Stoichiometry: typical mistakes

Stoichiometry is one of those topics that seems simple when you follow the lesson and becomes a disaster when you do the first exercise on your own. Even students who are very good at maths get stuck here — not because the calculations are hard, but because there is a sequence of steps that, if you get one wrong halfway, takes you very far from the right answer.

In this post I collect the mistakes I see most often in my students — both in high school (tests, maturità) and in preparation for the TOLC-MED, TOLC-F and TOLC-S (chemistry section; the TOLC being Italy’s university admission test) — and the way I teach to avoid them.

The basic scheme that always works

All basic stoichiometry problems follow the same 4-phase scheme:

Write the balanced reaction.
Convert the data to moles (from grams → moles, from volumes → moles, from concentration + volume → moles).
Use the coefficients of the balanced reaction as a “recipe” to go from the moles of one substance to the moles of another.
Convert back from the moles obtained to the unit the problem asks for (grams, litres, molarity, …).

If you get used to always doing these 4 steps in the same order, you stop making mistakes. Most of the mistakes I see come from skipping step 2 or step 4 and doing direct calculations between grams (which is wrong).

The fundamental relationship is

n = \frac{m}{M}

where $n$ is the number of moles, $m$ is the mass in grams and $M$ is the molar mass in g/mol. Always keep it in mind.

Mistake #1: confusing units of mass, amount of substance and volume

A classic example: “calculate how many grams of H₂O form when 32 g of H₂ react with excess O₂.”

Reaction: 2 H₂ + O₂ → 2 H₂O.

The average student does: “32 g of H₂ → 32 g of H₂O” because the coefficients are equal. Wrong. The coefficients are equal in moles, not in grams: one mole of H₂ weighs 2 g, one mole of H₂O weighs 18 g.

Correct procedure:

32 g of H₂ $\div$ 2 g/mol = 16 mol of H₂.
Equal coefficients → 16 mol of H₂O.
16 mol $\times$ 18 g/mol = 288 g of H₂O.

The difference between 32 and 288 is enormous. Mistakes like this cost you the test.

Mistake #2: forgetting to balance first

“How many grams of NH₃ form from 14 g of N₂ + 3 g of H₂?”

Reaction: N₂ + H₂ → NH₃ (NOT balanced).

The student who forgets to balance writes a 1:1:1 proportion and gets wrong results. The balanced reaction is N₂ + 3 H₂ → 2 NH₃: you need three times the amount of hydrogen and twice as much ammonia forms as there is nitrogen.

Practical rule: before touching the numbers, balance. If the reaction in the question is not balanced, rewrite it yourself.

Mistake #3: ignoring the limiting reagent

“I mix 4 g of H₂ with 16 g of O₂. How many grams of H₂O form?”

The student who skips this step does the calculation on only one reagent (usually the one mentioned first) and gets the wrong result because there may not be enough of the other.

Correct procedure:

4 g H₂ $\div$ 2 g/mol = 2 mol H₂.
16 g O₂ $\div$ 32 g/mol = 0.5 mol O₂.
Reaction 2 H₂ + O₂ → 2 H₂O: every mole of O₂ requires 2 moles of H₂.
For 0.5 mol O₂ you need 1 mol H₂ → I have 2, so O₂ is the limiting reagent.
Moles of H₂O produced = $2 \times 0.5 = 1$ mol H₂O = 18 g.

Mental trick: I divide the moles of each reagent by its own coefficient. The one with the smaller ratio is the limiting reagent.

Worked exercise: limiting reagent + percentage yield

This is the December test, the one where it is not enough to find the limiting reagent: you also have to compare how much you “should” get with how much you really get in the laboratory.

Question: “By reacting 28 g of N₂ with 12 g of H₂, 30 g of NH₃ are obtained in the laboratory. What is the percentage yield of the reaction?”

Balanced reaction: N₂ + 3 H₂ → 2 NH₃.

Convert to moles.
- $n (N_{2}) = 28 g \div 28 g/mol = 1 mol$ .
- $n (H_{2}) = 12 g \div 2 g/mol = 6 mol$ .
Find the limiting reagent by dividing the moles by the coefficient:
- N₂: $1 \div 1 = 1$ .
- H₂: $6 \div 3 = 2$ .
- The smallest ratio is that of N₂: it is the limiting reagent (H₂ is left over in excess — a frequent mistake is to “trust” the reagent that looks scarce by eye, here H₂, whereas you must always do the calculation).
Calculate the theoretical yield from the limiting reagent: 1 mol of N₂ produces, by the 2:1 ratio of the balanced reaction, 2 mol of NH₃.
- $m_{theoretical} = 2 mol \times 17 g/mol = 34 g$ .
Calculate the percentage yield by comparing the mass actually obtained with the theoretical one:

η = \frac{m _{actual}}{m _{theoretical}} \times 100 = \frac{30}{34} \times 100 \approx 88.2%

A yield of 88.2% is a data point, not a mistake: in the laboratory you never reach 100% (losses, side reactions, incomplete equilibrium). What counts in the exam is having worked out the theoretical yield on the limiting reagent and not on the first reagent in the question.

Worked exercise: from grams to molecules in three steps

Another type of conversion that is often intertwined with stoichiometry: from mass to number of particles (molecules, atoms, ions), always going through moles.

Question: “How many molecules are contained in 9 g of water?”

Mass → moles: $n = m / M = 9 g \div 18 g/mol = 0.5 mol$ .
Moles → particles, using Avogadro’s number $N_{A} = 6.022 \times 1 0^{23} mol^{- 1}$ :

N = n \times N_{A} = 0.5 \times 6.022 \times 1 0^{23} = 3.011 \times 1 0^{23} molecules

The same scheme (mass → moles → particles) applies identically to atoms in an element or to ions in an ionic compound: only what you are counting at the last step changes.

Solutions and molarity: the step after stoichiometry

As soon as stoichiometry is solid, the next lesson is almost always on solutions — and that is where mistakes #4 and #5 of this post (mL vs L, concentration vs amount) come back under another name. The basic formula is

M = \frac{n}{V}

where $M$ is the molarity in mol/L, $n$ the moles of solute and $V$ the volume of the solution in litres (not of the solvent: if you dissolve a solid in water and make up to volume, $V$ is the final volume of the solution).

Minimal example: if you have $0.5$ mol of solute dissolved in $2$ L of solution, $M = 0.5/2 = 0.25 mol/L$ . From here on, all the dilution, mixing and titration exercises are variations on the same 4-phase scheme seen at the start of the post, applied to molarity as well as to moles.

Mistake #4: confusing mol and mmol, M and mM

Typical in solution problems:

“I add 10 mL of 0.1 M HCl to 20 mL of 0.05 M NaOH. What is left?”

The student confuses mL with L and gets a result that is 1000 times wrong.

Rule: ALWAYS convert to SI units before doing the calculation. Mol and M (mol/L) are compatible; mmol and mM (mmol/L) are compatible with each other; mixing them without converting is the #1 cause of disasters.

Correct solution:

Moles of HCl:

10 mL \times 0.1 mol/L = 0.001 L \times 0.1 mol/L = 1 \times 1 0^{- 3} mol = 1 mmol

Moles of NaOH:

20 mL \times 0.05 mol/L = 0.020 L \times 0.05 mol/L = 1 \times 1 0^{- 3} mol = 1 mmol

The reaction is HCl + NaOH → NaCl + H₂O, ratio 1:1: the two amounts are equal, so they are exactly neutralised and the final solution is neutral.

In the maturità or the TOLC, “exactly neutralised” is a full-mark answer; a calculation seen at a glance shows mastery.

Mistake #5: not understanding the difference between concentration and amount

Confusing “concentration” (intensive: mol/L) and “amount” (extensive: mol) blocks a great many students in dilution problems.

“I dilute 100 mL of 0.5 M NaCl by adding 400 mL of water. What is the final concentration?”

The student who thinks directly in concentration does “0.5 / 5” without understanding why. The correct way is to think in moles:

Initial moles = $0.1 L \times 0.5 mol/L = 0.05 mol$ . The moles do not change when you add only water.
Final volume = 100 + 400 = 500 mL = 0.5 L.
Final concentration = $0.05 mol /0.5 L = 0.1 M$ .

When you get used to thinking “the moles are conserved, the volume changes, the concentration is recalculated”, you never get a dilution wrong again.

Mistakes specific to the TOLC-MED / TOLC-F

The TOLC for the medical-biological area has a chemistry section of 7-10 questions where stoichiometry is practically always present. Typical mistakes against the clock:

Calculating the molar mass from memory and getting it wrong. For H₂SO₄: $2 (1) + 32 + 4 (16) = 98 g/mol$ . Frequent mistakes: confusing the 32 of S with the 16 of O, or forgetting one of the 4 oxygens. It is better to redo the calculation underneath, not to trust memory.
Skipping the balancing because “it’s easy anyway”. In the quick questions they make you put coefficients like 4, 5, 6 — and the question is precisely “find the coefficient of X”. No improvising.
Confusing “moles that react” with “total moles”. When the problem talks about equilibrium, the reacting moles must be subtracted from the initial ones to find those remaining.

Realistic timescales

For a well-done review:

High school, ahead of a test: 3-4 targeted lessons on the mistakes above + 30-40 exercises between one lesson and the next. Doable in 2-3 weeks.
Maturità with a chemistry oral: 5-6 consolidation lessons + practice on typical questions for the second paper (for those schools where it is included).
TOLC-MED or TOLC-F: stoichiometry is inside a wider syllabus. Generally 2-3 lessons devoted to pure stoichiometry, integrated into the plan of 8-12 lessons in total for the whole chemistry section.

How we work in lessons

The first step is to understand where you stumble: I have you solve one or two typical exercises aloud, and generally after 5 minutes we understand whether the problem is balancing, units of measurement or the limiting reagent. Then we work only on that specific problem for a few lessons, until it becomes automatic.

For high school students ahead of a test we use the exercises from your book or those the teacher gave — not generic ones from other books. For the TOLC, official CISIA quizzes from previous years.

The aim is not to “learn chemistry”: it is to make the 4-phase scheme an automatic reflex, so that in the test/exam you can concentrate on the hard things instead of the basic steps.

See also the page dedicated to science tutoring with the full programme of general and inorganic chemistry, biology, earth sciences and astronomical geography, or the admission test preparation with the TOLC-MED / TOLC-F section. If you want to understand honestly where to restart before the test or the TOLC, start from the free introductory call.

To go further

If before balancing you still struggle to write the formulas of compounds, start from inorganic chemical nomenclature: IUPAC and traditional — it is the step before stoichiometry.
To review atoms and bonds: chemistry, atoms and bonds for Year 3 human sciences.
For balancing proper, step by step: balancing chemical equations in Year 2.
If you meet stoichiometry again in the first semester of a degree with limited places: chemistry in the medicine filter semester: what to study.
When the mole meets gases: Gas laws in General Chemistry: worked exercises.
To train on the format: the medicine filter semester practice test and the TOLC-F practice test.

Stoichiometry: typical mistakes and how to avoid them