One Number Per Element Generates Every Formula in Chemistry

Because a formula is nothing more than the ratio that balances two combining capacities.

Calcium has valency $2$ and chlorine has valency $1$. So one calcium can hold two chlorines, and the formula must be ${\text{CaCl}}_2$. Nothing was looked up and nothing was memorised — the numbers decided it.

This is why valency is worth learning properly. Get the two valencies right and the formula follows mechanically, for compounds you have never seen. This page covers the first part of the ICSE Class 8 Chemistry chapter on the language of chemistry: radicals and their valencies, writing formulae, reading valencies back out, and calculating molecular mass.

A radical is an atom or a group of atoms that carries a charge and behaves as a single unit in a chemical reaction.

Basic radicals are positive — they come from metals and from the ammonium group. Acidic radicals are negative — they come from non-metals and from acid groups.

Basic radicals, by valency:

- Valency 1 — sodium ${\text{Na}}^{+}$, potassium ${\text{K}}^{+}$, hydrogen ${\text{H}}^{+}$, silver ${\text{Ag}}^{+}$, ammonium ${\text{NH}}_4^{+}$, copper(I) ${\text{Cu}}^{+}$
- Valency 2 — magnesium ${\text{Mg}}^{2+}$, calcium ${\text{Ca}}^{2+}$, zinc ${\text{Zn}}^{2+}$, barium ${\text{Ba}}^{2+}$, lead ${\text{Pb}}^{2+}$, iron(II) ${\text{Fe}}^{2+}$, copper(II) ${\text{Cu}}^{2+}$
- Valency 3 — aluminium ${\text{Al}}^{3+}$, iron(III) ${\text{Fe}}^{3+}$

Acidic radicals, by valency:

- Valency 1 — chloride ${\text{Cl}}^{-}$, bromide ${\text{Br}}^{-}$, iodide ${\text{I}}^{-}$, hydroxide ${\text{OH}}^{-}$, nitrate ${\text{NO}}_3^{-}$, bicarbonate ${\text{HCO}}_3^{-}$
- Valency 2 — oxide ${\text{O}}^{2-}$, sulphide ${\text{S}}^{2-}$, sulphate ${\text{SO}}_4^{2-}$, carbonate ${\text{CO}}_3^{2-}$, sulphite ${\text{SO}}_3^{2-}$
- Valency 3 — phosphate ${\text{PO}}_4^{3-}$, nitride ${\text{N}}^{3-}$

Simple and compound radicals. A simple radical is a single atom — ${\text{Na}}^{+}$, ${\text{Cl}}^{-}$, ${\text{O}}^{2-}$. A compound radical is a group — ${\text{NH}}_4^{+}$, ${\text{SO}}_4^{2-}$, ${\text{CO}}_3^{2-}$, ${\text{OH}}^{-}$, ${\text{NO}}_3^{-}$. Compound radicals must be treated as one unbreakable unit and enclosed in brackets when more than one is needed.

The two exceptions worth marking. Ammonium ${\text{NH}}_4^{+}$ is the only common positive compound radical — everything else in the positive list is a metal. And iron and copper appear twice each, with two different valencies, which is why their compounds are named iron(II) and iron(III) to say which is meant.

Where these numbers come from. They are not arbitrary. Sodium's valency of $1$ is the single valence electron of the previous chapter's $2,8,1$; oxygen's valency of $2$ is the two electrons its $2,6$ needs. Compound radicals carry the leftover charge of a group, which is why ${\text{SO}}_4$ behaves as a unit of valency $2$ rather than as separate sulphur and oxygen.

Write the two radicals side by side with their valencies above them, then swap the valencies so each becomes the other's subscript.

The steps:

- Write the positive radical first, then the negative one.
- Write each valency above its radical.
- Criss-cross them down as subscripts.
- Simplify the subscripts if they share a common factor.
- Do not write a subscript of $1$.
- Put brackets around a compound radical whose subscript is more than $1$.

Worked example 1 — calcium chloride. Calcium is $2$, chlorine is $1$. Crossing over gives ${\text{Ca}}_1{\text{Cl}}_2$, so

$${\text{CaCl}}_2$$

Worked example 2 — aluminium oxide. Aluminium is $3$, oxygen is $2$:

$${\text{Al}}_2{\text{O}}_3$$

Worked example 3 — sodium sulphate. Sodium is $1$, sulphate is $2$:

$${\text{Na}}_2{\text{SO}}_4$$

No brackets needed here, because the sulphate subscript is $1$.

Worked example 4 — aluminium sulphate. Aluminium is $3$, sulphate is $2$. Crossing over needs three sulphate groups, so brackets are essential:

$${\text{Al}}_2({\text{SO}}_4{)}_3$$

Writing ${\text{Al}}_2{\text{SO}}_{43}$ would be meaningless, which is exactly why the bracket rule exists.

Worked example 5 — where you must simplify. Calcium carbonate: calcium is $2$, carbonate is $2$. Crossing over gives ${\text{Ca}}_2({\text{CO}}_3{)}_2$, and both subscripts share the factor $2$:

$${\text{CaCO}}_3$$

Similarly magnesium oxide, both valency $2$, gives ${\text{Mg}}_2{\text{O}}_2$ which simplifies to

$$\text{MgO}$$

Worked example 6 — a compound radical on both sides. Ammonium sulphate: ammonium is $1$, sulphate is $2$:

$$({\text{NH}}_4{)}_2{\text{SO}}_4$$

The ammonium needs brackets because its subscript is $2$.

The step students skip. Simplification is not optional. ${\text{Ca}}_2({\text{CO}}_3{)}_2$ is wrong, not merely ugly, because a formula must give the simplest whole-number ratio. Forgetting it is the single commonest mistake when both valencies are equal.

Use the radical whose valency you already know, total its charge, and divide by the number of the unknown atoms.

The principle is that a compound is electrically neutral, so the total positive charge must equal the total negative charge.

**Worked example 1 — iron in ${\text{Fe}}_2{\text{O}}_3$.** Oxygen has valency $2$, and there are three of them:

$$\text{total negative}=3\times 2=6$$

That must be balanced by two iron atoms, so

$$\text{valency of Fe}=\frac{6}{2}=3$$

Iron is therefore iron(III) here.

**Worked example 2 — nitrogen in ${\text{N}}_2{\text{O}}_5$.** Oxygen $2$, five of them:

$$\text{total}=5\times 2=10,\text{valency of N}=\frac{10}{2}=5$$

**Worked example 3 — a compound radical, carbonate in ${\text{K}}_2{\text{CO}}_3$.** Potassium has valency $1$, and there are two:

$$\text{total positive}=2\times 1=2$$

There is one carbonate group, so

$${\text{valency of CO}}_3=\frac{2}{1}=2$$

**Worked example 4 — phosphorus in ${\text{PH}}_3$.** Hydrogen has valency $1$, three of them:

$$\text{valency of P}=\frac{3\times 1}{1}=3$$

Worked example 5 — checking against the first section. In ${\text{FeCl}}_3$, chlorine is $1$ and there are three, so iron's valency is $3÷1=3$. In ${\text{FeCl}}_2$ the same working gives $2$. Both compounds exist, and this is precisely why iron needs the (II) and (III) labels.

Why this is the same skill as the criss-cross, reversed. The criss-cross turns valencies into subscripts; this turns subscripts back into valencies. If you can do one you can do the other, and checking a formula by deducing its valencies back out is the fastest way to catch an error you have just made.