### Recovery Check Calculations

```Result validation
Exercise 1
• You’ve done an analysis to the best of your ability using the correct
• possibly, hopefully
• you can’t ever be sure that the answer is correct because the sample is an
unknown
• all you can do is provide evidence that the answer “should” be correct
• calibration
• blanks
• controls
• spikes
• retesting
Calibration
• standards provide a day-to-day measure of an instrument response
• allow you to determine concentration
• do not necessarily provide a measurement of the ongoing performance of
the instrument
• also means the running of special standards which do not relate to sample
analysis, but to instrument performance
• also known as Performance Qualification
Blanks
•
•
•
•
“samples” with no analyte
field blanks – to check on contamination from the collection procedure
transport blanks – kept in the transport vehicle at the site
container blanks – prepared at the laboratory from a container which is
identical to those used for field collection
• matrix blanks – materials of similar matrix to the samples
• reagent blanks – some analytical methods require a number of sample
treatment steps involving addition of different chemicals
Example 1
• 10 mL of conc. HCl added to river water sample (25 ug/L Pb)
• the solution made up to 100 mL with ultra-pure water
• acid contains 0.1 mg/L (0.00001 %w/v) of lead
• the additional lead from the acid is 10 mL x 0.1 mg/L ÷ 100 mL = 10 ug/L
Exercise 2
How can blanks be used?
• a zero (eg spectro std) (only if a low value)
• an early sample to check level
• if high – find a different source of the reagent
Controls
• sources of error other than contamination:
• matrix interference
• method error
• instrument malfunction
• operator error
• for certain sample types, it is possible to have standard samples, known as
controls or certified reference materials (CRM)
• samples of the same matrix, which have known levels of the analytes
• purchased or prepared in the laboratory.
• taken through the same procedure as the normal sample
• the result compared with the “true value”
• used to judge the accuracy of the real samples
Example 2
• standard sample of flour known to contain 450 mg/kg of sodium
• analysed together with “real” samples of flour
• found to contain 425 mg/kg
• standard sample is reading 5.6% low [(425-450)/450],
• reasonable to assume that their results are similarly low
• you would not simply subtract 5.6% from their results!!
Spikes (recovery checks)
• the addition of a known amount of analyte (known as a spike) to the
sample, which is then analysed as normal
• increase in concentration compared to how much in sample and how much
• if a procedure works perfectly, then 100% of the added analyte will be
recovered
• e.g. sample contains 5 mg of analyte, added spike contains 5 mg, so spiked
sample should contain 10
• if not, then the value for the sample is wrong
• 90-110% recovery generally considered OK
• the percent recovered is a measure of how accurate the answer for your
sample is
• if 50% of the added analyte is recovered, it suggest that 50% of the analyte
in the sample is being missed as well
• you don’t then adjust the sample answer accordingly, just report both
results.
• recovery check process is similar in performance to standard addition
• done for totally different reasons
• standard addition is done to determine the sample concentration
• an analytical procedure
• recovery check requires that you know the concentration in the sample
• a check on the analytical procedure
Retesting
• analysing a duplicate sample placed in the batch
• re-running a calibration standard after a certain number of samples (known
as a re-slope)
• checks for instrumental drift
• analysing the sample by an alternative method
• another person (or laboratory) analysing the sample
The equation itself
 SPS  S 
% RC  100 x 

SP


•
•
•
•
•
•
•
•
SPS is amount found in spiked sample
S is amount found in sample
SP is amount added as spike
found means in analysis, added means as standard
amount can refer to mass, volume or concentration.
the final calculation is easy
getting the three numbers to plug into the equation is not easy,
it is easy to put the wrong number in when there are so many numbers to
choose from
Two basic rules & something to remember
• All three values must be in exactly the same unit, eg mg, g/L, %w/w
• All three values must be from the same stage of the analysis, eg the original
sample, the first solution, the analysed solution
• there is no one way to do these, just one correct answer
• 1 mL of 1000 mg/L contains 1 mg of analyte
Example 4




20 mL sample of river water is diluted to 100 mL
found to contain an average of 2 mg/L of Na
another 20 mL aliquot spiked with 2 mL of 50 mg/L Na,
found to contain 2.9 mg/L
Sample
2 mg/L
Spike
2 mL of 50 mg/L
Spiked sample
2.9 mg/L
Procedure
20 mL to 100 mL
Example 4
• the best unit to choose is mg/L
• the only one in common
• this procedure has only two stages:
• the original river water
• the diluted solution which is analysed
• Which stage do these mg/L numbers belong to?
• S : 2 – diluted
• SP: 50 – neither (it is a standard in another bottle)
• SPS: 2.9 – diluted
Example 4
•
•
•
•
simplest to work out what the mg/L of SP in the diluted solution is
use the dilution equation C1V1 = C2V2.
50 x 2 = ? x 100
SP = 1 mg/L
 2 .9  2 
% RC  100 x 
  90 %
1


Example 4
•
•
•
•
less obvious approach – mass
S: 20 mL of 10 mg/L = 0.2 mg
SPS: 100 mL of 2.9 mg/L = 0.29
SP: 2 mL of 50 mg/L = 0.05
• % = 100
0.29 −0.2
0.1
= 90%
• why even consider this way?
• each of the three values has to be worked out
• the “mass in the sample portion” method works regardless of the type of
sample (solid or liquid) and whether there are any dilutions
Example 5
•
•
•
•
potato chips contain 5.0%w/w of Na
1 g sample is spiked with 5 mL of 5000 mg/L Na
dissolved in 500 mL
contains 140 mg/L of Na
S
5.0 %w/w
SP
5 mL of 5000 mg/L
SPS
140 mg/L
Procedure
1 g, 500 mL
Example 5
Approach 1 – concentration based
• Use mg/L as the common unit - SPS is already done
• For S, how many mg are in 1 g of 5%w/w? The conc. in 500 mL?
• 5g/100g means 0.05 g/1 g or 50 mg
• dissolved in 500 mL => 100 mg/L
• For SP, what is the final concentration?
• 5000 mg/L x 5 mL = ? x 500 mL
• ? = 50 mg/L
 140  100 
% RC  100 x 
  80 %
50


Example 5
Approach 2 – mass based
• Use mg – all 3 have to be worked out
• SPS: 500 mL of 140 mg/L = 70 mg
• S: done in Approach 1: 50 mg
• SP: 5 mL of 5000 mg/L = 25 mg
 70  50 
% RC  100 x 
  80 %
25


Exercise 3(a)
• 10 mL of sample (known concentration 100 mg/L) is spiked with 0.5 mL of
1000 mg/L and diluted to 250 mL. This solution, when analysed, has a
concentration of 6.1 mg/L.
Sample
100 mg/L
Spike
500 uL of 1000 mg/L
Spiked sample 6.1 mg/L
Procedure
10 mL, 250 mL
Exercise 3(a)
Approach 1 – diluted soln, mg/L
• S: 10 mL of 100 mg/L to 250 mL of 4
• SP: 0.5 mL of 1000 mg/L to 250 mL of 2
 6 .1  4 
% RC  100 x 
  105 %
2


Exercise 3(a)
Approach 2 – mass, mg
• S: 10 mL of 100 mg/L is 1 mg
• SP: 0.5 mL of 1000 mg/L is 0.5 mg
• SPS: 250 mL of 6.1 mg/L is 1.525 mg
 1 . 525  1 
% RC  100 x 
  105 %
0 .5


Exercise 3(b)
• 5 g of sample (known concentration 50 mg/kg) is spiked with 5 mL of 50
mg/L, and made to 100 mL. The concentration of this solution is 4.4 mg/L.
Sample
50 mg/kg
Spike
5 mL of 50 mg/L
Spiked sample 4.4 mg/L
Procedure
5 g, 100 mL
Exercise 8(b)
Approach 1 – soln, mg/L
• S: 5 g of 50 mg/kg is 0.25 mg, in 100 mL is 2.5
• SP: 5 mL of 50 mg/L to 100 mL of 2.5
 4 .4  2 .5 
% RC  100 x 
  76 %
2 .5


Exercise 8(b)
Approach 2 – mass, mg
• S only known from calcs in #1: 0.25
• SPS: 100 mL of 4.4 mg/L is 0.44
• SP: 5 mL of 40 mg/L is 0.25
 0 . 44  0 . 25 
% RC  100 x 
  76 %
0 . 25


Exercise 3(c)
• The concentration in the beer is found to be 52 mg/L. A 50 mL aliquot of the
beer is spiked with 2 mL of 1000 mg/L Na, made up to 100 mL and this
solution found to contain 47 mg/L Na.
Sample
50 mg/L
Spike
2 mL of 1000 mg/L
Spiked sample 40 mg/L
Procedure
50 mL to 100 mL
Exercise 8(c)
Approach 1 – analysed soln, mg/L
• S: 50 mL of 52 diluted to 100 mL of 26
• SP: 2 mL of 1000 mg/L to 100 mL of 20
 47  26
% RC  100 x 
 20

  105 %

Exercise 8(c)
Approach 2 – mass, mg
• S: 50 mL of 52 mg/L: 2.6
• SPS: 100 mL of 47 mg/L is 4.7
• SP: 2 mL of 1000 mg/L is 2
 4 .7  2 .6 
% RC  100 x 
  105 %
2


```