Extraction recovery

Extraction recovery is the fraction of an analyte amount recovered after a specified extraction stage. For input amount $n_0$ and recovered amount $n_1$, write $r=n_1/n_0$ when $n_0>0$. The same quantity basis must be used in numerator and denominator. IUPAC distinguishes this preparation yield from apparent recovery, an observed/reference comparison involving a calibrated analytical result.^[ref]

Concentration alone is insufficient. In an original example, $10\mathrm{mL}$ at $10\mu \mathrm{mol}{\mathrm{L}}^{-1}$ contains $100\mathrm{nmol}$ of analyte. A final extract of $2\mathrm{mL}$ at $30\mu \mathrm{mol}{\mathrm{L}}^{-1}$ contains $60\mathrm{nmol}$. Recovery is $60/100=0.60$, or $60\%$, even though concentration increased threefold. Shrinking the volume can enrich a sample while losing material.

Measured response also need not track recovered amount with the same proportionality in different sample matrices. A method study explicitly distinguished matrix response from recovery in quantitative liquid chromatography–mass spectrometry.^[ref] In a matched comparison, adding equal nominal analyte amounts before preparation and after preparation can isolate preparation loss only if final volumes, acquisition and response conditions are adequately controlled.

A recovery correction divides an observed amount by an appropriate recovery estimate. A measured $6\mathrm{nmol}$ with $r=0.60$ gives $10\mathrm{nmol}$ before that loss, under the model. This is an inference, not recovered physical material. If the estimated recovery changes with sample composition or analyte level, applying one constant can introduce error. A measured apparent recovery above 100% need not represent creation of analyte during extraction; background, calibration, chemical conversion and the reference basis require examination.

A recovery correction adjusts an analytical result to account for an estimated loss of analyte during a specified preparation stage. If the recovered amount $y$ follows $y=rq$, with original amount $q$ and extraction recovery $r>0$, the corrected estimate is $\hat{q}=y/r$. Its validity depends on whether the recovery estimate applies to that analyte, sample matrix and procedure.

For an original example, $y=6\mathrm{nmol}$ and $r=0.60$ give $\hat{q}=10\mathrm{nmol}$. The calculation estimates the amount before loss; it does not restore molecules to the sample. It also does not separately correct a mass spectrometry matrix effect unless that effect has been included consistently in the measurement model.

Recovery-correction uncertainty follows from the quotient, not just from the measured amount. First-order propagation of uncertainty givesIf the two inputs are independent, the covariance term vanishes. With stipulated standard uncertainties $u_y=0.3\mathrm{nmol}$ and $u_r=0.06$, the two variance contributions are $0.25$ and $1.00\mathrm{nmo}{\mathrm{l}}^2$, giving $u(\hat{q})\approx 1.12\mathrm{nmol}$. The recovery estimate dominates this example; dividing by recovery has not created extra precision.

A recovery inferred from a spiked preparation need not reproduce the behavior of analyte originally embedded in a sample. Likewise, applying a single recovery at every concentration assumes a common model. Near-zero or very uncertain recovery can make quotient inference unreliable beyond the linear uncertainty approximation.

Reporting whether a correction was applied, its size and how it was derived makes results comparable. These reporting provisions appear in the harmonised recovery guidelines and their Codex adoption text.^[ref]

Recovery-correction uncertainty quantifies uncertainty in an amount inferred through a recovery correction. If $\hat{q}=y/r$ uses measured recovered amount $y$ and recovery estimate $r>0$, both inputs contribute. The uncertainty of the corrected result is not obtained merely by copying the uncertainty attached to $y$.

The first-order sensitivity coefficients are $\partial \hat{q}/\partial y=1/r$ and $\partial \hat{q}/\partial r=-y/r^2$. Combining them givesThis is the linear propagation of uncertainty law with correlated inputs.^[ref] With positive nominal inputs and correlation coefficient $\rho$, the relative form isRecovery-estimate covariance can matter because an amount and its estimated recovery may share measurements or common response changes.

For original values $y=6\mathrm{nmol}$, $r=0.60$, $u_y=0.30\mathrm{nmol}$ and $u_r=0.06$, the corrected amount is $10\mathrm{nmol}$. Independence gives relative variance $0.05^2+0.10^2=0.0125$ and standard uncertainty about $1.12\mathrm{nmol}$. With $\rho =0.8$, the variance becomes $0.0045$ and uncertainty about $0.67\mathrm{nmol}$. Correlation cannot be chosen to improve the answer; it must describe the actual joint measurements.

The formula is an approximation near the nominal values. A recovery near zero or with large relative uncertainty can make a quotient strongly asymmetric and poorly represented by a symmetric standard uncertainty. Model inadequacy—such as applying a recovery from an unrepresentative sample matrix—is a separate issue that careful numerical propagation alone does not resolve.

Recovery-estimate covariance describes joint variation between an estimated extraction recovery and another input used with it, such as the measured recovered amount. It matters when a recovery correction divides one estimate by the other: shared data can make their uncertainties correlated.

For corrected amount $q=y/r$, first-order uncertainty propagation givesThe sign of the covariance contribution follows from the opposite signs of the quotient's sensitivity coefficients. A positive covariance reduces the propagated variance relative to a calculation assuming independence; a negative covariance increases it. Neither sign is a discretionary choice.^[ref]

An original shared-data example sets $r=y/z$, where independent measured amounts have nominal values $y=6\mathrm{nmol}$ and $z=10\mathrm{nmol}$, with standard uncertainties $u_y=0.30\mathrm{nmol}$ and $u_z=0.40\mathrm{nmol}$. Then $r=0.60$ andSubstitution gives $u_q^2=0.25+0.41-0.50=0.16\mathrm{nmo}{\mathrm{l}}^2$, hence $u_q=0.40\mathrm{nmol}$. This agrees with the exact algebraic cancellation $y/(y/z)=z$. Ignoring the shared numerator would instead give $\sqrt{0.66}\approx 0.81\mathrm{nmol}$.

The cancellation also reveals a validation trap: using the same recovered amount both to estimate its recovery and to correct itself does not independently demonstrate accurate recovery of an unknown sample. Build the joint measurement model from the underlying observations. Shared calibration standards, preparations or instrument drift may create other covariance structures; matched replicate data or an explicit common-input model can estimate them. Do not assume independence merely because a table lists two separate estimates.