R2.1 How much? The amount of chemical change

Chemical equations as ratios

A chemical equation summarizes a reaction with formulas: reactants on the left, products on the right, and an arrow between them. In a balanced equation every atom on the left appears on the right, because atoms are neither created nor destroyed in a chemical reaction; they are only rearranged into new substances. This is the law of conservation of mass at the level of atoms.

The numbers in front of the formulas are the coefficients. They state how many particles of each substance react or form. In the equation for the synthesis of water, two molecules of hydrogen react with one molecule of oxygen to give two molecules of water:

Because a mole is a fixed number of particles ($6.02\times10^{23}$), the same ratio holds for any number of particles, including moles. Two moles of hydrogen react with one mole of oxygen to give two moles of water. The ratio of the coefficients is the mole ratio, and it is the central idea of every calculation in reacting quantities, an area of chemistry known as stoichiometry.

The three rows of the diagram of the synthesis of water show what the coefficients do and do not mean. They give the ratio of particles and of amounts in moles. They do not give the ratio of masses: 2 mol of hydrogen has a mass of only $4.04\ \text{g}$, while 1 mol of oxygen has a mass of $32.00\ \text{g}$. Masses are linked through molar masses, which is why almost every calculation starts by converting masses into moles.

State symbols

A state symbol after each formula shows the physical state of the substance under the conditions of the reaction. State symbols carry real chemical information: they show whether a product is a gas that escapes, a solid that precipitates from solution or a substance dissolved in water. They also matter for energy, since forming liquid water releases more energy than forming water vapour.

Balancing equations by inspection

An equation is balanced by changing coefficients only. The formulas themselves are fixed by the identity of the substances: changing ${{\mathrm{H}{\vphantom{X}}_{\smash[t]{2}}\mathrm{O}}}$ to ${{\mathrm{H}{\vphantom{X}}_{\smash[t]{2}}\mathrm{O}{\vphantom{X}}_{\smash[t]{2}}}}$ to make the oxygen atoms balance would turn water into hydrogen peroxide. A reliable order of working is:

1. Write the correct formula of every reactant and product. Elements that exist as diatomic molecules are written as such: ${{\mathrm{H}{\vphantom{X}}_{\smash[t]{2}}}}$, ${{\mathrm{N}{\vphantom{X}}_{\smash[t]{2}}}}$, ${{\mathrm{O}{\vphantom{X}}_{\smash[t]{2}}}}$, ${{\mathrm{F}{\vphantom{X}}_{\smash[t]{2}}}}$, ${{\mathrm{Cl}{\vphantom{X}}_{\smash[t]{2}}}}$, ${{\mathrm{Br}{\vphantom{X}}_{\smash[t]{2}}}}$, ${{\mathrm{I}{\vphantom{X}}_{\smash[t]{2}}}}$.

2. Balance the elements that appear in only one reactant and one product first, usually the metals and carbon.

3. Keep polyatomic ions such as ${{\mathrm{SO}{\vphantom{X}}_{\smash[t]{4}}{\vphantom{X}}^{2-}}}$, ${{\mathrm{NO}{\vphantom{X}}_{\smash[t]{3}}{\vphantom{X}}^{-}}}$

4. Balance hydrogen and then oxygen last, because they often appear in several substances.

5. If a fraction is needed (often for ${{\mathrm{O}{\vphantom{X}}_{\smash[t]{2}}}}$), use it, then multiply every coefficient by the denominator to give whole numbers.

6. Count every atom on each side as a final check, and check that the coefficients are in the smallest whole-number ratio.

Writing equations from a description

Many equations are written from a description in words. Most reactions met in introductory chemistry follow a small number of patterns, so recognizing the type of reaction gives the products; balancing then follows. The table summarizes the common patterns.

Ionic equations and spectator ions

When ionic compounds dissolve, their ions separate and move independently in the solution. In many reactions in solution only some of these ions change. The others are spectator ions: they are present in the same form before and after the reaction. An ionic equation leaves out the spectator ions and shows only the species that react.

Mixing silver nitrate and sodium chloride solutions gives a white precipitate of silver chloride. The full equation is ${{\mathrm{AgNO}{\vphantom{X}}_{\smash[t]{3}}\mskip2mu (\mathrm{aq}) {}+{} \mathrm{NaCl}\mskip2mu (\mathrm{aq})\xrightarrow{}\mathrm{AgCl}\mskip2mu (\mathrm{s}) {}+{} \mathrm{NaNO}{\vphantom{X}}_{\smash[t]{3}}\mskip2mu (\mathrm{aq})}}$, but the sodium and nitrate ions stay dissolved throughout, as the particle diagram of the mixture shows.

Writing every aqueous ionic compound as separate ions and then crossing out the ions that appear unchanged on both sides gives the ionic equation:

The ionic equation is more general than the full equation. It describes the reaction of any soluble silver salt with any soluble chloride, which is why acidified silver nitrate is used as a test for chloride ions. In the same way, the ionic equation for every neutralization of a strong acid by a strong base in solution is the same:

An ionic equation must balance in two ways: the number of each kind of atom, and the total charge. In ${{\mathrm{Mg}\mskip2mu (\mathrm{s}) {}+{} 2\,\mathrm{H}{\vphantom{X}}^{+}\mskip2mu (\mathrm{aq})\xrightarrow{}\mathrm{Mg}{\vphantom{X}}^{2+}\mskip2mu (\mathrm{aq}) {}+{} \mathrm{H}{\vphantom{X}}_{\smash[t]{2}}\mskip2mu (\mathrm{g})}}$ the total charge is $2+$ on each side. Solids, liquids, gases and molecular substances in solution (such as ethanoic acid, which is only slightly ionized) are written as complete formulas, not as ions.

Redox reactions involving ions need extra care, because the charges only balance when the electrons lost and gained are equal. For these reactions a method based on half-equations, one for oxidation and one for reduction, gives the balanced ionic equation reliably; it is developed with electron transfer reactions.

Calculating reacting masses

A balanced equation links amounts in moles, not masses. A calculation of reacting masses therefore has three steps: convert the known mass into moles, use the mole ratio from the equation to find the moles of the substance asked about, then convert those moles into the quantity asked for. The same three steps work whether the data are masses, gas volumes or solution concentrations, as the map of the routes between quantities shows.

In symbols, for a reaction $a\,\text{A} + b\,\text{B} \rightarrow c\,\text{C}$, where $a$, $b$ and $c$ are the coefficients:

Molar masses are calculated from the relative atomic masses in the data booklet to two decimal places: $M({{\mathrm{Fe}{\vphantom{X}}_{\smash[t]{2}}\mathrm{O}{\vphantom{X}}_{\smash[t]{3}}}}) = 2(55.85) + 3(16.00) = 159.70\ \text{g mol}^{-1}$. Intermediate values are kept unrounded in the calculator, and only the final answer is rounded, to the same number of significant figures as the least precise data.

A fuel calculation follows the same steps. The combustion of $1.00\ \text{kg}$ of propane, ${{\mathrm{C}{\vphantom{X}}_{\smash[t]{3}}\mathrm{H}{\vphantom{X}}_{\smash[t]{8}}\mskip2mu (\mathrm{g}) {}+{} 5\,\mathrm{O}{\vphantom{X}}_{\smash[t]{2}}\mskip2mu (\mathrm{g})\xrightarrow{}3\,\mathrm{CO}{\vphantom{X}}_{\smash[t]{2}}\mskip2mu (\mathrm{g}) {}+{} 4\,\mathrm{H}{\vphantom{X}}_{\smash[t]{2}}\mathrm{O}\mskip2mu (\mathrm{l})}}$, uses $1000/44.11 = 22.67$ mol of propane. Each mole gives 3 mol of carbon dioxide, so $68.01$ mol of ${{\mathrm{CO}{\vphantom{X}}_{\smash[t]{2}}}}$ forms, with a mass of $68.01 \times 44.01 = 2.99\times10^{3}\ \text{g}$, or about $3.0\ \text{kg}$. The mass of carbon dioxide is three times the mass of the fuel because each carbon atom (12.01) leaves combined with two oxygen atoms taken from the air.

Problems with an extra step

Real data rarely arrive as a pure mass of one reactant. A percentage composition, a mass change or a hydrated formula adds a step before the mole ratio can be used. The order of operations is the same: first find the mass (or amount) of the substance that actually reacts.

Reacting volumes of gases

Avogadro's law states that equal volumes of all gases, measured at the same temperature and pressure, contain equal numbers of molecules. It follows that, for gases at the same temperature and pressure, the volume of a gas is proportional to its amount in moles. The coefficients of a balanced equation therefore give the ratio of the volumes of the gases that react and form, with no need to convert to moles at all.

For the combustion of propane, ${{\mathrm{C}{\vphantom{X}}_{\smash[t]{3}}\mathrm{H}{\vphantom{X}}_{\smash[t]{8}}\mskip2mu (\mathrm{g}) {}+{} 5\,\mathrm{O}{\vphantom{X}}_{\smash[t]{2}}\mskip2mu (\mathrm{g})\xrightarrow{}3\,\mathrm{CO}{\vphantom{X}}_{\smash[t]{2}}\mskip2mu (\mathrm{g}) {}+{} 4\,\mathrm{H}{\vphantom{X}}_{\smash[t]{2}}\mathrm{O}\mskip2mu (\mathrm{l})}}$, $50\ \text{cm}^3$ of propane needs $5\times50 = 250\ \text{cm}^3$ of oxygen and gives $3\times50 = 150\ \text{cm}^3$ of carbon dioxide, all measured at the same temperature and pressure. The water is a liquid at room temperature, so it has a negligible volume and does not count as a gas.

Gas volumes from masses: the molar volume

Avogadro's law also means that one mole of any ideal gas occupies the same volume at a given temperature and pressure. This molar volume, $V_\text{m}$, is $22.7\ \text{dm}^3\ \text{mol}^{-1}$ at STP ($273\ \text{K}$ and $100\ \text{kPa}$) (data booklet). At STP, a gas volume and an amount are linked by:

Car airbags show how much gas a small mass of solid can produce. Early airbag inflators were triggered by a crash sensor to decompose sodium azide almost instantly: ${{2\,\mathrm{NaN}{\vphantom{X}}_{\smash[t]{3}}\mskip2mu (\mathrm{s})\xrightarrow{}2\,\mathrm{Na}\mskip2mu (\mathrm{s}) {}+{} 3\,\mathrm{N}{\vphantom{X}}_{\smash[t]{2}}\mskip2mu (\mathrm{g})}}$. About $130\ \text{g}$ of sodium azide produces enough nitrogen to fill a typical driver's airbag, and the decomposition is complete in a few hundredths of a second.

Sodium azide is highly toxic, and the sodium metal formed is dangerously reactive, so inflators also contained compounds such as potassium nitrate and silica to convert the sodium into a harmless glassy solid. Most modern inflators use non-azide gas generators instead, many of them based on nitrogen-rich compounds such as guanidine nitrate, but the calculation of gas volume from the equation is the same.

Molar volume under other conditions

The value $22.7\ \text{dm}^3\ \text{mol}^{-1}$ applies only at STP. Rearranging the ideal gas equation gives $V_\text{m} = V/n = RT/P$, so the molar volume is proportional to the absolute temperature and inversely proportional to the pressure. At $298\ \text{K}$ and $100\ \text{kPa}$ it is $24.8\ \text{dm}^3\ \text{mol}^{-1}$. For any conditions other than STP, $PV = nRT$ is used directly.

Avogadro's law is a description: it states a relationship that measurements of gas volumes confirm, but it does not say why the relationship holds. The explanation comes from the kinetic theory of gases. Gas particles are so far apart that their own volume is negligible, so the volume a gas occupies depends on the number of particles, their speed (set by temperature) and the pressure, not on what the particles are. This is also why the law fails for real gases at high pressure or low temperature, when the particles' own volume and the attractions between them are no longer negligible.

Reactions in solution and titration

Many reactions are carried out in solution, where amounts are measured by volume rather than by mass. The amount of solute in a volume of solution follows from its molar concentration:

Concentrations are written with square brackets: $[{{\mathrm{NaOH}}}] = 0.150\ \text{mol dm}^{-3}$. Volumes measured in $\text{cm}^3$ are divided by 1000 before use. With $n = CV$ the mole map applies to solutions exactly as it applies to masses.

Precipitation reactions link a solution to a mass of solid. Adding excess sodium sulfate solution to $25.0\ \text{cm}^3$ of $0.200\ \text{mol dm}^{-3}$ barium chloride precipitates all $5.00\times10^{-3}$ mol of the barium ions as barium sulfate, ${{\mathrm{Ba}{\vphantom{X}}^{2+}\mskip2mu (\mathrm{aq}) {}+{} \mathrm{SO}{\vphantom{X}}_{\smash[t]{4}}{\vphantom{X}}^{2-}\mskip2mu (\mathrm{aq})\xrightarrow{}\mathrm{BaSO}{\vphantom{X}}_{\smash[t]{4}}\mskip2mu (\mathrm{s})}}$. The mass of precipitate expected is $5.00\times10^{-3} \times 233.40 = 1.17\ \text{g}$. Filtering, drying and weighing a precipitate in this way is called gravimetric analysis.

What a titration is

A titration measures the volume of one solution that reacts exactly with a known volume of another. One solution, the titrant, is added from a burette to a measured volume of the other, the analyte, until the reaction is just complete. If the concentration of one solution is known, the mole ratio gives the concentration of the other.

The standard procedure, used for both acid–base and redox titrations, is:

1. Rinse the burette with a little of the titrant, then fill it, removing the air bubble below the tap. Rinse the pipette with a little of the analyte.

2. Use the pipette and a pipette filler to transfer an exact volume of the analyte (for example $25.00\ \text{cm}^3$) into a conical flask, then add two or three drops of indicator.

3. Record the initial burette reading, reading the bottom of the meniscus at eye level.

4. Add the titrant while swirling the flask. Do a rough titration first to find the approximate end point, then repeat, adding the titrant dropwise near the end point.

5. Record the final burette reading. The difference between the two readings is the titre.

6. Repeat until at least two titres are concordant (agree within $\pm0.10\ \text{cm}^3$), and use the mean of the concordant titres in the calculation.

The equipment is chosen for precision. A volumetric pipette and a burette measure volumes far more precisely than a measuring cylinder: a burette is read to $\pm0.05\ \text{cm}^3$, and since each titre is the difference between two readings, its uncertainty is $\pm0.10\ \text{cm}^3$. The white tile shows the colour change clearly, and swirling mixes each addition so that the colour change is not seen too early.

Rinsing matters for the same reason. Water left in a burette dilutes the titrant, so a larger titre is needed and the calculated concentration of the analyte comes out too high. Water left in the conical flask, by contrast, does no harm: it changes the concentration of the analyte but not the amount of it that the pipette delivered, and the amount is what the titration measures.

Acid–base titrations: the ethanoic acid in vinegar

Vinegar is a dilute solution of ethanoic acid, typically around 5 g per $100\ \text{cm}^3$. Its concentration is found by titration with standard sodium hydroxide solution and phenolphthalein indicator, which is colourless in acid and turns pink at the end point. Because vinegar is too concentrated to titrate conveniently with dilute alkali, it is first diluted accurately with a pipette and volumetric flask.

A dilution adds solvent but not solute, so the amount of solute is the same before and after. Since $n = CV$, the concentrations and volumes before (1) and after (2) dilution are linked by:

Any volume unit can be used here, as long as both volumes are in the same unit. Diluting $10.00\ \text{cm}^3$ of vinegar to $100.0\ \text{cm}^3$ divides its concentration by 10.

Redox titrations

In a redox titration the titrant is an oxidizing or reducing agent. A common example uses acidified potassium manganate(VII) to find the amount of iron(II) in a sample:

The manganate(VII) ion is deep purple and the manganese(II) ion is almost colourless, so the titration is self-indicating: each drop of titrant is decolourized while iron(II) remains, and the end point is the first permanent pale pink colour. The mole ratio ${{\mathrm{MnO}{\vphantom{X}}_{\smash[t]{4}}{\vphantom{X}}^{-}}} : {{\mathrm{Fe}{\vphantom{X}}^{2+}}}$ is $1 : 5$, which comes from the electrons transferred: each manganese gains five electrons, while each iron(II) ion loses only one.

Back titration

Some substances cannot be titrated directly: an insoluble solid such as calcium carbonate reacts too slowly with acid for a sharp end point, and a volatile base such as ammonia escapes from an open flask. In a back titration, a known amount of a reagent (more than enough) is added to the sample, and once the reaction is complete the amount of reagent left over is found by titration. The amount that reacted with the sample is the difference.

Limiting and excess reactants

Reactants are seldom mixed in exactly the ratio of the equation. When they are not, one reactant is used up first and the reaction stops, however much of the others remains. The reactant that runs out is the limiting reactant; the others are in excess. A sandwich shop with 20 slices of bread and 15 slices of cheese can make only 10 cheese sandwiches (two slices of bread and one of cheese each): bread limits the number, and 5 slices of cheese are left over.

Identifying the limiting reactant

The limiting reactant cannot be found by comparing masses, or even by comparing moles directly, because the reactants do not react 1 : 1 in general. Two equivalent methods work:

• Divide the amount of each reactant by its coefficient. The reactant with the smallest result is limiting. For ${{2\,\mathrm{H}{\vphantom{X}}_{\smash[t]{2}} {}+{} \mathrm{O}{\vphantom{X}}_{\smash[t]{2}}\xrightarrow{}2\,\mathrm{H}{\vphantom{X}}_{\smash[t]{2}}\mathrm{O}}}$ with 4 mol ${{\mathrm{H}{\vphantom{X}}_{\smash[t]{2}}}}$ and 3 mol ${{\mathrm{O}{\vphantom{X}}_{\smash[t]{2}}}}$: $4/2 = 2$ and $3/1 = 3$, so hydrogen is limiting.

• Calculate the amount of product each reactant could form. The reactant that gives the smaller amount is limiting, and that smaller amount is the amount of product formed.

Once the limiting reactant is known, every other quantity is calculated from it: the amount of product, and the amount of each excess reactant that reacts. The amount of excess reactant left over is the amount present minus the amount that reacts.

The same reaction shows how the limiting reactant changes as the proportions change. Adding increasing masses of magnesium to a fixed $50.0\ \text{cm}^3$ of the acid gives more hydrogen only while the magnesium is limiting. Once $0.608\ \text{g}$ ($0.0250$ mol) has been added, the acid becomes limiting and the volume of hydrogen stops rising.

The thermite reaction, used to weld railway lines on site, is a further example in which both reactants are solids: ${{\mathrm{Fe}{\vphantom{X}}_{\smash[t]{2}}\mathrm{O}{\vphantom{X}}_{\smash[t]{3}}\mskip2mu (\mathrm{s}) {}+{} 2\,\mathrm{Al}\mskip2mu (\mathrm{s})\xrightarrow{}\mathrm{Al}{\vphantom{X}}_{\smash[t]{2}}\mathrm{O}{\vphantom{X}}_{\smash[t]{3}}\mskip2mu (\mathrm{s}) {}+{} 2\,\mathrm{Fe}\mskip2mu (\mathrm{l})}}$. A mixture of $100.0\ \text{g}$ of iron(III) oxide ($0.6262$ mol) and $30.0\ \text{g}$ of aluminium ($1.112$ mol) contains less aluminium than needed, since $1.112/2 = 0.556$ is smaller than $0.6262/1$. Aluminium is limiting, so $1.112$ mol of molten iron forms ($62.1\ \text{g}$), and $0.070$ mol ($11.2\ \text{g}$) of iron(III) oxide is left over.

Limiting reactants in gas reactions

With gases at the same temperature and pressure, volumes can be compared directly. If $30\ \text{cm}^3$ of methane is mixed with $50\ \text{cm}^3$ of oxygen, ${{\mathrm{CH}{\vphantom{X}}_{\smash[t]{4}}\mskip2mu (\mathrm{g}) {}+{} 2\,\mathrm{O}{\vphantom{X}}_{\smash[t]{2}}\mskip2mu (\mathrm{g})\xrightarrow{}\mathrm{CO}{\vphantom{X}}_{\smash[t]{2}}\mskip2mu (\mathrm{g}) {}+{} 2\,\mathrm{H}{\vphantom{X}}_{\smash[t]{2}}\mathrm{O}\mskip2mu (\mathrm{l})}}$, the methane would need $60\ \text{cm}^3$ of oxygen. Oxygen is limiting: $50\ \text{cm}^3$ of oxygen reacts with $25\ \text{cm}^3$ of methane, giving $25\ \text{cm}^3$ of carbon dioxide and leaving $5\ \text{cm}^3$ of methane unreacted.

In practice a chemist often chooses which reactant to put in excess. The cheaper or less hazardous reactant is usually added in excess, so that all of the expensive reactant is converted into product. Excess air ensures complete combustion of a fuel; excess acid ensures that all of a carbonate in a sample reacts. The excess has to be separated from the product afterwards, which is one reason why a large excess is avoided.

Theoretical, experimental and percentage yield

The amount of product calculated from the limiting reactant is the most that the reaction can give. The amount actually obtained in a laboratory or a factory is almost always smaller.

The two are compared as the percentage yield, with both yields in the same units (both masses or both amounts):

Why experimental yields are lower

The causes of a low yield fall into two groups. Chemical causes mean that less product forms than the equation predicts; physical (practical) causes mean that product forms but is not all collected.

An experimental yield can also appear higher than the theoretical yield. This always points to an error: the product still contains water or solvent because it was not dried to constant mass, or it contains impurities such as unreacted reactant or a by-product. A percentage yield above 100% is impossible for a pure, dry product, so it is a signal to look for such an error rather than a result to report.

A gravimetric yield experiment: burning magnesium

Heating magnesium ribbon in a crucible is a classic way to measure a yield by mass: ${{2\,\mathrm{Mg}\mskip2mu (\mathrm{s}) {}+{} \mathrm{O}{\vphantom{X}}_{\smash[t]{2}}\mskip2mu (\mathrm{g})\xrightarrow{}2\,\mathrm{MgO}\mskip2mu (\mathrm{s})}}$. The cleaned ribbon is weighed in a crucible with its lid, heated strongly until it no longer glows, cooled and weighed again. Heating, cooling and weighing are repeated until two masses agree, which shows that the reaction is complete; this is called heating to constant mass.

The lid is lifted briefly from time to time with tongs to let air in, but kept on most of the time to stop the white magnesium oxide smoke from escaping. The crucible is allowed to cool before each weighing: a hot crucible sets up air currents around the balance and gives a falsely low reading.

Yields in multistep synthesis

Medicines are usually made in a sequence of reactions, each with its own yield. The overall yield is the product of the yields of the individual steps, so it falls quickly as the number of steps grows. A six-step synthesis in which every step gives 90% has an overall yield of $0.9^6 = 0.53$, or 53%; a ten-step route at 80% per step gives only 11%.

For this reason pharmaceutical chemists put great effort into shortening routes and raising the yield of each step, especially the late steps, where the material carries the cost of every earlier step. Shorter routes also use fewer reagents and solvents, which links yield to waste and to the next idea, atom economy.

Atom economy

Percentage yield measures how much of the possible product is actually obtained. It says nothing about the other products of the reaction. A reaction with a 100% yield may still turn most of the mass of its reactants into substances nobody wants. Atom economy measures this second kind of efficiency from the balanced equation alone.

The equation is given in the data booklet. Because mass is conserved, the total molar mass of the reactants equals the total molar mass of the products, so the denominator can be calculated from either side. Every atom that does not end up in the desired product ends up in a by-product, which is waste unless a use is found for it. The higher the atom economy, the less waste a process produces for each kilogram of product: atom economy and waste are inversely related.

Atom economy and the type of reaction

The type of reaction largely fixes the atom economy. In an addition reaction all the reactants combine into one product, so the atom economy is 100%. Addition polymerization is an addition reaction repeated thousands of times: every ethene molecule becomes part of the poly(ethene) chain, so the atom economy is 100%. Substitution and elimination reactions always produce a second product and so have lower atom economies.

Atom economies calculated from the relative atomic masses in the data booklet (2025).

Case study: two syntheses of ibuprofen

Ibuprofen, one of the most widely used painkillers, was developed by the Boots Company in the United Kingdom in the 1960s. The original Boots synthesis used six steps. Each step added reagents whose atoms did not end up in ibuprofen, and about 514 g of reactants were used for every mole (206 g) of ibuprofen made, an atom economy of about 40%. More than half of the mass of the reactants became waste, and with millions of kilograms of ibuprofen made every year, that waste had to be treated and disposed of.

In the 1990s the BHC Company developed a three-step route. Its overall equation, from 2-methylpropylbenzene (${{\mathrm{C}{\vphantom{X}}_{\smash[t]{10}}\mathrm{H}{\vphantom{X}}_{\smash[t]{14}}}}$), ethanoic anhydride, hydrogen and carbon monoxide, is:

With ibuprofen (${{\mathrm{C}{\vphantom{X}}_{\smash[t]{13}}\mathrm{H}{\vphantom{X}}_{\smash[t]{18}}\mathrm{O}{\vphantom{X}}_{\smash[t]{2}}}}$, $206.31\ \text{g mol}^{-1}$) as the desired product, the atom economy is:

The only by-product is ethanoic acid, which is recovered and sold, so in practice nearly all the reactant atoms end up in useful products. Published figures for the process vary with the reactants counted: the US Environmental Protection Agency quotes about 80% atom utilization, or about 99% when the recovered ethanoic acid is included. The BHC process also uses catalysts that are recovered and reused (including hydrogen fluoride, recycled with better than 99.9% efficiency) instead of reagents that are used up. It received a US Presidential Green Chemistry Challenge Award in 1997.

The comparison shows why atom economy is calculated at the design stage: it depends only on the chosen reactions, so a route with a poor atom economy cannot be rescued by running each step more carefully. Percentage yield, by contrast, can be raised by better practice and conditions.

Atom economy compared with percentage yield

A process can score well on one measure and badly on the other. The chlorohydrin route to ethylene oxide could run with a good percentage yield and still waste three-quarters of its reactant mass; a direct oxidation with 100% atom economy is of little use if most of the ethene is lost by burning to carbon dioxide. Both measures are needed to judge efficiency.

Green chemistry: judging the efficiency of a process

Green chemistry is the design of chemical products and processes that reduce or eliminate the use and generation of hazardous substances. It was set out as twelve principles by Paul Anastas and John Warner in 1998. Atom economy and percentage yield are two of the measures it uses; a full assessment of a process also looks at the following factors.

These factors can pull in different directions. Fermentation uses a renewable feedstock and runs near room temperature, but has an atom economy of only 51%, since a third of the carbon atoms leave as carbon dioxide. Weighing such factors against each other, and against cost and safety, is part of choosing a process. The speed and extent of a reaction, which also affect efficiency, depend on the factors that control rates and equilibria.

Applying the ideas in exam questions

Paper 1A: the limiting reactant

Paper 1A: reacting gas volumes

Paper 1B: collecting carbon dioxide from calcium carbonate

A student adds $0.250\ \text{g}$ ($\pm0.001\ \text{g}$) of powdered calcium carbonate to $20.0\ \text{cm}^3$ of $1.00\ \text{mol dm}^{-3}$ hydrochloric acid in the apparatus shown, inserting the bung immediately. The gas collected at $20.0\ ^\circ\text{C}$ and $100\ \text{kPa}$ has a final volume of $54.0\ \text{cm}^3$ ($\pm0.5\ \text{cm}^3$).

Paper 2: standardizing hydrochloric acid

Paper 2: extracting titanium

Paper 2: two routes to hydrogen

Self-check

• To do:

  I can write balanced equations, with state symbols, from descriptions of reactions and from given reactants and products.

• To do:

  I can balance equations by inspection and give the sum of the smallest whole-number coefficients.

• To do:

  I can write ionic equations, identify spectator ions and check that charges balance.

• To do:

  I can explain why the coefficients of an equation give mole ratios but not mass ratios.

• To do:

  I can calculate reacting masses through the mole ratio, including problems with a percentage composition or a mass difference.

• To do:

  I can use volume ratios for gases at the same temperature and pressure, and deduce a formula from combustion volumes.

• To do:

  I can calculate gas volumes at STP with 22.7 dm³ mol⁻¹ and under other conditions with PV = nRT in SI units.

• To do:

  I can calculate amounts, concentrations and volumes of solutions that react, including precipitation reactions.

• To do:

  I can describe how to carry out a titration and explain the choice of glassware, concordant titres and rinsing.

• To do:

  I can process acid–base, redox and back titration data to find a concentration, a mass or a percentage purity.

• To do:

  I can identify the limiting and excess reactants and calculate the amount of excess reactant remaining.

• To do:

  I can distinguish theoretical yield from experimental yield and calculate a percentage yield.

• To do:

  I can explain why an experimental yield is lower, or appears higher, than the theoretical yield.

• To do:

  I can process gravimetric data from heating to constant mass, including the uncertainty in a mass difference.

• To do:

  I can calculate the overall yield of a multistep synthesis.

• To do:

  I can calculate atom economy from a balanced equation and explain why addition reactions have 100% atom economy.

• To do:

  I can explain the difference between atom economy and percentage yield, and the link between atom economy and waste.

• To do:

  I can discuss specific green chemistry factors, beyond atom economy and yield, that affect the efficiency of a process.

The answer is B.

Oxygen last. The products contain $4\times2 + 5\times1 = 13$ O atoms, which needs $\tfrac{13}{2}$ ${{\mathrm{O}{\vphantom{X}}_{\smash[t]{2}}}}$:

Clear the fraction by doubling every coefficient:

Check: C 8 = 8; H 20 = 20; O 26 = 8 × 2 + 10 = 26. The sum of the coefficients is $2 + 13 + 8 + 10 = 33$.

Step 2: mole ratio. 1 mol ${{\mathrm{Fe}{\vphantom{X}}_{\smash[t]{2}}\mathrm{O}{\vphantom{X}}_{\smash[t]{3}}}}$ gives 2 mol Fe:

Step 3: convert to mass.

Check: the mass of iron is less than the mass of the oxide, as it must be, since the oxygen is removed. The proportion $2(55.85)/159.70 = 0.699$ is also the mass fraction of iron in ${{\mathrm{Fe}{\vphantom{X}}_{\smash[t]{2}}\mathrm{O}{\vphantom{X}}_{\smash[t]{3}}}}$.

Mole ratio 1 : 1, so $n({{\mathrm{SO}{\vphantom{X}}_{\smash[t]{2}}}}) = 779.5\ \text{mol}$, and $M({{\mathrm{SO}{\vphantom{X}}_{\smash[t]{2}}}}) = 32.07 + 2(16.00) = 64.07\ \text{g mol}^{-1}$:

The answer is given to two significant figures, matching the 2.5% in the data. Burning one tonne of this coal releases about 50 kg of a gas that forms acid rain, which is why coal-fired power stations remove sulfur dioxide from their flue gases.

Volume ratio = mole ratio. For $\mathrm{C}_x\mathrm{H}_y + (x + \tfrac{y}{4})\,{{\mathrm{O}{\vphantom{X}}_{\smash[t]{2}}}} \rightarrow x\,{{\mathrm{CO}{\vphantom{X}}_{\smash[t]{2}}}} + \tfrac{y}{2}\,{{\mathrm{H}{\vphantom{X}}_{\smash[t]{2}}\mathrm{O}}}$:

The hydrocarbon is ${{\mathrm{C}{\vphantom{X}}_{\smash[t]{3}}\mathrm{H}{\vphantom{X}}_{\smash[t]{8}}}}$, propane. Check: ${{\mathrm{C}{\vphantom{X}}_{\smash[t]{3}}\mathrm{H}{\vphantom{X}}_{\smash[t]{8}} {}+{} 5\,\mathrm{O}{\vphantom{X}}_{\smash[t]{2}}\xrightarrow{}3\,\mathrm{CO}{\vphantom{X}}_{\smash[t]{2}} {}+{} 4\,\mathrm{H}{\vphantom{X}}_{\smash[t]{2}}\mathrm{O}}}$; $20\ \text{cm}^3$ needs $100\ \text{cm}^3$ of ${{\mathrm{O}{\vphantom{X}}_{\smash[t]{2}}}}$ and gives $60\ \text{cm}^3$ of ${{\mathrm{CO}{\vphantom{X}}_{\smash[t]{2}}}}$, as measured.

Mole ratio. 2 mol ${{\mathrm{NaN}{\vphantom{X}}_{\smash[t]{3}}}}$ gives 3 mol ${{\mathrm{N}{\vphantom{X}}_{\smash[t]{2}}}}$: $n({{\mathrm{N}{\vphantom{X}}_{\smash[t]{2}}}}) = 1.999\times\tfrac{3}{2} = 2.999\ \text{mol}$.

(a) At STP: $V = 2.999 \times 22.7 = 68.1\ \text{dm}^3$.

(b) At 300 K and 100 kPa, use $PV = nRT$ in SI units ($P$ in Pa, $V$ in $\text{m}^3$):

The volume is larger at the higher temperature, as expected for a gas at constant pressure.

Volume: $V = n/C = 1.875\times10^{-3} / 0.100 = 0.01875\ \text{dm}^3 = 18.8\ \text{cm}^3$.

Moles of NaOH: $n = 0.100 \times 0.02085 = 2.085\times10^{-3}\ \text{mol}$.

Mole ratio 1 : 1: $n({{\mathrm{CH}{\vphantom{X}}_{\smash[t]{3}}\mathrm{COOH}}}) = 2.085\times10^{-3}\ \text{mol}$ in $25.00\ \text{cm}^3$ of diluted vinegar.

Undo the dilution. The vinegar was diluted by a factor of $100.0/10.00 = 10$, so the original concentration is $0.834\ \text{mol dm}^{-3}$.

Mass concentration: $M({{\mathrm{CH}{\vphantom{X}}_{\smash[t]{3}}\mathrm{COOH}}}) = 60.06\ \text{g mol}^{-1}$, so $0.834 \times 60.06 = 50.1\ \text{g dm}^{-3}$, which is 5.0 g per $100\ \text{cm}^3$.

Scale up to the whole tablet (the $100.0\ \text{cm}^3$ flask held four portions): $n = 4 \times 2.910\times10^{-4} = 1.164\times10^{-3}\ \text{mol}$.

Mass: $1.164\times10^{-3} \times 55.85 = 0.0650\ \text{g} = 65.0\ \text{mg}$. This matches the iron content stated for a common supplement tablet containing 200 mg of dried iron(II) sulfate.

Acid that reacted with the carbonate: $2.500\times10^{-2} - 6.020\times10^{-3} = 1.898\times10^{-2}\ \text{mol}$.

Mole ratio 2 : 1: $n({{\mathrm{CaCO}{\vphantom{X}}_{\smash[t]{3}}}}) = 9.490\times10^{-3}\ \text{mol}$; $m = 9.490\times10^{-3} \times 100.09 = 0.950\ \text{g}$.

Percentage: $0.950/1.000 \times 100 = 95.0\%$. The rest of the shell is mainly protein and other minerals, which do not react with the acid; any that did react would make the result too high.

Divide by the coefficients: Mg $0.1000/1 = 0.1000$; HCl $0.0500/2 = 0.0250$. The acid gives the smaller value, so HCl is limiting and magnesium is in excess.

Hydrogen formed (2 mol HCl gives 1 mol ${{\mathrm{H}{\vphantom{X}}_{\smash[t]{2}}}}$): $n({{\mathrm{H}{\vphantom{X}}_{\smash[t]{2}}}}) = 0.0250\ \text{mol}$; $V = 0.0250 \times 22.7 = 0.568\ \text{dm}^3 = 568\ \text{cm}^3$ at STP.

Magnesium left over: $0.0250$ mol reacts, so $0.1000 - 0.0250 = 0.0750\ \text{mol}$ remains, a mass of $0.0750 \times 24.31 = 1.82\ \text{g}$.

Theoretical yield: the ratio is 1 : 1 and $M({{\mathrm{C}{\vphantom{X}}_{\smash[t]{9}}\mathrm{H}{\vphantom{X}}_{\smash[t]{8}}\mathrm{O}{\vphantom{X}}_{\smash[t]{4}}}}) = 180.17\ \text{g mol}^{-1}$, so the theoretical mass of aspirin is $1.448\times10^{-2} \times 180.17 = 2.61\ \text{g}$.

Mass of product $= 25.981 - 25.412 = 0.569\ \text{g}$.

(b) The mass of product is the difference of two readings, so its absolute uncertainty is $2 \times 0.001 = \pm0.002\ \text{g}$. The percentage uncertainty is $0.002/0.569 \times 100 = 0.35\%$, about $0.4\%$.

(c) Yes. With nitrogen, ${{3\,\mathrm{Mg}\mskip2mu (\mathrm{s}) {}+{} \mathrm{N}{\vphantom{X}}_{\smash[t]{2}}\mskip2mu (\mathrm{g})\xrightarrow{}\mathrm{Mg}{\vphantom{X}}_{\smash[t]{3}}\mathrm{N}{\vphantom{X}}_{\smash[t]{2}}\mskip2mu (\mathrm{s})}}$; if all $1.481\times10^{-2}$ mol of Mg formed ${{\mathrm{Mg}{\vphantom{X}}_{\smash[t]{3}}\mathrm{N}{\vphantom{X}}_{\smash[t]{2}}}}$ ($M = 100.95\ \text{g mol}^{-1}$), the mass would be $\tfrac{1}{3} \times 1.481\times10^{-2} \times 100.95 = 0.498\ \text{g}$. Each magnesium atom gains less mass as nitride than as oxide, so a mixture of the two gives a mass between 0.498 g and 0.597 g, which is consistent with the 0.569 g obtained. The difference from 100% (4.7%) is much larger than the 0.4% uncertainty, so it is a real effect, not random error.

Chlorohydrin route:

Direct oxidation (using the coefficients):

The old route turned three-quarters of the mass of its reactants into calcium chloride and water; the direct route, an addition reaction, has no by-product at all.

(b)

• $n({{\mathrm{CO}{\vphantom{X}}_{\smash[t]{2}}}}) = 2.50\times10^{-3}$ «mol» ✓

• $V = \dfrac{nRT}{P} = \dfrac{2.498\times10^{-3} \times 8.31 \times 293}{1.00\times10^{5}} = 6.08\times10^{-5}\ \text{m}^3 = 60.8$