4.3 Quantitative chemistry — revision question pack

13 specification points · notes, questions, answers and worked methods

Checked against AQA 8462 section 4.3. Review basis: the qualification registry sourced from the AQA GCSE Chemistry (8462) specification; registry verification recorded 17 July 2026.

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4.3.1.1 · Conservation of mass and balanced chemical equations

Explanation

The law of conservation of mass states that atoms are neither created nor destroyed in a chemical reaction, so the total mass is unchanged in a closed system.
Balance a symbol equation by placing whole-number coefficients before formulae until each element has the same number of atoms on both sides.
For example, $2\mathrm{Mg}+\mathrm{O}_2\rightarrow2\mathrm{MgO}$ shows that two magnesium atoms and two oxygen atoms are present on each side.
Never change a subscript to balance an equation: that changes the identity of the substance rather than the quantity reacting.
Exam calculations should show the relationship used, substitution with units and a suitably rounded result.

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.

Exam tip

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

Tier 1 · Easy

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
Complete the balanced equation by inserting the missing coefficient: $2\mathrm{Na}+\mathrm{Cl}_2\rightarrow\square\mathrm{NaCl}$ .
[1 mark]
Total for this question: 1

Tier 2 · Standard

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
A sealed flask contains $7.0\,\mathrm{g}$ of iron and $4.0\,\mathrm{g}$ of sulfur. The elements react completely to make one product. Calculate the mass of product and state the law used.
[2 marks]
Total for this question: 2
Choose the balanced equation for the reaction between lithium and oxygen. A: $\mathrm{Li}+\mathrm{O}_2\rightarrow\mathrm{Li}_2\mathrm{O}$ ; B: $2\mathrm{Li}+\mathrm{O}_2\rightarrow\mathrm{Li}_2\mathrm{O}$ ; C: $4\mathrm{Li}+\mathrm{O}_2\rightarrow2\mathrm{Li}_2\mathrm{O}$ . Give the atom totals that support your choice.
[3 marks]
Total for this question: 3

Tier 3 · Hard

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
Balance $\mathrm{NH}_3+\mathrm{O}_2\rightarrow\mathrm{N}_2+\mathrm{H}_2\mathrm{O}$ . Verify the result by stating the number of nitrogen, hydrogen and oxygen atoms on each side.
[4 marks]
Total for this question: 4
Complete $\square\mathrm{Fe}+\square\mathrm{H}_2\mathrm{O}\rightarrow\mathrm{Fe}_3\mathrm{O}_4+\square\mathrm{H}_2$ using the smallest whole-number coefficients. Explain the different meanings of the coefficient before $\mathrm{H}_2\mathrm{O}$ and the subscript $4$ in $\mathrm{Fe}_3\mathrm{O}_4$ .
[4 marks]
Total for this question: 4
An oxide is represented by $\mathrm{X}_a\mathrm{O}_b$ . Its balanced formation equation is $4\mathrm{X}+3\mathrm{O}_2\rightarrow2\mathrm{X}_a\mathrm{O}_b$ . Determine $a$ and $b$ , write the formula of the oxide and verify the atom totals on both sides.
[5 marks]
Total for this question: 5
Hydrogen peroxide decomposes to water and oxygen. Balance $\mathrm{H}_2\mathrm{O}_2\rightarrow\mathrm{H}_2\mathrm{O}+\mathrm{O}_2$ . A sealed vessel initially contains $34.0\,\mathrm{g}$ of hydrogen peroxide and, after complete decomposition, contains $16.0\,\mathrm{g}$ of oxygen. Calculate the water mass and use both atom counts and masses to verify conservation.
[6 marks]
Total for this question: 6

4.3.1.2 · Relative formula mass

Explanation

Relative formula mass, $M_r$ , is the sum of the relative atomic masses of every atom shown in a formula.
Multiply each $A_r$ value by the number of that atom, including multipliers outside brackets, before adding the contributions.
Percentage by mass of an element is $\dfrac{\text{total }A_r\text{ of that element}}{M_r\text{ of the compound}}\times100$ .
A coefficient in an equation multiplies an entire formula; ignoring it when comparing equation masses is a common error.
Exam calculations should show the relationship used, substitution with units and a suitably rounded result.

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 $M_r$ .

Exam tip

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

Tier 1 · Easy

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
A student calculates $M_r$ of $\mathrm{(NH}_4)_2\mathrm{SO}_4$ as $2(14)+4(1)+32+4(16)=128$ . Identify the counting error and give the correct $M_r$ . Use $A_r$ : $\mathrm{N}=14$ , $\mathrm{H}=1$ , $\mathrm{S}=32$ , $\mathrm{O}=16$ .
[2 marks]
Total for this question: 2

Tier 2 · Standard

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
Compare the percentage by mass of oxygen in magnesium oxide, $\mathrm{MgO}$ , and sulfur dioxide, $\mathrm{SO}_2$ . Identify which compound has the greater oxygen percentage. Use $A_r$ : $\mathrm{Mg}=24$ , $\mathrm{S}=32$ , $\mathrm{O}=16$ .
[4 marks]
Total for this question: 4
A magnesium chloride compound has formula $\mathrm{MgCl}_x$ and relative formula mass $95$ . Determine the value of $x$ and write the completed formula. Use $A_r$ : $\mathrm{Mg}=24$ , $\mathrm{Cl}=35.5$ .
[3 marks]
Total for this question: 3

Tier 3 · Hard

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
In $2\mathrm{NaHCO}_3\rightarrow\mathrm{Na}_2\mathrm{CO}_3+\mathrm{CO}_2+\mathrm{H}_2\mathrm{O}$ , calculate the percentage of the equation's reactant mass that becomes carbon dioxide. Hence calculate the carbon dioxide mass from $42.0\,\mathrm{g}$ of sodium hydrogencarbonate. Use unrounded values in your working and $A_r$ : $\mathrm{Na}=23$ , $\mathrm{H}=1$ , $\mathrm{C}=12$ , $\mathrm{O}=16$ .
[5 marks]
Total for this question: 5
An oxide has formula $\mathrm{M}_2\mathrm{O}_3$ and relative formula mass $160$ . Calculate its percentage by mass of oxygen and determine $A_r$ of element $\mathrm{M}$ . Use $A_r(\mathrm{O})=16$ .
[4 marks]
Total for this question: 4
An oxide has formula $\mathrm{MO}$ and is $20.0\%$ oxygen by mass. Determine $A_r$ of M. Then calculate the mass of M in a $15.0\,\mathrm{g}$ sample of the oxide. Use $A_r(\mathrm{O})=16$ .
[5 marks]
Total for this question: 5
An oxide is $60.0\%$ oxygen by mass. Choose its formula from $\mathrm{CO}_2$ , $\mathrm{SO}_3$ and $\mathrm{NO}_2$ by calculating the oxygen percentage of each. Then calculate the oxygen mass in $40.0\,\mathrm{g}$ of the oxide. Use $A_r$ : $\mathrm{C}=12$ , $\mathrm{N}=14$ , $\mathrm{S}=32$ , $\mathrm{O}=16$ .
[5 marks]
Total for this question: 5

4.3.1.3 · Mass changes when a reactant or product is a gas

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

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
A metal has a mass of $7.6\,\mathrm{g}$ . After heating in air, its oxide has a mass of $9.5\,\mathrm{g}$ . Calculate the mass of oxygen that reacted.
[1 mark]
Total for this question: 1

Tier 2 · Standard

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
Copper carbonate is heated completely in an open container. The solid mass falls from $12.35\,\mathrm{g}$ to $7.95\,\mathrm{g}$ . Calculate the decrease and explain why the measured mass changes.
[3 marks]
Total for this question: 3
A reaction starts with $15.6\,\mathrm{g}$ of substances in a flask fitted with a gas bag. After the reaction, the flask contents have a mass of $13.9\,\mathrm{g}$ . Calculate the gas mass collected in the bag and explain why the mass of the complete apparatus remains $15.6\,\mathrm{g}$ .
[3 marks]
Total for this question: 3

Tier 3 · Hard

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
Equal masses of acid and carbonate react in two flasks. Flask X has a cotton-wool plug and its recorded mass falls by $1.10\,\mathrm{g}$ . Flask Y is sealed. Predict the mass change for flask Y and explain the difference between the readings.
[4 marks]
Total for this question: 4
Copper(II) nitrate decomposes in an open tube: $2\mathrm{Cu(NO}_3)_2\rightarrow2\mathrm{CuO}+4\mathrm{NO}_2+\mathrm{O}_2$ . A $18.8\,\mathrm{g}$ sample leaves $8.0\,\mathrm{g}$ of solid. Calculate the total mass of gases released. Show that the data are consistent using $A_r$ : $\mathrm{Cu}=64$ , $\mathrm{N}=14$ , $\mathrm{O}=16$ .
[4 marks]
Total for this question: 4
A metal sample has mass $9.60\,\mathrm{g}$ . After complete heating in air, its oxide has mass $12.0\,\mathrm{g}$ . Calculate the oxygen mass in the oxide and the percentage by mass of oxygen and metal. Use the results to explain the measured mass increase.
[5 marks]
Total for this question: 5
A $3.00\,\mathrm{g}$ sample of metal A is heated in an open crucible to constant mass, forming $3.75\,\mathrm{g}$ of oxide. A $3.00\,\mathrm{g}$ sample of metal B, heated the same way, forms $4.50\,\mathrm{g}$ of oxide. Calculate the mass of oxygen gained per gram of metal for each and identify which metal combines with more oxygen. Explain, in terms of particles, why both crucibles gain mass.
[5 marks]
Total for this question: 5

4.3.1.4 · Chemical measurements

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

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
A temperature rise is reported as $(12.6\pm0.2)\,^\circ\mathrm{C}$ . State the lowest and highest values in the uncertainty interval.
[2 marks]
Total for this question: 2

Tier 2 · Standard

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
A digital balance has a resolution of $0.01\,\mathrm{g}$ , so take its uncertainty for each mass result as $\pm0.01\,\mathrm{g}$ . Repeat measurements of a product's mass are $6.42$ , $6.48$ , $6.45$ and $6.43\,\mathrm{g}$ . Calculate the half-range uncertainty. Determine whether the balance resolution or the repeat spread gives the larger uncertainty, and state the uncertainty that should be used.
[4 marks]
Total for this question: 4
Thermometer P has $1.0\,^\circ\mathrm{C}$ divisions and thermometer Q has $0.2\,^\circ\mathrm{C}$ divisions. Take the reading uncertainty as half one division. Calculate the uncertainty for each thermometer and identify which gives the more precise temperature reading.
[3 marks]
Total for this question: 3

Tier 3 · Hard

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
Experiment A gives $(5.42\pm0.03)\,\mathrm{g}$ and experiment B gives $(5.50\pm0.04)\,\mathrm{g}$ . Calculate both uncertainty intervals. Determine whether the intervals overlap and explain what the comparison suggests.
[4 marks]
Total for this question: 4
A student records reaction times of $20.1$ , $20.5$ and $20.3\,\mathrm{s}$ , then adds repeats of $20.2$ and $20.4\,\mathrm{s}$ . Calculate the mean and range before and after the extra repeats. Evaluate the claim that taking more repeats must reduce the range.
[4 marks]
Total for this question: 4
The accepted temperature change is $25.0\,^\circ\mathrm{C}$ . Method A gives $24.6$ , $24.8$ and $24.7\,^\circ\mathrm{C}$ . Method B gives $24.9$ , $25.3$ and $25.1\,^\circ\mathrm{C}$ . For each method, calculate the mean and half-range uncertainty. Evaluate which method is more precise and which has a mean closer to the accepted value.
[6 marks]
Total for this question: 6
Five mass readings have a mean of $7.45\,\mathrm{g}$ . Four readings are $7.42$ , $7.46$ , $7.44$ and $7.48\,\mathrm{g}$ . Calculate the missing reading, then report the mean with a half-range uncertainty. Determine whether an expected value of $7.50\,\mathrm{g}$ lies within the resulting interval.
[6 marks]
Total for this question: 6

4.3.2.1 · Moles (HT only)

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

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
A helium sample contains $1.204\times10^{24}$ atoms. Calculate the amount in moles. Use the Avogadro constant $6.02\times10^{23}\,\mathrm{mol}^{-1}$ .
[2 marks]
Total for this question: 2

Tier 2 · Standard

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
A sample of calcium carbonate, $\mathrm{CaCO}_3$ , contains $3.01\times10^{23}$ formula units. Calculate its mass. Use the Avogadro constant $6.02\times10^{23}\,\mathrm{mol}^{-1}$ and $M_r(\mathrm{CaCO}_3)=100$ .
[3 marks]
Total for this question: 3
A sample contains $0.0400\,\mathrm{mol}$ of magnesium atoms. Calculate the total number of electrons in the neutral atoms. Each magnesium atom contains $12$ electrons. Use the Avogadro constant $6.02\times10^{23}\,\mathrm{mol}^{-1}$ .
[3 marks]
Total for this question: 3

Tier 3 · Hard

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
Calculate the number of oxygen atoms in $7.10\,\mathrm{g}$ of sodium sulfate, $\mathrm{Na}_2\mathrm{SO}_4$ . Give the answer to three significant figures. Use $M_r(\mathrm{Na}_2\mathrm{SO}_4)=142$ and the Avogadro constant $6.02\times10^{23}\,\mathrm{mol}^{-1}$ .
[4 marks]
Total for this question: 4
A calcium chloride sample contains $1.7458\times10^{23}$ chloride ions. Calculate the mass of calcium chloride, $\mathrm{CaCl}_2$ . Use the Avogadro constant $6.02\times10^{23}\,\mathrm{mol}^{-1}$ and $M_r(\mathrm{CaCl}_2)=111$ .
[4 marks]
Total for this question: 4
Higher Tier: A mixture contains $8.80\,\mathrm{g}$ of carbon dioxide and $2.80\,\mathrm{g}$ of nitrogen gas. Calculate the total number of molecules and the total number of atoms in the mixture. Use $M_r(\mathrm{CO}_2)=44.0$ , $M_r(\mathrm{N}_2)=28.0$ and the Avogadro constant $6.02\times10^{23}\,\mathrm{mol}^{-1}$ .
[6 marks]
Total for this question: 6
Higher Tier: A mixture of carbon monoxide and carbon dioxide contains $0.250\,\mathrm{mol}$ of molecules in total and $0.400\,\mathrm{mol}$ of oxygen atoms. Determine the amount and mass of each gas. Use $M_r(\mathrm{CO})=28.0$ and $M_r(\mathrm{CO}_2)=44.0$ .
[6 marks]
Total for this question: 6

4.3.2.2 · Amounts of substances in equations (HT only)

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

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
Methane burns in $\mathrm{CH}_4+2\mathrm{O}_2\rightarrow\mathrm{CO}_2+2\mathrm{H}_2\mathrm{O}$ . Calculate the amount of carbon dioxide formed when $0.300\,\mathrm{mol}$ of oxygen reacts with excess methane.
[2 marks]
Total for this question: 2

Tier 2 · Standard

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
Calcium reacts by $\mathrm{Ca}+2\mathrm{H}_2\mathrm{O}\rightarrow\mathrm{Ca(OH)}_2+\mathrm{H}_2$ . Calculate the calcium mass needed to form $0.600\,\mathrm{g}$ of hydrogen. Use $A_r(\mathrm{Ca})=40$ and $M_r(\mathrm{H}_2)=2$ .
[3 marks]
Total for this question: 3
Aluminium reacts by $4\mathrm{Al}+3\mathrm{O}_2\rightarrow2\mathrm{Al}_2\mathrm{O}_3$ . Choose the maximum aluminium oxide mass made from $5.40\,\mathrm{g}$ of aluminium: $6.40\,\mathrm{g}$ , $10.2\,\mathrm{g}$ or $20.4\,\mathrm{g}$ . Show your calculation. Use $A_r$ : $\mathrm{Al}=27$ , $\mathrm{O}=16$ .
[4 marks]
Total for this question: 4

Tier 3 · Hard

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
Iron is extracted in $\mathrm{Fe}_2\mathrm{O}_3+3\mathrm{CO}\rightarrow2\mathrm{Fe}+3\mathrm{CO}_2$ . Calculate the maximum iron mass made from $24.0\,\mathrm{g}$ of iron(III) oxide. Use $M_r(\mathrm{Fe}_2\mathrm{O}_3)=160$ and $A_r(\mathrm{Fe})=56$ .
[4 marks]
Total for this question: 4
A $12.0\,\mathrm{g}$ sample of impure calcium carbonate is heated completely. It produces $4.40\,\mathrm{g}$ of carbon dioxide. Use $\mathrm{CaCO}_3\rightarrow\mathrm{CaO}+\mathrm{CO}_2$ to calculate the mass of calcium carbonate in the sample and its percentage by mass. Use $M_r(\mathrm{CaCO}_3)=100$ and $M_r(\mathrm{CO}_2)=44$ .
[5 marks]
Total for this question: 5
Higher Tier: Aluminium reduces iron(III) oxide: $2\mathrm{Al}+\mathrm{Fe}_2\mathrm{O}_3\rightarrow\mathrm{Al}_2\mathrm{O}_3+2\mathrm{Fe}$ . Calculate the aluminium mass required to make $33.6\,\mathrm{g}$ of iron and the aluminium oxide mass formed at the same time. Use $A_r(\mathrm{Al})=27$ , $A_r(\mathrm{Fe})=56$ and $M_r(\mathrm{Al}_2\mathrm{O}_3)=102$ .
[6 marks]
Total for this question: 6
Higher Tier: Calcium carbonate is converted to calcium chloride in two reactions: $\mathrm{CaCO}_3\rightarrow\mathrm{CaO}+\mathrm{CO}_2$ and $\mathrm{CaO}+2\mathrm{HCl}\rightarrow\mathrm{CaCl}_2+\mathrm{H}_2\mathrm{O}$ . Calculate the maximum calcium chloride mass and the hydrogen chloride mass required when starting with $15.0\,\mathrm{g}$ of calcium carbonate. Use $M_r$ : $\mathrm{CaCO}_3=100$ , $\mathrm{CaCl}_2=111$ , $\mathrm{HCl}=36.5$ .
[6 marks]
Total for this question: 6

4.3.2.3 · Using moles to balance equations (HT only)

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

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
Hydrogen, oxygen and water occur in amounts $0.30$ , $0.15$ and $0.30\,\mathrm{mol}$ respectively. Use the mole ratio to balance $\mathrm{H}_2+\mathrm{O}_2\rightarrow\mathrm{H}_2\mathrm{O}$ .
[2 marks]
Total for this question: 2

Tier 2 · Standard

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
Magnesium reacts with nitrogen to form magnesium nitride, $\mathrm{Mg}_3\mathrm{N}_2$ . An experiment uses $7.20\,\mathrm{g}$ of magnesium and $2.80\,\mathrm{g}$ of nitrogen, forming $10.0\,\mathrm{g}$ of magnesium nitride. Determine the balanced equation. Use $A_r(\mathrm{Mg})=24$ , $M_r(\mathrm{N}_2)=28$ and $M_r(\mathrm{Mg}_3\mathrm{N}_2)=100$ .
[4 marks]
Total for this question: 4
A student converts reacting masses to $0.12$ , $0.18$ and $0.12\,\mathrm{mol}$ , then uses $0.12:0.18:0.12$ as equation coefficients. Explain the error and determine the smallest whole-number ratio.
[3 marks]
Total for this question: 3

Tier 3 · Hard

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
Iron reacts with oxygen to make $\mathrm{Fe}_3\mathrm{O}_4$ . A reaction uses $16.8\,\mathrm{g}$ of iron and $6.40\,\mathrm{g}$ of oxygen, forming $23.2\,\mathrm{g}$ of product. Determine the balanced equation. Use $A_r(\mathrm{Fe})=56$ , $M_r(\mathrm{O}_2)=32$ and $M_r(\mathrm{Fe}_3\mathrm{O}_4)=232$ .
[5 marks]
Total for this question: 5
A reaction uses $10.6\,\mathrm{g}$ of $\mathrm{Na}_2\mathrm{CO}_3$ and $7.30\,\mathrm{g}$ of $\mathrm{HCl}$ , forming $11.7\,\mathrm{g}$ of $\mathrm{NaCl}$ , $1.80\,\mathrm{g}$ of $\mathrm{H}_2\mathrm{O}$ and $4.40\,\mathrm{g}$ of $\mathrm{CO}_2$ . Determine the balanced equation. Use $M_r$ : $\mathrm{Na}_2\mathrm{CO}_3=106$ , $\mathrm{HCl}=36.5$ , $\mathrm{NaCl}=58.5$ , $\mathrm{H}_2\mathrm{O}=18.0$ , $\mathrm{CO}_2=44.0$ .
[5 marks]
Total for this question: 5
Higher Tier: Phosphorus burns to form $\mathrm{P}_4\mathrm{O}_{10}$ . A reaction uses $12.4\,\mathrm{g}$ of $\mathrm{P}_4$ and $16.0\,\mathrm{g}$ of $\mathrm{O}_2$ , forming $28.4\,\mathrm{g}$ of $\mathrm{P}_4\mathrm{O}_{10}$ . Use the experimental masses to determine the balanced equation. Use $M_r$ : $\mathrm{P}_4=124$ , $\mathrm{O}_2=32.0$ , $\mathrm{P}_4\mathrm{O}_{10}=284$ .
[5 marks]
Total for this question: 5
Higher Tier: Experimental masses for $\mathrm{Al}+\mathrm{Cl}_2\rightarrow\mathrm{AlCl}_3$ give amounts of $0.198$ , $0.303$ and $0.201\,\mathrm{mol}$ respectively. Determine the smallest whole-number coefficients and evaluate why the amounts do not give an exact whole-number ratio. Your answer should use the unrounded ratio before choosing coefficients.
[5 marks]
Total for this question: 5

4.3.2.4 · Limiting reactants (HT only)

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

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
State why the reactant with the smaller starting mass is not always the limiting reactant.
[1 mark]
Total for this question: 1

Tier 2 · Standard

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
Zinc reacts by $\mathrm{Zn}+2\mathrm{HCl}\rightarrow\mathrm{ZnCl}_2+\mathrm{H}_2$ . A mixture contains $6.50\,\mathrm{g}$ of zinc and $3.65\,\mathrm{g}$ of hydrogen chloride. Determine the limiting reactant and the mass of zinc left after reaction. Use $A_r$ : $\mathrm{Zn}=65$ , $\mathrm{H}=1$ , $\mathrm{Cl}=35.5$ .
[4 marks]
Total for this question: 4
Hydrogen reacts by $2\mathrm{H}_2+\mathrm{O}_2\rightarrow2\mathrm{H}_2\mathrm{O}$ . Mixture A contains $0.50\,\mathrm{mol}$ of hydrogen and $0.20\,\mathrm{mol}$ of oxygen. Mixture B contains $0.36\,\mathrm{mol}$ of hydrogen and $0.30\,\mathrm{mol}$ of oxygen. Determine the limiting reactant and water amount for each mixture, then identify which mixture makes more water.
[4 marks]
Total for this question: 4

Tier 3 · Hard

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
Propane burns by $\mathrm{C}_3\mathrm{H}_8+5\mathrm{O}_2\rightarrow3\mathrm{CO}_2+4\mathrm{H}_2\mathrm{O}$ . A vessel contains $8.80\,\mathrm{g}$ of propane and $40.0\,\mathrm{g}$ of oxygen. Determine the limiting reactant. Calculate the maximum propane mass that this oxygen supply could burn and hence the additional propane that could be added. Use $M_r$ : $\mathrm{C}_3\mathrm{H}_8=44$ , $\mathrm{O}_2=32$ .
[5 marks]
Total for this question: 5
Ammonia forms by $\mathrm{N}_2+3\mathrm{H}_2\rightarrow2\mathrm{NH}_3$ . A reactor contains $1.00\,\mathrm{mol}$ of nitrogen and $2.40\,\mathrm{mol}$ of hydrogen. The operator can add either $0.20\,\mathrm{mol}$ of nitrogen or $0.30\,\mathrm{mol}$ of hydrogen. Determine which addition increases the maximum ammonia amount and calculate the increase.
[5 marks]
Total for this question: 5
Higher Tier: Zinc reacts by $\mathrm{Zn}+2\mathrm{HCl}\rightarrow\mathrm{ZnCl}_2+\mathrm{H}_2$ . A reaction starts with $13.0\,\mathrm{g}$ of zinc and stops when all the hydrogen chloride has been used, leaving $3.25\,\mathrm{g}$ of zinc. Calculate the starting hydrogen chloride mass and the amount of hydrogen formed. Use $A_r(\mathrm{Zn})=65$ and $M_r(\mathrm{HCl})=36.5$ .
[6 marks]
Total for this question: 6
Higher Tier: A $12.5\,\mathrm{g}$ limestone sample is $80.0\%$ calcium carbonate by mass. It reacts with $5.475\,\mathrm{g}$ of hydrogen chloride. One mole of calcium carbonate reacts with two moles of hydrogen chloride and forms one mole of carbon dioxide. Determine the limiting reactant, the carbon dioxide mass formed and the calcium carbonate mass left. Use $M_r$ : $\mathrm{CaCO}_3=100$ , $\mathrm{HCl}=36.5$ , $\mathrm{CO}_2=44.0$ .
[6 marks]
Total for this question: 6

4.3.2.5 · Concentration of solutions

Explanation

Mass concentration in $\mathrm{g\,dm}^{-3}$ is the mass of dissolved solute divided by the solution volume: $c=m/V$ . Convert volumes before calculating: $1000\,\mathrm{cm}^3=1\,\mathrm{dm}^3$ .
For example, $6.0\,\mathrm{g}$ in $0.20\,\mathrm{dm}^3$ has concentration $6.0/0.20=30\,\mathrm{g\,dm}^{-3}$ .
Use the volume of the final solution, not the volume of solvent added or the mass of the whole solution.
Higher tier: explain how the mass of solute and volume of solution are related to concentration.
Exam calculations should show the relationship used, substitution with units and a suitably rounded result.

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.

Exam tip

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

Tier 1 · Easy

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
Two solutions contain equal masses of the same solute. Solution P has the smaller total volume. State which solution has the greater mass concentration.
[1 mark]
Total for this question: 1

Tier 2 · Standard

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
A student dissolves $7.50\,\mathrm{g}$ of a solid and makes the solution up to $250\,\mathrm{cm}^3$ . Calculate the mass concentration in $\mathrm{g\,dm}^{-3}$ .
[3 marks]
Total for this question: 3
A student mixes $100\,\mathrm{cm}^3$ of a solution containing $4.0\,\mathrm{g}$ of salt with $400\,\mathrm{cm}^3$ of a solution containing $6.0\,\mathrm{g}$ of the same salt. Calculate the mass concentration of the mixture, assuming the volumes are additive.
[3 marks]
Total for this question: 3

Tier 3 · Hard

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
A salt solution initially has volume $400\,\mathrm{cm}^3$ and concentration $18.0\,\mathrm{g\,dm}^{-3}$ . Water evaporates until the volume is $240\,\mathrm{cm}^3$ ; no salt is lost. Calculate the final concentration.
[4 marks]
Total for this question: 4
A salt solution has volume $300\,\mathrm{cm}^3$ and concentration $24.0\,\mathrm{g\,dm}^{-3}$ . A $100\,\mathrm{cm}^3$ sample is removed, then water is added to the remaining solution to make $500\,\mathrm{cm}^3$ . Calculate the final mass concentration.
[4 marks]
Total for this question: 4
A student mixes $200\,\mathrm{cm}^3$ of a $15.0\,\mathrm{g\,dm}^{-3}$ solution with an unknown volume of a $30.0\,\mathrm{g\,dm}^{-3}$ solution of the same solute. The mixture has concentration $24.0\,\mathrm{g\,dm}^{-3}$ . Calculate the unknown volume, assuming volumes are additive.
[5 marks]
Total for this question: 5
A solution has volume $500\,\mathrm{cm}^3$ and mass concentration $40.0\,\mathrm{g\,dm}^{-3}$ . Water evaporates and $5.00\,\mathrm{g}$ of the solute crystallises out. The remaining solution has volume $300\,\mathrm{cm}^3$ . Calculate the mass of solute still dissolved and its final mass concentration.
[5 marks]
Total for this question: 5

4.3.3.1 · Percentage yield (chemistry only)

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: calculate the theoretical mass of product from a given mass of reactant.
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

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
A process has a theoretical product mass of $12.0\,\mathrm{g}$ and a percentage yield of $75.0\%$ . Calculate the actual product mass.
[2 marks]
Total for this question: 2

Tier 2 · Standard

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
Method A has a theoretical yield of $20.0\,\mathrm{g}$ and gives $15.0\,\mathrm{g}$ . Method B has a theoretical yield of $12.0\,\mathrm{g}$ and gives $9.60\,\mathrm{g}$ . Calculate both percentage yields and identify the more efficient method by this measure.
[3 marks]
Total for this question: 3
A preparation has a theoretical yield of $10.0\,\mathrm{g}$ . A student first collects $6.20\,\mathrm{g}$ of crystals, then recovers another $1.40\,\mathrm{g}$ of the same pure product from the solution. Calculate the total actual mass and the overall percentage yield.
[3 marks]
Total for this question: 3

Tier 3 · Hard

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
Higher Tier: Magnesium burns by $2\mathrm{Mg}+\mathrm{O}_2\rightarrow2\mathrm{MgO}$ . A student heats $6.00\,\mathrm{g}$ of magnesium with excess oxygen and obtains $8.30\,\mathrm{g}$ of magnesium oxide. Calculate the theoretical product mass and the percentage yield. Use $A_r$ : $\mathrm{Mg}=24$ , $\mathrm{O}=16$ .
[5 marks]
Total for this question: 5
Method P gives a $84.0\%$ yield and method Q gives a $70.0\%$ yield. Each method would make $1.00\,\mathrm{kg}$ of product theoretically from $1.00\,\mathrm{kg}$ of its feed material. Calculate the feed mass each method needs to make $42.0\,\mathrm{kg}$ of actual product. Determine which method needs less feed.
[4 marks]
Total for this question: 4
A two-stage process would make $50.0\,\mathrm{g}$ of the final product if both stages had $100\%$ yield. Stage 1 has an $80.0\%$ yield. Each gram recovered from stage 1 could form one gram of final product, but stage 2 converts only $75.0\%$ of it. Calculate the final mass and the overall percentage yield.
[5 marks]
Total for this question: 5
A reaction has a theoretical yield of $18.0\,\mathrm{g}$ . The collected solid has mass $20.0\,\mathrm{g}$ , but analysis shows that $14.5\%$ of this mass is trapped solvent. Calculate the mass of pure product and its percentage yield. Evaluate the student's claim that the reaction has a yield above $100\%$ .
[5 marks]
Total for this question: 5

4.3.3.2 · Atom economy (chemistry only)

Explanation

Atom economy measures the proportion of reactant atoms that become the desired product in the balanced equation. Calculate it using $\dfrac{M_r\text{ of desired product from the equation}}{\text{sum of }M_r\text{ of all reactants from the equation}}\times100$ .
A high atom economy reduces unwanted by-products, conserves resources and can lower disposal costs.
Include equation coefficients when totaling formula masses; atom economy is not the same as percentage yield.
Higher tier: explain why a reaction pathway is chosen using atom economy, yield, rate, equilibrium position and usefulness of by-products.
Exam calculations should show the relationship used, substitution with units and a suitably rounded result.

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.

Exam tip

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

Tier 1 · Easy

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
Reaction 1 makes only the desired product. Reaction 2 makes the desired product and an unwanted product. Identify which reaction has $100\%$ atom economy and give one reason.
[2 marks]
Total for this question: 2

Tier 2 · Standard

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
Calcium chloride is the desired product in $\mathrm{CaCO}_3+2\mathrm{HCl}\rightarrow\mathrm{CaCl}_2+\mathrm{H}_2\mathrm{O}+\mathrm{CO}_2$ . Calculate the atom economy to three significant figures. Use $M_r$ : $\mathrm{CaCO}_3=100$ , $\mathrm{HCl}=36.5$ , $\mathrm{CaCl}_2=111$ .
[3 marks]
Total for this question: 3
A reaction has an atom economy of $62.5\%$ . The relative formula mass of the desired product shown in the equation is $80$ . Determine the sum of the relative formula masses of all reactants shown in the equation.
[3 marks]
Total for this question: 3

Tier 3 · Hard

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
Higher Tier: Route A has $78.0\%$ atom economy and $90.0\%$ yield. Route B has $92.0\%$ atom economy and $72.0\%$ yield. For $100\,\mathrm{kg}$ of reactants in each route, calculate the actual desired-product mass. Choose the route that maximises product mass and give one environmental advantage of the other route.
[5 marks]
Total for this question: 5
Copper is the desired product in two reactions. Route 1: $\mathrm{CuO}+\mathrm{H}_2\rightarrow\mathrm{Cu}+\mathrm{H}_2\mathrm{O}$ . Route 2: $2\mathrm{CuO}+\mathrm{C}\rightarrow2\mathrm{Cu}+\mathrm{CO}_2$ . Calculate the atom economy of each route and identify which route forms the smaller proportion of unwanted products. Use $A_r$ : $\mathrm{Cu}=63.5$ , $\mathrm{O}=16$ , $\mathrm{H}=1$ , $\mathrm{C}=12$ .
[5 marks]
Total for this question: 5
A factory processes $60.0\,\mathrm{kg}$ of reactants in the balanced-equation proportions for Route A, which has $80.0\%$ atom economy. It processes another $40.0\,\mathrm{kg}$ in the balanced-equation proportions for Route B, which has $65.0\%$ atom economy. Calculate the total desired-product mass, the total unwanted-product mass and the atom economy of the combined production.
[6 marks]
Total for this question: 6
Copper is the desired product in $2\mathrm{NH}_3+3\mathrm{CuO}\rightarrow\mathrm{N}_2+3\mathrm{Cu}+3\mathrm{H}_2\mathrm{O}$ . A student ignores the coefficients and calculates atom economy using $63.5/(17.0+79.5)$ . Calculate the student's value and the correct atom economy, then explain the effect of the error. Use $M_r$ : $\mathrm{NH}_3=17.0$ , $\mathrm{CuO}=79.5$ ; $A_r(\mathrm{Cu})=63.5$ .
[6 marks]
Total for this question: 6

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

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

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
Calculate the volume of a $0.300\,\mathrm{mol\,dm}^{-3}$ solution containing $0.0600\,\mathrm{mol}$ of solute. Give the volume in $\mathrm{cm}^3$ .
[2 marks]
Total for this question: 2

Tier 2 · Standard

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
$5.85\,\mathrm{g}$ of sodium chloride is dissolved to make $500\,\mathrm{cm}^3$ of solution. Calculate the concentration in $\mathrm{mol\,dm}^{-3}$ . Use $M_r(\mathrm{NaCl})=58.5$ .
[4 marks]
Total for this question: 4
$100\,\mathrm{cm}^3$ of a $0.800\,\mathrm{mol\,dm}^{-3}$ solution is diluted with water to a final volume of $250\,\mathrm{cm}^3$ . Calculate the final concentration.
[3 marks]
Total for this question: 3

Tier 3 · Hard

$25.0\,\mathrm{cm}^3$ 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
Higher Tier: $24.0\,\mathrm{cm}^3$ of $0.175\,\mathrm{mol\,dm}^{-3}$ nitric acid reacts exactly with potassium hydroxide solution of concentration $0.140\,\mathrm{mol\,dm}^{-3}$ . Use $\mathrm{HNO}_3+\mathrm{KOH}\rightarrow\mathrm{KNO}_3+\mathrm{H}_2\mathrm{O}$ to calculate the volume of potassium hydroxide solution in $\mathrm{cm}^3$ .
[5 marks]
Total for this question: 5
$25.0\,\mathrm{cm}^3$ portions of sodium hydroxide are titrated with $0.100\,\mathrm{mol\,dm}^{-3}$ hydrochloric acid. The titres are $23.60$ , $23.54$ and $25.10\,\mathrm{cm}^3$ . Identify the anomalous titre, calculate the mean of the consistent titres and determine the sodium hydroxide concentration. Use $\mathrm{HCl}+\mathrm{NaOH}\rightarrow\mathrm{NaCl}+\mathrm{H}_2\mathrm{O}$ and unrounded values.
[5 marks]
Total for this question: 5
Higher Tier: $100\,\mathrm{cm}^3$ of $0.200\,\mathrm{mol\,dm}^{-3}$ silver nitrate is mixed with $150\,\mathrm{cm}^3$ of $0.100\,\mathrm{mol\,dm}^{-3}$ sodium chloride. They react $1:1$ : $\mathrm{AgNO}_3+\mathrm{NaCl}\rightarrow\mathrm{AgCl}+\mathrm{NaNO}_3$ . Determine the limiting reactant and the concentration of the excess reactant in the final mixture. Assume volumes are additive.
[6 marks]
Total for this question: 6
Higher Tier: A student dilutes $16.0\,\mathrm{cm}^3$ of hydrochloric acid to $200\,\mathrm{cm}^3$ . A $20.0\,\mathrm{cm}^3$ portion of the diluted acid reacts exactly with $14.4\,\mathrm{cm}^3$ of $0.100\,\mathrm{mol\,dm}^{-3}$ sodium hydroxide. Use $\mathrm{HCl}+\mathrm{NaOH}\rightarrow\mathrm{NaCl}+\mathrm{H}_2\mathrm{O}$ to calculate the concentration of the original acid.
[6 marks]
Total for this question: 6

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

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

Calculate the volume occupied by $0.350\,\mathrm{mol}$ of a gas at room temperature and pressure.
[2 marks]
Total for this question: 2
Calculate the amount in moles of a gas occupying $3.60\,\mathrm{dm}^3$ at room temperature and pressure.
[2 marks]
Total for this question: 2

Tier 2 · Standard

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
Nitrogen and hydrogen react by $\mathrm{N}_2+3\mathrm{H}_2\rightarrow2\mathrm{NH}_3$ . Calculate the ammonia volume formed from $9.00\,\mathrm{dm}^3$ of hydrogen when nitrogen is in excess and all gases are measured under the same conditions.
[3 marks]
Total for this question: 3
A $0.960\,\mathrm{g}$ sample of a gas occupies $0.480\,\mathrm{dm}^3$ at room temperature and pressure. Calculate the relative formula mass of the gas. Use a molar gas volume of $24\,\mathrm{dm}^3$ .
[3 marks]
Total for this question: 3

Tier 3 · Hard

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
Calcium carbonate decomposes by $\mathrm{CaCO}_3\rightarrow\mathrm{CaO}+\mathrm{CO}_2$ . Heating $10.0\,\mathrm{g}$ produces $2.04\,\mathrm{dm}^3$ of carbon dioxide at room temperature and pressure. Calculate the theoretical gas volume and the percentage yield. Use $M_r(\mathrm{CaCO}_3)=100$ and a molar gas volume of $24\,\mathrm{dm}^3$ .
[5 marks]
Total for this question: 5
Ammonia decomposes by $2\mathrm{NH}_3\rightarrow\mathrm{N}_2+3\mathrm{H}_2$ . The nitrogen and hydrogen produced have a combined volume of $48.0\,\mathrm{dm}^3$ at room temperature and pressure. Calculate the volume that the ammonia would occupy under the same conditions and the mass of ammonia that decomposed. Use $M_r(\mathrm{NH}_3)=17.0$ and a molar gas volume of $24\,\mathrm{dm}^3$ .
[5 marks]
Total for this question: 5
Higher Tier: A $60.0\,\mathrm{cm}^3$ sample containing methane mixed with an unreactive gas is burned in excess oxygen. It produces $48.0\,\mathrm{cm}^3$ of carbon dioxide under the same conditions. Use $\mathrm{CH}_4+2\mathrm{O}_2\rightarrow\mathrm{CO}_2+2\mathrm{H}_2\mathrm{O}$ to calculate the percentage of methane in the sample and the oxygen volume used.
[5 marks]
Total for this question: 5
Higher Tier: A $60.0\,\mathrm{dm}^3$ mixture of nitrogen and hydrogen reacts according to $\mathrm{N}_2+3\mathrm{H}_2\rightarrow2\mathrm{NH}_3$ . Assume the reaction goes to completion. After reaction, the ammonia and excess nitrogen occupy $32.0\,\mathrm{dm}^3$ in total under the same conditions. Determine the starting volume of each reactant and the final volume of nitrogen.
[6 marks]
Total for this question: 6

Answer key

Answers begin on a new printed page so the question pack can be completed without the solutions alongside it.

4.3.1.1 · Conservation of mass and balanced chemical equations

Tier 1 · Easy

Mark scheme for 4.3.1.1 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	$4\mathrm{Al}+3\mathrm{O}_2\rightarrow2\mathrm{Al}_2\mathrm{O}_3$	Make the oxygen total even by placing $2$ before $\mathrm{Al}_2\mathrm{O}_3$ . This gives six oxygen atoms, so place $3$ before $\mathrm{O}_2$ . There are now four aluminium atoms on the right, so place $4$ before $\mathrm{Al}$ .	1
Total Question 1			1
02.1	Coefficient $=2$	The left side contains two sodium atoms and two chlorine atoms. A coefficient of $2$ gives two $\mathrm{NaCl}$ formula units and therefore the same atom totals on the right.	1
Total Question 2			1

Tier 2 · Standard

Mark scheme for 4.3.1.1 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	Changing the subscript changes magnesium oxide into a different substance $2\mathrm{Mg}+\mathrm{O}_2\rightarrow2\mathrm{MgO}$	Formulae must not be changed when balancing because their subscripts identify the substances. Place a coefficient of $2$ before both $\mathrm{Mg}$ and $\mathrm{MgO}$ . This gives two magnesium atoms and two oxygen atoms on each side: $2\mathrm{Mg}+\mathrm{O}_2\rightarrow2\mathrm{MgO}$ .	2
Total Question 1			2
02.1	Mass of product $=11.0\,\mathrm{g}$ The law of conservation of mass	No material enters or leaves the sealed flask. The product mass is therefore the total starting mass, $7.0+4.0=11.0\,\mathrm{g}$ , by conservation of mass.	2
Total Question 2			2
03.1	Equation C: $4\mathrm{Li}+\mathrm{O}_2\rightarrow2\mathrm{Li}_2\mathrm{O}$ Each side contains four lithium atoms and two oxygen atoms	Count each element without changing any formula. Equation C has four lithium atoms and two oxygen atoms on the left. Its two lithium oxide formula units contain four lithium atoms and two oxygen atoms, so both elements balance.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.3.1.1 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	Mass of oxygen used $=40.0\,\mathrm{g}$ Reactants and products each total $51.0\,\mathrm{g}$	The products have total mass $33.0+18.0=51.0\,\mathrm{g}$ . Conservation requires the reactants to have the same total mass, so the oxygen mass is $51.0-11.0=40.0\,\mathrm{g}$ . Checking gives $11.0+40.0=51.0\,\mathrm{g}$ on the reactant side and $33.0+18.0=51.0\,\mathrm{g}$ on the product side.	3
Total Question 1			3
02.1	$4\mathrm{NH}_3+3\mathrm{O}_2\rightarrow2\mathrm{N}_2+6\mathrm{H}_2\mathrm{O}$ Each side has $4$ nitrogen atoms, $12$ hydrogen atoms and $6$ oxygen atoms	Place $2$ before $\mathrm{N}_2$ , requiring $4$ before $\mathrm{NH}_3$ . Twelve hydrogen atoms then require $6$ water molecules. These contain six oxygen atoms, so place $3$ before $\mathrm{O}_2$ .	4
Total Question 2			4
03.1	$3\mathrm{Fe}+4\mathrm{H}_2\mathrm{O}\rightarrow\mathrm{Fe}_3\mathrm{O}_4+4\mathrm{H}_2$ The coefficient $4$ means four water molecules take part The subscript $4$ means each formula unit of $\mathrm{Fe}_3\mathrm{O}_4$ contains four oxygen atoms	The product formula requires three iron atoms and four oxygen atoms, so place $3$ before iron and $4$ before water. Four water molecules also contain eight hydrogen atoms, requiring $4$ hydrogen molecules. A coefficient changes the number of particles; a subscript is part of a substance's formula.	4
Total Question 3			4
04.1	The four X atoms on the left are shared between two formula units, so $a=2$ The six oxygen atoms on the left are shared between two formula units, so $b=3$ The oxide formula is $\mathrm{X}_2\mathrm{O}_3$ Each side contains four X atoms Each side contains six oxygen atoms	The coefficient $2$ means that two formula units of the oxide are formed. Four X atoms divided between those units gives two X atoms per formula unit. Three oxygen molecules contain six oxygen atoms, so each product formula unit contains three oxygen atoms. The completed equation is $4\mathrm{X}+3\mathrm{O}_2\rightarrow2\mathrm{X}_2\mathrm{O}_3$ , with four X atoms and six oxygen atoms on each side.	5
Total Question 4			5
05.1	$2\mathrm{H}_2\mathrm{O}_2\rightarrow2\mathrm{H}_2\mathrm{O}+\mathrm{O}_2$ The balanced equation has four hydrogen atoms on each side The balanced equation has four oxygen atoms on each side Water mass $=34.0-16.0=18.0\,\mathrm{g}$ Total product mass $=16.0+18.0=34.0\,\mathrm{g}$ The product mass equals the initial reactant mass because the vessel is sealed	Placing $2$ before hydrogen peroxide and water gives four hydrogen atoms and four oxygen atoms on each side. Because no substance can leave the sealed vessel, the water mass is the initial mass minus the oxygen mass: $34.0-16.0=18.0\,\mathrm{g}$ . The product masses total $16.0+18.0=34.0\,\mathrm{g}$ , matching the starting mass and confirming conservation at both the atom and mass levels.	6
Total Question 5			6

4.3.1.2 · Relative formula mass

Tier 1 · Easy

Mark scheme for 4.3.1.2 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	$M_r(\mathrm{Ca(OH)}_2)=74$	The bracket is multiplied by $2$ , so $M_r=40+2(16+1)=40+34=74$ .	2
Total Question 1			2
02.1	There are eight hydrogen atoms, not four $M_r=132$	The outside $2$ multiplies both atoms inside the bracket, giving two nitrogen atoms and eight hydrogen atoms. Therefore $M_r=2(14)+8(1)+32+4(16)=132$ .	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.3.1.2 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	Total for the reactants $=36$ Total for the products $=36$	$M_r(\mathrm{H}_2)=2$ and $M_r(\mathrm{O}_2)=32$ , so the reactant total is $2(2)+32=36$ . $M_r(\mathrm{H}_2\mathrm{O})=18$ , so the product total is $2(18)=36$ . The totals are equal.	3
Total Question 1			3
02.1	Oxygen in $\mathrm{MgO}=40.0\%$ Oxygen in $\mathrm{SO}_2=50.0\%$ $\mathrm{SO}_2$ has the greater oxygen percentage	$M_r(\mathrm{MgO})=40$ , so its oxygen percentage is $(16/40)\times100=40.0\%$ . $M_r(\mathrm{SO}_2)=64$ and oxygen contributes $32$ , giving $(32/64)\times100=50.0\%$ . Therefore sulfur dioxide has the greater value.	4
Total Question 2			4
03.1	$x=2$ $\mathrm{MgCl}_2$	The chlorine atoms contribute $95-24=71$ . Dividing by $A_r(\mathrm{Cl})$ gives $x=71/35.5=2$ , so the completed formula is $\mathrm{MgCl}_2$ .	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.3.1.2 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	Mass of oxygen $=9.60\,\mathrm{g}$	$M_r=2(27)+3[32+4(16)]=342$ . The twelve oxygen atoms contribute $12(16)=192$ , so the oxygen fraction is $192/342$ . The oxygen mass is $17.1\times(192/342)=9.60\,\mathrm{g}$ .	4
Total Question 1			4
02.1	Reactant equation mass $=168$ and carbon dioxide equation mass $=44$ Carbon dioxide percentage $=26.2\%$ Carbon dioxide mass $=11.0\,\mathrm{g}$	$M_r(\mathrm{NaHCO}_3)=84$ , so the equation has reactant mass $2(84)=168$ . Carbon dioxide contributes $44$ , giving $(44/168)\times100=26.190\ldots\%=26.2\%$ . Using the unrounded fraction, $42.0\times(44/168)=11.0\,\mathrm{g}$ .	5
Total Question 2			5
03.1	Percentage by mass of oxygen $=30.0\%$ $A_r(\mathrm{M})=56$	The oxygen contribution is $3(16)=48$ , so the oxygen percentage is $(48/160)\times100=30.0\%$ . The two M atoms contribute $160-48=112$ , giving $A_r(\mathrm{M})=112/2=56$ .	4
Total Question 3			4
04.1	Oxygen contributes $16$ out of the relative formula mass $M_r(\mathrm{MO})=16/0.200=80$ $A_r(\mathrm{M})=80-16=64$ M is $80.0\%$ of the oxide by mass Mass of M $=15.0\times0.800=12.0\,\mathrm{g}$	Write the oxygen fraction as $0.200=16/M_r$ . Rearranging gives $M_r=16/0.200=80$ . The metal therefore contributes $80-16=64$ , so $A_r(\mathrm{M})=64$ . The remaining $80.0\%$ of the oxide is M, giving a mass of $15.0\times0.800=12.0\,\mathrm{g}$ .	5
Total Question 4			5
05.1	$\mathrm{CO}_2$ contains $(32/44)\times100=72.7\%$ oxygen $\mathrm{SO}_3$ contains $(48/80)\times100=60.0\%$ oxygen $\mathrm{NO}_2$ contains $(32/46)\times100=69.6\%$ oxygen The oxide is $\mathrm{SO}_3$ Oxygen mass $=40.0\times0.600=24.0\,\mathrm{g}$	Work out the oxygen mass units in each formula: $32$ in $\mathrm{CO}_2$ , $48$ in $\mathrm{SO}_3$ and $32$ in $\mathrm{NO}_2$ . The unrounded oxygen percentages are $(32/44)\times100=72.727\ldots\%$ for $\mathrm{CO}_2$ , $(48/80)\times100=60.0\%$ for $\mathrm{SO}_3$ and $(32/46)\times100=69.565\ldots\%$ for $\mathrm{NO}_2$ . Only sulfur trioxide matches $60.0\%$ . Its oxygen fraction is $0.600$ , so the oxygen mass is $40.0\times0.600=24.0\,\mathrm{g}$ .	5
Total Question 5			5

4.3.1.3 · Mass changes when a reactant or product is a gas

Tier 1 · Easy

Mark scheme for 4.3.1.3 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	Oxygen from the air has combined with the magnesium Mass added $=4.0\,\mathrm{g}$	The magnesium is not the only reactant: oxygen gas enters from the surroundings and becomes part of the solid. The mass gained is $10.0-6.0=4.0\,\mathrm{g}$ , which is the oxygen mass taken in.	2
Total Question 1			2
02.1	Mass of oxygen $=1.9\,\mathrm{g}$	The added oxygen accounts for the mass gain: $9.5-7.6=1.9\,\mathrm{g}$ .	1
Total Question 2			1

Tier 2 · Standard

Mark scheme for 4.3.1.3 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	A gas is produced and gas particles escape from the open beaker The total mass would remain constant in a sealed flask	The reaction produces a gas. In the open beaker, gas particles leave the measured system, so the balance reading decreases. In a sealed flask no particles can enter or leave, so the mass of the flask and its contents remains constant.	3
Total Question 1			3
02.1	Mass decrease $=4.40\,\mathrm{g}$ Carbon dioxide gas forms and escapes from the open container	The decrease is $12.35-7.95=4.40\,\mathrm{g}$ . Copper carbonate decomposes to copper oxide and carbon dioxide. The carbon dioxide leaves the measured system, although mass is conserved when the gas is included.	3
Total Question 2			3
03.1	Mass of gas collected $=1.7\,\mathrm{g}$ The gas has moved into the bag but no particles have left the complete apparatus	The mass transferred from the flask contents to the bag is $15.6-13.9=1.7\,\mathrm{g}$ . The flask and bag form a closed system, so every particle remains within the measured apparatus and its total mass is conserved.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.3.1.3 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	Crucible and contents gain $3.2\,\mathrm{g}$ because oxygen enters the oxide The complete sealed chamber has no mass change	Oxygen particles with mass $3.2\,\mathrm{g}$ move from the chamber gas into the solid, so the crucible and contents alone gain that mass. No particles cross the chamber boundary, so the gas loses exactly $3.2\,\mathrm{g}$ and the total chamber mass remains constant.	4
Total Question 1			4
02.1	Flask Y has no change in total mass Carbon dioxide escapes through the cotton wool from flask X but remains inside sealed flask Y	Both reactions make carbon dioxide. The gas can pass through the cotton wool, so flask X loses $1.10\,\mathrm{g}$ from the measured system. In sealed flask Y every particle remains inside; gas formation changes its location and state but not the total mass.	4
Total Question 2			4
03.1	Total mass of gases $=10.8\,\mathrm{g}$ The equation-mass ratio predicts $8.0\,\mathrm{g}$ of copper(II) oxide from $18.8\,\mathrm{g}$ of copper(II) nitrate	$M_r(\mathrm{Cu(NO}_3)_2)=188$ and $M_r(\mathrm{CuO})=80$ . The equation shows $2(188)=376$ relative mass units of copper(II) nitrate forming $2(80)=160$ units of copper(II) oxide. The predicted solid mass is $18.8\times(160/376)=8.0\,\mathrm{g}$ . The remaining $18.8-8.0=10.8\,\mathrm{g}$ is the combined nitrogen dioxide and oxygen mass, so the stated data conserve mass when the gases are included.	4
Total Question 3			4
04.1	Oxygen mass $=12.0-9.60=2.40\,\mathrm{g}$ Oxygen percentage $=(2.40/12.0)\times100=20.0\%$ Metal percentage $=(9.60/12.0)\times100=80.0\%$ The two percentages total $100.0\%$ The mass increases because oxygen from the air becomes part of the solid oxide	The metal accounts for $9.60\,\mathrm{g}$ of the oxide, so the oxygen gained from the air has mass $12.0-9.60=2.40\,\mathrm{g}$ . Relative to the $12.0\,\mathrm{g}$ oxide, oxygen contributes $20.0\%$ and metal contributes $80.0\%$ . Their total is $100.0\%$ , showing that the apparent gain is the oxygen now incorporated into the solid.	5
Total Question 4			5
05.1	Metal A gains $3.75-3.00=0.750\,\mathrm{g}$ of oxygen, which is $0.750/3.00=0.250\,\mathrm{g}$ per gram of metal Metal B gains $4.50-3.00=1.50\,\mathrm{g}$ of oxygen, which is $1.50/3.00=0.500\,\mathrm{g}$ per gram of metal Metal B combines with twice as much oxygen per gram of metal as metal A Oxygen particles from the air enter each open crucible and combine with metal atoms, becoming part of the solid oxide No atoms are created or destroyed, so total mass is conserved when the air and crucible contents are considered together	Subtract each starting mass from its oxide mass: $3.75-3.00=0.750\,\mathrm{g}$ for A and $4.50-3.00=1.50\,\mathrm{g}$ for B. Divide by the $3.00\,\mathrm{g}$ of metal to compare fairly: $0.250$ and $0.500\,\mathrm{g}$ of oxygen per gram, so B combines with twice as much. Because the crucibles are open, oxygen particles from the surrounding air can enter and react; those atoms become part of each solid oxide. The atoms have only been rearranged, so including the oxygen removed from the air the total mass is unchanged.	5
Total Question 5			5

4.3.1.4 · Chemical measurements

Tier 1 · Easy

Mark scheme for 4.3.1.4 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	Mean $=12.5\,\mathrm{cm}^3$ Uncertainty $=\pm0.1\,\mathrm{cm}^3$	The mean is $(12.4+12.6+12.5)/3=12.5\,\mathrm{cm}^3$ . The range is $12.6-12.4=0.2\,\mathrm{cm}^3$ , so half the range is $0.1\,\mathrm{cm}^3$ . Report $(12.5\pm0.1)\,\mathrm{cm}^3$ .	3
Total Question 1			3
02.1	Lowest value $=12.4\,^\circ\mathrm{C}$ Highest value $=12.8\,^\circ\mathrm{C}$	Subtract the uncertainty for the lower limit: $12.6-0.2=12.4\,^\circ\mathrm{C}$ . Add it for the upper limit: $12.6+0.2=12.8\,^\circ\mathrm{C}$ .	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.3.1.4 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	Student A: uncertainty $=\pm3\,\mathrm{s}$ Student B: uncertainty $=\pm1\,\mathrm{s}$ Student B's results are more precise	Student A's range is $48-42=6\,\mathrm{s}$ , so the half-range uncertainty is $\pm3\,\mathrm{s}$ . Student B's range is $46-44=2\,\mathrm{s}$ , so the half-range uncertainty is $\pm1\,\mathrm{s}$ . Student B has the smaller spread and therefore the more precise results.	4
Total Question 1			4
02.1	Half-range uncertainty $=\pm0.03\,\mathrm{g}$ The repeat spread gives the larger uncertainty Use $\pm0.03\,\mathrm{g}$	The repeat range is $6.48-6.42=0.06\,\mathrm{g}$ , so the half-range uncertainty is $\pm0.03\,\mathrm{g}$ . This is larger than the $\pm0.01\,\mathrm{g}$ uncertainty due to the balance resolution, so the repeat spread dominates and $\pm0.03\,\mathrm{g}$ should be used.	4
Total Question 2			4
03.1	Thermometer P uncertainty $=\pm0.5\,^\circ\mathrm{C}$ Thermometer Q uncertainty $=\pm0.1\,^\circ\mathrm{C}$ Thermometer Q gives the more precise reading	Half of P's $1.0\,^\circ\mathrm{C}$ division is $0.5\,^\circ\mathrm{C}$ ; half of Q's $0.2\,^\circ\mathrm{C}$ division is $0.1\,^\circ\mathrm{C}$ . Q has the smaller uncertainty and therefore the greater precision.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.3.1.4 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	Anomalous reading $=36.9\,\mathrm{s}$ Mean of consistent readings $=31.45\,\mathrm{s}$ Reported result $=(31.45\pm0.15)\,\mathrm{s}$	$36.9\,\mathrm{s}$ lies far from the cluster and is anomalous. The other four readings have mean $(31.4+31.6+31.5+31.3)/4=31.45\,\mathrm{s}$ . Their range is $31.6-31.3=0.3\,\mathrm{s}$ , so the half-range uncertainty is $0.15\,\mathrm{s}$ .	4
Total Question 1			4
02.1	Experiment A interval: $5.39$ to $5.45\,\mathrm{g}$ Experiment B interval: $5.46$ to $5.54\,\mathrm{g}$ The intervals do not overlap, suggesting that the results are different beyond these estimated uncertainties	For A, the limits are $5.42-0.03=5.39\,\mathrm{g}$ and $5.42+0.03=5.45\,\mathrm{g}$ . For B, they are $5.50-0.04=5.46\,\mathrm{g}$ and $5.50+0.04=5.54\,\mathrm{g}$ . Since $5.45<5.46$ , the intervals do not overlap, providing evidence that the measured results differ.	4
Total Question 2			4
03.1	Before: mean $=20.3\,\mathrm{s}$ and range $=0.4\,\mathrm{s}$ After: mean $=20.3\,\mathrm{s}$ and range $=0.4\,\mathrm{s}$ The claim is not supported because the range is unchanged, although more repeats can improve confidence in the mean	The first total is $60.9\,\mathrm{s}$ , giving mean $60.9/3=20.3\,\mathrm{s}$ and range $20.5-20.1=0.4\,\mathrm{s}$ . All five readings total $101.5\,\mathrm{s}$ , giving mean $20.3\,\mathrm{s}$ and the same $0.4\,\mathrm{s}$ range. Extra repeats do not guarantee a smaller spread.	4
Total Question 3			4
04.1	Method A mean $=24.7\,^\circ\mathrm{C}$ Method A uncertainty $=\pm0.1\,^\circ\mathrm{C}$ Method B mean $=25.1\,^\circ\mathrm{C}$ Method B uncertainty $=\pm0.2\,^\circ\mathrm{C}$ Method A is more precise because it has the smaller spread Method B's mean is closer to the accepted value	Method A has mean $(24.6+24.8+24.7)/3=24.7\,^\circ\mathrm{C}$ and half-range $(24.8-24.6)/2=0.1\,^\circ\mathrm{C}$ . Method B has mean $(24.9+25.3+25.1)/3=25.1\,^\circ\mathrm{C}$ and half-range $(25.3-24.9)/2=0.2\,^\circ\mathrm{C}$ . A is more precise because its readings are less spread out, while B is closer to the accepted value because its mean differs by $0.1\,^\circ\mathrm{C}$ rather than $0.3\,^\circ\mathrm{C}$ .	6
Total Question 4			6
05.1	Required total for five readings $=5\times7.45=37.25\,\mathrm{g}$ The four known readings total $29.80\,\mathrm{g}$ Missing reading $=37.25-29.80=7.45\,\mathrm{g}$ Half-range uncertainty $=(7.48-7.42)/2=0.03\,\mathrm{g}$ Reported result $=(7.45\pm0.03)\,\mathrm{g}$ , with interval $7.42$ to $7.48\,\mathrm{g}$ $7.50\,\mathrm{g}$ is outside the interval	The five readings must total $5\times7.45=37.25\,\mathrm{g}$ . The four stated values total $29.80\,\mathrm{g}$ , so the missing reading is $7.45\,\mathrm{g}$ . The range is $7.48-7.42=0.06\,\mathrm{g}$ and the half-range is $0.03\,\mathrm{g}$ . The interval is therefore $7.42$ to $7.48\,\mathrm{g}$ , so the expected $7.50\,\mathrm{g}$ lies above it.	6
Total Question 5			6

4.3.2.1 · Moles (HT only)

Tier 1 · Easy

Mark scheme for 4.3.2.1 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	Amount of water $=0.50\,\mathrm{mol}$	Use $n=m/M_r$ : $n=9.0/18=0.50\,\mathrm{mol}$ .	2
Total Question 1			2
02.1	Amount of helium $=2.00\,\mathrm{mol}$	Divide the number of atoms by the Avogadro constant: $n=(1.204\times10^{24})/(6.02\times10^{23})=2.00\,\mathrm{mol}$ .	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.3.2.1 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	(a) Each sample contains $1.0\,\mathrm{mol}$ of molecules (b) The oxygen sample, with $2.0\,\mathrm{mol}$ of atoms against $1.0\,\mathrm{mol}$	Helium is monatomic, so its molecules are single atoms: amount $=4.0/4=1.0\,\mathrm{mol}$ . $M_r(\mathrm{O}_2)=32$ , so the oxygen amount $=32.0/32=1.0\,\mathrm{mol}$ of molecules. Equal amounts in moles means equal numbers of molecules. Each $\mathrm{O}_2$ molecule contains two atoms, so the oxygen sample holds $2.0\,\mathrm{mol}$ of atoms against $1.0\,\mathrm{mol}$ for helium.	4
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02.1	Mass of calcium carbonate $=50.0\,\mathrm{g}$	The amount is $(3.01\times10^{23})/(6.02\times10^{23})=0.500\,\mathrm{mol}$ . Therefore $m=nM_r=0.500\times100=50.0\,\mathrm{g}$ .	3
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03.1	Number of electrons $=2.89\times10^{23}$	The number of magnesium atoms is $0.0400\times6.02\times10^{23}=2.408\times10^{22}$ . Each neutral atom contains $12$ electrons, so the total is $12\times2.408\times10^{22}=2.8896\times10^{23}$ , or $2.89\times10^{23}$ to three significant figures.	3
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Tier 3 · Hard

Mark scheme for 4.3.2.1 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	$9.03\times10^{21}$ formula units $1.81\times10^{22}$ chloride ions	Formula units $=0.0150\times6.02\times10^{23}=9.03\times10^{21}$ . Each formula unit contains two chloride ions, so the ion count is $2\times9.03\times10^{21}=1.806\times10^{22}$ , or $1.81\times10^{22}$ to three significant figures.	4
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02.1	Number of oxygen atoms $=1.20\times10^{23}$	The amount of sodium sulfate is $7.10/142=0.0500\,\mathrm{mol}$ . This is $0.0500\times6.02\times10^{23}=3.01\times10^{22}$ formula units. Each formula unit has four oxygen atoms, giving $1.204\times10^{23}$ atoms, or $1.20\times10^{23}$ to three significant figures.	4
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03.1	Mass of calcium chloride $=16.1\,\mathrm{g}$	Each formula unit contains two chloride ions, so there are $8.729\times10^{22}$ formula units. The amount of calcium chloride is $(8.729\times10^{22})/(6.02\times10^{23})=0.145\,\mathrm{mol}$ . Its mass is $0.145\times111=16.095\,\mathrm{g}$ , or $16.1\,\mathrm{g}$ to three significant figures.	4
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04.1	Carbon dioxide amount $=8.80/44.0=0.200\,\mathrm{mol}$ Nitrogen amount $=2.80/28.0=0.100\,\mathrm{mol}$ Total amount of molecules $=0.300\,\mathrm{mol}$ Total number of molecules $=1.806\times10^{23}$ Amount of atoms $=3(0.200)+2(0.100)=0.800\,\mathrm{mol}$ Total number of atoms $=4.82\times10^{23}$ to three significant figures	The carbon dioxide amount is $8.80/44.0=0.200\,\mathrm{mol}$ and the nitrogen amount is $2.80/28.0=0.100\,\mathrm{mol}$ . The mixture therefore contains $0.300\times6.02\times10^{23}=1.806\times10^{23}$ molecules. Each carbon dioxide molecule contains three atoms and each nitrogen molecule contains two, so the atom amount is $3(0.200)+2(0.100)=0.800\,\mathrm{mol}$ . This corresponds to $0.800\times6.02\times10^{23}=4.816\times10^{23}$ atoms, or $4.82\times10^{23}$ .	6
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05.1	If carbon dioxide amount is $x$ and carbon monoxide amount is $y$ , then $x+y=0.250$ Counting oxygen atoms gives $2x+y=0.400$ Subtracting the equations gives $x=0.150\,\mathrm{mol}$ of carbon dioxide Carbon monoxide amount $=0.250-0.150=0.100\,\mathrm{mol}$ Carbon dioxide mass $=0.150\times44.0=6.60\,\mathrm{g}$ Carbon monoxide mass $=0.100\times28.0=2.80\,\mathrm{g}$	Let $x$ be the carbon dioxide amount and $y$ the carbon monoxide amount. The total molecule amount gives $x+y=0.250$ . Because carbon dioxide has two oxygen atoms per molecule and carbon monoxide has one, $2x+y=0.400$ . Subtracting the first equation from the second gives $x=0.150\,\mathrm{mol}$ and hence $y=0.100\,\mathrm{mol}$ . Multiplying each amount by its relative formula mass gives $6.60\,\mathrm{g}$ of carbon dioxide and $2.80\,\mathrm{g}$ of carbon monoxide.	6
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4.3.2.2 · Amounts of substances in equations (HT only)

Tier 1 · Easy

Mark scheme for 4.3.2.2 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	Mass of magnesium oxide $=8.0\,\mathrm{g}$	Moles of magnesium $=4.8/24=0.20\,\mathrm{mol}$ . The equation ratio $\mathrm{Mg}:\mathrm{MgO}$ is $2:2$ , so $0.20\,\mathrm{mol}$ of magnesium oxide forms. Its $M_r$ is $40$ , giving mass $0.20\times40=8.0\,\mathrm{g}$ .	3
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02.1	Amount of carbon dioxide $=0.150\,\mathrm{mol}$	The equation ratio $\mathrm{O}_2:\mathrm{CO}_2$ is $2:1$ . Therefore carbon dioxide amount $=0.300/2=0.150\,\mathrm{mol}$ .	2
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Tier 2 · Standard

Mark scheme for 4.3.2.2 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	Mass of hydrogen $=6.00\,\mathrm{g}$	Moles of nitrogen $=28.0/28=1.00\,\mathrm{mol}$ . The equation requires three moles of hydrogen for each mole of nitrogen, so hydrogen amount $=3.00\,\mathrm{mol}$ . Its mass is $3.00\times2=6.00\,\mathrm{g}$ .	3
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02.1	Mass of calcium $=12.0\,\mathrm{g}$	Hydrogen amount $=0.600/2=0.300\,\mathrm{mol}$ . The equation gives a $1:1$ ratio, so $0.300\,\mathrm{mol}$ of calcium is needed. Its mass is $0.300\times40=12.0\,\mathrm{g}$ .	3
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03.1	Maximum aluminium oxide mass $=10.2\,\mathrm{g}$	Aluminium amount $=5.40/27=0.200\,\mathrm{mol}$ . The $4:2$ ratio gives $0.100\,\mathrm{mol}$ of aluminium oxide. $M_r(\mathrm{Al}_2\mathrm{O}_3)=2(27)+3(16)=102$ , so the mass is $0.100\times102=10.2\,\mathrm{g}$ .	4
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Tier 3 · Hard

Mark scheme for 4.3.2.2 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	Mass of water $=16.2\,\mathrm{g}$	Moles of ammonia $=10.2/17=0.600\,\mathrm{mol}$ . The equation ratio is $4\mathrm{NH}_3:6\mathrm{H}_2\mathrm{O}$ , so water moles $=0.600\times(6/4)=0.900\,\mathrm{mol}$ . The water mass is $0.900\times18=16.2\,\mathrm{g}$ .	4
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02.1	Maximum mass of iron $=16.8\,\mathrm{g}$	Iron(III) oxide amount $=24.0/160=0.150\,\mathrm{mol}$ . Each mole forms two moles of iron, so iron amount $=0.300\,\mathrm{mol}$ . The mass is $0.300\times56=16.8\,\mathrm{g}$ .	4
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03.1	Mass of calcium carbonate $=10.0\,\mathrm{g}$ Calcium carbonate in the sample $=83.3\%$ by mass	Carbon dioxide amount $=4.40/44=0.100\,\mathrm{mol}$ . The equation ratio is $1:1$ , so the sample contained $0.100\,\mathrm{mol}$ of calcium carbonate, with mass $0.100\times100=10.0\,\mathrm{g}$ . Its percentage by mass is $(10.0/12.0)\times100=83.333\ldots\%=83.3\%$ .	5
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04.1	Iron amount $=33.6/56=0.600\,\mathrm{mol}$ The $2:2$ ratio requires $0.600\,\mathrm{mol}$ of aluminium Aluminium mass $=0.600\times27=16.2\,\mathrm{g}$ The $2:1$ ratio forms $0.300\,\mathrm{mol}$ of aluminium oxide Aluminium oxide mass $=0.300\times102=30.6\,\mathrm{g}$ Both results follow from the same $0.600\,\mathrm{mol}$ equation scale	The iron amount is $33.6/56=0.600\,\mathrm{mol}$ . The equation uses two aluminium moles for every two iron moles, so $0.600\,\mathrm{mol}$ of aluminium is required, with mass $0.600\times27=16.2\,\mathrm{g}$ . Two iron moles accompany one aluminium oxide mole, so $0.300\,\mathrm{mol}$ of aluminium oxide forms. Its mass is $0.300\times102=30.6\,\mathrm{g}$ .	6
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05.1	Calcium carbonate amount $=15.0/100=0.150\,\mathrm{mol}$ The first reaction forms $0.150\,\mathrm{mol}$ of calcium oxide The second reaction forms $0.150\,\mathrm{mol}$ of calcium chloride Maximum calcium chloride mass $=0.150\times111=16.65\,\mathrm{g}$ Hydrogen chloride amount required $=2\times0.150=0.300\,\mathrm{mol}$ Hydrogen chloride mass $=0.300\times36.5=10.95\,\mathrm{g}$	The starting amount is $15.0/100=0.150\,\mathrm{mol}$ of calcium carbonate. The first equation has a $1:1$ ratio to calcium oxide, and the second has a $1:1$ ratio from calcium oxide to calcium chloride, so $0.150\,\mathrm{mol}$ of calcium chloride can form. Its mass is $0.150\times111=16.65\,\mathrm{g}$ . The second equation needs twice as many moles of hydrogen chloride, so $0.300\,\mathrm{mol}$ is required, with mass $0.300\times36.5=10.95\,\mathrm{g}$ .	6
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4.3.2.3 · Using moles to balance equations (HT only)

Tier 1 · Easy

Mark scheme for 4.3.2.3 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	$\mathrm{N}_2+3\mathrm{H}_2\rightarrow2\mathrm{NH}_3$	Divide all amounts by the smallest, $0.20\,\mathrm{mol}$ : $0.20:0.60:0.40$ becomes $1:3:2$ . Use these as the coefficients.	2
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02.1	$2\mathrm{H}_2+\mathrm{O}_2\rightarrow2\mathrm{H}_2\mathrm{O}$	Divide every amount by $0.15\,\mathrm{mol}$ . The ratio is $2:1:2$ , so these are the smallest whole-number coefficients.	2
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Tier 2 · Standard

Mark scheme for 4.3.2.3 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	$2\mathrm{Cu}+\mathrm{O}_2\rightarrow2\mathrm{CuO}$	Convert each mass to moles: copper $=12.7/63.5=0.200$ , oxygen $=3.2/32.0=0.100$ , and copper(II) oxide $=15.9/79.5=0.200$ . Divide by the smallest amount, $0.100$ , to obtain the whole-number ratio $2:1:2$ . The balanced equation is $2\mathrm{Cu}+\mathrm{O}_2\rightarrow2\mathrm{CuO}$ .	4
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02.1	$3\mathrm{Mg}+\mathrm{N}_2\rightarrow\mathrm{Mg}_3\mathrm{N}_2$	The mole amounts are $7.20/24=0.300$ , $2.80/28=0.100$ and $10.0/100=0.100\,\mathrm{mol}$ . Divide by $0.100$ to obtain $3:1:1$ . The balanced equation is $3\mathrm{Mg}+\mathrm{N}_2\rightarrow\mathrm{Mg}_3\mathrm{N}_2$ .	4
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03.1	Mole amounts must be converted to a simple whole-number ratio Smallest whole-number ratio $=2:3:2$	Divide every amount by the smallest, $0.12$ , to obtain $1:1.5:1$ . These are not all whole numbers, so multiply every value by $2$ to get $2:3:2$ .	3
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Tier 3 · Hard

Mark scheme for 4.3.2.3 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	$2\mathrm{C}_2\mathrm{H}_6+7\mathrm{O}_2\rightarrow4\mathrm{CO}_2+6\mathrm{H}_2\mathrm{O}$	The mole amounts are $3.0/30=0.10$ , $11.2/32=0.35$ , $8.8/44=0.20$ and $5.4/18=0.30$ . Divide by $0.10$ to obtain $1:3.5:2:3$ . Multiply every term by $2$ to remove the half, giving $2:7:4:6$ .	5
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02.1	$3\mathrm{Fe}+2\mathrm{O}_2\rightarrow\mathrm{Fe}_3\mathrm{O}_4$	The mole amounts are $16.8/56=0.300$ , $6.40/32=0.200$ and $23.2/232=0.100\,\mathrm{mol}$ . Divide by $0.100$ to obtain $3:2:1$ . The equation is therefore $3\mathrm{Fe}+2\mathrm{O}_2\rightarrow\mathrm{Fe}_3\mathrm{O}_4$ .	5
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03.1	$\mathrm{Na}_2\mathrm{CO}_3+2\mathrm{HCl}\rightarrow2\mathrm{NaCl}+\mathrm{H}_2\mathrm{O}+\mathrm{CO}_2$	The mole amounts are $10.6/106=0.100$ , $7.30/36.5=0.200$ , $11.7/58.5=0.200$ , $1.80/18.0=0.100$ and $4.40/44.0=0.100$ . Divide by $0.100$ to obtain $1:2:2:1:1$ , giving the balanced equation shown.	5
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04.1	Phosphorus amount $=12.4/124=0.100\,\mathrm{mol}$ Oxygen amount $=16.0/32.0=0.500\,\mathrm{mol}$ Product amount $=28.4/284=0.100\,\mathrm{mol}$ Dividing by $0.100$ gives the ratio $1:5:1$ $\mathrm{P}_4+5\mathrm{O}_2\rightarrow\mathrm{P}_4\mathrm{O}_{10}$	Convert each measured mass to an amount: phosphorus is $0.100\,\mathrm{mol}$ , oxygen is $0.500\,\mathrm{mol}$ and the oxide is $0.100\,\mathrm{mol}$ . Dividing every amount by the smallest gives $1:5:1$ . These values are already whole numbers, so the balanced equation is $\mathrm{P}_4+5\mathrm{O}_2\rightarrow\mathrm{P}_4\mathrm{O}_{10}$ .	5
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05.1	Dividing by $0.198$ gives approximately $1:1.53:1.02$ This is close to $1:1.5:1$ rather than a whole-number ratio Multiplying all terms by $2$ gives $2:3:2$ $2\mathrm{Al}+3\mathrm{Cl}_2\rightarrow2\mathrm{AlCl}_3$ Small deviations can arise from measurement uncertainty, rounding or incomplete product recovery	Divide all three experimental amounts by $0.198\,\mathrm{mol}$ to obtain approximately $1:1.53:1.02$ . These values cluster around the simple ratio $1:1.5:1$ . Multiplying every term by $2$ removes the half and gives $2:3:2$ , so the balanced equation is $2\mathrm{Al}+3\mathrm{Cl}_2\rightarrow2\mathrm{AlCl}_3$ . Experimental values need not reproduce exact integers because measured masses have uncertainty, reported values are rounded and some product may be lost.	5
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4.3.2.4 · Limiting reactants (HT only)

Tier 1 · Easy

Mark scheme for 4.3.2.4 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	Chlorine is limiting $4.0\,\mathrm{mol}$ of hydrogen chloride forms	Hydrogen and chlorine react in a $1:1$ ratio. Only $2.0\,\mathrm{mol}$ of chlorine is present, so it uses $2.0\,\mathrm{mol}$ of hydrogen and is limiting. The $1:2$ ratio from chlorine to hydrogen chloride gives $4.0\,\mathrm{mol}$ of product.	2
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02.1	Limiting behaviour depends on the available mole amounts and the balanced-equation ratio, not mass alone	Different substances have different molar masses and may also have different coefficients in the equation. Convert masses to moles and compare those amounts with the required ratio.	1
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Tier 2 · Standard

Mark scheme for 4.3.2.4 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	Chlorine is the limiting reactant Amount of aluminium chloride $=0.133\,\mathrm{mol}$	Aluminium amount $=5.4/27=0.200\,\mathrm{mol}$ and chlorine amount $=14.2/71=0.200\,\mathrm{mol}$ . The equation needs $3\,\mathrm{mol}$ of chlorine for every $2\,\mathrm{mol}$ of aluminium, so the chlorine is limiting. The $3:2$ ratio gives aluminium chloride amount $=0.200\times(2/3)=0.133\,\mathrm{mol}$ to three significant figures.	4
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02.1	Hydrogen chloride is limiting Mass of zinc left $=3.25\,\mathrm{g}$	Zinc amount $=6.50/65=0.100\,\mathrm{mol}$ and hydrogen chloride amount $=3.65/36.5=0.100\,\mathrm{mol}$ . The $1:2$ ratio makes hydrogen chloride limiting and allows $0.0500\,\mathrm{mol}$ of zinc to react. Zinc left $=0.0500\,\mathrm{mol}$ , with mass $0.0500\times65=3.25\,\mathrm{g}$ .	4
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03.1	Mixture A: oxygen is limiting and $0.40\,\mathrm{mol}$ of water forms Mixture B: hydrogen is limiting and $0.36\,\mathrm{mol}$ of water forms Mixture A makes more water	In A, $0.20\,\mathrm{mol}$ of oxygen reacts with $0.40\,\mathrm{mol}$ of hydrogen, so oxygen limits and forms $0.40\,\mathrm{mol}$ of water. In B, $0.36\,\mathrm{mol}$ of hydrogen needs only $0.18\,\mathrm{mol}$ of oxygen, so hydrogen limits and forms $0.36\,\mathrm{mol}$ of water. Therefore A gives the greater amount.	4
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Tier 3 · Hard

Mark scheme for 4.3.2.4 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	Hydrogen is limiting Ammonia mass $=13.6\,\mathrm{g}$ Nitrogen left $=2.8\,\mathrm{g}$	The starting amounts are $14.0/28=0.500\,\mathrm{mol}$ of nitrogen and $2.40/2=1.20\,\mathrm{mol}$ of hydrogen. Using all the nitrogen would require $1.50\,\mathrm{mol}$ of hydrogen, so hydrogen is limiting. It forms $1.20\times(2/3)=0.800\,\mathrm{mol}$ of ammonia, with mass $0.800\times17=13.6\,\mathrm{g}$ . Nitrogen used is $1.20/3=0.400\,\mathrm{mol}$ , leaving $0.100\,\mathrm{mol}$ or $2.8\,\mathrm{g}$ .	5
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02.1	Propane is limiting Maximum propane mass for the oxygen supply $=11.0\,\mathrm{g}$ Additional propane $=2.20\,\mathrm{g}$	Propane amount $=8.80/44=0.200\,\mathrm{mol}$ and oxygen amount $=40.0/32=1.25\,\mathrm{mol}$ . The present propane needs $1.00\,\mathrm{mol}$ of oxygen, so propane is limiting. The full oxygen supply can burn $1.25/5=0.250\,\mathrm{mol}$ of propane, mass $0.250\times44=11.0\,\mathrm{g}$ . A further $11.0-8.80=2.20\,\mathrm{g}$ could be added.	5
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03.1	The original mixture can form $1.60\,\mathrm{mol}$ of ammonia Adding nitrogen gives no increase Adding hydrogen raises the ammonia amount to $1.80\,\mathrm{mol}$ , an increase of $0.20\,\mathrm{mol}$	Initially hydrogen is limiting, so ammonia amount $=2.40\times(2/3)=1.60\,\mathrm{mol}$ . Adding nitrogen leaves the hydrogen amount unchanged and still limiting, so the maximum stays $1.60\,\mathrm{mol}$ . Adding hydrogen gives $2.70\,\mathrm{mol}$ ; this is still below the $3.00\,\mathrm{mol}$ needed for all the nitrogen and forms $2.70\times(2/3)=1.80\,\mathrm{mol}$ , an increase of $0.20\,\mathrm{mol}$ .	5
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04.1	Zinc used $=13.0-3.25=9.75\,\mathrm{g}$ Zinc amount used $=9.75/65=0.150\,\mathrm{mol}$ Hydrogen chloride amount $=2\times0.150=0.300\,\mathrm{mol}$ Starting hydrogen chloride mass $=0.300\times36.5=10.95\,\mathrm{g}$ The $1:1$ zinc-to-hydrogen ratio forms $0.150\,\mathrm{mol}$ of hydrogen Hydrogen chloride is limiting because zinc remains	The zinc mass consumed is $13.0-3.25=9.75\,\mathrm{g}$ , which is $9.75/65=0.150\,\mathrm{mol}$ . The equation requires two hydrogen chloride moles per zinc mole, so the starting hydrogen chloride amount was $0.300\,\mathrm{mol}$ and its mass was $0.300\times36.5=10.95\,\mathrm{g}$ . Zinc and hydrogen have a $1:1$ ratio, so $0.150\,\mathrm{mol}$ of hydrogen forms. The leftover zinc confirms that hydrogen chloride was limiting.	6
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05.1	Calcium carbonate mass $=12.5\times0.800=10.0\,\mathrm{g}$ , or $0.100\,\mathrm{mol}$ Hydrogen chloride amount $=5.475/36.5=0.150\,\mathrm{mol}$ Hydrogen chloride is limiting because $0.100\,\mathrm{mol}$ of calcium carbonate would need $0.200\,\mathrm{mol}$ Carbon dioxide amount $=0.150/2=0.0750\,\mathrm{mol}$ Carbon dioxide mass $=0.0750\times44.0=3.30\,\mathrm{g}$ Calcium carbonate left $=(0.100-0.0750)\times100=2.50\,\mathrm{g}$	The sample contains $12.5\times0.800=10.0\,\mathrm{g}$ of calcium carbonate, equal to $0.100\,\mathrm{mol}$ . The hydrogen chloride amount is $5.475/36.5=0.150\,\mathrm{mol}$ , less than the $0.200\,\mathrm{mol}$ required for all the carbonate, so it is limiting. The $2:1$ ratio gives $0.0750\,\mathrm{mol}$ of carbon dioxide, mass $3.30\,\mathrm{g}$ . The same amount of carbonate reacts, leaving $0.0250\,\mathrm{mol}$ or $2.50\,\mathrm{g}$ .	6
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4.3.2.5 · Concentration of solutions

Tier 1 · Easy

Mark scheme for 4.3.2.5 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	Concentration $=30.0\,\mathrm{g\,dm}^{-3}$	Use $c=m/V$ : $c=12.0/0.400=30.0\,\mathrm{g\,dm}^{-3}$ .	2
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02.1	Solution P has the greater mass concentration	Mass concentration is solute mass divided by solution volume. With equal solute masses, dividing by the smaller volume gives the greater concentration.	1
Total Question 2			1

Tier 2 · Standard

Mark scheme for 4.3.2.5 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	Solution A: $25\,\mathrm{g\,dm}^{-3}$ Solution B: $20\,\mathrm{g\,dm}^{-3}$ Solution A is more concentrated	Convert the volumes to $0.200\,\mathrm{dm}^3$ and $0.400\,\mathrm{dm}^3$ . For A, $c=m/V=5.0/0.200=25\,\mathrm{g\,dm}^{-3}$ . For B, $c=8.0/0.400=20\,\mathrm{g\,dm}^{-3}$ . Solution A has the greater concentration.	3
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02.1	Mass concentration $=30.0\,\mathrm{g\,dm}^{-3}$	Convert the final solution volume: $250\,\mathrm{cm}^3=0.250\,\mathrm{dm}^3$ . Then $c=m/V=7.50/0.250=30.0\,\mathrm{g\,dm}^{-3}$ .	3
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03.1	Mass concentration of mixture $=20\,\mathrm{g\,dm}^{-3}$	The total salt mass is $4.0+6.0=10.0\,\mathrm{g}$ . The total volume is $500\,\mathrm{cm}^3=0.500\,\mathrm{dm}^3$ . Therefore $c=m/V=10.0/0.500=20\,\mathrm{g\,dm}^{-3}$ .	3
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Tier 3 · Hard

Mark scheme for 4.3.2.5 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	Volume of water added $=150\,\mathrm{cm}^3$	The solute mass stays $3.60\,\mathrm{g}$ . The required final volume is $V=m/c=3.60/12.0=0.300\,\mathrm{dm}^3=300\,\mathrm{cm}^3$ . The water added is therefore $300-150=150\,\mathrm{cm}^3$ .	4
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02.1	Final concentration $=30.0\,\mathrm{g\,dm}^{-3}$	The initial volume is $0.400\,\mathrm{dm}^3$ , so solute mass $=18.0\times0.400=7.20\,\mathrm{g}$ . The final volume is $0.240\,\mathrm{dm}^3$ . Therefore the final concentration is $7.20/0.240=30.0\,\mathrm{g\,dm}^{-3}$ .	4
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03.1	Final mass concentration $=9.60\,\mathrm{g\,dm}^{-3}$	The initial solute mass is $24.0\times0.300=7.20\,\mathrm{g}$ . The removed $0.100\,\mathrm{dm}^3$ sample contains $24.0\times0.100=2.40\,\mathrm{g}$ , leaving $4.80\,\mathrm{g}$ . After making the volume up to $0.500\,\mathrm{dm}^3$ , the concentration is $4.80/0.500=9.60\,\mathrm{g\,dm}^{-3}$ .	4
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04.1	Solute mass in the first solution $=15.0\times0.200=3.00\,\mathrm{g}$ If the unknown volume is $x\,\mathrm{dm}^3$ , its solute mass is $30.0x\,\mathrm{g}$ The mixture equation is $(3.00+30.0x)/(0.200+x)=24.0$ Solving gives $x=0.300\,\mathrm{dm}^3$ Unknown volume $=300\,\mathrm{cm}^3$	The first solution contains $15.0\times0.200=3.00\,\mathrm{g}$ of solute. Let the unknown volume be $x\,\mathrm{dm}^3$ ; it then contains $30.0x\,\mathrm{g}$ . Applying concentration to the combined mass and volume gives $(3.00+30.0x)/(0.200+x)=24.0$ . Therefore $3.00+30.0x=4.80+24.0x$ , so $6.00x=1.80$ and $x=0.300\,\mathrm{dm}^3=300\,\mathrm{cm}^3$ .	5
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05.1	Initial volume $=0.500\,\mathrm{dm}^3$ Initial dissolved mass $=40.0\times0.500=20.0\,\mathrm{g}$ Mass still dissolved $=20.0-5.00=15.0\,\mathrm{g}$ Final solution volume $=0.300\,\mathrm{dm}^3$ Final concentration $=15.0/0.300=50.0\,\mathrm{g\,dm}^{-3}$	Convert the starting volume to $0.500\,\mathrm{dm}^3$ , so the initial solute mass is $40.0\times0.500=20.0\,\mathrm{g}$ . Crystallisation removes $5.00\,\mathrm{g}$ from the dissolved solute, leaving $15.0\,\mathrm{g}$ . The final volume is $0.300\,\mathrm{dm}^3$ , giving concentration $15.0/0.300=50.0\,\mathrm{g\,dm}^{-3}$ .	5
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4.3.3.1 · Percentage yield (chemistry only)

Tier 1 · Easy

Mark scheme for 4.3.3.1 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	Percentage yield $=80\%$	Percentage yield $=(7.2/9.0)\times100=80\%$ .	2
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02.1	Actual product mass $=9.00\,\mathrm{g}$	Actual mass $=(75.0/100)\times12.0=9.00\,\mathrm{g}$ .	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.3.3.1 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	Percentage yield $=70\%$ Possible reasons include: the reaction is reversible and does not go to completion; product is lost during separation; side reactions form other products	Percentage yield $=(8.4/12.0)\times100=70\%$ . A yield below $100\%$ can result if a reversible reaction does not go to completion, if some product is lost while it is separated from the mixture, or if reactants form other products in side reactions.	4
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02.1	Method A yield $=75.0\%$ Method B yield $=80.0\%$ Method B has the higher percentage yield	For A, percentage yield $=(15.0/20.0)\times100=75.0\%$ . For B, it is $(9.60/12.0)\times100=80.0\%$ . Method B has the greater percentage yield.	3
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03.1	Total actual product mass $=7.60\,\mathrm{g}$ Overall percentage yield $=76.0\%$	The total actual yield is $6.20+1.40=7.60\,\mathrm{g}$ . Percentage yield $=(7.60/10.0)\times100=76.0\%$ .	3
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Tier 3 · Hard

Mark scheme for 4.3.3.1 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	Combined percentage yield $=74.0\%$	Add masses before finding the overall percentage. Total actual mass is $36.0+19.5=55.5\,\mathrm{g}$ and total theoretical mass is $45.0+30.0=75.0\,\mathrm{g}$ . The combined yield is $(55.5/75.0)\times100=74.0\%$ .	4
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02.1	Theoretical magnesium oxide mass $=10.0\,\mathrm{g}$ Percentage yield $=83.0\%$	Magnesium amount $=6.00/24=0.250\,\mathrm{mol}$ . The $2:2$ ratio gives $0.250\,\mathrm{mol}$ of magnesium oxide. Since $M_r(\mathrm{MgO})=40$ , the theoretical mass is $10.0\,\mathrm{g}$ . Percentage yield $=(8.30/10.0)\times100=83.0\%$ .	5
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03.1	Method P feed mass $=50.0\,\mathrm{kg}$ Method Q feed mass $=60.0\,\mathrm{kg}$ Method P needs less feed	Required theoretical product equals actual product divided by the fractional yield. For P, $42.0/0.840=50.0\,\mathrm{kg}$ . For Q, $42.0/0.700=60.0\,\mathrm{kg}$ . The stated one-to-one theoretical relationship makes these the feed masses, so P needs $10.0\,\mathrm{kg}$ less.	4
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04.1	Mass recovered after stage 1 $=50.0\times0.800=40.0\,\mathrm{g}$ Final mass $=40.0\times0.750=30.0\,\mathrm{g}$ The fractional overall yield is $0.800\times0.750=0.600$ Overall percentage yield $=60.0\%$ The same result follows from $(30.0/50.0)\times100$	Stage 1 recovers $50.0\times0.800=40.0\,\mathrm{g}$ . Stage 2 then produces $40.0\times0.750=30.0\,\mathrm{g}$ of final product. Relative to the original $50.0\,\mathrm{g}$ theoretical maximum, the overall yield is $(30.0/50.0)\times100=60.0\%$ . Equivalently, multiplying the two fractional yields gives $0.800\times0.750=0.600$ .	5
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05.1	Trapped solvent mass $=20.0\times0.145=2.90\,\mathrm{g}$ Pure product mass $=20.0-2.90=17.1\,\mathrm{g}$ Percentage yield $=(17.1/18.0)\times100=95.0\%$ The yield is below $100\%$ The student's claim treats solvent contamination as product mass and is therefore invalid	The solvent contributes $20.0\times0.145=2.90\,\mathrm{g}$ , so the actual pure-product mass is $17.1\,\mathrm{g}$ . Percentage yield must use this pure mass: $(17.1/18.0)\times100=95.0\%$ . The apparent value above $100\%$ arose because the wet solid included trapped solvent, not because more product formed than the theoretical maximum.	5
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4.3.3.2 · Atom economy (chemistry only)

Tier 1 · Easy

Mark scheme for 4.3.3.2 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	Atom economy $=56\%$	There is one mole of desired calcium oxide from one mole of reactant. Atom economy $=(56/100)\times100=56\%$ .	2
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02.1	Reaction 1 All reactant atoms become the desired product, so none form an unwanted product	Atom economy is $100\%$ only when every reactant atom is incorporated into the desired product. Reaction 2 assigns some atoms to its unwanted product.	2
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Tier 2 · Standard

Mark scheme for 4.3.3.2 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	$82\%$ of the starting materials, by mass, becomes the desired product A high atom economy produces less waste and uses raw materials more efficiently	An atom economy of $82\%$ means that $82\%$ of the starting materials, by mass in the balanced equation, forms the desired product. A high atom economy conserves raw materials, forms less unwanted product and can reduce separation or disposal costs.	3
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02.1	Atom economy $=64.2\%$	The reactant total is $100+2(36.5)=173$ . The desired calcium chloride contributes $111$ , so atom economy $=(111/173)\times100=64.161\ldots\%=64.2\%$ to three significant figures.	3
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03.1	Sum of reactant relative formula masses $=128$	Use $62.5=(80/\text{reactant total})\times100$ . Rearranging gives reactant total $=80\times100/62.5=128$ .	3
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Tier 3 · Hard

Mark scheme for 4.3.3.2 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	Atom economy $=17.0\%$ Titanium mass $=42.6\,\mathrm{kg}$	$M_r(\mathrm{TiCl}_4)=48+4(35.5)=190$ . The reactant total is $190+4(23)=282$ , while desired titanium contributes $48$ . Atom economy $=(48/282)\times100=17.0\%$ . The corresponding titanium mass is $250\times(48/282)=42.6\,\mathrm{kg}$ .	5
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02.1	Route A actual desired-product mass $=70.2\,\mathrm{kg}$ Route B actual desired-product mass $=66.2\,\mathrm{kg}$ Choose Route A to maximise product mass Route B has the higher atom economy, so a smaller proportion of reactant atoms forms unwanted products	For A, the theoretical desired product is $100\times0.780=78.0\,\mathrm{kg}$ and the actual mass is $78.0\times0.900=70.2\,\mathrm{kg}$ . For B, use unrounded values: $100\times0.920\times0.720=66.24\,\mathrm{kg}$ , or $66.2\,\mathrm{kg}$ to three significant figures. A gives more desired product, while B's higher atom economy means less material is assigned to by-products by the equation.	5
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03.1	Route 1 atom economy $=77.9\%$ Route 2 atom economy $=74.3\%$ Route 1 forms the smaller proportion of unwanted products	For Route 1, the desired copper mass is $63.5$ and the reactant total is $79.5+2=81.5$ , so atom economy $=(63.5/81.5)\times100=77.9\%$ . For Route 2, desired copper contributes $2(63.5)=127$ and the reactants total $2(79.5)+12=171$ , so atom economy $=(127/171)\times100=74.3\%$ . Route 1 has the higher atom economy and therefore the smaller unwanted-product proportion.	5
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04.1	Route A desired-product mass $=60.0\times0.800=48.0\,\mathrm{kg}$ Route B desired-product mass $=40.0\times0.650=26.0\,\mathrm{kg}$ Total desired-product mass $=74.0\,\mathrm{kg}$ Total reactant mass $=100.0\,\mathrm{kg}$ Total unwanted-product mass $=100.0-74.0=26.0\,\mathrm{kg}$ Combined atom economy $=(74.0/100.0)\times100=74.0\%$	Route A assigns $60.0\times0.800=48.0\,\mathrm{kg}$ to desired product, while Route B assigns $40.0\times0.650=26.0\,\mathrm{kg}$ . The combined desired mass is therefore $74.0\,\mathrm{kg}$ from $100.0\,\mathrm{kg}$ of reactants. The remaining $26.0\,\mathrm{kg}$ is unwanted product, and the combined atom economy is $(74.0/100.0)\times100=74.0\%$ .	6
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05.1	Student's value $=(63.5/96.5)\times100=65.8\%$ Correct reactant equation mass $=2(17.0)+3(79.5)=272.5$ Correct desired copper mass $=3(63.5)=190.5$ Correct atom economy $=(190.5/272.5)\times100=69.9\%$ The student's value is too low Coefficients multiply the quantities represented in the balanced equation and must be included for both reactants and desired product	The student's calculation gives $(63.5/96.5)\times100=65.803\ldots\%$ , or $65.8\%$ . The balanced equation contains two ammonia units and three copper(II) oxide units, so the reactant total is $2(17.0)+3(79.5)=272.5$ . It also forms three copper atoms, contributing $3(63.5)=190.5$ . The correct atom economy is $(190.5/272.5)\times100=69.908\ldots\%$ , or $69.9\%$ . Omitting unequal coefficients understates the proportion assigned to copper in this equation.	6
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4.3.4 · Using concentrations of solutions in mol/dm3 (chemistry only) (HT only)

Tier 1 · Easy

Mark scheme for 4.3.4 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	Concentration $=0.300\,\mathrm{mol\,dm}^{-3}$	Convert $250\,\mathrm{cm}^3$ to $0.250\,\mathrm{dm}^3$ . Then $c=n/V=0.0750/0.250=0.300\,\mathrm{mol\,dm}^{-3}$ .	2
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02.1	Volume $=200\,\mathrm{cm}^3$	Rearrange $c=n/V$ to $V=n/c$ . The volume is $0.0600/0.300=0.200\,\mathrm{dm}^3=200\,\mathrm{cm}^3$ .	2
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Tier 2 · Standard

Mark scheme for 4.3.4 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	Volume $=500\,\mathrm{cm}^3$	Moles of potassium hydroxide $=m/M_r=5.60/56.0=0.100\,\mathrm{mol}$ . Rearrange $c=n/V$ to $V=n/c$ . The volume is $0.100/0.200=0.500\,\mathrm{dm}^3=500\,\mathrm{cm}^3$ .	4
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02.1	Sodium chloride concentration $=0.200\,\mathrm{mol\,dm}^{-3}$	Sodium chloride amount $=5.85/58.5=0.100\,\mathrm{mol}$ . The solution volume is $0.500\,\mathrm{dm}^3$ , so $c=n/V=0.100/0.500=0.200\,\mathrm{mol\,dm}^{-3}$ .	4
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03.1	Final concentration $=0.320\,\mathrm{mol\,dm}^{-3}$	The original solute amount is $n=cV=0.800\times0.100=0.0800\,\mathrm{mol}$ . Dilution does not change this amount. The final volume is $0.250\,\mathrm{dm}^3$ , so $c=n/V=0.0800/0.250=0.320\,\mathrm{mol\,dm}^{-3}$ .	3
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Tier 3 · Hard

Mark scheme for 4.3.4 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	Sulfuric acid concentration $=0.0960\,\mathrm{mol\,dm}^{-3}$	Sodium hydroxide moles $=0.150\times0.0320=0.00480\,\mathrm{mol}$ . The equation ratio $\mathrm{H}_2\mathrm{SO}_4:\mathrm{NaOH}$ is $1:2$ , so acid moles $=0.00480/2=0.00240\,\mathrm{mol}$ . The acid volume is $0.0250\,\mathrm{dm}^3$ , hence $c=0.00240/0.0250=0.0960\,\mathrm{mol\,dm}^{-3}$ .	5
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02.1	Volume of potassium hydroxide solution $=30.0\,\mathrm{cm}^3$	Nitric acid amount $=0.175\times0.0240=0.00420\,\mathrm{mol}$ . The equation has a $1:1$ ratio, so $0.00420\,\mathrm{mol}$ of potassium hydroxide reacts. Its volume is $V=n/c=0.00420/0.140=0.0300\,\mathrm{dm}^3=30.0\,\mathrm{cm}^3$ .	5
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03.1	Anomalous titre $=25.10\,\mathrm{cm}^3$ Mean consistent titre $=23.57\,\mathrm{cm}^3$ Sodium hydroxide concentration $=0.0943\,\mathrm{mol\,dm}^{-3}$	$25.10\,\mathrm{cm}^3$ is far from the other two readings. Their mean is $(23.60+23.54)/2=23.57\,\mathrm{cm}^3$ . Hydrochloric acid amount $=0.100\times0.02357=0.002357\,\mathrm{mol}$ . The equation ratio is $1:1$ , so the sodium hydroxide portion contains the same amount. Its concentration is $0.002357/0.0250=0.09428\,\mathrm{mol\,dm}^{-3}$ , or $0.0943\,\mathrm{mol\,dm}^{-3}$ to three significant figures.	5
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04.1	Silver nitrate amount $=0.200\times0.100=0.0200\,\mathrm{mol}$ Sodium chloride amount $=0.100\times0.150=0.0150\,\mathrm{mol}$ Sodium chloride is limiting Silver nitrate left $=0.0200-0.0150=0.00500\,\mathrm{mol}$ Final volume $=0.100+0.150=0.250\,\mathrm{dm}^3$ Excess silver nitrate concentration $=0.00500/0.250=0.0200\,\mathrm{mol\,dm}^{-3}$	The solutions contain $0.200\times0.100=0.0200\,\mathrm{mol}$ of silver nitrate and $0.100\times0.150=0.0150\,\mathrm{mol}$ of sodium chloride. The ratio is $1:1$ , so sodium chloride is limiting and leaves $0.00500\,\mathrm{mol}$ of silver nitrate. The combined volume is $0.250\,\mathrm{dm}^3$ , giving an excess-reactant concentration of $0.00500/0.250=0.0200\,\mathrm{mol\,dm}^{-3}$ .	6
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05.1	Sodium hydroxide amount $=0.100\times0.0144=0.00144\,\mathrm{mol}$ The $1:1$ ratio means the acid portion contains $0.00144\,\mathrm{mol}$ Diluted acid concentration $=0.00144/0.0200=0.0720\,\mathrm{mol\,dm}^{-3}$ The full $200\,\mathrm{cm}^3$ diluted solution contains $0.0720\times0.200=0.0144\,\mathrm{mol}$ Dilution does not change the acid amount, so the original $16.0\,\mathrm{cm}^3$ contains $0.0144\,\mathrm{mol}$ Original acid concentration $=0.0144/0.0160=0.900\,\mathrm{mol\,dm}^{-3}$	The sodium hydroxide amount is $0.100\times0.0144=0.00144\,\mathrm{mol}$ . The equation ratio is $1:1$ , so the $20.0\,\mathrm{cm}^3$ acid portion contains the same amount and has concentration $0.00144/0.0200=0.0720\,\mathrm{mol\,dm}^{-3}$ . The full diluted solution contains $0.0720\times0.200=0.0144\,\mathrm{mol}$ . Those moles came from $0.0160\,\mathrm{dm}^3$ of original acid, whose concentration was therefore $0.0144/0.0160=0.900\,\mathrm{mol\,dm}^{-3}$ .	6
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4.3.5 · Use of amount of substance in relation to volumes of gases (chemistry only) (HT only)

Tier 1 · Easy

Mark scheme for 4.3.5 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	Gas volume $=8.40\,\mathrm{dm}^3$	At room temperature and pressure, $V=24n=24\times0.350=8.40\,\mathrm{dm}^3$ .	2
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02.1	Amount of gas $=0.150\,\mathrm{mol}$	At room temperature and pressure, one mole occupies $24\,\mathrm{dm}^3$ . Therefore $n=3.60/24=0.150\,\mathrm{mol}$ .	2
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Tier 2 · Standard

Mark scheme for 4.3.5 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	Mass of methane $=0.800\,\mathrm{g}$	Moles of methane $=1.20/24=0.0500\,\mathrm{mol}$ . Its mass is $m=nM_r=0.0500\times16.0=0.800\,\mathrm{g}$ .	3
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02.1	Ammonia volume $=6.00\,\mathrm{dm}^3$	At the same conditions, gas volumes follow the coefficient ratio. The $3:2$ ratio from hydrogen to ammonia gives volume $=9.00\times(2/3)=6.00\,\mathrm{dm}^3$ .	3
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03.1	Relative formula mass of the gas $=48.0$	The amount of gas is $0.480/24=0.0200\,\mathrm{mol}$ . Since $M_r=m/n$ , the relative formula mass is $0.960/0.0200=48.0$ .	3
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Tier 3 · Hard

Mark scheme for 4.3.5 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	Carbon dioxide volume $=36.0\,\mathrm{dm}^3$ Oxygen left $=6.0\,\mathrm{dm}^3$	Gas volumes follow the $2:1:2$ coefficient ratio. Reacting $36.0\,\mathrm{dm}^3$ of carbon monoxide needs $36.0/2=18.0\,\mathrm{dm}^3$ of oxygen and produces $36.0\,\mathrm{dm}^3$ of carbon dioxide. Oxygen is in excess, with $24.0-18.0=6.0\,\mathrm{dm}^3$ remaining.	4
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02.1	Theoretical carbon dioxide volume $=2.40\,\mathrm{dm}^3$ Percentage yield $=85.0\%$	Calcium carbonate amount $=10.0/100=0.100\,\mathrm{mol}$ . The $1:1$ ratio gives $0.100\,\mathrm{mol}$ of carbon dioxide, with theoretical volume $0.100\times24=2.40\,\mathrm{dm}^3$ . Percentage yield $=(2.04/2.40)\times100=85.0\%$ .	5
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03.1	Volume of ammonia decomposed $=24.0\,\mathrm{dm}^3$ Mass of ammonia decomposed $=17.0\,\mathrm{g}$	The equation converts two gas-volume parts of ammonia into four gas-volume parts of products. Therefore the starting ammonia volume is $48.0\times(2/4)=24.0\,\mathrm{dm}^3$ . This is $24.0/24=1.00\,\mathrm{mol}$ , so its mass is $1.00\times17.0=17.0\,\mathrm{g}$ .	5
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04.1	The $1:1$ methane-to-carbon-dioxide ratio means methane volume $=48.0\,\mathrm{cm}^3$ Methane percentage $=(48.0/60.0)\times100=80.0\%$ Unreactive gas volume $=60.0-48.0=12.0\,\mathrm{cm}^3$ The methane-to-oxygen ratio is $1:2$ Oxygen volume used $=2\times48.0=96.0\,\mathrm{cm}^3$	At the same temperature and pressure, gas volumes follow the equation coefficients. Methane and carbon dioxide have a $1:1$ ratio, so the sample contained $48.0\,\mathrm{cm}^3$ of methane. This is $(48.0/60.0)\times100=80.0\%$ , leaving $12.0\,\mathrm{cm}^3$ of unreactive gas. The $1:2$ methane-to-oxygen ratio means $96.0\,\mathrm{cm}^3$ of oxygen was consumed.	5
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05.1	Let the starting nitrogen volume be $x$ , so hydrogen volume is $60.0-x$ Hydrogen is limiting, so ammonia volume is $(2/3)(60.0-x)$ Excess nitrogen volume is $x-(60.0-x)/3$ Solving $(2/3)(60.0-x)+x-(60.0-x)/3=32.0$ gives $x=18.0\,\mathrm{dm}^3$ Starting volumes are $18.0\,\mathrm{dm}^3$ nitrogen and $42.0\,\mathrm{dm}^3$ hydrogen Nitrogen is in excess; $14.0\,\mathrm{dm}^3$ reacts and $4.0\,\mathrm{dm}^3$ remains	Let nitrogen start at $x\,\mathrm{dm}^3$ , leaving $60.0-x$ for hydrogen. If hydrogen limits, it forms $(2/3)(60.0-x)$ of ammonia and uses $(60.0-x)/3$ of nitrogen. Setting ammonia plus nitrogen left equal to $32.0$ gives $(2/3)(60.0-x)+x-(60.0-x)/3=32.0$ , so $x=18.0$ . Hydrogen therefore starts at $42.0\,\mathrm{dm}^3$ . It uses $14.0\,\mathrm{dm}^3$ of nitrogen and forms $28.0\,\mathrm{dm}^3$ of ammonia, leaving $4.0\,\mathrm{dm}^3$ of nitrogen; the final total is $32.0\,\mathrm{dm}^3$ .	6
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