Combustion stoichiometry — fuel + O₂ → CO₂ + H₂O: Worked examples

Before we start, three words used everywhere below, defined from zero:

The scenario matrix

Every combustion problem lives in one of these case classes. Each row is a distinct way the numbers can behave; the last column names the example that nails it.

#	Case class	What makes it tricky	Covered by
C1	Pure hydrocarbon, $z = 0$ , whole-number answer	the baseline — must be flawless	Ex 1
C2	Hydrocarbon giving a fraction $\frac{y}{4}$ not whole	must double to clear halves	Ex 2
C3	Oxygenated fuel, $z > 0$ (the $- \frac{z}{2}$ credit)	fuel brings its own oxygen	Ex 3
C4	Degenerate fuel: no carbon ( $x = 0$ ) or no hydrogen ( $y = 0$ )	one product vanishes	Ex 4
C5	Mass calculation with mass-conservation check	grams ↔ moles ↔ grams	Ex 5
C6	Limiting reagent: not enough O₂ supplied	leftover fuel; incomplete	Ex 6
C7	Real-world word problem (air, %, tank)	translate English → $x, y, z$	Ex 7
C8	Exam twist: unknown fuel found from product masses	run the logic backwards	Ex 8

Read the matrix as a checklist. Below, each example is tagged with the cell it fills.

Worked example Example 1 — Propane

$C_{3} H_{8}$ (Cell C1: clean hydrocarbon) Statement. Balance the complete combustion of propane (the gas in a barbecue cylinder).

Forecast: Cover the steps and guess the O₂ coefficient. Is it a whole number or a fraction?

Read off $x, y, z$ . $x = 3, y = 8, z = 0$ . Why this step? The whole method is "plug the fuel's atom counts into the master formula", so we must extract them first.
Carbon → CO₂: $b = x = 3$ . Why? Each of the 3 carbons must land in one CO₂, and no carbon may vanish (conservation of atoms, Law of Conservation of Mass).
Hydrogen → H₂O: $c = \frac{y}{2} = \frac{8}{2} = 4$ . Why? Each water molecule holds 2 H, so 8 H make 4 waters.
Oxygen last: $a = x + \frac{y}{4} - \frac{z}{2} = 3 + 2 - 0 = 5$ . Why last? O₂ is the only free oxygen source; its size is dictated by however much CO₂ and H₂O we just fixed. $C_{3} H_{8} + 5 O_{2} \to 3 CO_{2} + 4 H_{2} O$ Verify: Count O atoms. Left: $5 \times 2 = 10$ . Right: $3 \times 2 + 4 \times 1 = 6 + 4 = 10$ . ✓ Carbon: $3 = 3$ ✓. Hydrogen: $8 = 4 \times 2$ ✓.

Worked example Example 2 — Butane

$C_{4} H_{10}$ (Cell C2: fractional coefficient) Statement. Balance butane (lighter fluid) combustion in whole numbers.

Forecast: $\frac{y}{4} = \frac{10}{4}$ is not whole. Guess: will $a$ be a half? What do we do about it?

$x = 4, y = 10, z = 0$ .
$b = 4$ , $c = \frac{10}{2} = 5$ .
$a = 4 + \frac{10}{4} - 0 = 4 + 2.5 = 6.5 = \frac{13}{2}$ . Why a fraction is OK for now? Coefficients only need to make atoms balance; a "half a molecule" is a bookkeeping placeholder we clean up next. $C_{4} H_{10} + \frac{13}{2} O_{2} \to 4 CO_{2} + 5 H_{2} O$
Clear the fraction: multiply every coefficient by 2. Why? You can't burn half a molecule in real life; multiplying the whole equation by 2 keeps ratios identical but makes all counts whole (see Balancing Chemical Equations). $2 C_{4} H_{10} + 13 O_{2} \to 8 CO_{2} + 10 H_{2} O$ Verify: O left $= 13 \times 2 = 26$ ; O right $= 8 \times 2 + 10 = 26$ ✓. C: $8 = 8$ ✓. H: $20 = 10 \times 2$ ✓.

Worked example Example 4 — Degenerate fuels (Cell C4: a product disappears)

Statement. Balance (a) hydrogen gas $H_{2}$ and (b) carbon $C$ (graphite) burning completely.

Forecast: Hydrogen has no carbon. Carbon has no hydrogen. So one product should vanish in each. Which one, for which fuel?

(a) Hydrogen $H_{2}$ — treat as $C_{0} H_{2} O_{0}$ , so $x = 0, y = 2, z = 0$ .

$b = x = 0$ → no CO₂ at all. Why? No carbon atoms exist to fill any CO₂.
$c = \frac{2}{2} = 1$ → 1 H₂O.
$a = 0 + \frac{2}{4} - 0 = \frac{1}{2}$ . $H_{2} + \frac{1}{2} O_{2} \to H_{2} O ⟹ 2 H_{2} + O_{2} \to 2 H_{2} O$

(b) Carbon $C$ — $x = 1, y = 0, z = 0$ .

$b = 1$ → 1 CO₂.
$c = \frac{0}{2} = 0$ → no H₂O. Why? No hydrogen to build water.
$a = 1 + 0 - 0 = 1$ . $C + O_{2} \to CO_{2}$ Verify (a): O left $= 1$ ; O right $= 1$ ✓. Verify (b): O left $= 2$ ; O right $= 2$ ✓. The master formula survives the edge cases where a whole product term drops to zero.

Worked example Example 5 — Mass calculation with conservation check (Cell C5)

Statement. How many grams of CO₂ are produced when 44 g of propane burns completely? Also confirm mass is conserved.

Forecast: Guess the order of magnitude — will CO₂ mass be more or less than the fuel mass? (Hint: we ADD oxygen atoms.)

We use Ex 1's equation: $C_{3} H_{8} + 5 O_{2} \to 3 CO_{2} + 4 H_{2} O$ .

Molar mass of propane. $M (C_{3} H_{8}) = 3 (12) + 8 (1) = 36 + 8 = 44$ g/mol. Why? We need to turn grams into moles, and moles are what the coefficients count.
Moles of propane $= \frac{44 g}{44 g/mol} = 1$ mol.
Mole ratio. From the equation, $1$ propane makes $3$ CO₂. So $1 \times 3 = 3$ mol CO₂. Why? Coefficients ARE the mole ratio — that's the payoff of balancing.
Grams of CO₂. $M (CO_{2}) = 12 + 2 (16) = 44$ g/mol, so $3 \times 44 = 132$ g CO₂.
Water for the check. $4$ mol H₂O $\times 18$ g/mol $= 72$ g. O₂ used: $5$ mol $\times 32 = 160$ g. Verify (mass conservation, Law of Conservation of Mass): $fuel 44 + O_{2} 160 = 204 g in CO_{2} 132 + H_{2} O 72 = 204 g out ✓$ Answer: 132 g of CO₂. Heavier than the fuel because oxygen atoms were added on.

Worked example Example 6 — Limiting reagent (Cell C6: not enough O₂)

Statement. You mix 1 mol of methane $CH_{4}$ with only 1.5 mol of O₂ (less than the recipe wants). If we insist on complete combustion, how much CH₄ can actually burn, and how much is left over?

Forecast: The recipe wants $2$ O₂ per CH₄. We only have $1.5$ . Guess: does all the methane burn, or is some stranded?

Equation: $CH_{4} + 2 O_{2} \to CO_{2} + 2 H_{2} O$ .

O₂ demanded by all the fuel. $1$ mol CH₄ $\times 2 = 2$ mol O₂ needed. Why? Check whether supply meets demand.
Compare to supply. We have $1.5 < 2$ , so oxygen runs out first — O₂ is the limiting reagent (see Limiting Reagent).
How much CH₄ the O₂ can burn. $1.5$ mol O₂ $\div 2 (O_{2} per CH_{4}) = 0.75$ mol CH₄ burned. Why divide by 2? Each burned CH₄ consumes 2 O₂, so available O₂ divided by 2 gives burnable fuel.
Leftover fuel. $1 - 0.75 = 0.25$ mol CH₄ unburned.
Products (from the 0.75 mol that burned). CO₂ $= 0.75$ mol; H₂O $= 1.5$ mol. Verify: O₂ consumed by 0.75 mol CH₄ $= 0.75 \times 2 = 1.5$ mol = exactly what we supplied ✓. Carbon check: burned C $= 0.75$ into CO₂ $= 0.75$ ✓. In a real engine this stranded fuel + low O₂ is exactly when soot and CO appear — incomplete combustion.

Worked example Example 7 — Real-world word problem: burning in AIR (Cell C7)

Statement. A stove burns methane in air, not pure oxygen. Air is about $21%$ O₂ by moles ( $79%$ is inert nitrogen). How many moles of air are needed to burn $1$ mol of methane completely?

Forecast: We need $2$ mol O₂ (from Ex-1-style logic). Since O₂ is only a fifth of air, guess: roughly $5 \times$ more air than O₂?

Equation: $CH_{4} + 2 O_{2} \to CO_{2} + 2 H_{2} O$ .

O₂ required. $x + \frac{y}{4} - \frac{z}{2} = 1 + 1 - 0 = 2$ mol O₂ for 1 mol CH₄. Why? Same master formula; air just changes how we source that O₂.
Convert O₂ to air. Only $21%$ of air is O₂, so $n_{air} = \frac{n _{O_{2}}}{0.21} = \frac{2}{0.21} \approx 9.52 mol air .$ Why divide by 0.21? If oxygen is a fraction $0.21$ of the mixture, you must bring $1/0.21$ times as much total gas to obtain that much O₂.
Nitrogen along for the ride. $9.52 - 2 = 7.52$ mol N₂ passes through unchanged (it exits hot but chemically untouched). Verify: $9.52 \times 0.21 = 2.0$ mol O₂ ✓ — exactly the demand. Answer: about 9.52 mol of air. This "excess air" idea is why furnaces move huge volumes of gas for a little fuel.

Worked example Example 8 — Exam twist: find the unknown fuel (Cell C8: run it backwards)

Statement. A pure hydrocarbon $C_{x} H_{y}$ is burned completely. It produces 0.4 mol CO₂ and 0.5 mol H₂O, all from 0.1 mol of fuel. Find the formula.

Forecast: We normally go fuel → products. Here we invert. Guess: how do you get $x$ from CO₂, and $y$ from H₂O?

Carbon per fuel molecule. All carbon ends in CO₂, and each CO₂ carries 1 C: $x = \frac{mol CO _{2}}{mol fuel} = \frac{0.4}{0.1} = 4.$ Why divide by mol fuel? $x$ is carbons per molecule, so we scale total carbon down to one fuel molecule.
Hydrogen per fuel molecule. Each H₂O carries 2 H: $y = \frac{2 \times mol H _{2} O}{mol fuel} = \frac{2 \times 0.5}{0.1} = \frac{1.0}{0.1} = 10.$ Why the factor 2? Water has two hydrogens each, so total H $= 2 \times$ (mol H₂O).
The fuel is $C_{4} H_{10}$ — butane! (Same molecule as Ex 2.) Verify: Forward-check with the master formula: $a = 4 + \frac{10}{4} = 6.5$ , and $0.1$ mol fuel should give $0.1 \times 4 = 0.4$ mol CO₂ ✓ and $0.1 \times 5 = 0.5$ mol H₂O ✓. The backward logic reproduces the given data exactly.

Recall Which cell was which? (self-test)

Match each: whole-number hydrocarbon / fractional coefficient / oxygen credit / vanishing product / mass check / limiting O₂ / air percentage / unknown fuel. Cell C1 propane baseline ::: Example 1 Cell C2 butane fraction cleared by ×2 ::: Example 2 Cell C3 glucose with $- z /2$ credit ::: Example 3 Cell C4 H₂ (no CO₂) and C (no H₂O) ::: Example 4 Cell C5 propane 44 g → 132 g CO₂, mass balances 204 g ::: Example 5 Cell C6 O₂ limiting, 0.25 mol CH₄ left over ::: Example 6 Cell C7 methane in air needs ≈9.52 mol air ::: Example 7 Cell C8 products → unknown fuel C₄H₁₀ ::: Example 8

Connections

Mole Concept and Molar Mass — powers every gram ↔ mole step (Ex 5, 7).
Limiting Reagent — the engine of Example 6.
Law of Conservation of Mass — the 204 g in = 204 g out check.
Balancing Chemical Equations — clearing fractions (Ex 2, 4).
Enthalpy of Combustion — once atoms rearrange, energy is released.
Oxidation States — carbon climbs to $+ 4$ in every complete case.

Worked examples — Combustion stoichiometry — fuel + O₂ → CO₂ + H₂O

The scenario matrix

Connections

Scenario map