6.2.1 Concentration in mol dm⁻³ and titration calculations

Concentration in mol dm⁻³ counts the moles in each cubic decimetre

1. A concentration in ${mol\; dm}^{-3}$ states the amount of solute dissolved in each ${dm}^3$ of solution.
2. The relationship between amount, volume and concentration is: $$concentration=\frac{amount\; of\; solute\; in\; mol}{{volume\; in\; dm}^3}$$
3. Rearranged to give the amount present: $$amount\; in\; mol=concentration\times {volume\; in\; dm}^3$$
4. Volumes measured in ${cm}^3$ are divided by $1000$ to convert them to ${dm}^3$.
5. A more concentrated solution holds more solute in the same volume.

Converting between g dm⁻³ and mol dm⁻³

1. A concentration in ${g\; dm}^{-3}$ states the mass of solute in each ${dm}^3$ of solution.
2. Dividing by the relative formula mass converts a mass to an amount: $${concentration\; in\; mol\; dm}^{-3}=\frac{{concentration\; in\; g\; dm}^{-3}}{M_r}$$
3. Multiplying reverses the conversion: $${concentration\; in\; g\; dm}^{-3}={concentration\; in\; mol\; dm}^{-3}\times M_r$$
4. Sodium hydroxide has an $M_r$ of $40$, so $4.0{g\; dm}^{-3}$ is $0.10{mol\; dm}^{-3}$.
5. The conversion factor is the $M_r$ of the solute, so it changes from one solution to another.

A titration measures the volume that exactly reacts

1. A pipette delivers a fixed, accurately known volume of one solution into a conical flask.
2. A burette adds the second solution a little at a time, with its volume read to $0.05{cm}^3$.
3. A few drops of a suitable indicator show when the reaction is exactly complete.
4. The end point is the first permanent colour change, and the burette reading is taken there.
5. The volume added is the titre, and concordant titres are averaged before any calculation.

A worked titration calculation

1. $25.0{cm}^3$ of sodium hydroxide of concentration $0.100{mol\; dm}^{-3}$ needs $20.0{cm}^3$ of hydrochloric acid.
2. First find the amount of the solution whose concentration is known: $$n(NaOH)=0.100\times 0.0250=2.50\times 10^{-3}mol$$
3. The balancing numbers in the equation are one to one, so the acid supplies the same amount.
4. Then divide by the volume of acid, converted to ${dm}^3$: $$c(HCl)=\frac{2.50\times 10^{-3}}{0.0200}=0.125{mol\; dm}^{-3}$$
5. The three steps are always the same: amount, ratio, then divide by the other volume.

Finding an unknown volume instead

1. The same three steps give a volume when the two concentrations are known.
2. A titration of $25.0{cm}^3$ of $0.100{mol\; dm}^{-3}$ sodium hydroxide against $0.200{mol\; dm}^{-3}$ acid gives $n=2.50\times 10^{-3}mol$ of each.
3. Dividing the amount by the concentration gives the volume: $$V=\frac{2.50\times 10^{-3}}{0.200}=0.0125{dm}^3$$
4. Multiplying by $1000$ converts the answer back to $12.5{cm}^3$.
5. A more concentrated acid needs a smaller volume, which is a useful check on the answer.

6.2.2 Percentage yield and why actual yield falls short

Theoretical yield is the maximum mass a reaction could give

1. The maximum mass of product assumes every particle of the limiting reactant is converted.
2. It is calculated from the balanced equation, not measured in the laboratory.
3. Whichever reactant runs out first sets that ceiling, and it is the only one the calculation uses.
4. Any reactant in excess is left over and adds nothing to the mass of product.
5. No real reaction reaches this figure, which is why the calculated value is called theoretical.

Actual yield is the mass actually collected

1. The actual yield is the mass of pure, dry product weighed at the end.
2. It is measured on a balance after the product has been separated and dried.
3. Product still wet with solvent weighs too much, so the sample is dried to constant mass.
4. Product left in the apparatus or lost in filtering never reaches the balance.
5. Both yields are masses, so both are quoted in grams.

Percentage yield compares the two

1. The actual yield is expressed as a percentage of the theoretical yield: $$percentage\; yield=\frac{actual\; yield}{theoretical\; yield}\times 100$$
2. Both masses must be in the same unit, so the units cancel and the answer is a pure number.
3. The answer carries no unit, and the $\%$ sign is part of the number rather than a unit.
4. A value above $100\%$ is impossible and points to an impure or damp sample.
5. A high percentage yield means the process wastes little of the starting material.

A worked percentage yield

1. A reaction has a theoretical yield of $8.0g$ and gives $6.0g$ of dry product.
2. Substituting into the expression gives: $$\frac{6.0}{8.0}\times 100=75\%$$
3. Three quarters of the possible product was recovered, so a quarter was lost or never formed.
4. Rearranging finds a missing figure, so $75\%$ of $8.0g$ gives the actual yield.
5. Dividing the actual yield by the percentage and multiplying by $100$ recovers the theoretical yield.

Why the actual yield falls short

1. The reaction may be incomplete, leaving some reactant unreacted when the mixture is worked up.
2. A reversible reaction reaches equilibrium, so it can never convert all of the reactants.
3. Practical losses occur at every transfer, in filtering, and as product left on the glassware.
4. Side reactions use up reactants to make substances other than the one wanted.
5. Purifying the product removes impurities but removes some product along with them.

6.2.3 Atom economy of a reaction

Atom economy measures how much of the product mass is wanted

1. A reaction usually makes more than one product, and only one of them is the desired product.
2. Atom economy compares the mass of that product with the mass of everything formed: $$atom\; economy=\frac{M_{r}of\; the\; desired\; product}{sum\; of\; the{M}_{r}of\; all\; products}\times 100$$
3. The masses come from the balanced equation, with each $M_r$ multiplied by its balancing number.
4. A high atom economy means little of the starting material ends up as waste.
5. The answer is a percentage and carries no unit.

Working out an atom economy step by step

1. Write the balanced equation and identify which product is wanted.
2. Find the $M_r$ of that product and multiply it by its balancing number.
3. Do the same for every other product, then add all the products together.
4. Divide the desired figure by the total and multiply by $100$.
5. The reactants never enter the calculation, since conservation of mass makes their total the same.

A worked atom economy: extracting iron

1. The equation for reducing iron(III) oxide with carbon monoxide is: $${Fe}_2{O}_3+3CO\to 2Fe+3{CO}_2$$
2. The desired product is iron, of $A_r$ $56$: $$2\times 56=112$$
3. The only by-product is carbon dioxide, of $M_r$ $44$: $$3\times 44=132$$
4. The atom economy follows from the two totals: $$\frac{112}{112+132}\times 100=45.9\%$$
5. Less than half the mass formed is iron, and the rest leaves the process as carbon dioxide.

Why a high atom economy matters

1. A reaction with a low atom economy converts much of its raw material into waste.
2. That waste has to be separated, treated or disposed of, which costs money.
3. Raw materials are often finite, so wasting them shortens the supply.
4. A route with a higher atom economy is usually more sustainable for the same product.
5. A useful by-product softens the objection, because the waste can be sold rather than discarded.

Comparing two routes to the same product

1. Two reactions that give the same product can have very different atom economies.
2. The route that forms fewer by-products keeps more of the mass in the wanted product.
3. An addition reaction, which joins everything into one product, scores highest of all.
4. A route with a lower atom economy may still be chosen if it is faster or cheaper to run.
5. Comparing routes fairly means quoting both figures and the reason for preferring one.

6.2.4 Choosing a reaction pathway

A product can usually be made by more than one route

1. Different reactions can lead to the same product from different starting materials.
2. Choosing between them means comparing the routes on several measures at once.
3. The measures used are atom economy, percentage yield, rate, position of equilibrium and the use made of by-products.
4. No single measure decides the choice, because a route can score well on one and badly on another.
5. The cost of running the process is what all of these feed into.

Atom economy and percentage yield measure different things

1. The equation alone fixes atom economy, which says how much of the mass formed is the wanted product.
2. Measuring the product gives the percentage yield, which says how much of the possible product was recovered.
3. A route can have a high atom economy and a low percentage yield, or the reverse.
4. A low percentage yield can sometimes be improved by changing the conditions, while the atom economy cannot.
5. Both matter, because raw material is wasted either by becoming a by-product or by never being converted.

Rate and equilibrium position decide how fast and how far

1. A faster reaction makes more product per hour, so it needs smaller plant for the same output.
2. A very slow route may be rejected however good its atom economy.
3. A reversible reaction in a closed system settles at an equilibrium that limits how much product forms.
4. A position lying well to the right means a large proportion of the reactants is converted.
5. A route whose equilibrium lies to the left needs its product removing continually to be worth running.

Useful by-products change the economics

1. A by-product that can be sold offsets the cost of the raw materials.
2. A by-product that must be treated as waste adds a disposal cost instead.
3. A by-product that is fed back into the process reduces what has to be bought in.
4. Steam produced by an exothermic route can be used to heat another part of the plant.
5. A modest atom economy can still be acceptable when the other product has a market.

Weighing the factors against one another

1. A judgement begins by stating which route wins on each measure separately.
2. The measures are then weighed, and the weighting depends on what the product is worth.
3. Where the raw material is expensive, atom economy and yield carry the most weight.
4. Where demand is high and the material cheap, rate matters more than either.
5. A conclusion is only sound when the reason for the weighting is given alongside it.

6.2.5 Molar volume of gases and Avogadro's law

One mole of any gas occupies the same volume

1. At room temperature and pressure, one mole of molecules of any gas occupies the same volume.
2. That volume is $24{dm}^3$, which is the same as $24000{cm}^3$.
3. Room temperature and pressure means about $20{}^{\circ }C$ and normal atmospheric pressure.
4. The identity of the gas makes no difference, so a mole of hydrogen and a mole of carbon dioxide fill the same space.
5. For a gas such as helium the particles are single atoms, and a mole of them occupies $24{dm}^3$ as well.

Converting between amount and volume of gas

1. The volume occupied by a known amount of gas is: $${volume\; in\; dm}^3=amount\; in\; mol\times 24$$
2. Rearranged to give the amount from a measured volume: $$amount\; in\; mol=\frac{{volume\; in\; dm}^3}{24}$$
3. A volume in ${cm}^3$ is divided by $24000$ instead, or converted to ${dm}^3$ first.
4. The amount scales the volume directly, so $0.5mol$ fills $12{dm}^3$ and $2mol$ fills $48{dm}^3$.
5. The figure $24{dm}^3$ is given whenever a calculation needs it.

Linking the mass of a solid to the volume of a gas

1. A mass of solid is converted to an amount by dividing by its relative formula mass.
2. The balancing numbers in the equation convert that amount into the amount of gas.
3. Multiplying the amount of gas by $24$ gives its volume in ${dm}^3$.
4. Heating $10g$ of calcium carbonate, of $M_r$ $100$, gives: $$n({CaCO}_3)=\frac{10}{100}=0.10mol$$
5. The equation is one to one, so the volume of carbon dioxide is: $$0.10\times 24=2.4{dm}^3$$

Avogadro's law relates volumes of gases directly

1. Equal volumes of gases at the same temperature and pressure hold equal numbers of particles.
2. The ratio of volumes of reacting gases is therefore the ratio of the balancing numbers.
3. No relative formula masses are needed, so the volumes are compared directly.
4. In the formation of hydrogen chloride the balancing numbers are one, one and two: $$H_2(g)+{Cl}_2(g)\to 2HCl(g)$$
5. $30{cm}^3$ of hydrogen therefore reacts with $30{cm}^3$ of chlorine to give $60{cm}^3$ of hydrogen chloride.

Common slips in gas calculations

1. Dividing by $24$ when the volume is in ${cm}^3$ makes the answer a thousand times too large.
2. Applying the molar volume to a solid or liquid gives a meaningless figure.
3. Skipping the ratio from the equation is the most frequent error in a mass to volume question.
4. Applying the volume ratio to a solid or liquid in the equation is the error, since only the gases have a volume to compare.
5. Checking that the final unit is ${dm}^3$ or ${cm}^3$ catches most of these.