Chemical Formula of Common Compounds

A chemical formula records two things at once: which elements are present, and the fixed ratio in which their atoms combine. That ratio is not arbitrary — it is fixed by each element's valency, the combining capacity that decides how many bonds (or charge units) an atom typically forms.

Step-by-step recipe. Write the two combining species side by side with their valency (or ionic charge magnitude, ignoring the sign) shown above each symbol. Criss-cross: the valency of the first becomes the subscript of the second, and vice versa. Simplify the resulting subscripts to their smallest whole-number ratio by cancelling any common factor. Finally, if a polyatomic ion ends up needing a subscript greater than 1, enclose the whole ion in brackets before writing that subscript — the bracket signals "this many units of the entire group," not of one atom inside it.

where: $\text{A}$ and $\text{B}$ are the two combining elements or ions, $a$ is the valency (or charge magnitude) of A, and $b$ is the valency (or charge magnitude) of B — each becomes the other symbol's subscript, which is exactly what "criss-cross" means.

Two families to keep separate. Oxides and hydrides combine an element directly with oxygen (valency 2) or hydrogen (valency 1) — an element of valency 2 gives an oxide of type XO, one of valency 3 gives X$_2$O$_3$ (as in aluminium oxide, Al$_2$O$_3$). Compounds built from polyatomic ions — hydroxide $\text{OH}^-$, ammonium $\text{NH}_4^+$, sulphate $\text{SO}_4^{2-}$, carbonate $\text{CO}_3^{2-}$, nitrate $\text{NO}_3^-$, phosphate $\text{PO}_4^{3-}$ — always need the extra bracket step once their subscript exceeds 1, because the whole ion must move as one indivisible unit through the criss-cross.

Formulae worth recognising instantly (each is nonetheless derivable, not just memorised): $\text{H}_2\text{O}$, $\text{NH}_3$, $\text{CO}_2$, $\text{CH}_4$, $\text{NaCl}$, $\text{CaO}$.

Why the method actually works. A compound must be electrically neutral overall — the positive charge contributed by one species has to exactly cancel the negative charge contributed by the other. Criss-crossing the valencies is simply a shortcut for finding the smallest whole numbers of each species that make the total charge zero: if A carries charge $+a$ and B carries charge $-b$, then $b$ units of A and $a$ units of B give a total charge of $b(+a) + a(-b) = 0$, which is exactly the criss-crossed subscript pattern. Seen this way, the method is not an arbitrary rule to memorise but a direct consequence of charge balance, which is also why simplifying to the lowest ratio never changes the chemistry — $\text{Ca}_2\text{O}_2$ and $\text{CaO}$ describe the same 1:1 combination, just scaled differently.

Problem. Derive the chemical formula of calcium phosphate, given calcium ion $\text{Ca}^{2+}$ and phosphate ion $\text{PO}_4^{3-}$.

Step 1. Write the ions with their charge magnitudes above each: $\overset{2}{\text{Ca}^{2+}} \quad \overset{3}{\text{PO}_4^{3-}}$.

Step 2. Criss-cross the magnitudes: calcium's charge (2) becomes phosphate's subscript, and phosphate's charge (3) becomes calcium's subscript, giving $\text{Ca}_3(\text{PO}_4)_2$.

Step 3. Check whether it simplifies further. The subscripts 3 and 2 share no common factor other than 1, so this is already the lowest whole-number ratio — no further reduction is possible.

Step 4. Bracket the polyatomic ion. Phosphate needs a subscript of 2, so it must be enclosed in brackets; calcium's subscript of 3 needs no bracket since it is a single atom, not a group.

Answer: the formula of calcium phosphate is $\mathbf{\text{Ca}_3(\text{PO}_4)_2}$ — three calcium ions (total charge $+6$) exactly balancing two phosphate ions (total charge $-6$).

Competitive stretch

For aluminium sulphate, criss-crossing $\text{Al}^{3+}$ (charge 3) with $\text{SO}_4^{2-}$ (charge 2) gives $\text{Al}_2(\text{SO}_4)_3$ directly, since 2 and 3 share no common factor — a useful check: whenever the two charge magnitudes are coprime, the criss-crossed subscripts are already the final answer, with no simplification step needed.

• Writing subscripts for single atoms. A subscript of 1 is never shown explicitly — write NaCl, not "Na$_1$Cl$_1$".
• Forgetting brackets for a repeating polyatomic ion. Ca(OH)$_2$ and (NH$_4$)$_2$SO$_4$ both need brackets around the ion group whenever its subscript exceeds 1; dropping the brackets changes what the formula means.
• Not simplifying to the lowest ratio. Criss-crossing valencies 2 and 4 gives X$_4$Y$_2$, which must be reduced to X$_2$Y by dividing both subscripts by their common factor 2.
• Mixing up which valency goes where. The valency of A becomes the subscript of B, and vice versa — reversing this gives a wrong formula, such as writing Al$_3$O$_2$ instead of the correct Al$_2$O$_3$.
• Using the ionic charge sign instead of its magnitude. Only the size of the charge (1, 2, 3...) is criss-crossed into a subscript; the plus or minus sign only tells you the two species attract, it never appears as part of the formula itself.

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