4.3 Quantitative chemistry — 8462 revision notes

4.3.1.1

Conservation of mass and balanced chemical equations

Notes

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Definition

The law of conservation of mass states that no atoms are lost or made during a chemical reaction, so the total mass of the products equals the total mass of the reactants.

Explanation

Because atoms are only rearranged, a chemical reaction can always be written as a symbol equation balanced by the number of atoms of each element on both sides. Two kinds of number appear and they mean different things:
a multiplier in front of a formula multiplies the whole formula, while a subscript inside fixes that substance's composition and may never be altered. Balancing means adjusting only the front multipliers.
A reliable order is the metal or least common element first, then the other non-metals, then hydrogen, then oxygen last, because oxygen usually appears in more than one product.
Apparent mass changes in non-enclosed systems are the standard twist: mass appears to decrease when a gas escapes, and to increase when a gas from the air, usually oxygen, is taken in; in a closed system mass is unchanged.
To show mass is conserved, add the relative formula masses on each side, including multipliers, and state that the totals are equal.

Mass is conserved in both flasks; only the open flask appears to lose mass, because the gas produced leaves the system.

Worked example

Methane reacts completely with oxygen in a sealed vessel. The reactants have masses $4.0\,\mathrm{g}$ and $16.0\,\mathrm{g}$ . One product is $11.0\,\mathrm{g}$ of carbon dioxide. Calculate the mass of water formed.

1.The sealed vessel contains every reactant and product, so total mass is conserved. The reactant mass is $4.0+16.0=20.0\,\mathrm{g}$ . Therefore the water mass is $20.0-11.0=9.0\,\mathrm{g}$ .

Answer: Mass of water $=9.0\,\mathrm{g}$

Common mistakes

Don't change a subscript while balancing, which changes the substance itself.
Don't balance one element and fail to recount every element on both sides.

Mark-scheme wording to use

“left = 232 + 6 = 238. 1 mark: right = 184 + 54 = 238 and equality shown” — AQA 8462, June 2025 Paper 1F, Q10.5
“2C₄H₆O₂ + 9O₂ → 8CO₂ + 6H₂O; allow multiples … Ignore state symbols” — AQA 8462, June 2024 Paper 2H, Q08.4
“allow 1 mark for correct reactant formulae with missing/incorrect coefficients and 1 mark for correct product formulae with missing/incorrect coefficients” — AQA 8462, June 2024 Paper 2H, Q08.4

What the examiner wants

Both totals written out and the equality stated in words. Getting the right formulae still earns partial credit when the coefficients are wrong, so always write the formulae down even if the balancing defeats you.

Exam tip

For a balancing question, alter coefficients only and finish by recounting every element on both sides.

Tier 1 · Easy

ORIGINAL

Insert the smallest whole-number coefficients to balance

\mathrm{Al}+\mathrm{O}_2\rightarrow\mathrm{Al}_2\mathrm{O}_3

.

[1 mark]

Total for this question: 1

Tier 2 · Standard

ORIGINAL

A student tries to balance the equation

\mathrm{Mg}+\mathrm{O}_2\rightarrow\mathrm{MgO}

by changing it to

\mathrm{Mg}+\mathrm{O}_2\rightarrow\mathrm{MgO}_2

. Explain why the student's equation is incorrect. Write the correctly balanced equation.

[2 marks]

Total for this question: 2

Tier 3 · Hard

ORIGINAL

Propane burns according to

\mathrm{C}_3\mathrm{H}_8+5\mathrm{O}_2\rightarrow3\mathrm{CO}_2+4\mathrm{H}_2\mathrm{O}

. A sealed reaction uses

11.0\,\mathrm{g}

of propane and forms

33.0\,\mathrm{g}

of carbon dioxide plus

18.0\,\mathrm{g}

of water. Determine the mass of oxygen used and show that the masses obey conservation.

[3 marks]

Total for this question: 3

4.3.1.2

Relative formula mass

Notes

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Definition

The relative formula mass, Mr, of a compound is the sum of the relative atomic masses of all the atoms in the numbers shown in its formula.

Explanation

In a balanced equation, the sum of the relative formula masses of the reactants in the quantities shown equals the sum for the products — conservation of mass expressed in Mr. Two skills are examined: computing Mr from a formula, including brackets, and using it to find a percentage by mass.
For a formula with brackets such as Ca(OH)₂, multiply everything inside the bracket by the outside subscript before adding. Percentage by mass of an element is the total relative atomic mass contributed by that element, counting every atom of it, divided by the Mr of the compound, multiplied by 100.
Two habits protect marks. Write the substitution line out in full, because the scheme usually awards the expression separately from the final number.
And do not round Mr partway through: carry the full value and round only at the end.
The scheme routinely allows error carried forward, so a wrong Mr used correctly still earns the later marks.

Worked example

Ammonium nitrate is $\mathrm{NH}_4\mathrm{NO}_3$ . Calculate its percentage by mass of nitrogen. Use $A_r$ : $\mathrm{N}=14$ , $\mathrm{H}=1$ , $\mathrm{O}=16$ .

1. $M_r(\mathrm{NH}_4\mathrm{NO}_3)=2(14)+4(1)+3(16)=80$ . Nitrogen contributes $2(14)=28$ , so its percentage by mass is $(28/80)\times100=35.0\%$ .

Answer: Percentage nitrogen $=35.0\%$

Common mistakes

Don't forget to multiply every atom inside brackets by the outside subscript.
Don't use one atom’s relative mass as the denominator in a percentage-by-mass calculation instead of the compound’s $M_r$ .

Mark-scheme wording to use

“(Mr =) 63.5 + 56 + (2 × 32) = 183.5. 1 mark: (% of copper =) 63.5 / 183.5 × 100. 1 mark: = 34.6 (%) — allow correct use of an incorrectly determined Mr” — AQA 8462, June 2023 Paper 2H, Q05.2
“allow 34.60490 correctly rounded to at least 2 significant figures” — AQA 8462, June 2023 Paper 2H, Q05.2
“Mr(Fe₃O₄) = 232. 1 mark: 168/232 x 100, allowing correct use of an incorrectly determined Mr” — AQA 8462, June 2024 Paper 1H, Q07.5

What the examiner wants

The Mr on its own line, then the substituted fraction, then the answer. Error carried forward means a single arithmetic slip costs one mark rather than three — but only if the working is written down.

Exam tip

Write each atom’s contribution to $M_r$ before calculating a percentage by mass.

Tier 1 · Easy

ORIGINAL

Calculate the relative formula mass of

\mathrm{Ca(OH)}_2

. Use

A_r

:

\mathrm{Ca}=40

,

\mathrm{O}=16

,

\mathrm{H}=1

.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

Use the equation

2\mathrm{H}_2+\mathrm{O}_2\rightarrow2\mathrm{H}_2\mathrm{O}

to show that the total relative formula mass of the reactants equals the total relative formula mass of the products. Use

A_r

:

\mathrm{H}=1

,

\mathrm{O}=16

.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

A

17.1\,\mathrm{g}

sample contains only aluminium sulfate,

\mathrm{Al}_2(\mathrm{SO}_4)_3

. Calculate the mass of oxygen in the sample. Use

A_r

:

\mathrm{Al}=27

,

\mathrm{S}=32

,

\mathrm{O}=16

.

[4 marks]

Total for this question: 4

4.3.1.3

Mass changes when a reactant or product is a gas

Notes

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

An apparent mass change can occur in an open system when a gaseous reactant enters or a gaseous product escapes.
A metal can gain mass while reacting because oxygen particles from the air become part of the solid metal oxide.
A metal carbonate can lose measured mass on heating because carbon dioxide leaves, while the solid metal oxide remains.
Conservation of mass still holds when the gas is included; claiming that atoms or mass have disappeared is the common error.
Exam calculations should show the relationship used, substitution with units and a suitably rounded result.

Worked example

A student heats $12.5\,\mathrm{g}$ of calcium carbonate in an open tube. The equation is $\mathrm{CaCO}_3\rightarrow\mathrm{CaO}+\mathrm{CO}_2$ . The solid left has a mass of $7.0\,\mathrm{g}$ . Calculate the mass change and explain it using particles.

1.The decrease is $12.5-7.0=5.5\,\mathrm{g}$ . The equation shows that carbon dioxide gas is formed. Its particles leave the open tube, so the balance records only the calcium oxide; including the escaped gas would restore the original total mass.

Answer: The measured mass decreases by $5.5\,\mathrm{g}$ Carbon dioxide particles escape from the open tube

Common mistakes

Don't claim mass has been destroyed when a gaseous product escapes from an open container.
Don't explain a metal’s mass gain without including oxygen from the air as a reactant.

Exam tip

For an apparent mass change, name the gas and state whether its particles enter or leave the measured system.

Tier 1 · Easy

ORIGINAL

A strip of magnesium has a mass of

6.0\,\mathrm{g}

before heating in air and the magnesium oxide has a mass of

10.0\,\mathrm{g}

. Explain the increase and find the mass added.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

An effervescent tablet reacts with water in an open beaker. The mass recorded on a balance decreases during the reaction. Explain the decrease in mass. Predict what would happen to the total mass if the reaction took place in a sealed flask.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

A crucible and its contents gain

3.2\,\mathrm{g}

while a metal is converted fully into its oxide. Predict the change in the combined mass of the crucible, contents and surrounding sealed chamber, and account for both observations.

[4 marks]

Total for this question: 4

4.3.1.4

Chemical measurements

Notes

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

Every measured result has uncertainty because instruments have limited resolution and repeated readings vary. For repeats, calculate the mean after checking whether any result is anomalous and should be investigated.
A useful estimate is half the range, written as $\pm\dfrac{\text{maximum}-\text{minimum}}{2}$ about the mean.
Do not quote an uncertainty without a unit or keep unjustified extra decimal places in the reported result.
Repeated measurements reveal the spread and improve confidence in the mean.
Exam calculations should show the relationship used, substitution with units and a suitably rounded result.

Worked example

Five mass-loss results are $8.21$ , $8.25$ , $8.23$ , $8.20$ and $8.26\,\mathrm{g}$ . Represent their distribution by calculating the mean, range and half-range uncertainty.

1.The readings total $41.15\,\mathrm{g}$ , so the mean is $41.15/5=8.23\,\mathrm{g}$ . The range is $8.26-8.20=0.06\,\mathrm{g}$ . Half the range is $0.03\,\mathrm{g}$ , giving $(8.23\pm0.03)\,\mathrm{g}$ .

Answer: Mean $=8.23\,\mathrm{g}$ Range $=0.06\,\mathrm{g}$ Result $=(8.23\pm0.03)\,\mathrm{g}$

Common mistakes

Don't include an anomalous reading in a mean without first investigating it.
Don't report uncertainty without a unit or with precision inconsistent with the measurements.

Exam tip

A calculation answer should show the mean, range and half-range with units and justified precision.

Tier 1 · Easy

ORIGINAL

Three titre readings are

12.4

,

12.6

and

12.5\,\mathrm{cm}^3

. Calculate their mean and estimate the uncertainty as half the range.

[3 marks]

Total for this question: 3

Tier 2 · Standard

ORIGINAL

Two students measure the time for the same reaction. Student A obtains

42

,

48

and

45\,\mathrm{s}

. Student B obtains

44

,

46

and

45\,\mathrm{s}

. Estimate the uncertainty in each set of results using half the range. Which student's results are more precise?

[4 marks]

Total for this question: 4

Tier 3 · Hard

ORIGINAL

A reaction-time experiment gives

31.4

,

31.6

,

31.5

,

36.9

and

31.3\,\mathrm{s}

. Identify the anomalous reading, then report the mean of the consistent readings with a half-range uncertainty.

[4 marks]

Total for this question: 4

4.3.2.1

Moles (HT only)

Notes

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

Higher tier: the mole, symbol $\mathrm{mol}$ , measures amount of substance; one mole contains $6.02\times10^{23}$ stated particles. The mass of one mole in grams is numerically equal to its $M_r$ , so $n=\dfrac{m}{M_r}$ and $m=nM_r$ .
For example, $9.0\,\mathrm{g}$ of water with $M_r=18$ is $9.0/18=0.50\,\mathrm{mol}$ .
State the particle type carefully: ionic substances have formula units and ions, not molecules.
The Avogadro constant converts between amount in moles and number of stated particles.
Exam calculations should show the relationship used, substitution with units and a suitably rounded result.

Worked example

Find the mass of $0.250\,\mathrm{mol}$ of sodium carbonate, $\mathrm{Na}_2\mathrm{CO}_3$ . Use $A_r$ : $\mathrm{Na}=23$ , $\mathrm{C}=12$ , $\mathrm{O}=16$ .

1. $M_r(\mathrm{Na}_2\mathrm{CO}_3)=2(23)+12+3(16)=106$ . Then $m=nM_r=0.250\times106=26.5\,\mathrm{g}$ .

Answer: Mass $=26.5\,\mathrm{g}$

Common mistakes

Don't divide $M_r$ by mass instead of using $n=m/M_r$ .
Don't call the particles in an ionic compound molecules instead of formula units or ions.

Exam tip

Write $n=m/M_r$ before substitution and name the particle type if a particle count is required.

Tier 1 · Easy

ORIGINAL

Calculate the amount in moles in

9.0\,\mathrm{g}

of water,

\mathrm{H}_2\mathrm{O}

. Use

M_r=18

.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

A sample contains

4.0\,\mathrm{g}

of helium,

\mathrm{He}

. Another sample contains

32.0\,\mathrm{g}

of oxygen,

\mathrm{O}_2

. (a) Show that each sample contains the same number of molecules. (b) Determine which sample contains more atoms. Use

A_r

:

\mathrm{He}=4

,

\mathrm{O}=16

.

[4 marks]

Total for this question: 4

Tier 3 · Hard

ORIGINAL

A sample contains

0.0150\,\mathrm{mol}

of magnesium chloride,

\mathrm{MgCl}_2

. Calculate the number of formula units and the number of chloride ions. Use the Avogadro constant

6.02\times10^{23}\,\mathrm{mol}^{-1}

.

[4 marks]

Total for this question: 4

4.3.2.2

Amounts of substances in equations (HT only)

Notes

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

Higher tier: balanced-equation coefficients give mole ratios, which can be used to connect masses of different substances.
Convert the known mass to moles, apply the coefficient ratio, then convert the required moles back to mass.
In $\mathrm{CaCO}_3\rightarrow\mathrm{CaO}+\mathrm{CO}_2$ , one mole of calcium carbonate produces one mole of carbon dioxide.
Do not use a coefficient ratio directly on masses unless the molar masses happen to be equal.
Exam calculations should show the relationship used, substitution with units and a suitably rounded result.

Worked example

Calcium carbonate decomposes as $\mathrm{CaCO}_3\rightarrow\mathrm{CaO}+\mathrm{CO}_2$ . Calculate the mass of carbon dioxide made from $25.0\,\mathrm{g}$ of calcium carbonate. Use $M_r(\mathrm{CaCO}_3)=100$ and $M_r(\mathrm{CO}_2)=44$ .

1.Moles of calcium carbonate $=25.0/100=0.250\,\mathrm{mol}$ . The equation gives a $1:1$ mole ratio, so $0.250\,\mathrm{mol}$ of carbon dioxide forms. Its mass is $0.250\times44=11.0\,\mathrm{g}$ .

Answer: Mass of carbon dioxide $=11.0\,\mathrm{g}$

Common mistakes

Don't apply equation coefficients directly to masses instead of converting mass to moles first.
Don't use the inverse mole ratio when moving from the known substance to the required substance.

Exam tip

Use the sequence mass → moles → equation ratio → moles → mass.

Tier 1 · Easy

ORIGINAL

For

2\mathrm{Mg}+\mathrm{O}_2\rightarrow2\mathrm{MgO}

, calculate the mass of magnesium oxide made from

4.8\,\mathrm{g}

of magnesium when oxygen is in excess. Use

A_r

:

\mathrm{Mg}=24

,

\mathrm{O}=16

.

[3 marks]

Total for this question: 3

Tier 2 · Standard

ORIGINAL

Nitrogen reacts with hydrogen:

\mathrm{N}_2+3\mathrm{H}_2\rightarrow2\mathrm{NH}_3

. Calculate the mass of hydrogen needed to react completely with

28.0\,\mathrm{g}

of nitrogen. Use

M_r

:

\mathrm{N}_2=28

,

\mathrm{H}_2=2

.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

Ammonia is oxidised by

4\mathrm{NH}_3+5\mathrm{O}_2\rightarrow4\mathrm{NO}+6\mathrm{H}_2\mathrm{O}

. Calculate the mass of water formed from

10.2\,\mathrm{g}

of ammonia when oxygen is in excess. Use

M_r(\mathrm{NH}_3)=17

and

M_r(\mathrm{H}_2\mathrm{O})=18

.

[4 marks]

Total for this question: 4

4.3.2.3

Using moles to balance equations (HT only)

Notes

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

Higher tier: experimental masses can reveal balancing coefficients after each mass is converted into moles.
Divide all mole amounts by the smallest value to obtain a simple ratio, then multiply every value if fractions remain.
For mole amounts $0.20:0.60:0.40$ , division by $0.20$ gives the whole-number ratio $1:3:2$ .
Rounding a ratio too early can produce incorrect coefficients; keep enough significant figures until the ratio is clear.
Exam calculations should show the relationship used, substitution with units and a suitably rounded result.

Worked example

Iron and oxygen form iron(III) oxide. The reacting masses are $11.2\,\mathrm{g}$ of $\mathrm{Fe}$ , $4.8\,\mathrm{g}$ of $\mathrm{O}_2$ and $16.0\,\mathrm{g}$ of $\mathrm{Fe}_2\mathrm{O}_3$ . Determine the balanced equation. Use $M_r$ : $\mathrm{Fe}=56$ , $\mathrm{O}_2=32$ , $\mathrm{Fe}_2\mathrm{O}_3=160$ .

1.Convert each mass to moles: iron $=11.2/56=0.20$ , oxygen $=4.8/32=0.15$ , and iron(III) oxide $=16.0/160=0.10$ . Divide by $0.10$ to get $2:1.5:1$ , then multiply all terms by $2$ to get $4:3:2$ .

Answer: $4\mathrm{Fe}+3\mathrm{O}_2\rightarrow2\mathrm{Fe}_2\mathrm{O}_3$

Common mistakes

Don't use experimental masses directly as balancing coefficients.
Don't round a mole ratio too early and miss a simple whole-number ratio.

Exam tip

Keep unrounded mole values until every amount has been divided by the smallest.

Tier 1 · Easy

ORIGINAL

Nitrogen, hydrogen and ammonia occur in amounts

0.20

,

0.60

and

0.40\,\mathrm{mol}

respectively. Use these amounts to balance

\mathrm{N}_2+\mathrm{H}_2\rightarrow\mathrm{NH}_3

.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

Copper reacts with oxygen to form copper(II) oxide.

12.7\,\mathrm{g}

of copper reacts with

3.2\,\mathrm{g}

of oxygen to form

15.9\,\mathrm{g}

of copper(II) oxide. A student writes

\mathrm{Cu}+\mathrm{O}_2\rightarrow\mathrm{CuO}

. Use the masses to determine the correctly balanced equation. Use

M_r

:

\mathrm{Cu}=63.5

,

\mathrm{O}_2=32.0

,

\mathrm{CuO}=79.5

.

[4 marks]

Total for this question: 4

Tier 3 · Hard

ORIGINAL

Ethane burns in oxygen. A complete reaction uses

3.0\,\mathrm{g}

of

\mathrm{C}_2\mathrm{H}_6

and

11.2\,\mathrm{g}

of

\mathrm{O}_2

, producing

8.8\,\mathrm{g}

of

\mathrm{CO}_2

and

5.4\,\mathrm{g}

of

\mathrm{H}_2\mathrm{O}

. Determine the balanced equation. Use

M_r

:

30

,

32

,

44

and

18

in the same order.

[5 marks]

Total for this question: 5

4.3.2.4

Limiting reactants (HT only)

Notes

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

Higher tier: the limiting reactant is used up completely and therefore fixes the maximum amount of product that can form.
Compare available moles with the balanced-equation ratio; the smaller mass is not necessarily the limiting amount.
Once the limiting reactant is known, use its moles and the coefficient ratio to calculate the product amount.
An excess reactant remains after the reaction, so using its full starting amount to calculate product overestimates the yield.
Exam calculations should show the relationship used, substitution with units and a suitably rounded result.

Worked example

Magnesium reacts by $2\mathrm{Mg}+\mathrm{O}_2\rightarrow2\mathrm{MgO}$ . A vessel contains $7.2\,\mathrm{g}$ of magnesium and $6.4\,\mathrm{g}$ of oxygen. Determine the limiting reactant and the mass of magnesium oxide. Use $A_r$ : $\mathrm{Mg}=24$ , $\mathrm{O}=16$ .

1.Magnesium moles $=7.2/24=0.300$ and oxygen moles $=6.4/32=0.200$ . The $2:1$ ratio means $0.300\,\mathrm{mol}$ of magnesium needs only $0.150\,\mathrm{mol}$ of oxygen, so magnesium is limiting. The $\mathrm{Mg}:\mathrm{MgO}$ ratio is $1:1$ , giving $0.300\,\mathrm{mol}$ of magnesium oxide. Its $M_r$ is $40$ , so its mass is $0.300\times40=12.0\,\mathrm{g}$ .

Answer: Magnesium is limiting Mass of magnesium oxide $=12.0\,\mathrm{g}$

Common mistakes

Don't assume the reactant with the smaller mass is automatically limiting.
Don't calculate product from the full amount of an excess reactant.

Exam tip

Test each reactant against the coefficient ratio before calculating product from the limiting reactant.

Tier 1 · Easy

ORIGINAL

For

\mathrm{H}_2+\mathrm{Cl}_2\rightarrow2\mathrm{HCl}

, a mixture contains

3.0\,\mathrm{mol}

of hydrogen and

2.0\,\mathrm{mol}

of chlorine. Identify the limiting reactant and calculate the amount of hydrogen chloride formed.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

Aluminium reacts with chlorine:

2\mathrm{Al}+3\mathrm{Cl}_2\rightarrow2\mathrm{AlCl}_3

. A mixture contains

5.4\,\mathrm{g}

of aluminium and

14.2\,\mathrm{g}

of chlorine. Determine the limiting reactant and the amount in moles of aluminium chloride formed. Use

A_r(\mathrm{Al})=27

and

M_r(\mathrm{Cl}_2)=71

.

[4 marks]

Total for this question: 4

Tier 3 · Hard

ORIGINAL

Ammonia forms by

\mathrm{N}_2+3\mathrm{H}_2\rightarrow2\mathrm{NH}_3

. A reactor receives

14.0\,\mathrm{g}

of nitrogen and

2.40\,\mathrm{g}

of hydrogen. Calculate the ammonia mass and the mass of excess reactant left. Use

M_r

:

\mathrm{N}_2=28

,

\mathrm{H}_2=2

,

\mathrm{NH}_3=17

.

[5 marks]

Total for this question: 5

4.3.2.5

Concentration of solutions

Notes

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Definition

The concentration of a solution is the amount of solute in a given volume of solution. At GCSE it is measured as mass per given volume, usually grams per dm³ (g/dm³), and at Higher tier also in mol/dm³.

Explanation

The volume in a concentration is always the volume of the solution, not the volume of solvent added, and it must be in dm³ before dividing. Since 1 dm³ = 1000 cm³, divide a volume in cm³ by 1000 — the mark scheme often awards a mark for that conversion, so write it out.
Concentration in g/dm³ is mass of solute divided by volume of solution; rearranged, mass of solute is concentration multiplied by volume. Higher tier candidates also work in mol/dm³, converting via the relative formula mass.
Concentration rises if more solute dissolves in the same volume, or the same mass in a smaller volume.
Higher tier: AQA restricts "explain how the mass of a solute and the volume of a solution is related to the concentration of the solution" to Higher tier; calculating the mass of solute in a given volume is examined on both tiers.
Concentration is not strength: a strong acid is completely ionised.

Concentration is the mass of dissolved solute divided by the volume of the whole solution in dm cubed.

Worked example

A fertiliser solution has concentration $18.0\,\mathrm{g\,dm}^{-3}$ . Calculate the solute mass in $250\,\mathrm{cm}^3$ of solution.

1.Convert the volume: $250\,\mathrm{cm}^3=0.250\,\mathrm{dm}^3$ . Rearrange to $m=cV$ , then $m=18.0\times0.250=4.50\,\mathrm{g}$ .

Answer: Mass of solute $=4.50\,\mathrm{g}$

Common mistakes

Don't use $\mathrm{cm}^3$ in a formula requiring volume in $\mathrm{dm}^3$ .
Don't use the volume of solvent added instead of the final solution volume.

Mark-scheme wording to use

“25.0/1000=0.025 dm³. 1 mark: 0.025x₄.00, allowing correct use of an incorrect/unconverted volume. 1 mark: 0.1 g” — AQA 8462, June 2023 Paper 1F, Q02.7
“the mass of acid dissolved is doubled and the volume of the solution is halved” — AQA 8462, June 2023 Paper 1H, Q05.2
“0.025 dm³. 1 mark: 6.75/0.025, allowing correct use of an incorrectly determined volume. 1 mark: 270 g/dm³” — AQA 8462, June 2024 Paper 1F, Q03.6

What the examiner wants

The cm^3−to−dm³ conversion written as its own step. It carries its own mark in almost every scheme, and it is the single most common reason an otherwise correct concentration is out by a factor of 1000.

Exam tip

Convert $\mathrm{cm}^3$ to $\mathrm{dm}^3$ before substituting into $c=m/V$ .

Tier 1 · Easy

ORIGINAL

A solution contains

12.0\,\mathrm{g}

of solute in

0.400\,\mathrm{dm}^3

. Calculate its concentration in

\mathrm{g\,dm}^{-3}

.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

Solution A contains

5.0\,\mathrm{g}

of solute in

200\,\mathrm{cm}^3

of solution. Solution B contains

8.0\,\mathrm{g}

of solute in

400\,\mathrm{cm}^3

of solution. Determine which solution is more concentrated.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

A beaker initially contains

3.60\,\mathrm{g}

of dissolved salt in

150\,\mathrm{cm}^3

of solution. Water is added until the concentration is

12.0\,\mathrm{g\,dm}^{-3}

. Calculate the volume of water added, assuming volumes are additive.

[4 marks]

Total for this question: 4

4.3.3.1

Percentage yield (chemistry only)

Notes

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

Percentage yield compares the actual product obtained with the maximum theoretical product: $\dfrac{\text{actual yield}}{\text{theoretical yield}}\times100$ . Yield may be below $100\%$ because a reversible reaction is incomplete, side reactions occur, or product is lost during separation.
For example, an actual mass of $8.0\,\mathrm{g}$ from a theoretical $10.0\,\mathrm{g}$ gives an $80\%$ yield.
Use actual over theoretical, not the reverse, and compare quantities in the same unit.
Higher tier: AQA restricts "calculate the theoretical mass of a product from a given mass of reactant and the balanced equation for the reaction" to Higher tier; calculating percentage yield from an actual and a theoretical yield is examined on both tiers.
Exam calculations should show the relationship used, substitution with units and a suitably rounded result.

Worked example

A process makes $20.4\,\mathrm{kg}$ of product at a percentage yield of $68.0\%$ . Calculate the theoretical product mass.

1.Write $68.0=(20.4/\text{theoretical mass})\times100$ . Rearranging gives theoretical mass $=20.4\times100/68.0=30.0\,\mathrm{kg}$ .

Answer: Theoretical mass $=30.0\,\mathrm{kg}$

Common mistakes

Don't divide theoretical yield by actual yield, producing a percentage above $100\%$ .
Don't use actual and theoretical quantities expressed in different units.

Exam tip

State actual ÷ theoretical × $100$ before substituting, then give the result as a percentage.

Tier 1 · Easy

ORIGINAL

A preparation has a theoretical product mass of

9.0\,\mathrm{g}

and an actual product mass of

7.2\,\mathrm{g}

. Calculate the percentage yield.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

A reaction has a theoretical product mass of

12.0\,\mathrm{g}

. The actual product mass is

8.4\,\mathrm{g}

. Calculate the percentage yield. Give two reasons why the yield may be less than

100\%

.

[4 marks]

Total for this question: 4

Tier 3 · Hard

ORIGINAL

Two batches have theoretical product masses of

45.0\,\mathrm{g}

and

30.0\,\mathrm{g}

. Their actual masses are

36.0\,\mathrm{g}

and

19.5\,\mathrm{g}

. Determine the combined percentage yield.

[4 marks]

Total for this question: 4

4.3.3.2

Atom economy (chemistry only)

Notes

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Definition

The atom economy of a reaction is a measure of how much of the starting material ends up as the useful product. It is the relative formula mass of the desired product from the equation, divided by the sum of the relative formula masses of all the reactants from the equation, multiplied by 100.

Explanation

Atom economy is calculated from the balanced equation alone, so the multipliers matter: Mr of the desired product times its coefficient on top, over the sum of every reactant Mr times its coefficient.
Percentage yield is different — it compares the mass actually obtained with the maximum possible and depends on how the reaction was carried out; atom economy depends only on the equation.
A reaction with a single product has an atom economy of 100 per cent, and that sentence is often all a one-mark question wants.
High atom economy matters for sustainable development and for economic reasons: less starting material becomes waste, so raw-material, separation, disposal and environmental costs all fall.
Higher tier: AQA restricts "explain why a particular reaction pathway is chosen to produce a specified product given appropriate data such as atom economy (if not calculated), yield, rate, equilibrium position and usefulness of by-products" to Higher tier; calculating atom economy is examined on both tiers.

Worked example

Chlorine is the desired product in $2\mathrm{NaCl}+2\mathrm{H}_2\mathrm{O}\rightarrow\mathrm{Cl}_2+\mathrm{H}_2+2\mathrm{NaOH}$ . Calculate the atom economy. Use $M_r$ : $\mathrm{NaCl}=58.5$ , $\mathrm{H}_2\mathrm{O}=18.0$ , $\mathrm{Cl}_2=71.0$ .

1.The reactant total from the equation is $2(58.5)+2(18.0)=153.0$ . The desired chlorine contributes $71.0$ . Atom economy $=(71.0/153.0)\times100=46.4\%$ .

Answer: Atom economy $=46.4\%$

Common mistakes

Don't omit balanced-equation coefficients when totaling relative formula masses.
Don't confuse atom economy, which comes from the equation, with percentage yield, which uses the actual product obtained.

Mark-scheme wording to use

“total Mr 140. 1 mark: 112/140 x 100. 1 mark: 80%” — AQA 8462, June 2025 Paper 1H, Q03.3
“there is one product” — AQA 8462, June 2022 Paper 2F, Q05.3
“for sustainable development … for economic reasons … to produce a high(er) percentage of useful product … to reduce waste” — AQA 8462, November 2020 Paper 1H, Q03.6

What the examiner wants

The full substituted fraction before the percentage, and every reactant included on the bottom line. Omitting a reactant is the standard error, and it changes the answer without any visible sign that something is wrong.

Exam tip

For atom economy, use equation quantities and identify the desired product before totaling masses.

Tier 1 · Easy

ORIGINAL

Calcium oxide is the desired product in

\mathrm{CaCO}_3\rightarrow\mathrm{CaO}+\mathrm{CO}_2

. Calculate the atom economy using

M_r(\mathrm{CaCO}_3)=100

and

M_r(\mathrm{CaO})=56

.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

A reaction has an atom economy of

82\%

. Explain what this value means. Give two advantages of a reaction having a high atom economy.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

Titanium is the desired product in

\mathrm{TiCl}_4+4\mathrm{Na}\rightarrow\mathrm{Ti}+4\mathrm{NaCl}

. Calculate the atom economy and the theoretical titanium mass represented by

250\,\mathrm{kg}

of reactants in this ratio. Use

A_r

:

\mathrm{Ti}=48

,

\mathrm{Cl}=35.5

,

\mathrm{Na}=23

.

[5 marks]

Total for this question: 5

4.3.4

Using concentrations of solutions in mol/dm3 (chemistry only) (HT only)

Notes

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

Higher tier: molar concentration is amount of solute per solution volume: $c=n/V$ , with $c$ in $\mathrm{mol\,dm}^{-3}$ and $V$ in $\mathrm{dm}^3$ . Use $n=cV$ to find moles, then use $m=nM_r$ when a solute mass is required.
Reacting-solution calculations use the balanced-equation mole ratio between the two dissolved substances.
Convert $\mathrm{cm}^3$ to $\mathrm{dm}^3$ by dividing by $1000$ before multiplying by concentration.
The requested concentration follows only after the reacting mole ratio has been applied.
Exam calculations should show the relationship used, substitution with units and a suitably rounded result.

Worked example

Calculate the sodium hydroxide mass in $150\,\mathrm{cm}^3$ of a $0.400\,\mathrm{mol\,dm}^{-3}$ solution. Use $M_r(\mathrm{NaOH})=40.0$ .

1.The volume is $0.150\,\mathrm{dm}^3$ . Moles of sodium hydroxide $=cV=0.400\times0.150=0.0600\,\mathrm{mol}$ . Its mass is $nM_r=0.0600\times40.0=2.40\,\mathrm{g}$ .

Answer: Mass of sodium hydroxide $=2.40\,\mathrm{g}$

Common mistakes

Don't multiply concentration by a volume still expressed in $\mathrm{cm}^3$ .
Don't ignore the balanced-equation mole ratio between reacting solutions.

Exam tip

A reacting-solutions calculation needs volume conversion, $n=cV$ , the mole ratio and then the requested concentration.

Tier 1 · Easy

ORIGINAL

A solution contains

0.0750\,\mathrm{mol}

of solute in

250\,\mathrm{cm}^3

. Calculate its concentration in

\mathrm{mol\,dm}^{-3}

.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

Calculate the volume of

0.200\,\mathrm{mol\,dm}^{-3}

potassium hydroxide solution that contains

5.60\,\mathrm{g}

of potassium hydroxide. Give your answer in

\mathrm{cm}^3

. Use

M_r(\mathrm{KOH})=56.0

.

[4 marks]

Total for this question: 4

Tier 3 · Hard

ORIGINAL

25.0\,\mathrm{cm}^3

of sulfuric acid reacts exactly with

32.0\,\mathrm{cm}^3

of

0.150\,\mathrm{mol\,dm}^{-3}

sodium hydroxide. Use

\mathrm{H}_2\mathrm{SO}_4+2\mathrm{NaOH}\rightarrow\mathrm{Na}_2\mathrm{SO}_4+2\mathrm{H}_2\mathrm{O}

to calculate the acid concentration.

[5 marks]

Total for this question: 5

4.3.5

Use of amount of substance in relation to volumes of gases (chemistry only) (HT only)

Notes

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

Higher tier: equal mole amounts of gases occupy equal volumes at the same temperature and pressure. At room temperature and pressure, one mole of any gas occupies $24\,\mathrm{dm}^3$ , so $V=24n$ .
Balanced coefficients give gas-volume ratios directly when all gaseous substances are compared under the same conditions.
Keep units consistent: $1000\,\mathrm{cm}^3=1\,\mathrm{dm}^3$ , and do not use $24\,\mathrm{dm}^3$ before converting mass to moles.
Mass-based gas questions therefore require an initial mole calculation.
Exam calculations should show the relationship used, substitution with units and a suitably rounded result.

Worked example

Calculate the volume of carbon dioxide at room temperature and pressure produced by $8.80\,\mathrm{g}$ of the gas. Use $M_r(\mathrm{CO}_2)=44.0$ .

1.Moles of carbon dioxide $=8.80/44.0=0.200\,\mathrm{mol}$ . Its volume is $0.200\times24=4.80\,\mathrm{dm}^3$ .

Answer: Carbon dioxide volume $=4.80\,\mathrm{dm}^3$

Common mistakes

Don't use $24\,\mathrm{dm}^3$ before converting a given mass to moles.
Don't apply a gas-volume ratio to substances that are not all gases under the stated conditions.

Exam tip

Gas volumes follow balanced coefficients directly only at the same temperature and pressure.

Tier 1 · Easy

ORIGINAL

Calculate the volume occupied by

0.350\,\mathrm{mol}

of a gas at room temperature and pressure.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

A sample of methane occupies

1.20\,\mathrm{dm}^3

at room temperature and pressure. Calculate the mass of methane in the sample. The volume of one mole of gas is

24\,\mathrm{dm}^3

at room temperature and pressure. Use

M_r(\mathrm{CH}_4)=16.0

.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

Carbon monoxide reacts by

2\mathrm{CO}+\mathrm{O}_2\rightarrow2\mathrm{CO}_2

. A mixture contains

36.0\,\mathrm{dm}^3

of carbon monoxide and

24.0\,\mathrm{dm}^3

of oxygen under the same conditions. Calculate the carbon dioxide volume and the volume of reactant gas left in excess after complete reaction.

[4 marks]

Total for this question: 4

Conservation of mass and balanced chemical equations

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Relative formula mass

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Mass changes when a reactant or product is a gas

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Chemical measurements

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Moles (HT only)

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Amounts of substances in equations (HT only)

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Using moles to balance equations (HT only)

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Limiting reactants (HT only)

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Concentration of solutions

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Percentage yield (chemistry only)

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Atom economy (chemistry only)

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Using concentrations of solutions in mol/dm3 (chemistry only) (HT only)

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Use of amount of substance in relation to volumes of gases (chemistry only) (HT only)

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

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