4.3 Quantitative chemistry — printable coverage pack

4.3.1.1 · Conservation of mass and balanced chemical equations

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.

Tier 1 · Easy

1. Insert the smallest whole-number coefficients to balance $\mathrm{Al}+\mathrm{O}_2\rightarrow\mathrm{Al}_2\mathrm{O}_3$ .[1 mark]
Answer
- $4\mathrm{Al}+3\mathrm{O}_2\rightarrow2\mathrm{Al}_2\mathrm{O}_3$
Method: 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 $\mathrm{Al}$ .

Tier 2 · Standard

1. 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.[2 marks]
Answer
- Mass of water $=9.0\,\mathrm{g}$
Method: 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 $20.0-11.0=9.0\,\mathrm{g}$ .

Tier 3 · Hard

1. 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]
Answer
- Mass of oxygen used $=40.0\,\mathrm{g}$
- Reactants and products each total $51.0\,\mathrm{g}$
Method: 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.

4.3.1.2 · Relative formula mass

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.

Tier 1 · Easy

1. Calculate the relative formula mass of $\mathrm{Ca(OH)}_2$ . Use $A_r$ : $\mathrm{Ca}=40$ , $\mathrm{O}=16$ , $\mathrm{H}=1$ .[2 marks]
Answer
- $M_r(\mathrm{Ca(OH)}_2)=74$
Method: The bracket is multiplied by $2$ , so $M_r=40+2(16+1)=40+34=74$ .

Tier 2 · Standard

1. Ammonium nitrate is $\mathrm{NH}_4\mathrm{NO}_3$ . Calculate its percentage by mass of nitrogen. Use $A_r$ : $\mathrm{N}=14$ , $\mathrm{H}=1$ , $\mathrm{O}=16$ .[3 marks]
Answer
- Percentage nitrogen $=35.0\%$
Method: $M_r(\mathrm{NH}_4\mathrm{NO}_3)=2(14)+4(1)+3(16)=80$ . Nitrogen contributes $2(14)=28$ , so its percentage by mass is $(28/80)\times100=35.0\%$ .

Tier 3 · Hard

1. 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]
Answer
- Mass of oxygen $=9.60\,\mathrm{g}$
Method: $M_r=2(27)+3[32+4(16)]=342$ . The twelve oxygen atoms contribute $12(16)=192$ , so the oxygen fraction is $192/342$ . The oxygen mass is $17.1\times(192/342)=9.60\,\mathrm{g}$ .

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

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.

Tier 1 · Easy

1. 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]
Answer
- Oxygen from the air has combined with the magnesium
- Mass added $=4.0\,\mathrm{g}$
Method: 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.

Tier 2 · Standard

1. 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.[3 marks]
Answer
- The measured mass decreases by $5.5\,\mathrm{g}$
- Carbon dioxide particles escape from the open tube
Method: 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.

Tier 3 · Hard

1. 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]
Answer
- Crucible and contents gain $3.2\,\mathrm{g}$ because oxygen enters the oxide
- The complete sealed chamber has no mass change
Method: 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.

4.3.1.4 · Chemical measurements

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.

Tier 1 · Easy

1. 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]
Answer
- Mean $=12.5\,\mathrm{cm}^3$
- Uncertainty $=\pm0.1\,\mathrm{cm}^3$
Method: 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 $(12.5\pm0.1)\,\mathrm{cm}^3$ .

Tier 2 · Standard

1. 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.[4 marks]
Answer
- Mean $=8.23\,\mathrm{g}$
- Range $=0.06\,\mathrm{g}$
- Result $=(8.23\pm0.03)\,\mathrm{g}$
Method: 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 $(8.23\pm0.03)\,\mathrm{g}$ .

Tier 3 · Hard

1. 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]
Answer
- Anomalous reading $=36.9\,\mathrm{s}$
- Mean of consistent readings $=31.45\,\mathrm{s}$
- Reported result $=(31.45\pm0.15)\,\mathrm{s}$
Method: $36.9\,\mathrm{s}$ 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 $0.15\,\mathrm{s}$ .

4.3.2.1 · Moles (HT only)

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.

Tier 1 · Easy

1. Calculate the amount in moles in $9.0\,\mathrm{g}$ of water, $\mathrm{H}_2\mathrm{O}$ . Use $M_r=18$ .[2 marks]
Answer
- Amount of water $=0.50\,\mathrm{mol}$
Method: Use $n=m/M_r$ : $n=9.0/18=0.50\,\mathrm{mol}$ .

Tier 2 · Standard

1. Find the mass of $0.250\,\mathrm{mol}$ of sodium carbonate, $\mathrm{Na}_2\mathrm{CO}_3$ . Use $A_r$ : $\mathrm{Na}=23$ , $\mathrm{C}=12$ , $\mathrm{O}=16$ .[3 marks]
Answer
- Mass $=26.5\,\mathrm{g}$
Method: $M_r(\mathrm{Na}_2\mathrm{CO}_3)=2(23)+12+3(16)=106$ . Then $m=nM_r=0.250\times106=26.5\,\mathrm{g}$ .

Tier 3 · Hard

1. 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]
Answer
- $9.03\times10^{21}$ formula units
- $1.81\times10^{22}$ chloride ions
Method: 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.

4.3.2.2 · Amounts of substances in equations (HT only)

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.

Tier 1 · Easy

1. 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]
Answer
- Mass of magnesium oxide $=8.0\,\mathrm{g}$
Method: 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 $0.20\times40=8.0\,\mathrm{g}$ .

Tier 2 · Standard

1. 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 $M_r(\mathrm{CO}_2)=44$ .[3 marks]
Answer
- Mass of carbon dioxide $=11.0\,\mathrm{g}$
Method: 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 $0.250\times44=11.0\,\mathrm{g}$ .

Tier 3 · Hard

1. 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]
Answer
- Mass of water $=16.2\,\mathrm{g}$
Method: 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 $0.900\times18=16.2\,\mathrm{g}$ .

4.3.2.3 · Using moles to balance equations (HT only)

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.

Tier 1 · Easy

1. 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]
Answer
- $\mathrm{N}_2+3\mathrm{H}_2\rightarrow2\mathrm{NH}_3$
Method: 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.

Tier 2 · Standard

1. 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 $M_r$ : $\mathrm{Fe}=56$ , $\mathrm{O}_2=32$ , $\mathrm{Fe}_2\mathrm{O}_3=160$ .[4 marks]
Answer
- $4\mathrm{Fe}+3\mathrm{O}_2\rightarrow2\mathrm{Fe}_2\mathrm{O}_3$
Method: 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 $4:3:2$ .

Tier 3 · Hard

1. 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]
Answer
- $2\mathrm{C}_2\mathrm{H}_6+7\mathrm{O}_2\rightarrow4\mathrm{CO}_2+6\mathrm{H}_2\mathrm{O}$
Method: 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 $2:7:4:6$ .

4.3.2.4 · Limiting reactants (HT only)

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.

Tier 1 · Easy

1. 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]
Answer
- Chlorine is limiting
- $4.0\,\mathrm{mol}$ of hydrogen chloride forms
Method: 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.

Tier 2 · Standard

1. 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 $A_r$ : $\mathrm{Mg}=24$ , $\mathrm{O}=16$ .[4 marks]
Answer
- Magnesium is limiting
- Mass of magnesium oxide $=12.0\,\mathrm{g}$
Method: 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 $0.300\times40=12.0\,\mathrm{g}$ .

Tier 3 · Hard

1. 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]
Answer
- Hydrogen is limiting
- Ammonia mass $=13.6\,\mathrm{g}$
- Nitrogen left $=2.8\,\mathrm{g}$
Method: 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 $2.8\,\mathrm{g}$ .

4.3.2.5 · Concentration of solutions

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.

Tier 1 · Easy

1. 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]
Answer
- Concentration $=30.0\,\mathrm{g\,dm}^{-3}$
Method: Use $c=m/V$ : $c=12.0/0.400=30.0\,\mathrm{g\,dm}^{-3}$ .

Tier 2 · Standard

1. A fertiliser solution has concentration $18.0\,\mathrm{g\,dm}^{-3}$ . Calculate the solute mass in $250\,\mathrm{cm}^3$ of solution.[3 marks]
Answer
- Mass of solute $=4.50\,\mathrm{g}$
Method: Convert the volume: $250\,\mathrm{cm}^3=0.250\,\mathrm{dm}^3$ . Rearrange to $m=cV$ , then $m=18.0\times0.250=4.50\,\mathrm{g}$ .

Tier 3 · Hard

1. 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]
Answer
- Volume of water added $=150\,\mathrm{cm}^3$
Method: 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 $300-150=150\,\mathrm{cm}^3$ .

4.3.3.1 · Percentage yield (chemistry only)

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.

Tier 1 · Easy

1. 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]
Answer
- Percentage yield $=80\%$
Method: Percentage yield $=(7.2/9.0)\times100=80\%$ .

Tier 2 · Standard

1. A process makes $20.4\,\mathrm{kg}$ of product at a percentage yield of $68.0\%$ . Calculate the theoretical product mass.[3 marks]
Answer
- Theoretical mass $=30.0\,\mathrm{kg}$
Method: Write $68.0=(20.4/\text{theoretical mass})\times100$ . Rearranging gives theoretical mass $=20.4\times100/68.0=30.0\,\mathrm{kg}$ .

Tier 3 · Hard

1. 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]
Answer
- Combined percentage yield $=74.0\%$
Method: 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 $(55.5/75.0)\times100=74.0\%$ .

4.3.3.2 · Atom economy (chemistry only)

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.

Tier 1 · Easy

1. 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]
Answer
- Atom economy $=56\%$
Method: There is one mole of desired calcium oxide from one mole of reactant. Atom economy $=(56/100)\times100=56\%$ .

Tier 2 · Standard

1. 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 $M_r$ : $\mathrm{NaCl}=58.5$ , $\mathrm{H}_2\mathrm{O}=18.0$ , $\mathrm{Cl}_2=71.0$ .[3 marks]
Answer
- Atom economy $=46.4\%$
Method: The reactant total from the equation is $2(58.5)+2(18.0)=153.0$ . The desired chlorine contributes $71.0$ . Atom economy $=(71.0/153.0)\times100=46.4\%$ .

Tier 3 · Hard

1. 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]
Answer
- Atom economy $=17.0\%$
- Titanium mass $=42.6\,\mathrm{kg}$
Method: $M_r(\mathrm{TiCl}_4)=48+4(35.5)=190$ . 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 $250\times(48/282)=42.6\,\mathrm{kg}$ .

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

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.

Tier 1 · Easy

1. A solution contains $0.0750\,\mathrm{mol}$ of solute in $250\,\mathrm{cm}^3$ . Calculate its concentration in $\mathrm{mol\,dm}^{-3}$ .[2 marks]
Answer
- Concentration $=0.300\,\mathrm{mol\,dm}^{-3}$
Method: Convert $250\,\mathrm{cm}^3$ to $0.250\,\mathrm{dm}^3$ . Then $c=n/V=0.0750/0.250=0.300\,\mathrm{mol\,dm}^{-3}$ .

Tier 2 · Standard

1. Calculate the sodium hydroxide mass in $150\,\mathrm{cm}^3$ of a $0.400\,\mathrm{mol\,dm}^{-3}$ solution. Use $M_r(\mathrm{NaOH})=40.0$ .[4 marks]
Answer
- Mass of sodium hydroxide $=2.40\,\mathrm{g}$
Method: The volume is $0.150\,\mathrm{dm}^3$ . Moles of sodium hydroxide $=cV=0.400\times0.150=0.0600\,\mathrm{mol}$ . Its mass is $nM_r=0.0600\times40.0=2.40\,\mathrm{g}$ .

Tier 3 · Hard

1. $25.0\,\mathrm{cm}^3$ 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]
Answer
- Sulfuric acid concentration $=0.0960\,\mathrm{mol\,dm}^{-3}$
Method: 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 $c=0.00240/0.0250=0.0960\,\mathrm{mol\,dm}^{-3}$ .

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

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.

Tier 1 · Easy

1. Calculate the volume occupied by $0.350\,\mathrm{mol}$ of a gas at room temperature and pressure.[2 marks]
Answer
- Gas volume $=8.40\,\mathrm{dm}^3$
Method: At room temperature and pressure, $V=24n=24\times0.350=8.40\,\mathrm{dm}^3$ .

Tier 2 · Standard

1. Calculate the volume of carbon dioxide at room temperature and pressure produced by $8.80\,\mathrm{g}$ of the gas. Use $M_r(\mathrm{CO}_2)=44.0$ .[3 marks]
Answer
- Carbon dioxide volume $=4.80\,\mathrm{dm}^3$
Method: Moles of carbon dioxide $=8.80/44.0=0.200\,\mathrm{mol}$ . Its volume is $0.200\times24=4.80\,\mathrm{dm}^3$ .

Tier 3 · Hard

1. 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]
Answer
- Carbon dioxide volume $=36.0\,\mathrm{dm}^3$
- Oxygen left $=6.0\,\mathrm{dm}^3$
Method: 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.