Stoichiometric calculations

A recipe card tells you exactly how much flour, sugar and butter combine to bake a cake — and a balanced chemical equation does the same job for a reaction. Its coefficients are a mole recipe: they say precisely how many moles of each reactant combine and how many moles of each product form. Stoichiometry is the arithmetic that turns this recipe into real quantities — grams, litres of solution, numbers of moles — that a chemist can actually weigh out or measure on a burette. This topic is the culminating quantitative skill of the ISC Class 11 chapter "Some Basic Concepts of Chemistry." It pulls together molar mass and the mole concept into one method: convert what you are given into moles, apply the mole ratio from the balanced equation, then convert back into the unit the question asks for. When two reactants are supplied and their amounts do not exactly match the recipe, one of them runs out first — the limiting reagent — and it alone decides how much product can form.

The core insight

The coefficients of a balanced equation fix a mole ratio between reactants and products, and every mass-to-mass, mass-to-volume or limiting-reagent calculation is just that ratio applied after converting the given quantity to moles and before converting the answer back.

The core idea

A balanced equation is a mole ratio, not a mass ratio. For a general reaction

aA + bB \rightarrow cC + dD

the coefficients $a, b, c, d$ give the relative number of moles reacting or forming — never the relative masses, since different substances have different molar masses. Any two species in the equation are locked together by the ratio of their coefficients:

\frac{n_A}{a} = \frac{n_B}{b} = \frac{n_C}{c} = \frac{n_D}{d}

where: $n_A, n_B, n_C, n_D$ are the moles of each species actually reacting or forming, and $a, b, c, d$ are the balancing coefficients read straight off the equation.

The four-step calculation method. Every stoichiometric problem, however it is dressed up, reduces to the same chain: (1) write and balance the equation, (2) convert the given quantity of the known substance into moles — using $n = m/M$ for a mass or $n = MV$ for a solution of known molarity, (3) apply the mole ratio from the balanced equation to get moles of the desired substance, (4) convert those moles back into the unit the question wants.

Limiting reagent. When two reactants are supplied in amounts that are not in the exact ratio the equation demands, one is used up first — the limiting reagent — while some of the other, the excess reagent, is left unreacted. The amount of product actually formed is always fixed by the limiting reagent alone, never by the reactant supplied in excess. To find it, work out how much product each reactant could give if it reacted completely by itself; whichever gives the smaller amount of product is limiting.

Figure

A balance scale with reactant moles on the left pan converted through a ratio arrow into product moles on the right pan

Derivation ATurning a balanced equation into a mass of product

The goal is to show, with nothing beyond the mole concept already covered in this chapter, exactly how a coefficient ratio between two species in a balanced equation becomes an actual mass of product you can weigh, starting from a given mass of a single reactant and passing through moles at every stage rather than jumping straight between masses.

Step 1. Start from the balanced equation $\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2$ . Reading the coefficients (all $1$ ), the mole ratio of calcium carbonate consumed to carbon dioxide formed is fixed at $1:1$ , whatever mass of $\text{CaCO}_3$ is actually taken.

Step 2. Convert the given mass of $\text{CaCO}_3$ into moles using $n = m/M$ , where $M = 100\ \text{g mol}^{-1}$ is its molar mass. This step is unavoidable because the coefficients relate moles, not grams — skipping it and dividing masses directly would silently assume both substances have the same molar mass, which they never do.

Step 3. Apply the mole ratio read off in Step 1 to convert moles of $\text{CaCO}_3$ into moles of $\text{CO}_2$ : since the ratio is $1:1$ , the moles of $\text{CO}_2$ formed equal the moles of $\text{CaCO}_3$ consumed exactly. For a ratio other than $1:1$ , this step would multiply by $c/a$ , the ratio of the product's coefficient to the reactant's coefficient.

Step 4. Convert the moles of $\text{CO}_2$ back into a mass using $m = n \times M$ with $M = 44\ \text{g mol}^{-1}$ for $\text{CO}_2$ . This final conversion is what makes the number weighable — moles are a counting unit, useful for the ratio arithmetic, but the chemist ultimately wants grams. The full numerical run of this chain — starting from $5.0\ \text{g}$ of $\text{CaCO}_3$ and ending at $2.2\ \text{g}$ of $\text{CO}_2$ — is carried out in the worked example below, reusing the same four steps in the same order.

Worked example

$5.6\ \text{g}$ of iron reacts with $80\ \text{mL}$ of $1.0\ \text{mol L}^{-1}$ hydrochloric acid solution according to

\text{Fe} + 2\text{HCl} \rightarrow \text{FeCl}_2 + \text{H}_2

(molar mass of $\text{Fe} = 56\ \text{g mol}^{-1}$ ). Identify the limiting reagent and calculate the mass of $\text{FeCl}_2$ (molar mass $= 127\ \text{g mol}^{-1}$ ) formed, and how much of the excess reagent is left over.

Step 1. Convert both given quantities to moles. $n_{\text{Fe}} = 5.6/56 = 0.100\ \text{mol}$ . For the acid solution, $n_{\text{HCl}} = M \times V = 1.0\ \text{mol L}^{-1} \times 0.080\ \text{L} = 0.080\ \text{mol}$ .

Step 2. Compare against the required mole ratio. The equation demands $\text{Fe} : \text{HCl} = 1:2$ , so the $0.100\ \text{mol}$ Fe available would need $2 \times 0.100 = 0.200\ \text{mol}$ HCl to react completely — but only $0.080\ \text{mol}$ HCl is present, far short of that.

Step 3. Confirm the limiting reagent from the other side. The $0.080\ \text{mol}$ HCl available can consume only $0.080/2 = 0.040\ \text{mol}$ Fe. Since $0.040\ \text{mol}$ is less than the $0.100\ \text{mol}$ Fe on hand, HCl runs out first — HCl is the limiting reagent, and Fe is in excess.

Step 4. Use only the limiting reagent's stoichiometry to find product. From $\text{HCl} : \text{FeCl}_2 = 2:1$ , $n_{\text{FeCl}_2} = n_{\text{HCl}}/2 = 0.080/2 = 0.040\ \text{mol}$ , so $m_{\text{FeCl}_2} = n \times M = 0.040 \times 127 = 5.08\ \text{g}$ .

Step 5. Find the leftover excess reagent. Fe actually consumed $= n_{\text{HCl}}/2 = 0.040\ \text{mol}$ (matching Step 3), so Fe remaining $= 0.100 - 0.040 = 0.060\ \text{mol}$ , i.e. $0.060 \times 56 = 3.36\ \text{g}$ of unreacted iron.

Limiting reagent = HCl; $m_{\text{FeCl}_2} \approx 5.1\ \text{g}$ , with $3.36\ \text{g}$ of Fe left unreacted.

Common traps

Comparing the raw masses of two reactants to spot the limiting reagent instead of converting both to moles first and checking against the balanced equation's mole ratio.
Forgetting to balance the equation before reading off coefficients — an unbalanced equation gives the wrong mole ratio and every downstream number is wrong.
Basing the product calculation on whichever reactant happens to be listed or given first, rather than actually checking which one is limiting.
Confusing "moles supplied" of the excess reagent with "moles that actually reacted" — only the limiting reagent's full amount reacts; the excess reagent's reacted amount is fixed by the limiting reagent's stoichiometry, not by how much of it was originally there.

Ask the tutor: want to work through another limiting-reagent problem step by step, or check your own mass-to-mass or mass-to-volume stoichiometry calculation with guided help?

Why can't I just scale the masses in the equation directly, the way I would scale a recipe by weight?

Because the coefficients in a balanced equation count particles (moles), not grams. A recipe scaled by weight only works if every ingredient has the same "weight per unit," which is never true for different chemical substances — one mole of $\text{CaCO}_3$ weighs $100\ \text{g}$ while one mole of $\text{CO}_2$ weighs only $44\ \text{g}$ , even though they react in a $1:1$ mole ratio. Converting to moles first removes this mismatch, since moles count identical "recipe units" regardless of what the substance is.

How do I know which reactant to pick first when checking for a limiting reagent?

It does not matter which one you start with, as long as you check consistently. Convert both given quantities to moles, then either (a) ask how much of the second reactant the first would need to react completely and compare that to what is actually available, or (b) calculate the moles of product each reactant could form alone and take the smaller value. Both methods must agree — use whichever feels more natural, but always finish by stating which reactant is limiting and why, in one sentence.

If a reactant is in excess, does none of it react at all?

No — the excess reagent still reacts, just not completely. Whatever amount is needed to consume the limiting reagent according to the mole ratio does react; only the leftover portion, beyond what the limiting reagent could combine with, remains unreacted. In the worked example, all $0.080\ \text{mol}$ of the limiting HCl reacts, consuming exactly $0.040\ \text{mol}$ of the $0.100\ \text{mol}$ Fe supplied — so Fe does react, just partially, and $0.060\ \text{mol}$ of it is left over.

Stoichiometric calculations

The core idea

Worked example

Common traps

Before this

Where this leads