Decomposition reactions

A decomposition reaction is a chemical reaction in which one compound breaks down into two or more simpler substances. It is written in general form as

where: $AB$ is the single starting compound, and $A$, $B$ are the products, which may be elements or compounds themselves. Read backwards, this is exactly the pattern of a combination reaction — one reactant on the left instead of many — which is why the two reaction types are always taught side by side and are easy to mix up if you do not first count how many substances sit on the reactant side. Most decompositions in this chapter are driven by heat, so a $\Delta$ symbol sits above the arrow to mark a thermal decomposition. The benchmark example is calcium carbonate breaking down on strong heating:

Beyond carbonates, two more compound families decompose thermally and are equally examinable: metal nitrates, which release nitrogen dioxide and oxygen, and metal sulphates, which release sulfur dioxide and sulfur trioxide. In every case the metal ends up as an oxide, and the family the salt belongs to — not the individual salt — is what tells you which gas or gases to expect.

• Carbonates → metal oxide + carbon dioxide (colourless gas, turns limewater milky).
• Nitrates → metal oxide + nitrogen dioxide (brown gas) + oxygen (relights a glowing splint).
• Sulphates → metal oxide + sulfur dioxide + sulfur trioxide (both pungent, choking gases).

There is no formula in this topic beyond bookkeeping the atoms, so the "grounded rule" behind every decomposition equation is conservation of mass: every atom present in the reactant must reappear, unaltered, among the products. This single constraint is powerful enough to determine every coefficient you will ever need to write in this chapter, without any table of reactivity or energy values, and it is the reason the hit-and-trial balancing method taught in class always terminates in a correct equation as long as the products themselves are correctly identified from the compound family. When a nitrate or sulphate releases more than one gas, the atom count on both sides must match separately for every element present — metal, nonmetal, and oxygen — which is exactly why lead nitrate needs a coefficient of 2 while calcium carbonate needs none at all.

Step 1. Identify the reactant and write down how many atoms of each element it contains, since every one of those atoms must survive somewhere in the products. For $\text{Pb(NO}_3)_2$, that is 1 lead, 2 nitrogen, and 6 oxygen atoms.

Step 2. Identify the products from the compound family — a nitrate always gives a metal oxide, nitrogen dioxide, and oxygen — and write their unbalanced skeleton equation: $\text{Pb(NO}_3)_2 \rightarrow \text{PbO} + \text{NO}_2 + \text{O}_2$.

Step 3. Balance nitrogen first, since it appears in only one product. Two nitrogen atoms on the left force a coefficient of 2 on $\text{NO}_2$, and to keep lead and oxygen consistent with two nitrate units reacting, the lead side needs a coefficient of 2 as well, giving $2\text{Pb(NO}_3)_2 \rightarrow 2\text{PbO} + 4\text{NO}_2 + \text{O}_2$.

Step 4. Check oxygen last, because it is the element most likely to appear in every term. The left side now has $2 \times 6 = 12$ oxygen atoms; the right side has $2$ (in PbO) $+ 8$ (in $4\text{NO}_2$) $+ 2$ (in $\text{O}_2$) $= 12$ — the equation balances, confirming no atom was lost or created.

Question: Green crystals of ferrous sulphate ($\text{FeSO}_4$) are strongly heated in a dry test tube. A reddish-brown solid residue remains, and two pungent, choking gases are given off. Identify the reaction type, the products, and write the balanced equation.

Step 1. Count the reactants: there is only $\text{FeSO}_4$ being heated, one substance breaking apart under heat, so this must be a decomposition reaction, not a displacement or combination.

Step 2. Apply the sulphate family rule: a metal sulphate decomposes on heating into the metal oxide plus sulfur dioxide and sulfur trioxide. The reddish-brown solid is therefore ferric oxide ($\text{Fe}_2\text{O}_3$), and the two pungent gases are $\text{SO}_2$ and $\text{SO}_3$.

Step 3. Balance by atom conservation. Two $\text{FeSO}_4$ units supply 2 Fe, 2 S, and 8 O on the left. On the right, $\text{Fe}_2\text{O}_3$ needs both iron atoms, so the skeleton is $2\text{FeSO}_4 \rightarrow \text{Fe}_2\text{O}_3 + \text{SO}_2 + \text{SO}_3$; checking oxygen, the right side gives $3 + 2 + 3 = 8$, matching the left side exactly, so no further coefficients are needed.

Equation: $2\text{FeSO}_4(s) \xrightarrow{\Delta} \text{Fe}_2\text{O}_3(s) + \text{SO}_2(g) + \text{SO}_3(g)$

Conclusion: This is a decomposition reaction. The reddish-brown solid is ferric oxide ($\text{Fe}_2\text{O}_3$), and the two gases are sulfur dioxide ($\text{SO}_2$) and sulfur trioxide ($\text{SO}_3$).

Competitive stretch

The same family logic answers a follow-up question instantly: if copper sulphate ($\text{CuSO}_4$) were heated instead, you would predict a black solid residue of copper(II) oxide plus the same pair of gases, $\text{SO}_2$ and $\text{SO}_3$, without needing to look up a new equation — the sulphate pattern, not the specific metal, decides the gas products.

• Confusing decomposition with displacement. Decomposition has exactly one reactant breaking apart; displacement always has two reactants, an element and a compound — count the substances on the left before naming the reaction.
• Forgetting the heat symbol. Almost every decomposition in this chapter is thermal, so a complete, fully correct equation always shows $\Delta$ above the arrow.
• Dropping a product. Nitrates give two gases ($\text{NO}_2$ and $\text{O}_2$), and sulphates give two gases ($\text{SO}_2$ and $\text{SO}_3$) — leaving out the second gas is the most common exam slip.
• Balancing metal atoms before checking the multi-element product. Balance the element that appears in only one compound first (like nitrogen in the nitrate example), then work outward to oxygen, which usually appears everywhere and is easiest to check last.

Ask the tutor: Want to practise predicting the products for a metal carbonate, nitrate, or sulphate you have not seen before, or get a step-by-step check on your own balancing attempt?