioxide are based on a single carbon
atom (12 amu) and 2 oxygen atoms (16 amu each). Thus carbon dioxide has
a total of 44 amu. There are 44 amu of carbon dioxide for every 12 amu of
carbon. Therefore, the conversion factor to calculate CO2 from C is 44/12.
Document No: Accutest BT1 Rev 1.0
Effective Date: 01/31/2012
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Example 1:
The analyst performed a test that converted the total sulfur content of a
sample to sulfate (which was then captured in an aqueous solution). The
solution had a total sulfate content of 3000 mg/L. What was the total sulfur
content (expressed as S)?
Answer:
First ascertain the atomic masses of the components from the periodic table.
Sulfur has 32.06 amu, and oxygen has 15.999 (normally rounded to 16) amu.
The formula of the sulfate ion is SO4, so the sulfate ion has 96.06 amu (32.06
for the S, 16 each for the O times 4). The conversion factor is S/SO4. The
problem should be set up as follows:
32.06 amu S
3000 mg SO4

= 1001.25 mg/L S  1000 mg/L S
L
96.06 amu SO4
Example 2:
The analyst performed a test that measured the total ammonia content of a
sample as nitrogen (N). The sample had a total ammonia-N content of 25.6
mg/L. What was the total ammonia content (expressed as NH3)?
Answer:
First ascertain the atomic masses of the components from the periodic table.
Nitrogen has 14.00067 amu (normally rounded to 14.00), and hydrogen has
1.008 (normally rounded to 1.01) amu. The formula of ammonia is NH3, so
the ammonia molecule has 17.03 amu (14.0 for the S, 1.001 each for the H
times 3). The conversion factor is NH3/N. The problem should be set up as
follows:
25.6 mg N 17.03 amu NH3
= 31.1406 mg/L NH3  31 mg/L NH3

L
14.00 amu N