Recovery Check Calculations

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. • 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.

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.

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.

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