5.1.1 The technique of titration

A titration measures the volume that reacts completely

Definition

Titration

A technique that finds the volume of one solution needed to react completely with a measured volume of another solution.

Titrations are most often used for neutralisation reactions between an acid and an alkali.
A fixed volume of one solution, often the alkali, is measured into a conical flask.
The other solution is added gradually from a burette until the reaction is complete.
A few drops of indicator in the flask change colour when neutralisation is complete.
1. The volume added from the burette to reach this point is called the titre.
If the concentration of one solution is known, the titre can later be used to work out the unknown concentration of the other.

Key Idea

Every step of a titration is designed to measure two volumes accurately: the fixed volume in the flask and the titre from the burette.

Pipettes measure fixed volumes and burettes measure variable volumes

Definition

Pipette

Glassware that measures and transfers one fixed volume of liquid accurately, such as 25.0 cm³.

Definition

Burette

A long graduated glass tube with a tap, used to add a variable volume of solution and measure it accurately.

Rinse the pipette with a little of the solution it will measure, so that no water is left inside to dilute it.
Use a pipette filler to draw the solution up until the bottom of the meniscus sits on the line, reading it at eye level.
1. Let the solution run into the conical flask and touch the tip on the inside of the flask, but do not blow out the last drop.
Rinse the burette with the solution it will hold, then fill it using a small funnel and remove the funnel before reading it.
Open the tap briefly so that the jet below the tap fills and contains no air bubble.
Read the burette at eye level from the bottom of the meniscus, recording each reading to the nearest $0.05 cm30.05\,\text{cm}^3$ $0.05 cm^{3}$ .
1. Readings are written to two decimal places, so a reading halfway between $22.40$ and $22.50$ is recorded as $cm322.45\,\text{cm}^3$ .

Common Mistake

Rinse the conical flask with distilled water only, because water in the flask does not change the amount of alkali the pipette delivered, whereas rinsing it with alkali would add extra.

An indicator's sharp colour change shows the end point

Definition

Indicator

A substance that has different colours in acidic and alkaline solutions, so its colour change shows when neutralisation is complete.

Definition

End point

The point in a titration at which the indicator changes colour, showing that the reaction is complete.

Add only a few drops of indicator, because the indicator is itself a weak acid and too much can affect the result.
Phenolphthalein is colourless in acid and pink in alkali.
Methyl orange is red in acid and yellow in alkali.
The colour change you see depends on which solution starts in the flask, so describe it in the direction the titration runs.
1. With alkali in the flask and acid in the burette, phenolphthalein turns from pink to colourless and methyl orange turns from yellow to red.
Universal indicator is not used, because it changes colour gradually through many shades instead of at one sharp point.
Standing the flask on a white tile makes the colour change easier to see.

Example

Hydrochloric acid is added from a burette to $cm325.0\,\text{cm}^3$ of sodium hydroxide solution containing phenolphthalein, and the end point is reached when one drop turns the pink solution permanently colourless.

A rough titration comes before accurate dropwise titrations

The first rough titration adds the solution quickly while swirling, to find roughly where the end point is.
In each accurate titration, add the solution quickly until you are about $2 cm32\,\text{cm}^3$ $2 cm^{3}$ short of the rough titre.
1. Then add it drop by drop, swirling after each drop, so that you do not overshoot the end point.
Wash any drops on the inside of the flask down with a little distilled water, so all the solution added can react.
Stop at the first permanent colour change and record the final burette reading.
Calculate the titre by subtracting the initial burette reading from the final reading.

Practical

Aim: find the volume of acid that reacts completely with $cm325.0\,\text{cm}^3$ of alkali.
Apparatus: a $cm350\,\text{cm}^3$ burette, a clamp and stand, a small funnel, a $cm325.0\,\text{cm}^3$ pipette with a pipette filler, a $cm3250\,\text{cm}^3$ conical flask, a white tile, an indicator such as phenolphthalein or methyl orange, the acid, the alkali and a wash bottle of distilled water.

Part 1: setting up

Rinse the burette with a little of the acid, then clamp it upright.
Fill the burette with acid through a funnel held below eye level, then remove the funnel.
Run a little acid out so that the tip has no air bubble.
Use the pipette and pipette filler to transfer $cm325.0\,\text{cm}^3$ of alkali into the conical flask, then add a few drops of indicator.
Stand the flask on the white tile under the burette.

Part 2: titrating

Record the initial burette reading at eye level, from the bottom of the meniscus.
Do a rough titration first, adding acid quickly while swirling until the indicator changes colour.
Refill the flask with fresh alkali and indicator, and run in acid to about $cm31\,\text{cm}^3$ less than the rough titre.
Add acid drop by drop while swirling, until one drop gives a permanent colour change.
Record the final reading to the nearest $cm30.05\,\text{cm}^3$ and work out the titre.
Repeat until at least two titres are concordant, within $cm30.10\,\text{cm}^3$ of each other, and use only those for the mean.

Accuracy and safety

Rinse the burette with the acid and the pipette with the alkali, so that leftover water does not dilute them.
Swirl the flask all the time and rinse its sides with distilled water near the end, so that all the acid added can react.
Read every volume at eye level, from the bottom of the meniscus.
Wear eye protection, because acids and alkalis are irritant or corrosive.

Concordant titres are averaged to give a reliable titre

Definition

Concordant results

Repeat titres that agree closely with each other, usually within 0.10 cm³.

Repeat the accurate titration until at least two titres are concordant, which usually means within $cm30.10\,\text{cm}^3$ of each other.
Leave the rough titre out of the mean, because the solution was added too quickly for it to be accurate.
Also leave out any anomalous titre, such as one where the end point was overshot.
Calculate the mean titre from the concordant results only.
Concordant results show that the method is repeatable, so the mean titre is a trustworthy value.

Example

Titres of $cm323.10\,\text{cm}^3$ (rough), $cm322.50\,\text{cm}^3$ , $cm322.40\,\text{cm}^3$ and $cm322.45\,\text{cm}^3$ are recorded; find the mean titre.
Leave out the rough titre of $cm323.10\,\text{cm}^3$ .
The other three titres all lie within $cm30.10\,\text{cm}^3$ of each other, so they are concordant.
Mean titre $\dfrac{22.50 + 22.40 + 22.45}{3} = 22.45\,\text{cm}^3$ .

Self review

Why is the pipette rinsed with the solution it will measure, but the conical flask rinsed only with distilled water?
What colour change does phenolphthalein show when acid is added to alkali in the flask?
Why is universal indicator not used in a titration?
Why is the solution added dropwise near the end point?
Which titres are used to calculate the mean titre?

5.1.2 Theoretical mass, percentage yield and atom economy

Scaling reacting masses gives the theoretical yield

Definition

Theoretical yield

The maximum mass of product that could form from a given mass of reactant, calculated from the balanced equation.

Definition

Relative formula mass

The mass of one formula unit of a substance compared with one-twelfth of the mass of a carbon-12 atom, given the symbol Mr.

A balanced equation shows the reacting masses of each substance when you replace each formula with its $M_{r}$ .
1. Mass is conserved, so these reacting masses always add up to the same total on both sides.
The reacting masses stay in the same ratio whatever mass of reactant is used.
To find the theoretical yield, multiply the mass of reactant by the mass ratio of product to reactant.
This gives the greatest mass of product possible, because it assumes all the reactant turns into product.

Example

Calculate the theoretical yield of calcium oxide when $g10.0\,\text{g}$ of calcium carbonate is heated: $CaCO3→CaO+CO2\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2$ ( $A_r$ : Ca = 40.1, C = 12.0, O = 16.0).
Work out the relative formula masses: $Mr(CaCO3)=40.1+12.0+(3×16.0)=100.1M_r(\text{CaCO}_3) = 40.1 + 12.0 + (3 \times 16.0) = 100.1$ and $Mr(CaO)=40.1+16.0=56.1M_r(\text{CaO}) = 40.1 + 16.0 = 56.1$ .
Write the reacting masses: $g100.1\,\text{g}$ of $CaCO3\text{CaCO}_3$ gives $g56.1\,\text{g}$ of $CaO\text{CaO}$ .
Scale to the mass used: $10.0 \times \dfrac{56.1}{100.1} = 5.60\,\text{g}$ of calcium oxide.

Multiply relative formula masses by the balancing numbers

A balancing number in front of a formula multiplies the whole formula, so it also multiplies that substance's reacting mass.
In $2Mg+O2→2MgO2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}$ $2 Mg + O_{2} \to 2 MgO$ , the reacting masses are $2×24.3=48.62 \times 24.3 = 48.6$ $2 \times 24.3 = 48.6$ for magnesium and $2×40.3=80.62 \times 40.3 = 80.6$ $2 \times 40.3 = 80.6$ for magnesium oxide.
1. So $g6.00\,\text{g}$ of magnesium gives a theoretical yield of $6.00 \times \dfrac{80.6}{48.6} = 9.95\,\text{g}$ of magnesium oxide.
Only the reactant you are given and the product you want are needed, so other substances in the equation can be ignored.
Round only at the end, and give the answer with a unit of mass, because relative formula masses themselves have no units.

Example

Calculate the theoretical yield of iron from $g80.0\,\text{g}$ of iron(III) oxide: $Fe2O3+3CO→2Fe+3CO2\text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2$ ( $A_r$ : Fe = 55.8, O = 16.0).
Work out the reacting masses: $Fe2O3=(2×55.8)+(3×16.0)=159.6\text{Fe}_2\text{O}_3 = (2 \times 55.8) + (3 \times 16.0) = 159.6$ and $2Fe=2×55.8=111.62\text{Fe} = 2 \times 55.8 = 111.6$ .
Scale to the mass used: $80.0 \times \dfrac{111.6}{159.6} = 55.9\,\text{g}$ of iron.

Percentage yield compares actual yield with theoretical yield

Definition

Actual yield

The mass of product that is collected when a reaction is carried out.

Definition

Percentage yield

The actual yield of a product expressed as a percentage of its theoretical yield.

Percentage yield is calculated with this equation: $yield×100\text{percentage yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100$
The actual yield is almost always less than the theoretical yield.
1. The reaction may be reversible or may not go to completion, so some reactant never turns into product.
2. Side reactions can turn some of the reactant into other, unwanted products.
3. Some product is lost when it is filtered, transferred between containers or purified.
A percentage yield above $100%100\%$ suggests the product was still wet or impure when it was weighed.

Example

The theoretical yield of calcium oxide is $g5.60\,\text{g}$ , but only $g4.48\,\text{g}$ is collected; calculate the percentage yield.
Substitute into the equation: $4.485.60×100\dfrac{4.48}{5.60} \times 100$ .
Percentage yield $80.0\%$ .

Atom economy shows how much reactant mass becomes product

Definition

Atom economy

The percentage of the total mass of reactants that ends up in the desired product, calculated from the balanced equation.

Definition

Desired product

The useful product that a reaction is carried out to make.

Atom economy is calculated from the balanced equation: $products×100\text{atom economy} = \frac{M_r \text{ of desired product}}{\text{sum of } M_r \text{ of all products}} \times 100$
Multiply each $M_{r}$ by its balancing number before adding, as for reacting masses.
1. The total mass of the products equals the total mass of the reactants, so the bottom line can also be the sum of the reactants' $M_r$ values.
A reaction with only one product has an atom economy of $100%100\%$ , because every atom ends up in the desired product.
A low atom economy means much of the reactant mass becomes waste products, which wastes raw materials unless the by-products can be used or sold.

Example

Calculate the atom economy for making calcium oxide: $CaCO3→CaO+CO2\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2$ ( $M_r$ : CaO = 56.1, $CO2\text{CO}_2$ = 44.0).
Add the $M_r$ values of all the products: $56.1 + 44.0 = 100.1$ .
Atom economy $\dfrac{56.1}{100.1} \times 100 = 56.0\%$ .

Percentage yield and atom economy measure different things

Percentage yield depends on how well the reaction is carried out, so it can change from one attempt to the next.
Atom economy depends only on the balanced equation, so it is fixed for a given reaction.
A reaction can have a high atom economy but a low percentage yield.
1. Making ethanol from ethene and steam has an atom economy of $100%100\%$ , yet only a small fraction of the gases react each time they pass through the reactor.
Chemists want both values to be high, because together they mean less waste and more product from the same raw materials.

Exam technique

When a question asks for percentage yield, look for a measured mass of product, but when it asks for atom economy, use only the $M_r$ values from the balanced equation.
Show the reacting masses or $M_r$ values you used, then give the final answer with its unit or percentage sign.

Self review

What is the theoretical yield of a reaction?
What mass of calcium oxide could form from $g20.0\,\text{g}$ of calcium carbonate ( $M_r$ : $CaCO3\text{CaCO}_3$ = 100.1, CaO = 56.1)?
Give two reasons why the actual yield is usually less than the theoretical yield.
Why does a reaction with only one product have an atom economy of $100%100\%$ ?
Which of percentage yield and atom economy can be calculated without doing the experiment?

5.1.3 Concentration of a solution in mol/dm3

Concentration measures solute per volume of solution

Definition

Concentration

The amount of solute dissolved in a stated volume of solution, measured in g/dm³ or mol/dm³.

Definition

Solute

The substance that dissolves in a solvent to make a solution.

A solution forms when a solute dissolves in a solvent, such as salt dissolving in water.
A concentrated solution has a lot of solute in a given volume, while a dilute solution has only a little.
Concentration can be measured in grams per cubic decimetre, written $g/dm3\text{g/dm}^3$ $g/dm^{3}$ .
1. One cubic decimetre is $cm31000\,\text{cm}^3$ , which is the same as one litre.
The volume used is the volume of the whole solution, not the volume of water that was added.
Concentration can also be given in $mol/dm3\text{mol/dm}^3$ , which uses the amount in moles of solute instead of its mass.

Key Idea

Concentration depends on two things only: how much solute there is and how much solution it is spread through.

Concentration equals mass of solute over volume

The relationship is: $(dm3)\text{concentration (g/dm}^3\text{)} = \frac{\text{mass of solute (g)}}{\text{volume of solution (dm}^3\text{)}}$
Convert a volume in $cm3\text{cm}^3$ $cm^{3}$ into $dm3\text{dm}^3$ $dm^{3}$ by dividing by 1000 before using the equation.
1. So $cm3250\,\text{cm}^3$ is $dm30.250\,\text{dm}^3$ .
Rearranging gives mass of solute $=$ concentration $×\times$ $\times$ volume, which tells you how much solid to weigh out.
1. Making $dm32.00\,\text{dm}^3$ of a $g/dm315.0\,\text{g/dm}^3$ solution needs $15.0 \times 2.00 = 30.0\,\text{g}$ of solute.
Give every answer with its unit, because a number alone does not say whether it is a mass, a volume or a concentration.

Example

Calculate the concentration of a solution made by dissolving $g5.00\,\text{g}$ of sodium chloride in water to make $cm3250\,\text{cm}^3$ of solution.
Convert the volume: $dm3250\,\text{cm}^3 \div 1000 = 0.250\,\text{dm}^3$ .
Divide mass by volume: $g/dm3\dfrac{5.00\,\text{g}}{0.250\,\text{dm}^3} = 20.0\,\text{g/dm}^3$ .

How solute mass and volume affect concentration

At a fixed volume, concentration is directly proportional to the mass of solute.
1. Dissolving twice the mass of solute in the same volume doubles the concentration.
For a fixed mass of solute, a larger volume of solution gives a lower concentration.
1. Spreading the same mass through twice the volume halves the concentration.
Doubling both the mass and the volume leaves the concentration unchanged.
A sample of any size taken from a well-mixed solution has the same concentration as the rest of it.

Example

Adding more squash to the same glass of water makes a stronger drink, while topping the glass up with water makes it weaker, although the amount of squash in the glass has not changed.

Diluting a solution lowers its concentration

When water is added, the mass of solute stays the same.
The volume of solution increases, so the concentration falls.
Evaporating some of the water has the opposite effect, raising the concentration because the volume decreases while the solute stays.
To find a new concentration, first work out the mass of solute present, then divide it by the new volume.

Example

Water is added to $cm350.0\,\text{cm}^3$ of a $g/dm340.0\,\text{g/dm}^3$ solution until its volume is $cm3200\,\text{cm}^3$ ; find the new concentration.
Mass of solute $40.0\,\text{g/dm}^3 \times 0.0500\,\text{dm}^3 = 2.00\,\text{g}$ .
New concentration $\dfrac{2.00\,\text{g}}{0.200\,\text{dm}^3} = 10.0\,\text{g/dm}^3$ .

Making a solution of known concentration

Weigh the solid accurately on a balance in a weighing boat or small beaker.
Dissolve it in a small volume of distilled water in a beaker, stirring until no solid remains.
Pour the solution into a volumetric flask and rinse the beaker and stirring rod into the flask, so no solute is left behind.
Add distilled water until the bottom of the meniscus sits on the line, then stopper and invert the flask to mix.
Divide the mass weighed by the volume of the flask in $dm3\text{dm}^3$ to find the concentration.

Exam technique

When a question asks how a change affects concentration, say what happens to the mass of solute and what happens to the volume of solution, then state the effect.

Self review

What does a concentration of $g/dm320.0\,\text{g/dm}^3$ mean?
What is the concentration of $g3.00\,\text{g}$ of solute in $cm3150\,\text{cm}^3$ of solution?
What happens to the concentration if the same mass of solute is dissolved in twice the volume?
Why does adding water lower the concentration of a solution?
What mass of solute is needed to make $cm3500\,\text{cm}^3$ of a $g/dm312.0\,\text{g/dm}^3$ solution?

5.1.4 Concentration calculations and titration results

Concentration is moles of solute per cubic decimetre

Definition

Concentration

The amount of solute dissolved in a stated volume of solution, measured in g/dm³ or mol/dm³.

The relationship is: $(dm3)\text{concentration (mol/dm}^3\text{)} = \frac{\text{amount of solute (mol)}}{\text{volume of solution (dm}^3\text{)}}$
Convert a volume in $cm3\text{cm}^3$ into $dm3\text{dm}^3$ by dividing by 1000.
Rearranging gives amount in moles $=$ concentration $×\times$ $\times$ volume in $dm3\text{dm}^3$ $dm^{3}$ .
1. So $cm320.0\,\text{cm}^3$ of a $mol/dm30.100\,\text{mol/dm}^3$ solution contains $0.100 \times 0.0200 = 0.00200\,\text{mol}$ of solute.
Doubling the amount of solute in the same volume doubles the concentration, while doubling the volume for the same amount halves it.

Example

A solution contains $mol0.0500\,\text{mol}$ of sodium chloride in $cm3250\,\text{cm}^3$ ; calculate its concentration.
Convert the volume: $250 \div 1000 = 0.250\,\text{dm}^3$ .
Concentration $\dfrac{0.0500}{0.250} = 0.200\,\text{mol/dm}^3$ .

Relative formula mass links solute mass and concentration

Definition

Relative formula mass

The mass of one formula unit of a substance compared with one-twelfth of the mass of a carbon-12 atom, given the symbol Mr.

Convert a mass of solute into moles with amount $\dfrac{\text{mass (g)}}{M_r}$ .
Then divide the amount by the volume of solution in $dm3\text{dm}^3$ to find the concentration in $mol/dm3\text{mol/dm}^3$ .
Working backwards, mass of solute $=$ concentration $×\times$ $\times$ volume $×Mr\times M_r$ $\times M_{r}$ .
1. Making $cm3500\,\text{cm}^3$ of $mol/dm30.100\,\text{mol/dm}^3$ copper(II) sulfate, $M_r = 159.6$ , needs $0.100 \times 0.500 \times 159.6 = 7.98\,\text{g}$ of solid.
To change a concentration in $mol/dm3\text{mol/dm}^3$ into $g/dm3\text{g/dm}^3$ , multiply by $M_r$ , and to go the other way, divide by $M_r$ .

Example

Calculate the concentration in $mol/dm3\text{mol/dm}^3$ of a solution made by dissolving $g4.00\,\text{g}$ of sodium hydroxide to make $cm3250\,\text{cm}^3$ of solution ( $A_r$ : Na = 23.0, O = 16.0, H = 1.0).
$Mr(NaOH)=23.0+16.0+1.0=40.0M_r(\text{NaOH}) = 23.0 + 16.0 + 1.0 = 40.0$ .
Amount $\dfrac{4.00}{40.0} = 0.100\,\text{mol}$ .
Concentration $\dfrac{0.100}{0.250} = 0.400\,\text{mol/dm}^3$ .

The mole ratio links reacting volumes and concentrations

Definition

Mole ratio

The ratio of the amounts in moles of substances that react or form, given by the balancing numbers in the equation.

At the end point, the two substances have reacted completely, so their amounts in moles are in the ratio shown by the balanced equation.
The amount of the substance with a known concentration comes from concentration $×\times$ volume.
In $HCl(aq)+NaOH(aq)→NaCl(aq)+H2O(l)\text{HCl(aq)} + \text{NaOH(aq)} \rightarrow \text{NaCl(aq)} + \text{H}_2\text{O(l)}$ $HCl(aq) + NaOH(aq) \to NaCl(aq) + H_{2} O(l)$ , the ratio is 1 : 1, so equal volumes react when the concentrations are equal.
1. If the sodium hydroxide were twice as concentrated, only half the volume would be needed, because each $cm3\text{cm}^3$ would hold twice as many moles.
In $H2SO4(aq)+2NaOH(aq)→Na2SO4(aq)+2H2O(l)\text{H}_2\text{SO}_4\text{(aq)} + 2\text{NaOH(aq)} \rightarrow \text{Na}_2\text{SO}_4\text{(aq)} + 2\text{H}_2\text{O(l)}$ $H_{2} SO_{4} (aq) + 2 NaOH(aq) \to Na_{2} SO_{4} (aq) + 2 H_{2} O(l)$ , each mole of acid reacts with two moles of alkali.
1. So for solutions of equal concentration, the volume of sodium hydroxide needed is twice the volume of sulfuric acid.

Common Mistake

Do not assume every acid reacts with an alkali in a 1 : 1 ratio; always read the ratio from the balanced equation.
Do not put a volume in $cm3\text{cm}^3$ into concentration $×\times$ volume, because the concentration is per $dm3\text{dm}^3$ .

Titration results give an unknown concentration

Definition

Titration

A technique that finds the volume of one solution needed to react completely with a measured volume of another solution.

The pipette gives the fixed volume of one solution and the mean titre gives the volume of the other.
Write the balanced equation first, so you know the mole ratio.
Find the amount in moles of the solution whose concentration is known.
Use the mole ratio to find the amount in moles of the other substance.
Divide that amount by its volume in $dm3\text{dm}^3$ to find the unknown concentration.

Example

Concentration of sodium hydroxide from a titration

$cm325.0\,\text{cm}^3$ of sodium hydroxide solution is neutralised by a mean titre of $cm320.00\,\text{cm}^3$ of $mol/dm30.100\,\text{mol/dm}^3$ sulfuric acid; find the concentration of the sodium hydroxide.
Amount of $mol\text{H}_2\text{SO}_4 = 0.100 \times 0.02000 = 0.002000\,\text{mol}$ .
The ratio $H2SO4:NaOH\text{H}_2\text{SO}_4 : \text{NaOH}$ is $1 : 2$ , so the amount of $mol\text{NaOH} = 2 \times 0.002000 = 0.004000\,\text{mol}$ .
Concentration of $mol/dm3\text{NaOH} = \dfrac{0.004000}{0.0250} = 0.160\,\text{mol/dm}^3$ .

Finding required volumes and mass concentrations

To find a volume needed, divide the amount in moles required by the concentration of that solution.
1. $cm325.0\,\text{cm}^3$ of $mol/dm30.100\,\text{mol/dm}^3$ sodium hydroxide contains $mol0.00250\,\text{mol}$ , so it needs $mol0.00250\,\text{mol}$ of hydrochloric acid.
2. With $mol/dm30.200\,\text{mol/dm}^3$ acid, the volume needed is $dm3\dfrac{0.00250}{0.200} = 0.0125\,\text{dm}^3$ , which is $cm312.5\,\text{cm}^3$ .
Convert a volume in $dm3\text{dm}^3$ back to $cm3\text{cm}^3$ by multiplying by 1000.
To give an answer in $g/dm3\text{g/dm}^3$ $g/dm^{3}$ , multiply the concentration in $mol/dm3\text{mol/dm}^3$ $mol/dm^{3}$ by $M_{r}$ .
1. The $mol/dm30.160\,\text{mol/dm}^3$ sodium hydroxide above has a concentration of $0.160 \times 40.0 = 6.40\,\text{g/dm}^3$ .

Exam technique

When a question asks for a concentration from titration results, write out each link of the chain, volume to moles, moles to moles by the ratio, then moles to concentration, with a unit on every value.
Check that the answer fits the equation, so in a 1 : 2 reaction the amount in moles of the second substance must be twice the amount of the first.

Self review

What is the concentration of $mol0.600\,\text{mol}$ of solute in $cm3250\,\text{cm}^3$ of solution?
How many moles of solute are in $cm330.0\,\text{cm}^3$ of a $mol/dm30.200\,\text{mol/dm}^3$ solution?
How do you convert a concentration in $mol/dm3\text{mol/dm}^3$ into $g/dm3\text{g/dm}^3$ ?
Why does the mole ratio in the balanced equation matter in a titration calculation?
$cm320.0\,\text{cm}^3$ of $mol/dm30.150\,\text{mol/dm}^3$ hydrochloric acid neutralises $cm325.0\,\text{cm}^3$ of potassium hydroxide solution; what is the concentration of the potassium hydroxide?

5.1.5 Molar gas volume and volumes of gases in reactions

Equal moles of any gas occupy equal volumes

Definition

Molar gas volume

The volume occupied by one mole of any gas at a stated temperature and pressure, which is 24 dm³ at room temperature and pressure.

Gas particles are very far apart, so the volume of a gas depends on the number of particles and not on the size of each particle.
At the same temperature and pressure, the volume of a gas is directly proportional to its amount in moles.
1. Doubling the amount of gas doubles its volume, and halving it halves the volume.
At room temperature and pressure, about $20 ∘C20\ ^{\circ}\text{C}$ $20^{\circ} C$ and 1 atmosphere, one mole of any gas occupies $24 dm324\,\text{dm}^3$ $24 dm^{3}$ .
1. This is the same as $cm324\,000\,\text{cm}^3$ , whether the gas is hydrogen, oxygen or carbon dioxide.

Graph showing that the volume of a gas is directly proportional to the number of moles (V = constant × n), represented by a straight line passing through the origin.

Key Idea

One mole of hydrogen weighs $g2.0\,\text{g}$ and one mole of carbon dioxide weighs $g44.0\,\text{g}$ , yet at room temperature and pressure each fills the same $dm324\,\text{dm}^3$ .

Converting gas volume and moles using 24 dm³

To find a gas volume, multiply the amount in moles by $24$ : $(mol)×24\text{volume (dm}^3\text{)} = \text{amount (mol)} \times 24$
To find an amount in moles, divide the gas volume in $dm3\text{dm}^3$ $dm^{3}$ by $24$ .
1. For a volume in $cm3\text{cm}^3$ , divide by $00024\,000$ instead, or convert it into $dm3\text{dm}^3$ first.
So $mol0.25\,\text{mol}$ of carbon dioxide occupies $0.25 \times 24 = 6.0\,\text{dm}^3$ at room temperature and pressure.
The value of $dm324\,\text{dm}^3$ applies only at room temperature and pressure, so a different temperature or pressure gives a different volume.

Example

Calculate the amount in moles of oxygen in $cm3480\,\text{cm}^3$ of the gas at room temperature and pressure.
Convert the volume: $dm3480\,\text{cm}^3 \div 1000 = 0.480\,\text{dm}^3$ .
Amount $\dfrac{0.480}{24} = 0.0200\,\text{mol}$ .

Balancing numbers give the ratio of gas volumes

Definition

Mole ratio

The ratio of the amounts in moles of substances that react or form, given by the balancing numbers in the equation.

The balancing numbers give the mole ratio, and equal amounts of gases have equal volumes, so for gases they also give the volume ratio.
In $CH4(g)+2O2(g)→CO2(g)+2H2O(l)\text{CH}_4\text{(g)} + 2\text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)} + 2\text{H}_2\text{O(l)}$ $CH_{4} (g) + 2 O_{2} (g) \to CO_{2} (g) + 2 H_{2} O(l)$ , the volumes of methane, oxygen and carbon dioxide are in the ratio $1 : 2 : 1$ .
1. So $cm350\,\text{cm}^3$ of methane reacts with $cm3100\,\text{cm}^3$ of oxygen and makes $cm350\,\text{cm}^3$ of carbon dioxide.
Water is a liquid at room temperature, so it is left out of the gas volume ratio.
In $2H2(g)+O2(g)→2H2O(l)2\text{H}_2\text{(g)} + \text{O}_2\text{(g)} \rightarrow 2\text{H}_2\text{O(l)}$ , $cm340\,\text{cm}^3$ of hydrogen needs half its volume, $cm320\,\text{cm}^3$ , of oxygen.

Common Mistake

Do not use the balancing numbers as a volume ratio for solids, liquids or solutions, because the rule applies only to gases measured at the same temperature and pressure.

Finding gas volume from reactant mass via moles

First convert the mass of reactant into moles by dividing by its $M_r$ .
Next use the mole ratio from the balanced equation to find the amount in moles of gas formed.
Then multiply the amount of gas by $24$ to find its volume in $dm3\text{dm}^3$ at room temperature and pressure.
Give the answer to a sensible number of significant figures, usually matching the data in the question.

Example

Calculate the volume of carbon dioxide at room temperature and pressure when $g5.00\,\text{g}$ of calcium carbonate is heated: $CaCO3(s)→CaO(s)+CO2(g)\text{CaCO}_3\text{(s)} \rightarrow \text{CaO(s)} + \text{CO}_2\text{(g)}$ ( $A_r$ : Ca = 40.1, C = 12.0, O = 16.0).
$Mr(CaCO3)=40.1+12.0+(3×16.0)=100.1M_r(\text{CaCO}_3) = 40.1 + 12.0 + (3 \times 16.0) = 100.1$ , so the amount $\dfrac{5.00}{100.1} = 0.04995\,\text{mol}$ .
The ratio $CaCO3:CO2\text{CaCO}_3 : \text{CO}_2$ is $1 : 1$ , so $mol0.04995\,\text{mol}$ of carbon dioxide forms.
Volume $\times 24 = 1.20\,\text{dm}^3$ .

Working back from gas volume to reactant mass

Divide the gas volume in $dm3\text{dm}^3$ by $24$ to find the amount of gas in moles.
Use the mole ratio to find the amount in moles of the reactant.
Multiply by the reactant's $A_{r}$ or $M_{r}$ to find its mass in grams.
1. In $Mg(s)+2HCl(aq)→MgCl2(aq)+H2(g)\text{Mg(s)} + 2\text{HCl(aq)} \rightarrow \text{MgCl}_2\text{(aq)} + \text{H}_2\text{(g)}$ , $cm3180\,\text{cm}^3$ of hydrogen is $mol\dfrac{0.180}{24} = 0.00750\,\text{mol}$ .
2. The ratio $Mg:H2\text{Mg} : \text{H}_2$ is $1 : 1$ , so the mass of magnesium that reacted is $0.00750 \times 24.3 = 0.182\,\text{g}$ .

Exam technique

When a question gives a gas volume in $cm3\text{cm}^3$ , convert it into $dm3\text{dm}^3$ before dividing by $24$ , and show that step, because it is the step most often missed.

Self review

What volume does one mole of any gas occupy at room temperature and pressure?
What is the volume of $mol0.150\,\text{mol}$ of nitrogen at room temperature and pressure?
How many moles of gas are in $cm31200\,\text{cm}^3$ at room temperature and pressure?
In $CH4(g)+2O2(g)→CO2(g)+2H2O(l)\text{CH}_4\text{(g)} + 2\text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)} + 2\text{H}_2\text{O(l)}$ , what volume of oxygen reacts with $cm330\,\text{cm}^3$ of methane?
Why is water left out of the volume ratio in this equation?

5.1.6 Choosing a reaction pathway

One product can be made by different pathways

Definition

Reaction pathway

The route of one or more reactions, with its starting materials and conditions, used to make a particular product.

Ethanol can be made by fermentation of sugars with yeast, or by reacting ethene with steam, which is called hydration.
Each pathway uses different starting materials, conditions and numbers of steps.
Chemists compare the pathways using data rather than choosing the one that seems simplest.
The data usually cover yield and atom economy, rate, conditions and energy use, raw materials, by-products and how pure the product is.

Key Idea

The best pathway is the one that gives the most suitable overall balance of these factors for the product and its use, even if it is not the best on every one.

Yield and atom economy measure how much becomes product

Definition

By-product

A substance formed in a reaction alongside the desired product.

A high percentage yield means that less of the reactant is lost or left unreacted.
A high atom economy means that less of the reactant mass turns into by-products.
A pathway with fewer steps usually loses less product overall, because some product is lost at every step.
A low atom economy matters less if the by-product is useful or saleable, and more if it is harmful and must be treated.
Unreacted starting materials can sometimes be recycled through the reactor, which improves the overall yield.

Example

Compare the atom economies of the two pathways to ethanol: $C6H12O6→2C2H5OH+2CO2\text{C}_6\text{H}_{12}\text{O}_6 \rightarrow 2\text{C}_2\text{H}_5\text{OH} + 2\text{CO}_2$ and $C2H4+H2O→C2H5OH\text{C}_2\text{H}_4 + \text{H}_2\text{O} \rightarrow \text{C}_2\text{H}_5\text{OH}$ ( $M_r$ : glucose = 180.0, ethanol = 46.0).
For fermentation, atom economy $\dfrac{2 \times 46.0}{180.0} \times 100 = 51.1\%$ .
For hydration, ethanol is the only product, so the atom economy is $100%100\%$ .
Hydration wastes far less of the reactant mass, while almost half the mass in fermentation leaves as carbon dioxide.

Rate, conditions and energy decide a pathway's cost

A fast reaction makes more product in a given time, so less equipment is needed for the same output.
High temperatures and pressures need more energy and stronger, more expensive equipment.
A catalyst can let a pathway run fast enough at a lower temperature, which saves energy.
A continuous process, fed with reactants all the time, makes product faster than a batch that is started and stopped.
1. Hydration of ethene runs continuously and quickly, while fermentation is a slow batch process that takes several days.

Example

Hydration of ethene needs about $∘C300\ ^{\circ}\text{C}$ , a pressure about 60 times atmospheric pressure and a phosphoric acid catalyst, whereas fermentation runs at about $∘C30\ ^{\circ}\text{C}$ and normal pressure with enzymes in yeast.

Raw materials, purification and safety can outweigh yield

A pathway is more sustainable if its raw materials are renewable, cheap and easy to obtain.
1. Sugar for fermentation comes from crops such as sugar cane, which can be regrown, while ethene comes from crude oil, which is finite.
A pathway that gives a purer product needs less separation afterwards.
1. Fermentation gives a dilute solution of ethanol that must be separated by fractional distillation, while hydration gives almost pure ethanol.
Hazardous substances, flammable gases and harmful waste add safety and disposal costs to a pathway.

Note

Countries with large areas of farmland and little oil, such as Brazil, make much of their fuel ethanol by fermentation, while countries with a large petrochemical industry have used the hydration of ethene.

Justifying a pathway choice with data

State which pathway you would choose for the specified product.
Quote at least two pieces of data that support the choice.
Explain what each value means in practice, such as less waste, lower energy costs or more product per hour.
Say why the other pathway is less suitable, even if it has one advantage.
Finish with a judgement linked to the stated priority, such as low cost, speed or sustainability.

Exam technique

When a question asks you to choose a pathway from data, compare the pathways factor by factor using the numbers given, then say which factor matters most for this product and why it decides your choice.

Self review

What is meant by a reaction pathway?
Why does fermentation have a lower atom economy than the hydration of ethene?
Why can a pathway with fewer steps give a greater overall yield?
Give one advantage of fermentation over the hydration of ethene.
Why might a pathway with a lower percentage yield still be chosen?

5.1.1 The technique of titration

A titration measures the volume that reacts completely

Definition

Titration

A technique that finds the volume of one solution needed to react completely with a measured volume of another solution.

Titrations are most often used for neutralisation reactions between an acid and an alkali.
A fixed volume of one solution, often the alkali, is measured into a conical flask.
The other solution is added gradually from a burette until the reaction is complete.
A few drops of indicator in the flask change colour when neutralisation is complete.
1. The volume added from the burette to reach this point is called the titre.
If the concentration of one solution is known, the titre can later be used to work out the unknown concentration of the other.

Key Idea

Every step of a titration is designed to measure two volumes accurately: the fixed volume in the flask and the titre from the burette.

Pipettes measure fixed volumes and burettes measure variable volumes

Definition

Pipette

Glassware that measures and transfers one fixed volume of liquid accurately, such as 25.0 cm³.

Definition

Burette

A long graduated glass tube with a tap, used to add a variable volume of solution and measure it accurately.

Rinse the pipette with a little of the solution it will measure, so that no water is left inside to dilute it.
Use a pipette filler to draw the solution up until the bottom of the meniscus sits on the line, reading it at eye level.
1. Let the solution run into the conical flask and touch the tip on the inside of the flask, but do not blow out the last drop.
Rinse the burette with the solution it will hold, then fill it using a small funnel and remove the funnel before reading it.
Open the tap briefly so that the jet below the tap fills and contains no air bubble.
Read the burette at eye level from the bottom of the meniscus, recording each reading to the nearest $0.05 cm30.05\,\text{cm}^3$ $0.05 cm^{3}$ .
1. Readings are written to two decimal places, so a reading halfway between $22.40$ and $22.50$ is recorded as $cm322.45\,\text{cm}^3$ .

Common Mistake

Rinse the conical flask with distilled water only, because water in the flask does not change the amount of alkali the pipette delivered, whereas rinsing it with alkali would add extra.

An indicator's sharp colour change shows the end point

Definition

Indicator

A substance that has different colours in acidic and alkaline solutions, so its colour change shows when neutralisation is complete.

Definition

End point

The point in a titration at which the indicator changes colour, showing that the reaction is complete.

Add only a few drops of indicator, because the indicator is itself a weak acid and too much can affect the result.
Phenolphthalein is colourless in acid and pink in alkali.
Methyl orange is red in acid and yellow in alkali.
The colour change you see depends on which solution starts in the flask, so describe it in the direction the titration runs.
1. With alkali in the flask and acid in the burette, phenolphthalein turns from pink to colourless and methyl orange turns from yellow to red.
Universal indicator is not used, because it changes colour gradually through many shades instead of at one sharp point.
Standing the flask on a white tile makes the colour change easier to see.

Example

A rough titration comes before accurate dropwise titrations

The first rough titration adds the solution quickly while swirling, to find roughly where the end point is.
In each accurate titration, add the solution quickly until you are about $2 cm32\,\text{cm}^3$ $2 cm^{3}$ short of the rough titre.
1. Then add it drop by drop, swirling after each drop, so that you do not overshoot the end point.
Wash any drops on the inside of the flask down with a little distilled water, so all the solution added can react.
Stop at the first permanent colour change and record the final burette reading.
Calculate the titre by subtracting the initial burette reading from the final reading.

Practical

Aim: find the volume of acid that reacts completely with $cm325.0\,\text{cm}^3$ of alkali.
Apparatus: a $cm350\,\text{cm}^3$ burette, a clamp and stand, a small funnel, a $cm325.0\,\text{cm}^3$ pipette with a pipette filler, a $cm3250\,\text{cm}^3$ conical flask, a white tile, an indicator such as phenolphthalein or methyl orange, the acid, the alkali and a wash bottle of distilled water.

Part 1: setting up

Rinse the burette with a little of the acid, then clamp it upright.
Fill the burette with acid through a funnel held below eye level, then remove the funnel.
Run a little acid out so that the tip has no air bubble.
Use the pipette and pipette filler to transfer $cm325.0\,\text{cm}^3$ of alkali into the conical flask, then add a few drops of indicator.
Stand the flask on the white tile under the burette.

Part 2: titrating

Record the initial burette reading at eye level, from the bottom of the meniscus.
Do a rough titration first, adding acid quickly while swirling until the indicator changes colour.
Refill the flask with fresh alkali and indicator, and run in acid to about $cm31\,\text{cm}^3$ less than the rough titre.
Add acid drop by drop while swirling, until one drop gives a permanent colour change.
Record the final reading to the nearest $cm30.05\,\text{cm}^3$ and work out the titre.
Repeat until at least two titres are concordant, within $cm30.10\,\text{cm}^3$ of each other, and use only those for the mean.

Accuracy and safety

Rinse the burette with the acid and the pipette with the alkali, so that leftover water does not dilute them.
Swirl the flask all the time and rinse its sides with distilled water near the end, so that all the acid added can react.
Read every volume at eye level, from the bottom of the meniscus.
Wear eye protection, because acids and alkalis are irritant or corrosive.

Concordant titres are averaged to give a reliable titre

Definition

Concordant results

Repeat titres that agree closely with each other, usually within 0.10 cm³.

Repeat the accurate titration until at least two titres are concordant, which usually means within $cm30.10\,\text{cm}^3$ of each other.
Leave the rough titre out of the mean, because the solution was added too quickly for it to be accurate.
Also leave out any anomalous titre, such as one where the end point was overshot.
Calculate the mean titre from the concordant results only.
Concordant results show that the method is repeatable, so the mean titre is a trustworthy value.

Example

Titres of $cm323.10\,\text{cm}^3$ (rough), $cm322.50\,\text{cm}^3$ , $cm322.40\,\text{cm}^3$ and $cm322.45\,\text{cm}^3$ are recorded; find the mean titre.
Leave out the rough titre of $cm323.10\,\text{cm}^3$ .
The other three titres all lie within $cm30.10\,\text{cm}^3$ of each other, so they are concordant.
Mean titre $\dfrac{22.50 + 22.40 + 22.45}{3} = 22.45\,\text{cm}^3$ .

Self review

Why is the pipette rinsed with the solution it will measure, but the conical flask rinsed only with distilled water?
What colour change does phenolphthalein show when acid is added to alkali in the flask?
Why is universal indicator not used in a titration?
Why is the solution added dropwise near the end point?
Which titres are used to calculate the mean titre?

5.1.2 Theoretical mass, percentage yield and atom economy

Scaling reacting masses gives the theoretical yield

Definition

Theoretical yield

The maximum mass of product that could form from a given mass of reactant, calculated from the balanced equation.

Definition

Relative formula mass

The mass of one formula unit of a substance compared with one-twelfth of the mass of a carbon-12 atom, given the symbol Mr.

A balanced equation shows the reacting masses of each substance when you replace each formula with its $M_{r}$ .
1. Mass is conserved, so these reacting masses always add up to the same total on both sides.
The reacting masses stay in the same ratio whatever mass of reactant is used.
To find the theoretical yield, multiply the mass of reactant by the mass ratio of product to reactant.
This gives the greatest mass of product possible, because it assumes all the reactant turns into product.

Example

Calculate the theoretical yield of calcium oxide when $g10.0\,\text{g}$ of calcium carbonate is heated: $CaCO3→CaO+CO2\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2$ ( $A_r$ : Ca = 40.1, C = 12.0, O = 16.0).
Work out the relative formula masses: $Mr(CaCO3)=40.1+12.0+(3×16.0)=100.1M_r(\text{CaCO}_3) = 40.1 + 12.0 + (3 \times 16.0) = 100.1$ and $Mr(CaO)=40.1+16.0=56.1M_r(\text{CaO}) = 40.1 + 16.0 = 56.1$ .
Write the reacting masses: $g100.1\,\text{g}$ of $CaCO3\text{CaCO}_3$ gives $g56.1\,\text{g}$ of $CaO\text{CaO}$ .
Scale to the mass used: $10.0 \times \dfrac{56.1}{100.1} = 5.60\,\text{g}$ of calcium oxide.

Multiply relative formula masses by the balancing numbers

A balancing number in front of a formula multiplies the whole formula, so it also multiplies that substance's reacting mass.
In $2Mg+O2→2MgO2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}$ $2 Mg + O_{2} \to 2 MgO$ , the reacting masses are $2×24.3=48.62 \times 24.3 = 48.6$ $2 \times 24.3 = 48.6$ for magnesium and $2×40.3=80.62 \times 40.3 = 80.6$ $2 \times 40.3 = 80.6$ for magnesium oxide.
1. So $g6.00\,\text{g}$ of magnesium gives a theoretical yield of $6.00 \times \dfrac{80.6}{48.6} = 9.95\,\text{g}$ of magnesium oxide.
Only the reactant you are given and the product you want are needed, so other substances in the equation can be ignored.
Round only at the end, and give the answer with a unit of mass, because relative formula masses themselves have no units.

Example

Calculate the theoretical yield of iron from $g80.0\,\text{g}$ of iron(III) oxide: $Fe2O3+3CO→2Fe+3CO2\text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2$ ( $A_r$ : Fe = 55.8, O = 16.0).
Work out the reacting masses: $Fe2O3=(2×55.8)+(3×16.0)=159.6\text{Fe}_2\text{O}_3 = (2 \times 55.8) + (3 \times 16.0) = 159.6$ and $2Fe=2×55.8=111.62\text{Fe} = 2 \times 55.8 = 111.6$ .
Scale to the mass used: $80.0 \times \dfrac{111.6}{159.6} = 55.9\,\text{g}$ of iron.

Percentage yield compares actual yield with theoretical yield

Definition

Actual yield

The mass of product that is collected when a reaction is carried out.

Definition

Percentage yield

The actual yield of a product expressed as a percentage of its theoretical yield.

Percentage yield is calculated with this equation: $yield×100\text{percentage yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100$
The actual yield is almost always less than the theoretical yield.
1. The reaction may be reversible or may not go to completion, so some reactant never turns into product.
2. Side reactions can turn some of the reactant into other, unwanted products.
3. Some product is lost when it is filtered, transferred between containers or purified.
A percentage yield above $100%100\%$ suggests the product was still wet or impure when it was weighed.

Example

The theoretical yield of calcium oxide is $g5.60\,\text{g}$ , but only $g4.48\,\text{g}$ is collected; calculate the percentage yield.
Substitute into the equation: $4.485.60×100\dfrac{4.48}{5.60} \times 100$ .
Percentage yield $80.0\%$ .

Atom economy shows how much reactant mass becomes product

Definition

Atom economy

The percentage of the total mass of reactants that ends up in the desired product, calculated from the balanced equation.

Definition

Desired product

The useful product that a reaction is carried out to make.

Atom economy is calculated from the balanced equation: $products×100\text{atom economy} = \frac{M_r \text{ of desired product}}{\text{sum of } M_r \text{ of all products}} \times 100$
Multiply each $M_{r}$ by its balancing number before adding, as for reacting masses.
1. The total mass of the products equals the total mass of the reactants, so the bottom line can also be the sum of the reactants' $M_r$ values.
A reaction with only one product has an atom economy of $100%100\%$ , because every atom ends up in the desired product.
A low atom economy means much of the reactant mass becomes waste products, which wastes raw materials unless the by-products can be used or sold.

Example

Calculate the atom economy for making calcium oxide: $CaCO3→CaO+CO2\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2$ ( $M_r$ : CaO = 56.1, $CO2\text{CO}_2$ = 44.0).
Add the $M_r$ values of all the products: $56.1 + 44.0 = 100.1$ .
Atom economy $\dfrac{56.1}{100.1} \times 100 = 56.0\%$ .

Percentage yield and atom economy measure different things

Percentage yield depends on how well the reaction is carried out, so it can change from one attempt to the next.
Atom economy depends only on the balanced equation, so it is fixed for a given reaction.
A reaction can have a high atom economy but a low percentage yield.
1. Making ethanol from ethene and steam has an atom economy of $100%100\%$ , yet only a small fraction of the gases react each time they pass through the reactor.
Chemists want both values to be high, because together they mean less waste and more product from the same raw materials.

Exam technique

When a question asks for percentage yield, look for a measured mass of product, but when it asks for atom economy, use only the $M_r$ values from the balanced equation.
Show the reacting masses or $M_r$ values you used, then give the final answer with its unit or percentage sign.

Self review

What is the theoretical yield of a reaction?
What mass of calcium oxide could form from $g20.0\,\text{g}$ of calcium carbonate ( $M_r$ : $CaCO3\text{CaCO}_3$ = 100.1, CaO = 56.1)?
Give two reasons why the actual yield is usually less than the theoretical yield.
Why does a reaction with only one product have an atom economy of $100%100\%$ ?
Which of percentage yield and atom economy can be calculated without doing the experiment?

5.1.3 Concentration of a solution in mol/dm3

Concentration measures solute per volume of solution

Definition

Concentration

The amount of solute dissolved in a stated volume of solution, measured in g/dm³ or mol/dm³.

Definition

Solute

The substance that dissolves in a solvent to make a solution.

A solution forms when a solute dissolves in a solvent, such as salt dissolving in water.
A concentrated solution has a lot of solute in a given volume, while a dilute solution has only a little.
Concentration can be measured in grams per cubic decimetre, written $g/dm3\text{g/dm}^3$ $g/dm^{3}$ .
1. One cubic decimetre is $cm31000\,\text{cm}^3$ , which is the same as one litre.
The volume used is the volume of the whole solution, not the volume of water that was added.
Concentration can also be given in $mol/dm3\text{mol/dm}^3$ , which uses the amount in moles of solute instead of its mass.

Key Idea

Concentration depends on two things only: how much solute there is and how much solution it is spread through.

Concentration equals mass of solute over volume

The relationship is: $(dm3)\text{concentration (g/dm}^3\text{)} = \frac{\text{mass of solute (g)}}{\text{volume of solution (dm}^3\text{)}}$
Convert a volume in $cm3\text{cm}^3$ $cm^{3}$ into $dm3\text{dm}^3$ $dm^{3}$ by dividing by 1000 before using the equation.
1. So $cm3250\,\text{cm}^3$ is $dm30.250\,\text{dm}^3$ .
Rearranging gives mass of solute $=$ concentration $×\times$ $\times$ volume, which tells you how much solid to weigh out.
1. Making $dm32.00\,\text{dm}^3$ of a $g/dm315.0\,\text{g/dm}^3$ solution needs $15.0 \times 2.00 = 30.0\,\text{g}$ of solute.
Give every answer with its unit, because a number alone does not say whether it is a mass, a volume or a concentration.

Example

Calculate the concentration of a solution made by dissolving $g5.00\,\text{g}$ of sodium chloride in water to make $cm3250\,\text{cm}^3$ of solution.
Convert the volume: $dm3250\,\text{cm}^3 \div 1000 = 0.250\,\text{dm}^3$ .
Divide mass by volume: $g/dm3\dfrac{5.00\,\text{g}}{0.250\,\text{dm}^3} = 20.0\,\text{g/dm}^3$ .

How solute mass and volume affect concentration

At a fixed volume, concentration is directly proportional to the mass of solute.
1. Dissolving twice the mass of solute in the same volume doubles the concentration.
For a fixed mass of solute, a larger volume of solution gives a lower concentration.
1. Spreading the same mass through twice the volume halves the concentration.
Doubling both the mass and the volume leaves the concentration unchanged.
A sample of any size taken from a well-mixed solution has the same concentration as the rest of it.

Example

Adding more squash to the same glass of water makes a stronger drink, while topping the glass up with water makes it weaker, although the amount of squash in the glass has not changed.

Diluting a solution lowers its concentration

When water is added, the mass of solute stays the same.
The volume of solution increases, so the concentration falls.
Evaporating some of the water has the opposite effect, raising the concentration because the volume decreases while the solute stays.
To find a new concentration, first work out the mass of solute present, then divide it by the new volume.

Example

Water is added to $cm350.0\,\text{cm}^3$ of a $g/dm340.0\,\text{g/dm}^3$ solution until its volume is $cm3200\,\text{cm}^3$ ; find the new concentration.
Mass of solute $40.0\,\text{g/dm}^3 \times 0.0500\,\text{dm}^3 = 2.00\,\text{g}$ .
New concentration $\dfrac{2.00\,\text{g}}{0.200\,\text{dm}^3} = 10.0\,\text{g/dm}^3$ .

Making a solution of known concentration

Weigh the solid accurately on a balance in a weighing boat or small beaker.
Dissolve it in a small volume of distilled water in a beaker, stirring until no solid remains.
Pour the solution into a volumetric flask and rinse the beaker and stirring rod into the flask, so no solute is left behind.
Add distilled water until the bottom of the meniscus sits on the line, then stopper and invert the flask to mix.
Divide the mass weighed by the volume of the flask in $dm3\text{dm}^3$ to find the concentration.

Exam technique

When a question asks how a change affects concentration, say what happens to the mass of solute and what happens to the volume of solution, then state the effect.

Self review

What does a concentration of $g/dm320.0\,\text{g/dm}^3$ mean?
What is the concentration of $g3.00\,\text{g}$ of solute in $cm3150\,\text{cm}^3$ of solution?
What happens to the concentration if the same mass of solute is dissolved in twice the volume?
Why does adding water lower the concentration of a solution?
What mass of solute is needed to make $cm3500\,\text{cm}^3$ of a $g/dm312.0\,\text{g/dm}^3$ solution?

5.1.4 Concentration calculations and titration results

Concentration is moles of solute per cubic decimetre

Definition

Concentration

The amount of solute dissolved in a stated volume of solution, measured in g/dm³ or mol/dm³.

The relationship is: $(dm3)\text{concentration (mol/dm}^3\text{)} = \frac{\text{amount of solute (mol)}}{\text{volume of solution (dm}^3\text{)}}$
Convert a volume in $cm3\text{cm}^3$ into $dm3\text{dm}^3$ by dividing by 1000.
Rearranging gives amount in moles $=$ concentration $×\times$ $\times$ volume in $dm3\text{dm}^3$ $dm^{3}$ .
1. So $cm320.0\,\text{cm}^3$ of a $mol/dm30.100\,\text{mol/dm}^3$ solution contains $0.100 \times 0.0200 = 0.00200\,\text{mol}$ of solute.
Doubling the amount of solute in the same volume doubles the concentration, while doubling the volume for the same amount halves it.

Example

A solution contains $mol0.0500\,\text{mol}$ of sodium chloride in $cm3250\,\text{cm}^3$ ; calculate its concentration.
Convert the volume: $250 \div 1000 = 0.250\,\text{dm}^3$ .
Concentration $\dfrac{0.0500}{0.250} = 0.200\,\text{mol/dm}^3$ .

Relative formula mass links solute mass and concentration

Definition

Relative formula mass

The mass of one formula unit of a substance compared with one-twelfth of the mass of a carbon-12 atom, given the symbol Mr.

Convert a mass of solute into moles with amount $\dfrac{\text{mass (g)}}{M_r}$ .
Then divide the amount by the volume of solution in $dm3\text{dm}^3$ to find the concentration in $mol/dm3\text{mol/dm}^3$ .
Working backwards, mass of solute $=$ concentration $×\times$ $\times$ volume $×Mr\times M_r$ $\times M_{r}$ .
1. Making $cm3500\,\text{cm}^3$ of $mol/dm30.100\,\text{mol/dm}^3$ copper(II) sulfate, $M_r = 159.6$ , needs $0.100 \times 0.500 \times 159.6 = 7.98\,\text{g}$ of solid.
To change a concentration in $mol/dm3\text{mol/dm}^3$ into $g/dm3\text{g/dm}^3$ , multiply by $M_r$ , and to go the other way, divide by $M_r$ .

Example

Calculate the concentration in $mol/dm3\text{mol/dm}^3$ of a solution made by dissolving $g4.00\,\text{g}$ of sodium hydroxide to make $cm3250\,\text{cm}^3$ of solution ( $A_r$ : Na = 23.0, O = 16.0, H = 1.0).
$Mr(NaOH)=23.0+16.0+1.0=40.0M_r(\text{NaOH}) = 23.0 + 16.0 + 1.0 = 40.0$ .
Amount $\dfrac{4.00}{40.0} = 0.100\,\text{mol}$ .
Concentration $\dfrac{0.100}{0.250} = 0.400\,\text{mol/dm}^3$ .

The mole ratio links reacting volumes and concentrations

Definition

Mole ratio

The ratio of the amounts in moles of substances that react or form, given by the balancing numbers in the equation.

At the end point, the two substances have reacted completely, so their amounts in moles are in the ratio shown by the balanced equation.
The amount of the substance with a known concentration comes from concentration $×\times$ volume.
In $HCl(aq)+NaOH(aq)→NaCl(aq)+H2O(l)\text{HCl(aq)} + \text{NaOH(aq)} \rightarrow \text{NaCl(aq)} + \text{H}_2\text{O(l)}$ $HCl(aq) + NaOH(aq) \to NaCl(aq) + H_{2} O(l)$ , the ratio is 1 : 1, so equal volumes react when the concentrations are equal.
1. If the sodium hydroxide were twice as concentrated, only half the volume would be needed, because each $cm3\text{cm}^3$ would hold twice as many moles.
In $H2SO4(aq)+2NaOH(aq)→Na2SO4(aq)+2H2O(l)\text{H}_2\text{SO}_4\text{(aq)} + 2\text{NaOH(aq)} \rightarrow \text{Na}_2\text{SO}_4\text{(aq)} + 2\text{H}_2\text{O(l)}$ $H_{2} SO_{4} (aq) + 2 NaOH(aq) \to Na_{2} SO_{4} (aq) + 2 H_{2} O(l)$ , each mole of acid reacts with two moles of alkali.
1. So for solutions of equal concentration, the volume of sodium hydroxide needed is twice the volume of sulfuric acid.

Common Mistake

Do not assume every acid reacts with an alkali in a 1 : 1 ratio; always read the ratio from the balanced equation.
Do not put a volume in $cm3\text{cm}^3$ into concentration $×\times$ volume, because the concentration is per $dm3\text{dm}^3$ .

Titration results give an unknown concentration

Definition

Titration

A technique that finds the volume of one solution needed to react completely with a measured volume of another solution.

The pipette gives the fixed volume of one solution and the mean titre gives the volume of the other.
Write the balanced equation first, so you know the mole ratio.
Find the amount in moles of the solution whose concentration is known.
Use the mole ratio to find the amount in moles of the other substance.
Divide that amount by its volume in $dm3\text{dm}^3$ to find the unknown concentration.

Example

Concentration of sodium hydroxide from a titration

$cm325.0\,\text{cm}^3$ of sodium hydroxide solution is neutralised by a mean titre of $cm320.00\,\text{cm}^3$ of $mol/dm30.100\,\text{mol/dm}^3$ sulfuric acid; find the concentration of the sodium hydroxide.
Amount of $mol\text{H}_2\text{SO}_4 = 0.100 \times 0.02000 = 0.002000\,\text{mol}$ .
The ratio $H2SO4:NaOH\text{H}_2\text{SO}_4 : \text{NaOH}$ is $1 : 2$ , so the amount of $mol\text{NaOH} = 2 \times 0.002000 = 0.004000\,\text{mol}$ .
Concentration of $mol/dm3\text{NaOH} = \dfrac{0.004000}{0.0250} = 0.160\,\text{mol/dm}^3$ .

Finding required volumes and mass concentrations

To find a volume needed, divide the amount in moles required by the concentration of that solution.
1. $cm325.0\,\text{cm}^3$ of $mol/dm30.100\,\text{mol/dm}^3$ sodium hydroxide contains $mol0.00250\,\text{mol}$ , so it needs $mol0.00250\,\text{mol}$ of hydrochloric acid.
2. With $mol/dm30.200\,\text{mol/dm}^3$ acid, the volume needed is $dm3\dfrac{0.00250}{0.200} = 0.0125\,\text{dm}^3$ , which is $cm312.5\,\text{cm}^3$ .
Convert a volume in $dm3\text{dm}^3$ back to $cm3\text{cm}^3$ by multiplying by 1000.
To give an answer in $g/dm3\text{g/dm}^3$ $g/dm^{3}$ , multiply the concentration in $mol/dm3\text{mol/dm}^3$ $mol/dm^{3}$ by $M_{r}$ .
1. The $mol/dm30.160\,\text{mol/dm}^3$ sodium hydroxide above has a concentration of $0.160 \times 40.0 = 6.40\,\text{g/dm}^3$ .

Exam technique

When a question asks for a concentration from titration results, write out each link of the chain, volume to moles, moles to moles by the ratio, then moles to concentration, with a unit on every value.
Check that the answer fits the equation, so in a 1 : 2 reaction the amount in moles of the second substance must be twice the amount of the first.

Self review

What is the concentration of $mol0.600\,\text{mol}$ of solute in $cm3250\,\text{cm}^3$ of solution?
How many moles of solute are in $cm330.0\,\text{cm}^3$ of a $mol/dm30.200\,\text{mol/dm}^3$ solution?
How do you convert a concentration in $mol/dm3\text{mol/dm}^3$ into $g/dm3\text{g/dm}^3$ ?
Why does the mole ratio in the balanced equation matter in a titration calculation?
$cm320.0\,\text{cm}^3$ of $mol/dm30.150\,\text{mol/dm}^3$ hydrochloric acid neutralises $cm325.0\,\text{cm}^3$ of potassium hydroxide solution; what is the concentration of the potassium hydroxide?

5.1.5 Molar gas volume and volumes of gases in reactions

Equal moles of any gas occupy equal volumes

Definition

Molar gas volume

The volume occupied by one mole of any gas at a stated temperature and pressure, which is 24 dm³ at room temperature and pressure.

Gas particles are very far apart, so the volume of a gas depends on the number of particles and not on the size of each particle.
At the same temperature and pressure, the volume of a gas is directly proportional to its amount in moles.
1. Doubling the amount of gas doubles its volume, and halving it halves the volume.
At room temperature and pressure, about $20 ∘C20\ ^{\circ}\text{C}$ $20^{\circ} C$ and 1 atmosphere, one mole of any gas occupies $24 dm324\,\text{dm}^3$ $24 dm^{3}$ .
1. This is the same as $cm324\,000\,\text{cm}^3$ , whether the gas is hydrogen, oxygen or carbon dioxide.

Graph showing that the volume of a gas is directly proportional to the number of moles (V = constant × n), represented by a straight line passing through the origin.

Key Idea

One mole of hydrogen weighs $g2.0\,\text{g}$ and one mole of carbon dioxide weighs $g44.0\,\text{g}$ , yet at room temperature and pressure each fills the same $dm324\,\text{dm}^3$ .

Converting gas volume and moles using 24 dm³

To find a gas volume, multiply the amount in moles by $24$ : $(mol)×24\text{volume (dm}^3\text{)} = \text{amount (mol)} \times 24$
To find an amount in moles, divide the gas volume in $dm3\text{dm}^3$ $dm^{3}$ by $24$ .
1. For a volume in $cm3\text{cm}^3$ , divide by $00024\,000$ instead, or convert it into $dm3\text{dm}^3$ first.
So $mol0.25\,\text{mol}$ of carbon dioxide occupies $0.25 \times 24 = 6.0\,\text{dm}^3$ at room temperature and pressure.
The value of $dm324\,\text{dm}^3$ applies only at room temperature and pressure, so a different temperature or pressure gives a different volume.

Example

Calculate the amount in moles of oxygen in $cm3480\,\text{cm}^3$ of the gas at room temperature and pressure.
Convert the volume: $dm3480\,\text{cm}^3 \div 1000 = 0.480\,\text{dm}^3$ .
Amount $\dfrac{0.480}{24} = 0.0200\,\text{mol}$ .

Balancing numbers give the ratio of gas volumes

Definition

Mole ratio

The ratio of the amounts in moles of substances that react or form, given by the balancing numbers in the equation.

The balancing numbers give the mole ratio, and equal amounts of gases have equal volumes, so for gases they also give the volume ratio.
In $CH4(g)+2O2(g)→CO2(g)+2H2O(l)\text{CH}_4\text{(g)} + 2\text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)} + 2\text{H}_2\text{O(l)}$ $CH_{4} (g) + 2 O_{2} (g) \to CO_{2} (g) + 2 H_{2} O(l)$ , the volumes of methane, oxygen and carbon dioxide are in the ratio $1 : 2 : 1$ .
1. So $cm350\,\text{cm}^3$ of methane reacts with $cm3100\,\text{cm}^3$ of oxygen and makes $cm350\,\text{cm}^3$ of carbon dioxide.
Water is a liquid at room temperature, so it is left out of the gas volume ratio.
In $2H2(g)+O2(g)→2H2O(l)2\text{H}_2\text{(g)} + \text{O}_2\text{(g)} \rightarrow 2\text{H}_2\text{O(l)}$ , $cm340\,\text{cm}^3$ of hydrogen needs half its volume, $cm320\,\text{cm}^3$ , of oxygen.

Common Mistake

Do not use the balancing numbers as a volume ratio for solids, liquids or solutions, because the rule applies only to gases measured at the same temperature and pressure.

Finding gas volume from reactant mass via moles

First convert the mass of reactant into moles by dividing by its $M_r$ .
Next use the mole ratio from the balanced equation to find the amount in moles of gas formed.
Then multiply the amount of gas by $24$ to find its volume in $dm3\text{dm}^3$ at room temperature and pressure.
Give the answer to a sensible number of significant figures, usually matching the data in the question.

Example

Calculate the volume of carbon dioxide at room temperature and pressure when $g5.00\,\text{g}$ of calcium carbonate is heated: $CaCO3(s)→CaO(s)+CO2(g)\text{CaCO}_3\text{(s)} \rightarrow \text{CaO(s)} + \text{CO}_2\text{(g)}$ ( $A_r$ : Ca = 40.1, C = 12.0, O = 16.0).
$Mr(CaCO3)=40.1+12.0+(3×16.0)=100.1M_r(\text{CaCO}_3) = 40.1 + 12.0 + (3 \times 16.0) = 100.1$ , so the amount $\dfrac{5.00}{100.1} = 0.04995\,\text{mol}$ .
The ratio $CaCO3:CO2\text{CaCO}_3 : \text{CO}_2$ is $1 : 1$ , so $mol0.04995\,\text{mol}$ of carbon dioxide forms.
Volume $\times 24 = 1.20\,\text{dm}^3$ .

Working back from gas volume to reactant mass

Divide the gas volume in $dm3\text{dm}^3$ by $24$ to find the amount of gas in moles.
Use the mole ratio to find the amount in moles of the reactant.
Multiply by the reactant's $A_{r}$ or $M_{r}$ to find its mass in grams.
1. In $Mg(s)+2HCl(aq)→MgCl2(aq)+H2(g)\text{Mg(s)} + 2\text{HCl(aq)} \rightarrow \text{MgCl}_2\text{(aq)} + \text{H}_2\text{(g)}$ , $cm3180\,\text{cm}^3$ of hydrogen is $mol\dfrac{0.180}{24} = 0.00750\,\text{mol}$ .
2. The ratio $Mg:H2\text{Mg} : \text{H}_2$ is $1 : 1$ , so the mass of magnesium that reacted is $0.00750 \times 24.3 = 0.182\,\text{g}$ .

Exam technique

When a question gives a gas volume in $cm3\text{cm}^3$ , convert it into $dm3\text{dm}^3$ before dividing by $24$ , and show that step, because it is the step most often missed.

Self review

What volume does one mole of any gas occupy at room temperature and pressure?
What is the volume of $mol0.150\,\text{mol}$ of nitrogen at room temperature and pressure?
How many moles of gas are in $cm31200\,\text{cm}^3$ at room temperature and pressure?
In $CH4(g)+2O2(g)→CO2(g)+2H2O(l)\text{CH}_4\text{(g)} + 2\text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)} + 2\text{H}_2\text{O(l)}$ , what volume of oxygen reacts with $cm330\,\text{cm}^3$ of methane?
Why is water left out of the volume ratio in this equation?

5.1.6 Choosing a reaction pathway

One product can be made by different pathways

Definition

Reaction pathway

The route of one or more reactions, with its starting materials and conditions, used to make a particular product.

Ethanol can be made by fermentation of sugars with yeast, or by reacting ethene with steam, which is called hydration.
Each pathway uses different starting materials, conditions and numbers of steps.
Chemists compare the pathways using data rather than choosing the one that seems simplest.
The data usually cover yield and atom economy, rate, conditions and energy use, raw materials, by-products and how pure the product is.

Key Idea

The best pathway is the one that gives the most suitable overall balance of these factors for the product and its use, even if it is not the best on every one.

Yield and atom economy measure how much becomes product

Definition

By-product

A substance formed in a reaction alongside the desired product.

A high percentage yield means that less of the reactant is lost or left unreacted.
A high atom economy means that less of the reactant mass turns into by-products.
A pathway with fewer steps usually loses less product overall, because some product is lost at every step.
A low atom economy matters less if the by-product is useful or saleable, and more if it is harmful and must be treated.
Unreacted starting materials can sometimes be recycled through the reactor, which improves the overall yield.

Example

Compare the atom economies of the two pathways to ethanol: $C6H12O6→2C2H5OH+2CO2\text{C}_6\text{H}_{12}\text{O}_6 \rightarrow 2\text{C}_2\text{H}_5\text{OH} + 2\text{CO}_2$ and $C2H4+H2O→C2H5OH\text{C}_2\text{H}_4 + \text{H}_2\text{O} \rightarrow \text{C}_2\text{H}_5\text{OH}$ ( $M_r$ : glucose = 180.0, ethanol = 46.0).
For fermentation, atom economy $\dfrac{2 \times 46.0}{180.0} \times 100 = 51.1\%$ .
For hydration, ethanol is the only product, so the atom economy is $100%100\%$ .
Hydration wastes far less of the reactant mass, while almost half the mass in fermentation leaves as carbon dioxide.

Rate, conditions and energy decide a pathway's cost

A fast reaction makes more product in a given time, so less equipment is needed for the same output.
High temperatures and pressures need more energy and stronger, more expensive equipment.
A catalyst can let a pathway run fast enough at a lower temperature, which saves energy.
A continuous process, fed with reactants all the time, makes product faster than a batch that is started and stopped.
1. Hydration of ethene runs continuously and quickly, while fermentation is a slow batch process that takes several days.

Example

Raw materials, purification and safety can outweigh yield

A pathway is more sustainable if its raw materials are renewable, cheap and easy to obtain.
1. Sugar for fermentation comes from crops such as sugar cane, which can be regrown, while ethene comes from crude oil, which is finite.
A pathway that gives a purer product needs less separation afterwards.
1. Fermentation gives a dilute solution of ethanol that must be separated by fractional distillation, while hydration gives almost pure ethanol.
Hazardous substances, flammable gases and harmful waste add safety and disposal costs to a pathway.

Note

Justifying a pathway choice with data

State which pathway you would choose for the specified product.
Quote at least two pieces of data that support the choice.
Explain what each value means in practice, such as less waste, lower energy costs or more product per hour.
Say why the other pathway is less suitable, even if it has one advantage.
Finish with a judgement linked to the stated priority, such as low cost, speed or sustainability.

Exam technique

Self review

What is meant by a reaction pathway?
Why does fermentation have a lower atom economy than the hydration of ethene?
Why can a pathway with fewer steps give a greater overall yield?
Give one advantage of fermentation over the hydration of ethene.
Why might a pathway with a lower percentage yield still be chosen?

5.1 Monitoring chemical reactions

5.1.1 The technique of titration

A titration measures the volume that reacts completely

Pipettes measure fixed volumes and burettes measure variable volumes

An indicator's sharp colour change shows the end point

A rough titration comes before accurate dropwise titrations

Part 1: setting up

Part 2: titrating

Accuracy and safety

Concordant titres are averaged to give a reliable titre

5.1.2 Theoretical mass, percentage yield and atom economy

Scaling reacting masses gives the theoretical yield

Multiply relative formula masses by the balancing numbers

Percentage yield compares actual yield with theoretical yield

Atom economy shows how much reactant mass becomes product

Percentage yield and atom economy measure different things

5.1.3 Concentration of a solution in mol/dm3

Concentration measures solute per volume of solution

Concentration equals mass of solute over volume

How solute mass and volume affect concentration

Diluting a solution lowers its concentration

Making a solution of known concentration

5.1.4 Concentration calculations and titration results

Concentration is moles of solute per cubic decimetre

Relative formula mass links solute mass and concentration

The mole ratio links reacting volumes and concentrations

Titration results give an unknown concentration

Concentration of sodium hydroxide from a titration

Finding required volumes and mass concentrations

5.1.5 Molar gas volume and volumes of gases in reactions

Equal moles of any gas occupy equal volumes

Converting gas volume and moles using 24 dm³

Balancing numbers give the ratio of gas volumes

Finding gas volume from reactant mass via moles

Working back from gas volume to reactant mass

5.1.6 Choosing a reaction pathway

One product can be made by different pathways

Yield and atom economy measure how much becomes product

Rate, conditions and energy decide a pathway's cost

Raw materials, purification and safety can outweigh yield

Justifying a pathway choice with data

Review 5.1 Monitoring chemical reactions by teaching Genie

Recap your knowledge with an interactive lesson

Put it into practice with exam-style questions

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5.1 Monitoring chemical reactions Revision Guide

5.1 Monitoring chemical reactions

5.1.1 The technique of titration

A titration measures the volume that reacts completely

Pipettes measure fixed volumes and burettes measure variable volumes

An indicator's sharp colour change shows the end point

A rough titration comes before accurate dropwise titrations

Part 1: setting up

Part 2: titrating

Accuracy and safety

Concordant titres are averaged to give a reliable titre

5.1.2 Theoretical mass, percentage yield and atom economy

Scaling reacting masses gives the theoretical yield

Multiply relative formula masses by the balancing numbers

Percentage yield compares actual yield with theoretical yield

Atom economy shows how much reactant mass becomes product

Percentage yield and atom economy measure different things

5.1.3 Concentration of a solution in mol/dm3

Concentration measures solute per volume of solution

Concentration equals mass of solute over volume

How solute mass and volume affect concentration

Diluting a solution lowers its concentration

Making a solution of known concentration

5.1.4 Concentration calculations and titration results

Concentration is moles of solute per cubic decimetre

Relative formula mass links solute mass and concentration

The mole ratio links reacting volumes and concentrations

Titration results give an unknown concentration

Concentration of sodium hydroxide from a titration

Finding required volumes and mass concentrations

5.1.5 Molar gas volume and volumes of gases in reactions

Equal moles of any gas occupy equal volumes

Converting gas volume and moles using 24 dm³

Balancing numbers give the ratio of gas volumes

Finding gas volume from reactant mass via moles