SME
Analytical Chemistry · Problem type
Spike recovery
Use this for “A river sample reads 3.6 ppb lead. A spike that adds 5 ppb is put into a second portion, which then reads 8.3 ppb. What is the recovery?”
Chart
sample: 12.4 ppb
found: 9.5 ppb
This is an example. Type your own values to start a new problem.
R (%)
=
Cₛ (ppb)
−
Cᵤ (ppb)
Cₐ (ppb)
×
100
ΔC
Concentration found from the spike
calculated · ppb
Formulas
ΔC = Cₛ − Cᵤ
9.5 = 21.9 − 12.4
R
=
100
×
ΔC
Cₐ
95
=
100
×
9.5
10
Assumptions
•
The spike goes into a second portion of the same sample, and the volume it adds is too small to matter.
•
Cₐ is the concentration the spike adds to the sample, not the strength of the spiking solution.
•
Recovery near 100% shows the matrix neither hides nor adds signal; many methods accept 80% to 120%.
Step-by-step
WE KNOW
Cᵤ = 12.4 ppb, Cₛ = 21.9 ppb, Cₐ = 10 ppb
FIND
concentration found from the spike (ΔC), recovery (R)
Step 1 · Find concentration found from the spike (ΔC)
Use ΔC = Cₛ − Cᵤ
Take what the sample held on its own from the spiked reading.
ΔC = 21.9 − 12.4
ΔC = 9.5 ppb
Step 2 · Find recovery (R)
Use R = 100 × ΔC ÷ Cₐ
The spike found as a percent of the spike added.
R
=
100
×
9.5
10
R
=
950
10
R
=
95%
→
Between
80%
and
120%:
the
method
recovers
the
analyte
well
in
this
matrix.
Check
Put the numbers back in. Both sides should match.
9.5 = 21.9 − 12.4 ✓
95
=
100
×
9.5
10
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.