= Gravimetric dilution factor
{title2=$F=(m_s+m_d)/m_s$}
The <gravimetric dilution factor> is the ratio of final mixture mass to the stock mass used in a <gravimetric dilution>:
$$
F=\frac{m_s+m_d}{m_s}=1+\frac{m_d}{m_s}=\frac1f.
$$
Here $m_d$ is added diluent mass and $f$ is the retained stock mass fraction. For a diluent containing none of the component being measured, <mass conservation> gives final component <mass fraction> $w_f=w_s/F$. This mass-based relation avoids assuming that liquid volumes are additive. Its uncertainty follows from the same input masses and their <covariance> as the <uncertainty of a gravimetric dilution>.<Guidelines for Evaluating and Expressing the Uncertainty of NIST Measurement Results by Barry N. Taylor and Chris E. Kuyatt>[Appendix A, equations A-1–A-3, measurement function and uncertainty propagation]{ref}
For an original two-step example, combine $2.000\,\mathrm{g}$ of stock with $18.000\,\mathrm{g}$ of diluent, giving $F_1=10$. After complete mixing, transfer $1.000\,\mathrm{g}$ of that mixture and add $4.000\,\mathrm{g}$ of diluent, giving $F_2=5$. The combined <gravimetric dilution factor> is
$$
\boxed{F_{\rm total}=F_1F_2=50.}
$$
If the original stock has component <mass fraction> $0.2000$, the first mixture has $0.02000$ and the final mixture $0.004000$. The transferred gram contains $0.02000\,\mathrm{g}$ of the component; dividing by the final $5.000\,\mathrm{g}$ checks the result. This calculation requires a representative, homogeneous transferred portion and no component loss.
State whether a dilution number denotes $F$ or its reciprocal $f$: here a tenfold dilution means $F=10$ and $f=0.1$, not that ten parts of diluent were added per part of stock. The latter recipe gives $F=11$.
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