🎯What you need to be able to do
- Deduce and balance chemical equations when reactants and products are specified, including state symbols.
- Use the mole ratio of an equation to calculate reacting masses, volumes and concentrations of reactants and products.
- Identify the limiting and excess reactants from given data, and use the limiting reactant to determine the theoretical yield.
- Distinguish between theoretical and experimental yield, and calculate percentage yield.
- Calculate atom economy from the stoichiometry of a reaction, and discuss its relationship to waste in industrial processes.
📚The chemistry
Equations and the mole ratio
A balanced equation is a statement about ratios of amounts. In
\[ \mathrm{N_2}(g) + 3\mathrm{H_2}(g) \rightarrow 2\mathrm{NH_3}(g) \]
the coefficients say that one mole of nitrogen reacts with three moles of hydrogen to give two
moles of ammonia. They say nothing about masses — that is why every stoichiometry calculation has
the same three-step shape:
- convert what you are given into moles
(\( n = m/M \), \( n = cV \), or \( n = V/22.7 \) at STP);
- use the mole ratio from the balanced equation to find the moles of what you
want;
- convert that amount back into the quantity asked for.
Include state symbols: (s), (l), (g), (aq). They are not decoration — they
decide whether a species appears in an equilibrium expression
(R2.3) and are frequently worth a mark of their
own.
Limiting and excess reactants
Reactants are rarely mixed in exactly the ratio the equation requires. The limiting
reactant is the one that runs out first; it determines the theoretical yield,
and everything else is in excess.
You cannot identify the limiting reactant by comparing masses, or even by comparing
moles directly. You must compare each amount against what the equation requires. The
reliable method:
\[ \text{for each reactant, compute } \frac{n}{\text{coefficient}} \ ; \ \text{the smallest value is limiting} \]
Then base every subsequent calculation on the limiting reactant. Deliberately using an
excess of the cheaper reagent is standard industrial practice, because it drives the equilibrium
towards products and makes sure the expensive reagent is fully consumed.
Theoretical, experimental and percentage yield
Theoretical yield
the maximum amount obtainable, calculated from the limiting reactant assuming the reaction goes to completion and nothing is lost
Experimental yield
the amount actually isolated and weighed in the laboratory
\[ \text{percentage yield} = \frac{\text{experimental yield}}{\text{theoretical yield}} \times 100 \]
Percentage yield is essentially always below 100%, and the reasons are examinable:
- the reaction may be reversible and reach equilibrium rather than
completion;
- side reactions consume reactant to give something other than the desired
product;
- losses during purification — product left on the filter paper, in the
reaction vessel, or dissolved in the mother liquor after recrystallization;
- the reactants may be impure.
A yield above 100% is not a triumph but a sign of error: the product is almost certainly still wet,
or contains an impurity.
Atom economy and green chemistry
Atom economy asks a different question from yield: not how much of the possible
product did I get, but what proportion of the atoms in my reactants ended up in the product I
wanted.
\[ \text{atom economy} = \frac{M_{\mathrm{r}}\,(\text{desired product})}{\Sigma\, M_{\mathrm{r}}\,(\text{reactants})} \times 100 \]
The equation is in the data booklet. Two points about using it:
- Use the stoichiometric coefficients — if the equation makes two moles of
the product, the numerator is \( 2 M_{\mathrm{r}} \).
- Atom economy is calculated from the balanced equation alone. It does not depend
on how well the reaction actually went, which is why a reaction can have 100% atom economy and a 40%
yield, or vice versa.
The relationship with waste is inverse: whatever fraction of the reactant mass does
not appear in the desired product becomes a by-product that must be separated, treated or disposed of.
A low atom economy therefore means a process is intrinsically wasteful even if it works
perfectly, which is why green chemistry treats atom economy as a design criterion
rather than an afterthought. Addition reactions and addition polymerization have
100% atom economy, because the product contains every atom of the reactants;
substitution, elimination and condensation reactions necessarily have less.
Confusing yield with atom economy. They answer different questions and can point in
opposite directions. Yield is about the efficiency of the process —
did the reaction go, and did you keep what it made? Atom economy is about the
design of the reaction — are the reactants intrinsically capable of producing
mostly the thing you want? A reaction with 100% atom economy performed badly gives a low yield; a
well-run reaction with 40% atom economy still throws away more than half of the atoms it started
with. A good industrial process needs both.
✏️Worked example
Iron is extracted from iron(III) oxide in a blast furnace:
\( \mathrm{Fe_2O_3}(s) + 3\mathrm{CO}(g) \rightarrow 2\mathrm{Fe}(l) + 3\mathrm{CO_2}(g) \).
80.0 g of \( \mathrm{Fe_2O_3} \) is heated with 42.0 g of carbon monoxide.
(a) Determine the limiting reactant.
(b) Calculate the theoretical yield of iron.
(c) 48.0 g of iron is obtained. Calculate the percentage yield and give two reasons why it is
below 100%.
(d) Calculate the atom economy of the process with respect to iron.
\( M_{\mathrm{r}} \): \( \mathrm{Fe_2O_3} \) 159.70, CO 28.01, Fe 55.85,
\( \mathrm{CO_2} \) 44.01.
(a) Convert both to moles, then divide by the coefficients.
\[ n(\mathrm{Fe_2O_3}) = \frac{80.0}{159.70} = 0.501\ \mathrm{mol} \qquad n(\mathrm{CO}) = \frac{42.0}{28.01} = 1.50\ \mathrm{mol} \]
\[ \frac{n(\mathrm{Fe_2O_3})}{1} = 0.501 \qquad\qquad \frac{n(\mathrm{CO})}{3} = \frac{1.50}{3} = 0.500 \]
The smaller value belongs to carbon monoxide, so CO is the
limiting reactant — only just, but decisively.
(b) Work from the limiting reactant. The ratio CO : Fe is 3 : 2, so
\[ n(\mathrm{Fe}) = 1.50 \times \frac{2}{3} = 1.00\ \mathrm{mol} \]
\[ m(\mathrm{Fe}) = 1.00 \times 55.85 = 55.9\ \mathrm{g} \]
(c)
\[ \text{percentage yield} = \frac{48.0}{55.9} \times 100 = 85.9\% \]
Reasons (any two): some product is lost during separation and purification, for
instance iron retained in the slag or on the walls of the furnace; the reaction may not
go to completion, since it is reversible and reaches an equilibrium; there may be
side reactions, such as the formation of iron carbide; and the
reactants may be impure, so less \( \mathrm{Fe_2O_3} \) is present than the 80.0 g
suggests.
(d) The desired product is iron, and the equation produces
two moles of it:
\[ \text{atom economy} = \frac{2(55.85)}{159.70 + 3(28.01)} \times 100 = \frac{111.7}{243.73} \times 100 = 45.8\% \]
Fewer than half the atoms fed into the furnace end up in the iron; the rest leave as carbon
dioxide. That is intrinsic to the reaction and cannot be improved by running it better — only by
choosing a different reaction.
Check it. Three checks. Limiting reactant: confirm by working the other
way — 0.501 mol of \( \mathrm{Fe_2O_3} \) would need \( 3 \times 0.501 = 1.50 \) mol of CO,
and there is 1.50 mol, so they are almost exactly matched with CO very slightly short. That
agreement is a good sign the moles are right. Percentage yield: must be below 100%; if it
is above, you have probably used the excess reactant to find the theoretical yield.
Atom economy: must also be below 100% whenever there is a by-product, and the two
percentages should be recognisably different numbers — if yield and atom economy come out the
same, check you have not calculated the same thing twice.
Deciding the limiting reactant by comparing moles without dividing by the
coefficients. Here there is 1.50 mol of CO against 0.501 mol of \( \mathrm{Fe_2O_3} \), so
the oxide “obviously” looks limiting — and it is not, because three moles of CO
are needed for every one of the oxide. Always divide each amount by its coefficient before
comparing. The second trap: after identifying the limiting reactant, students sometimes revert to
the other one for the yield calculation. Every subsequent step uses the limiting reactant, and
nothing else.
📝Practise
Work through these on paper, then reveal the answer.
1. Balance and add state symbols: (a) the reaction of aqueous sodium carbonate with dilute hydrochloric acid, (b) the combustion of ethanol, (c) the thermal decomposition of solid calcium carbonate.
(a) \( \mathbf{Na_2CO_3(aq) + 2HCl(aq) \rightarrow 2NaCl(aq) + H_2O(l) + CO_2(g)} \). The carbon dioxide is a gas, which is why the reaction effervesces, and this is the standard test for a carbonate. (b) \( \mathbf{C_2H_5OH(l) + 3O_2(g) \rightarrow 2CO_2(g) + 3H_2O(l)} \) — the water is liquid under standard conditions; write (g) only if the question specifies a temperature above 100 °C. (c) \( \mathbf{CaCO_3(s) \rightarrow CaO(s) + CO_2(g)} \). State symbols matter here for a reason beyond marks: only the \( \mathrm{CO_2}(g) \) appears in the equilibrium expression for this reaction, because the two solids have constant concentration.
2. 25.0 cm3 of 0.100 mol dm−3 NaOH is titrated against sulfuric acid, requiring 20.0 cm3. Calculate the concentration of the acid.
Equation: \( \mathrm{H_2SO_4}(aq) + 2\mathrm{NaOH}(aq) \rightarrow \mathrm{Na_2SO_4}(aq) + 2\mathrm{H_2O}(l) \) — note the 1 : 2 ratio, since sulfuric acid is diprotic. Step 1: \( n(\mathrm{NaOH}) = cV = 0.100 \times 0.0250 = 2.50 \times 10^{-3}\ \mathrm{mol} \). Step 2: the mole ratio \( \mathrm{H_2SO_4} : \mathrm{NaOH} \) is 1 : 2, so \( n(\mathrm{H_2SO_4}) = \dfrac{2.50 \times 10^{-3}}{2} = 1.25 \times 10^{-3}\ \mathrm{mol} \). Step 3: \( c = \dfrac{n}{V} = \dfrac{1.25 \times 10^{-3}}{0.0200} = \mathbf{0.0625\ mol\,dm^{-3}} \). The commonest error is omitting the factor of two, which doubles the answer to 0.125 — so always write the balanced equation before touching the numbers.
3. 5.00 g of magnesium is added to 100.0 cm3 of 2.00 mol dm−3 hydrochloric acid. Determine the limiting reactant and the volume of hydrogen produced at STP. (\( A_{\mathrm{r}}(\mathrm{Mg}) = 24.31 \); molar volume 22.7 dm3 mol−1.)
Equation: \( \mathrm{Mg}(s) + 2\mathrm{HCl}(aq) \rightarrow \mathrm{MgCl_2}(aq) + \mathrm{H_2}(g) \). Amounts: \( n(\mathrm{Mg}) = \dfrac{5.00}{24.31} = 0.206\ \mathrm{mol} \); \( n(\mathrm{HCl}) = 2.00 \times 0.1000 = 0.200\ \mathrm{mol} \). Divide by coefficients: \( \dfrac{0.206}{1} = 0.206 \) for Mg; \( \dfrac{0.200}{2} = 0.100 \) for HCl. The smaller belongs to the acid, so HCl is limiting and magnesium is in excess — some metal will remain undissolved. From the limiting reactant, the ratio HCl : H2 is 2 : 1, so \( n(\mathrm{H_2}) = \dfrac{0.200}{2} = 0.100\ \mathrm{mol} \). Volume at STP: \( V = 0.100 \times 22.7 = \mathbf{2.27\ dm^3} \). The visual confirmation in the laboratory is that effervescence stops while unreacted magnesium is still visible.
4. Aspirin is made by \( \mathrm{C_7H_6O_3} + \mathrm{C_4H_6O_3} \rightarrow \mathrm{C_9H_8O_4} + \mathrm{CH_3COOH} \). Starting from 10.0 g of salicylic acid (\( M = 138.13 \)) with ethanoic anhydride in excess, 11.2 g of aspirin (\( M = 180.17 \)) is obtained. Calculate the percentage yield and the atom economy. (\( M(\mathrm{C_4H_6O_3}) = 102.10 \).)
Percentage yield. Salicylic acid is limiting, since the anhydride is in excess: \( n = \dfrac{10.0}{138.13} = 0.0724\ \mathrm{mol} \). The ratio is 1 : 1, so the theoretical amount of aspirin is 0.0724 mol, i.e. \( 0.0724 \times 180.17 = 13.0\ \mathrm{g} \). Percentage yield \( = \dfrac{11.2}{13.0} \times 100 = \mathbf{85.9\%} \). Atom economy. \( \dfrac{180.17}{138.13 + 102.10} \times 100 = \dfrac{180.17}{240.23} \times 100 = \mathbf{75.0\%} \). Note that the two numbers describe different things. The 85.9% says the preparation was carried out reasonably well; the 75.0% says that even a perfect preparation would convert only three quarters of the reactant mass into aspirin, because a molecule of ethanoic acid is produced as an unavoidable by-product. Reducing the first requires better technique; reducing the second would require a different synthetic route.
5. Compare these two routes to 1,2-dibromoethane on atom economy: (i) \( \mathrm{C_2H_4} + \mathrm{Br_2} \rightarrow \mathrm{C_2H_4Br_2} \), (ii) \( \mathrm{C_2H_6} + 2\mathrm{Br_2} \rightarrow \mathrm{C_2H_4Br_2} + 2\mathrm{HBr} \). (\( M \): C2H4 28.05, C2H6 30.07, Br2 159.80, C2H4Br2 187.86, HBr 80.91.)
Route (i), addition: \( \dfrac{187.86}{28.05 + 159.80} \times 100 = \dfrac{187.86}{187.85} \times 100 = \mathbf{100\%} \). Every atom of both reactants appears in the product, because nothing else is formed — which is true of all addition reactions. Route (ii), substitution: \( \dfrac{187.86}{30.07 + 2(159.80)} \times 100 = \dfrac{187.86}{349.67} \times 100 = \mathbf{53.7\%} \). Almost half the reactant mass leaves as hydrogen bromide, which must then be separated, neutralised or recovered. Route (i) is therefore far preferable on green chemistry grounds: it generates no by-product at all, so there is nothing to dispose of and no separation step. This comparison is the general reason why the chemical industry prefers addition to substitution wherever a route exists — and it is worth remembering that atom economy says nothing about how fast, how safe or how cheap either route is, so it is one criterion among several.
6. A student obtains a percentage yield of 104% for a recrystallized solid product. Explain why this is impossible, give the most likely cause, and state how to correct it.
A yield above 100% would mean more product was formed than the limiting reactant could possibly supply, which violates the conservation of mass: the atoms in the product must all have come from the reactants, and the calculation already assumes complete conversion with no losses. So a figure above 100% indicates an error in the measurement, not in the chemistry. The most likely cause after recrystallization is that the product is not completely dry — residual solvent adds to the mass without adding product, and even a small amount of retained water or ethanol is enough to push a yield past 100%. Other possible causes: the product is contaminated with an impurity, such as unreacted starting material that crystallised out with it, or the mass of the filter paper or watch glass has not been correctly subtracted. Correction: dry the solid to constant mass — weigh it, dry it further in a desiccator or a low oven, reweigh, and repeat until two successive masses agree. Purity should then be confirmed by taking a melting point: a sharp melting point over a narrow range indicates a pure product, while a depressed and broadened range confirms contamination.
🔗Go deeper — other people’s work
These are external resources, not mine. If one stops working, tell me and
everything above it on this page still stands.
- RSC Learn Chemistry — the preparation of aspirin, which is the standard school synthesis
for practising yield, purification, drying to constant mass and melting point as a purity
check.
- The ACS Green Chemistry Institute — the twelve principles of green chemistry, of which
atom economy is the second; useful context for an Extended Essay or the collaborative sciences
project.
- PhET — Reactants, Products and Leftovers, a genuinely useful way to make the
limiting-reactant idea concrete before it becomes arithmetic.