Stoichiometric Calculations and Reaction Yield

The previous chapter established how to write a correctly balanced chemical equation and read its coefficients as a mole ratio. This chapter puts that mole ratio to work: given a measured mass of one reactant, how much of another reactant is needed, and how much product can be formed? Real laboratory practice immediately raises two further questions that a bare mole ratio cannot answer by itself. First, a reaction is rarely supplied with reactants in the exact mole ratio the equation demands, so one reactant runs out first and halts the reaction while the other is still present in excess — identifying which one is the limiting reagent is essential before any further calculation can proceed. Second, the mass of product actually collected in a real experiment is almost always less than the mass the equation predicts, so chemists distinguish a theoretical yield from an actual yield and report the difference as a percentage yield.

This chapter develops all three ideas — mass-to-mass stoichiometry, limiting reagent calculations, and reaction yield — together with atom economy, a related but distinct measure of how efficiently a reaction's atoms end up in the desired product rather than in waste. Nothing here requires revisiting how an equation is balanced; every worked example below starts from an equation already given in its correctly balanced form.

1. The Mole Ratio as a Conversion Factor

A balanced equation supplies a fixed conversion factor between the amounts, in mol, of any two species appearing in it. For the general equation

the coefficients give the ratio in which moles of any pair of species are related, for instance

Every stoichiometric calculation in this chapter runs the same three-stage path, regardless of which quantities are given or asked for:

The middle step — converting moles of A into moles of C using the mole ratio from the balanced equation — is the one genuinely new idea in this chain; the two outer steps are nothing more than the $n=\dfrac{m}{M}$ relationship from Chapter 1, applied once on each side.

Key distinction. The mole ratio $\dfrac{c}{a}$ converts moles of one substance into moles of another. It is never valid to apply this ratio directly to masses or to numbers of grams — mass must first be converted to moles, and only moles may be scaled by the equation's coefficients.

2. Mass–Mass Stoichiometry

Worked example. Calcium carbonate decomposes on heating to form calcium oxide (quicklime) and carbon dioxide:

Find the mass of calcium oxide produced when $50.0~\text{g}$ of calcium carbonate is fully decomposed.

Step 1 — moles of CaCO$_3$, using $M(\text{CaCO}_3) = 100.09~\text{g}\,\text{mol}^{-1}$ (Chapter 1):

Step 2 — moles of CaO, using the $1:1$ mole ratio read directly from the equation:

Step 3 — mass of CaO, using $M(\text{CaO}) = A_r(\text{Ca}) + A_r(\text{O}) = 40.08 + 16.00 = 56.08~\text{g}\,\text{mol}^{-1}$:

The sample mass was given to three significant figures, so the answer is quoted to three as well: $28.0~\text{g}$ of calcium oxide.

3. Limiting Reagent and Excess Reagent

When a reaction is carried out with specific masses of every reactant, those masses rarely happen to stand in exactly the mole ratio the balanced equation demands. One reactant is consumed completely first, at which point the reaction can proceed no further — this is the limiting reagent. Any reactant left over once the limiting reagent is exhausted is said to be in excess. Confusing "the reactant present in the smaller mass" with "the limiting reagent" is one of the most common stoichiometry errors; the two coincide only by coincidence, because the comparison that matters is between moles divided by coefficient, not between raw masses or even raw moles.

Method. For each reactant, divide the number of moles available by its coefficient in the balanced equation. The reactant with the smallest such value is the limiting reagent, because it is the one that runs out first as the reaction advances.

Worked example. In the Haber process, nitrogen and hydrogen react to form ammonia:

A reaction vessel is charged with $56.04~\text{g}$ of $\text{N}_2$ ($M = 28.02~\text{g}\,\text{mol}^{-1}$) and $6.06~\text{g}$ of $\text{H}_2$ ($M = 2.02~\text{g}\,\text{mol}^{-1}$). Identify the limiting reagent and find the mass of ammonia produced.

Step 1 — convert each mass to moles:

Step 2 — divide each by its coefficient in the equation:

The smaller value belongs to $\text{H}_2$, so hydrogen is the limiting reagent: all $3.00~\text{mol}$ of it will be consumed, while nitrogen remains partly unreacted.

Step 3 — use the limiting reagent to find the moles of product, via the mole ratio $\text{H}_2 : \text{NH}_3 = 3:2$:

Step 4 — convert to mass, using $M(\text{NH}_3) = A_r(\text{N}) + 3A_r(\text{H}) = 14.01 + 3(1.01) = 17.04~\text{g}\,\text{mol}^{-1}$:

Step 5 — find the mass of the excess reagent remaining, by first finding how much $\text{N}_2$ the limiting reagent actually consumed, via the ratio $\text{H}_2:\text{N}_2 = 3:1$:

Warning. Once the limiting reagent is identified, every further mole-ratio calculation in the problem must start from the limiting reagent, never from the reactant in excess — using the excess reagent would predict more product than the reaction can actually make, since the limiting reagent runs out first.

4. Theoretical Yield, Actual Yield and Percentage Yield

The mass of product calculated from the limiting reagent by the method above — $34.1~\text{g}$ of $\text{NH}_3$ in the example just completed — is the theoretical yield: the maximum mass of product the stoichiometry permits, assuming the reaction goes to completion with no losses whatsoever. The mass of product actually collected and weighed at the end of a real experiment is the actual yield, and it is essentially always smaller. The percentage yield compares the two:

Worked example. In the Haber-process reaction above, the theoretical yield of ammonia was $34.08~\text{g}$ (keeping an extra figure before final rounding). A real synthesis under these conditions produces only $29.8~\text{g}$ of ammonia once collected and purified. Find the percentage yield.

A percentage yield is always between $0\%$ and $100\%$; a calculated value above $100\%$ signals an error somewhere — most often an impure or wet product, which has been weighed as if it were pure, or an arithmetic slip in the theoretical-yield calculation, never a reaction that has somehow produced more product than stoichiometry allows.

5. Why Actual Yield Is Less Than Theoretical Yield

Several distinct, genuine physical and chemical causes — not merely "experimental error" — explain why $\%\,\text{yield} < 100\%$ in practice:

• The reaction does not go to completion. Many reactions are reversible and reach a dynamic equilibrium (developed in a later block) well before all of the limiting reagent is consumed, so some of it remains unreacted.
• Side reactions compete with the main reaction, consuming some of the reactant(s) to form by-products other than the intended one, so less limiting reagent remains available for the desired pathway.
• Mechanical losses during practical steps — transferring a solution between vessels, filtering a precipitate, recrystallizing a solid, evaporating a solvent — leave a small residue of product behind at every stage, none of which reaches the final weighed sample.
• Impure starting materials mean the actual amount of true reactant present is smaller than the mass weighed out suggests, so less product than calculated can ever form.

6. Atom Economy

Percentage yield measures how efficiently a given reaction was carried out in practice. Atom economy measures something different and purely theoretical: of all the atoms present in the reactants, what percentage end up in the desired product, according to the stoichiometry of the equation alone, assuming a perfect $100\%$ yield?

where each molar mass is multiplied by its coefficient in the balanced equation (e.g. $2\times M_r(\text{C}_2\text{H}_5\text{OH})$ in the fermentation example below).

A reaction with several products necessarily has an atom economy below $100\%$, since mass is conserved overall (Chapter 1) but only part of it leaves in the wanted product; the rest leaves as by-products, which are frequently waste. A reaction that forms only a single product automatically has an atom economy of exactly $100\%$, since every atom supplied by the reactants is necessarily incorporated into that one product.

Key distinction. A reaction can have a high percentage yield yet a low atom economy (a clean, efficient reaction that nonetheless produces a large mass of unwanted by-product), or a low percentage yield yet a high atom economy (an intrinsically efficient reaction performed poorly in practice). The two quantities are independent and answer different questions — practical efficiency versus theoretical, stoichiometric efficiency.

Worked example — comparing two routes to ethanol. Ethanol can be manufactured by the catalytic hydration of ethene,

or obtained from the fermentation of glucose,

Hydration route: using $M(\text{C}_2\text{H}_4) = 28.06$, $M(\text{H}_2\text{O}) = 18.02$ and $M(\text{C}_2\text{H}_5\text{OH}) = 46.08~\text{g}\,\text{mol}^{-1}$, and noting this reaction has only one product:

Fermentation route: using $M(\text{C}_6\text{H}_{12}\text{O}_6) = 180.18~\text{g}\,\text{mol}^{-1}$, with $2~\text{mol}$ of the desired product $\text{C}_2\text{H}_5\text{OH}$ formed per mole of glucose:

As a check, the mass of the co-product $\text{CO}_2$ accounts for exactly the remaining mass: $2\times 44.01 = 88.02~\text{g}$ per mole of glucose, and $92.16 + 88.02 = 180.18~\text{g}$, matching the glucose consumed, as conservation of mass requires. The hydration of ethene is an addition reaction with no by-product and therefore a perfect atom economy, while fermentation necessarily discards a large fraction of the glucose's mass as carbon dioxide — a central reason atom economy matters to green chemistry, which seeks reaction routes that minimise waste at the design stage, not merely maximise yield of a reaction already chosen.

7. Common Exam Pitfalls

• Scaling masses directly by the equation's coefficients, skipping the conversion to moles. The mole ratio from a balanced equation applies only to amounts in mol; masses must never be substituted into it directly, since molar masses differ between substances.
• Identifying the limiting reagent by comparing raw masses or raw moles, rather than moles divided by each reactant's coefficient. A reactant present in a larger mass, or even a larger number of moles, can still be the limiting reagent if the equation demands proportionally more of it.
• Continuing a calculation from the excess reagent after the limiting reagent has already been identified. Every subsequent mole-ratio step must be anchored to the limiting reagent, since it is what actually runs out and halts the reaction.
• Reporting a percentage yield above $100\%$ without recognising this as a sign of error (commonly an impure or still-wet product weighed as if pure), rather than evidence of an unusually efficient reaction.
• Treating percentage yield and atom economy as the same quantity. Percentage yield compares an actual experimental result to a theoretical prediction for one specific run; atom economy is a fixed property of the chosen reaction's stoichiometry and does not depend on how carefully the experiment was performed.
• Forgetting that atom economy only equals $100\%$ when there is exactly one product. Any reaction producing two or more products necessarily has an atom economy below $100\%$, however efficiently it is carried out.

Summary

• Every mass-to-mass stoichiometry calculation follows $m(\text{A}) \to n(\text{A}) \to n(\text{C}) \to m(\text{C})$, converting with $n = m/M$ on each side and with the mole ratio $c/a$ from the balanced equation in the middle.
• The limiting reagent is found by dividing the moles of each reactant by its coefficient in the balanced equation; the smallest resulting value identifies the reactant that runs out first and must be used for every further calculation in the problem.
• The excess reagent is any reactant left over once the limiting reagent is exhausted; the amount remaining is found by subtracting the amount actually consumed (via the mole ratio with the limiting reagent) from the amount originally supplied.
• Theoretical yield is the maximum product mass predicted from the limiting reagent assuming complete reaction; actual yield is the mass actually collected; $\%\,\text{yield} = \dfrac{\text{actual}}{\text{theoretical}}\times 100\%$ and can never legitimately exceed $100\%$.
• Real yields fall short of $100\%$ because of incomplete (reversible) reactions, competing side reactions, mechanical losses during purification, and impure starting materials.
• Atom economy, $\%\,\text{atom economy} = \dfrac{M_r(\text{desired product})}{\sum M_r(\text{reactants})}\times 100\%$, is a purely stoichiometric measure of efficiency, independent of percentage yield, and equals $100\%$ only for a reaction with a single product.
• The next chapter extends mole-ratio reasoning to reactions involving gaseous reactants and products, where volume, rather than mass, is usually the quantity measured directly.