Stoichiometry: one reaction taken from balancing through moles, the mole ratio, limiting reactant and percent yield

Stoichiometry is the arithmetic of a balanced equation: convert what you were given to moles, use the coefficients as a ratio to find moles of what you want, and convert back. Every stoichiometry problem, including limiting reactant and percent yield, is those three moves in some order. Below, one reaction (aluminium burning in oxygen) is taken through all five steps with every line shown, then the line students usually get wrong at each step, then a problem with new numbers to try cold.

What stoichiometry is, in one sentence

The coefficients of a balanced equation are a mole ratio and nothing else. They are not a mass ratio: 4 mol of aluminium does not weigh the same as 3 mol of oxygen. They are not a volume ratio for solids or liquids. They say only that 4 mol of one thing reacts with 3 mol of another to make 2 mol of a third.

Because the ratio lives in moles, every problem has the same three moves. Go to moles from whatever you were given. Cross the ratio to the substance you want. Come back from moles to whatever the question asks for, usually grams. Balancing comes before the three moves, and percent yield comes after them. That is the whole procedure, and the five steps below are those pieces in order.

In Ontario, SCH4U files this as Gas laws and stoichiometry (prerequisite review): the one idea the rest of the course's calculations assume rather than teach. The SCH4U course map shows where it sits; every chemistry idea worked, in course order is the section this page belongs to. Students in a US chemistry course or AP Chemistry meet the same five steps in the first unit on reactions.

The data in the example carry three significant figures, so every final answer is reported to three. Intermediate values keep a fourth digit so that rounding happens once, at the end.

Step 1: balance the equation

Our rule for how to balance chemical equations has three parts. Balance the element that appears in the fewest formulas first. Leave oxygen and hydrogen for last, because they usually appear in several formulas. Never change a subscript; only the coefficients in front of a formula move.

Aluminium and oxygen make aluminium oxide: Al + O₂ → Al₂O₃.

1. Aluminium first. It appears in two formulas, oxygen also in two, so start with the metal. There are 2 Al on the right, so write 2 Al on the left: 2 Al + O₂ → Al₂O₃.
2. Oxygen next. There are 2 O on the left and 3 on the right. The smallest number both divide is 6, so 3 O₂ on the left and 2 Al₂O₃ on the right: 2 Al + 3 O₂ → 2 Al₂O₃.
3. Return to aluminium. The right side now has 2 × 2 = 4 Al, so the left needs 4 Al: 4 Al + 3 O₂ → 2 Al₂O₃.
4. Count both sides. Al: 4 = 2 × 2 = 4. O: 3 × 2 = 6 = 2 × 3 = 6. Balanced.

A second balance, briefly, because combustion comes up on every test. Propane burns: C₃H₈ + O₂ → CO₂ + H₂O. Carbon first: 3 CO₂. Hydrogen next: 4 H₂O. Oxygen last: the right side has 3 × 2 + 4 × 1 = 10 O, so 5 O₂. Result: C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O. Count: C 3 = 3, H 8 = 8, O 10 = 6 + 4.

The wrong line. A student stuck on Al + O₂ → Al₂O₃ changes the product to AlO₂ so that Al + O₂ → AlO₂ "balances" with no coefficients at all. AlO₂ is not aluminium oxide; the product of this reaction is Al₂O₃, fixed by the charges of Al³⁺ and O²⁻. Changing a subscript changes what the reaction is; changing a coefficient changes how much of it there is. Only coefficients move.

Step 2: grams to moles with the molar mass

The problem: 13.5 g of aluminium is burned in 12.8 g of oxygen, and the lab collects 21.3 g of aluminium oxide. The problem states its molar masses: Al 26.98 g/mol, O₂ 32.00 g/mol, Al₂O₃ 101.96 g/mol (2 × 26.98 + 3 × 16.00 = 53.96 + 48.00 = 101.96).

Moles are grams divided by molar mass, one line per substance:

• n(Al) = 13.5 g ÷ 26.98 g/mol = 0.5004 mol
• n(O₂) = 12.8 g ÷ 32.00 g/mol = 0.4000 mol

The unit check is part of the line: g ÷ (g/mol) = mol. Keep four digits here; the data have three, and the rounding waits for the end.

The wrong line. Dividing 12.8 g by 16.00 g/mol, the atomic mass of an oxygen atom rather than the molar mass of the O₂ molecule, gives 0.800 mol, double the truth. Every line after it inherits the error, and the final mass comes out wrong without a single arithmetic slip. The molar mass is the mass of the formula as written in the equation, and the equation says O₂.

Step 3: the mole ratio and the limiting reactant

Here is the mole comparison that misleads. There are 0.5004 mol of aluminium and 0.4000 mol of oxygen. Aluminium has more moles, so a student concludes that oxygen, with fewer, runs out first. The ratio decides, not the count. The equation needs 4 Al for every 3 O₂, so aluminium is used up faster than oxygen is.

Ask how much oxygen the aluminium would need:

• O₂ needed to use all the Al = 0.5004 mol Al × (3 mol O₂ ÷ 4 mol Al) = 0.3753 mol O₂
• O₂ available = 0.4000 mol, which is more than 0.3753 mol

So there is oxygen to spare. Oxygen is the excess reactant and aluminium is the limiting reactant: it is gone first, and it decides how much product can form.

The other direction gives the same verdict, and doing it once is a good check:

• Al needed to use all the O₂ = 0.4000 mol O₂ × (4 mol Al ÷ 3 mol O₂) = 0.5333 mol Al
• Al available = 0.5004 mol, which is less than 0.5333 mol

Not enough aluminium to use the oxygen, so aluminium limits. Both directions agree.

The oxygen left over, if the question asks:

• excess O₂ = 0.4000 mol − 0.3753 mol = 0.02472 mol
• mass = 0.02472 mol × 32.00 g/mol = 0.791 g

The wrong line. "0.4000 is less than 0.5004, so O₂ is limiting." It is the most common error on this unit, and it is wrong because the two substances are not consumed one for one. Every product figure that follows from it comes out 6.6 % high (27.2 g instead of 25.5 g, as the next step shows), and a marker who sees 27.2 g knows exactly which line produced it.

Step 4: moles of product, then grams

Product is always computed from the limiting reactant, because the limiting reactant is what the reaction actually had to work with.

• n(Al₂O₃) = 0.5004 mol Al × (2 mol Al₂O₃ ÷ 4 mol Al) = 0.2502 mol
• m(Al₂O₃) = 0.2502 mol × 101.96 g/mol = 25.51 g

Reported to three significant figures, the theoretical yield is 25.5 g of aluminium oxide. That is the most the reaction can make from 13.5 g of aluminium, however much oxygen is present.

The wrong line. Starting from the excess reactant: 0.4000 mol O₂ × (2 ÷ 3) = 0.2667 mol, then 0.2667 mol × 101.96 g/mol = 27.2 g. The reaction cannot make 27.2 g, because the aluminium is gone before that much oxygen has reacted. Any product figure must trace back to the limiting reactant, and if yours traces back to the other one, the answer is too high.

Step 5: percent yield

The lab collected 21.3 g. Percent yield compares what was collected with what the stoichiometry says was possible:

• percent yield = actual ÷ theoretical × 100 = 21.3 g ÷ 25.51 g × 100 = 83.5 %

Use the unrounded theoretical yield in the division and round once at the end. A yield of 83.5 % is ordinary for a reaction run in a school lab: some product sticks to the glassware, some reaction is incomplete, some is lost in the transfer.

The wrong line. Theoretical over actual: 25.51 ÷ 21.3 × 100 = 119.8 %. A yield over 100 % is impossible for a dry, pure product, so the number itself tells you something went wrong. Three causes cover almost every case: the fraction is upside down, a slip earlier (a wrong molar mass or a ratio read backwards) made the theoretical yield too small, or the product was weighed wet. Computing the theoretical yield from the excess reactant is a different error: it makes the theoretical yield too large and the percent too small.

The whole thing on one screen

Step	What you compute	The line from this problem	The wrong line
1 Balance	Coefficients only	4 Al + 3 O₂ → 2 Al₂O₃	Changing Al₂O₃ to AlO₂
2 Moles	Grams ÷ molar mass	13.5 ÷ 26.98 = 0.5004 mol Al; 12.8 ÷ 32.00 = 0.4000 mol O₂	12.8 ÷ 16.00 = 0.800 mol
3 Limiting	Moles needed vs moles available, through the ratio	Al needs 0.3753 mol O₂; 0.4000 available; Al limits	"Fewer moles of O₂, so O₂ limits"
4 Product	From the limiting reactant	0.5004 × 2/4 = 0.2502 mol → 25.5 g	From O₂: 0.2667 mol → 27.2 g
5 Yield	Actual ÷ theoretical × 100	21.3 ÷ 25.51 × 100 = 83.5 %	25.51 ÷ 21.3 = 119.8 %

Three moves, two bookends. Moles, ratio, moles back; balance before, yield after. The worked example is the shape every page in this section follows, and the reason is in chapter 3 of the free playbook: Sweller and Cooper (1985) found that students who studied a worked example and then solved a matching problem learned the procedure faster than students who only solved problems. The matching problem is next.

Try one cold: magnesium in oxygen

Cover everything above. Magnesium burns in oxygen: 2 Mg + O₂ → 2 MgO. A student burns 6.00 g of magnesium (24.31 g/mol) in 3.00 g of oxygen (32.00 g/mol) and collects 6.80 g of magnesium oxide (40.31 g/mol). Which reactant is limiting, what is the theoretical yield, and what is the percent yield? Write every line with its units before you look.

Check

• n(Mg) = 6.00 g ÷ 24.31 g/mol = 0.2468 mol
• n(O₂) = 3.00 g ÷ 32.00 g/mol = 0.09375 mol
• O₂ needed for all the Mg = 0.2468 mol × (1 ÷ 2) = 0.1234 mol; only 0.09375 mol is available, so O₂ is limiting
• n(MgO) = 0.09375 mol O₂ × (2 ÷ 1) = 0.1875 mol
• m(MgO) = 0.1875 mol × 40.31 g/mol = 7.558 g → 7.56 g theoretical yield
• percent yield = 6.80 g ÷ 7.558 g × 100 = 90.0 %

Notice the twist. Here the reactant with fewer moles is the limiting one, which is the opposite of the aluminium case. You cannot pattern-match the limiting reactant from the mole counts; you have to run the ratio each time. If you got oxygen as limiting by reasoning "fewer moles", you got the right answer for the wrong reason, and the aluminium problem would have caught you.

Where this sits in the course

In the SCH4U pack this is Gas laws and stoichiometry (prerequisite review), the last of the 24 ideas, filed last because it is review, met first because every calculation chapter stands on it. The gas half of that review is its own page: the gas half of this review, PV = nRT, where a gas's amount comes from its pressure, volume and temperature instead of a mass.

What stoichiometry unlocks, by the pack's own titles: Enthalpy changes and calorimetry, where the heat released is per mole of reactant and the moles come from here; Dynamic equilibrium and the equilibrium constant, where the ICE table is stoichiometry written in a column; Buffers and acid–base titration, where the volume at the equivalence point is a mole ratio; and Electrolytic cells and Faraday's laws, where the moles of metal deposited come from the moles of electrons passed. The enthalpy, equilibrium and pH pages in this section each open with a moles line that is this page's step 2.

How to practise it

Tomorrow, redo the aluminium example cold: the problem statement only, every line written, then compare. Then the magnesium sibling the same way. Then, and this is the step most students skip, mixed problems where you do not know in advance whether it is a limiting-reactant question or a plain mass-to-mass one. Rohrer and Taylor (2007) found that students who practised problems shuffled across types did worse during practice and better on the test a week later than students who practised them in blocks, because the shuffled group practised the decision about which procedure to use. Why mixed practice beats twenty of the same has the reasoning; where practice problems come from and the order that teaches has the sources.

Keep an error log by step number. "Step 3: compared mole counts instead of running the ratio" is a line you can test yourself on; "got it wrong" is not. After a week of this, the five steps are one habit, and the test's twelve-mark stoichiometry question is twelve marks you can see.