1. Recovery Check Calculations

2. The equation itself 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

3. Two basic rules 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

4. Example 1 Samples of river water are diluted 10:50 and analysed for chloride and found to contain an average of 2 mg/L of Na. Another 10 mL aliquot of the water is spiked with 1 mL of 50 mg/L Na, made up to 50 mL and then analysed. It was found to contain 2.9 mg/L. Sample2 mg/L Spike1 mL of 50 mg/L Spiked sample2.9 mg/L Procedure10:50, 10 mL, 50 mL Samples of river water are diluted 10:50 and analysed for chloride and found to contain an average of 2 mg/L of Na. Another 10 mL aliquot of the water is spiked with 1 mL of 50 mg/L Na, made up to 50 mL and then analysed. It was found to contain 2.9 mg/L.

5. Example 1 the 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

6. Example 1 simplest to work out what the mg/L of SP in the diluted solution is use the dilution equation C1V1 = C2V2. 50 x 1 = ? x 50 SP mg/L = 1

7. Example 2 Potato chips are analysed and found to contain 5.0%w/w of Na. A 1 g sample is spiked with 5 mL of 5000 mg/L Na and analysed. It is found to contain 70 mg of Na. S5.0 %w/w SP5 mL of 5000 mg/L SPS70 mg Procedure1 g

8. Example 2 Approach 1 – mass based Use mg as the common unit - SPS is already done For S, how many mg are in 1 g of 5%w/w? 5g/100g means 0.05 g/1 g or 50 mg. For SP, how many mg are in 5 mL of 5000 mg/L? 1 mL of 1000 mg/L is 1 mg, so SP is 25.

9. Example 2 Approach 2 - concentration based Use %w/w as the common unit – S is already done For SPS, what is the % of 70 mg in 1 g 70 x 100 ÷ 1000 = 7% For SP, which has 25 mg, what is the % 25 x 100 ÷ 1000 = 2.5% same answer neither approach is better

10. Exercise 1 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. Sample100 mg/L Spike500 uL of 1000 mg/L Spiked sample6.1 mg/L Procedure10 mL, 250 mL

11. Exercise 1 Approach 1 – diluted soln, mg/L SPS already known: 6.1 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

12. Exercise 1 Approach 2 – original soln, mg/L S already known: 100 SPS: 250 mL of 6.1 mg/L to 10 mL of 152.5 SP: 0.5 mL of 1000 mg/L to 10 mL of 50

13. Exercise 1 Approach 3 – mass in 250 mL soln, 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

14. Exercise 1 Approach 4 – mass in 10 mL soln, mg S: 10 mL of 100 mg/L is 1 mg SP: doesn’t really make sense SPS: doesn’t really make sense Is there a preferred approach? I prefer #1, but that’s just my opinion

15. Exercise 2 5 g of sample (known concentration 50 mg/kg) is spiked with 5 mL of 40 mg/L, and made to 100 mL. The concentration of this solution is 4.4 mg/L. Sample50 mg/kg Spike5 mL of 40 mg/L Spiked sample4.4 mg/L Procedure5 g, 100 mL

16. Exercise 2 Approach 1 – soln, mg/L SPS already known: 4.4 S: 5 g of 50 mg/kg is 0.25 mg, in 100 mL is 2.5 SP: 5 mL of 40 mg/L to 100 mL of 2

17. Exercise 2 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 0.2

18. Exercise 2 Approach 3 – solid sample, mg/kg S already known: 50 SPS: 0.44 mg in 5 g is 88 SP: 0.2 mg in 5 g is 40 Preferred approach? #1 has less work #2 requires all 3 to be calc’d #3 needs calcs done in #1