Quantitative Chemistry

Turn every element into moles by dividing its mass (or its percentage read as a mass out of 100) by its $A_{\mathrm{r}}$, then divide through by the smallest amount to get the atom ratio; that ratio is the empirical formula. For a molecular formula, scale the whole empirical unit by $\frac{M_{\mathrm{r}}(\text{molecule})}{M_{\mathrm{r}}(\text{empirical unit})}$.

Trigger: a compound described by the masses or mass fractions of its elements, or a molecular mass stated as a multiple of the empirical mass.

Instances:
(i) mass data: subtract to recover any missing element, divide each mass by its $A_{\mathrm{r}}$, then take the simplest ratio (a $2:5$ ratio gives $\mathrm{B}{\mathrm{r}}_2{\mathrm{O}}_5$);
(ii) an awkward ratio such as $1.5:2$ is not an error: multiply the whole ratio up to whole numbers ($1.5:2=3:4$, giving $\mathrm{F}{\mathrm{e}}_3{\mathrm{O}}_4$), never round;
(iii) molecular from empirical: find the empirical mass, then multiply every subscript by $M_{\mathrm{r}}$ over that mass (${\mathrm{C}}_2{\mathrm{F}}_3$ has mass $81$, the molecule $162$, so ${\mathrm{C}}_4{\mathrm{F}}_6$);
(iv) if composition is given as fractions and one element's absolute mass is fixed by the chemistry (bromine adds as $\mathrm{B}{\mathrm{r}}_2$, two atoms per molecule), read $M_{\mathrm{r}}$ as that element's mass divided by its mass fraction.