Stoichiometry: Solving Mole Problems (A 4-Step Method)

Solving a stoichiometry problem takes four steps: balance the chemical equation, convert the known data into moles, apply the ratios between the coefficients of the balanced equation to find the moles of the substance you are after, then convert the result back into grams, litres or a number of particles. Balancing first is the only way to get the ratios right.

On our chemistry courses there is a moment I recognise straight away: the student has a problem with reactants and products in front of them, more or less knows what moles are, but gets stuck on "and now what do I do with these grams?". Almost always it isn't a problem of theory — it's that a fixed method is missing, always the same, to be applied to every exercise without reinventing it each time.

Stoichiometry is exactly that: the accounting of a chemical reaction. It tells you how much product you get from a given amount of reactant, or how much reactant you need to obtain what you want. And the currency of that accounting isn't grams, it's moles. In this article I give you the four-step method I teach in lessons, then apply it to three worked examples with real numbers and units — including the exact mistakes I see students fall into.

Why does everything revolve around the mole?

The mole is the unit that connects the world you weigh in the laboratory (grams) to the world of the particles that react with one another (atoms and molecules). A mole always contains the same number of particles — Avogadro's number, $N_A = 6.022 \times 10^{23}$ — whatever the substance. That is what makes it possible to "count" molecules by weighing them.

The coefficients of a balanced equation tell you nothing about grams: they tell you in what ratio the moles react. In the synthesis of ammonia

one mole of nitrogen reacts with three moles of hydrogen to give two moles of ammonia — not "one gram with three grams". That is why you can't work directly on the masses: you have to go through moles first. The three conversions you will always use are:

where $n$ is the number of moles, $m$ the mass in grams, $M$ the molar mass in g/mol and $N$ the number of particles.

The method in 4 steps

Every stoichiometry problem, from the simplest to the most tangled, is solved with this same sequence. Don't skip any of them, especially the first.

1. Balance the equation. Write out reactants and products and adjust the coefficients until every element has the same number of atoms on the left and on the right. Without this step, the ratios in step 3 are wrong and everything else collapses.
2. Convert the data into moles. Take the starting figure (a mass, a gas volume, a number of particles) and turn it into moles with the right relation. In most cases that is $n = \dfrac{m}{M}$.
3. Use the mole ratio. Look at the coefficients of the balanced equation and build the ratio between the substance you know and the one you want. Multiply the known moles by that ratio: you get the moles of the substance you are after.
4. Convert back into the unit required. The moles you have found become grams ($m = n \cdot M$), litres or a number of particles, depending on what the problem asks for.

Keep it in mind as a funnel: mass → moles → (ratio) → moles → mass. The moles are the bridge in the middle. Now let's see it work.

Worked example 1 — From grams to grams: how much oxide forms?

Aluminium burns in air to form aluminium oxide. How many grams of $Al_2O_3$ do you obtain from $5.40$ g of aluminium, with oxygen in excess?

Step 1 — Balance. The raw equation $Al + O_2 \rightarrow Al_2O_3$ is not balanced. Adjusting the coefficients:

Check: 4 Al atoms on the left and on the right, 6 O atoms on the left ($3 \times 2$) and on the right ($2 \times 3$). Good.

Step 2 — From grams to moles. The molar mass of aluminium is $M = 26.98$ g/mol:

Step 3 — Mole ratio. From the coefficients, the $Al : Al_2O_3$ ratio is $4 : 2$, that is $2 : 1$. Two moles of aluminium are needed for every mole of oxide:

Step 4 — From moles to grams. The molar mass of $Al_2O_3$ is $M = 2 \times 26.98 + 3 \times 16.00 = 101.96$ g/mol:

The mistake I see most often here: using a $1 : 1$ ratio because "there's one $Al$ and one $Al_2O_3$", ignoring the coefficients $4$ and $2$. Anyone who does that writes $0.200$ mol of oxide and gets $20.4$ g — twice the correct value. The coefficients of the balanced equation are the ratio: you don't look at the subscripts inside the formulae, you look at the numbers in front of them.

Worked example 2 — Limiting reactant: which one runs out first?

So far one reactant was in excess and it was enough to follow the other. But when the problem gives you the amounts of both reactants, you first have to work out which one runs out first: that is the one that decides how much product you get, the limiting reactant.

We react $14$ g of nitrogen ($N_2$) with $6.0$ g of hydrogen ($H_2$) to form ammonia. How many grams of $NH_3$ are formed?

Step 1 — Balance.

Step 2 — Moles of both reactants. With $M_{N_2} = 28.02$ g/mol and $M_{H_2} = 2.016$ g/mol:

Step 3 — Find the limiting reactant. You don't compare the grams, you compare the moles against the coefficients. To consume all $0.500$ mol of nitrogen you would need

and I have $2.98$ mol of it: the hydrogen is in excess, so nitrogen is the limiting reactant. From here on I follow the nitrogen only. Ratio $N_2 : NH_3 = 1 : 2$:

Step 4 — From moles to grams. With $M_{NH_3} = 17.03$ g/mol:

Here the mistake is almost automatic: look at the masses and conclude "hydrogen is only $6.0$ g against $14$ g of nitrogen, so hydrogen is the limiting reactant". Wrong. In moles the hydrogen is nearly six times the nitrogen, and with the $3:1$ ratio there is still some of it left over. The limiting reactant is always decided in moles, never in grams. Anyone who skips this check and works on the wrong reactant gets a result higher than the real one.

Worked example 3 — Percentage yield: how much do you really get?

In the laboratory you never recover all the product the theory promises: some is lost, some doesn't react. The percentage yield compares what you obtain with the theoretical maximum.

We precipitate silver chloride by adding sodium chloride in excess to $17.0$ g of silver nitrate ($AgNO_3$). We recover $12.9$ g of $AgCl$. What is the yield?

Step 1 — Balance. The equation is already balanced ($1:1:1:1$), but checking is part of the method:

Step 2 — Moles of the known reactant. With $M_{AgNO_3} = 169.88$ g/mol:

The sodium chloride is in excess, so the silver nitrate is the limiting reactant and sets the theoretical yield.

Step 3 — Mole ratio. The $AgNO_3 : AgCl$ ratio is $1 : 1$, so $n_{AgCl} = 0.100$ mol.

Step 4 — Theoretical yield and comparison. With $M_{AgCl} = 107.87 + 35.45 = 143.32$ g/mol:

A detail that is worth marks in an exam: the actual yield ($12.9$ g) is an experimental figure, and it is always compared against the theoretical one, never the other way round. And if the problem asked how many units of $AgCl$ you obtained, one last step would do it: $0.0900$ mol actually obtained $\times \, 6.022 \times 10^{23} = 5.42 \times 10^{22}$ formula units. Always the same bridge: the moles.

The mistakes that cost the most marks

Here are the traps seen in the worked examples, plus two that keep coming back in class tests:

• Not balancing (or balancing badly) before anything else. It is mistake number one: it makes every ratio that follows wrong. Balance and re-check the atoms before touching the grams.
• Confusing subscripts and coefficients. The mole ratio is the numbers in front of the formulae, not the subscripts inside them.
• Deciding the limiting reactant from the grams. Always in moles, compared against the coefficients.
• Getting the molar mass wrong. The classic slip is forgetting that oxygen, nitrogen, hydrogen and chlorine are diatomic: $M_{O_2} = 32.00$ g/mol, not $16.00$. And add up all the atoms in the formula, subscripts included.
• Ignoring significant figures. If the data have three significant figures, the result has three: writing $10.19587$ g only shows that you copied off the calculator without thinking about the uncertainty.

This is the same method you need in entrance tests: in the Chemistry exams of the Medicine filter semester stoichiometry problems have to be closed in a couple of minutes, and without a fixed method the clock eats you alive.

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