4.3 Quantitative chemistry — notes on one page

4.3.1.1 · Conservation of mass and balanced chemical equations

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

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

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

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

4.3.1.2 · Relative formula mass

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

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

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

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

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.
• Exam calculations should show the relationship used, substitution with units and a suitably rounded result.

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

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.
• 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.

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

4.3.2.1 · Moles (HT only)

• 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.

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

4.3.2.2 · Amounts of substances in equations (HT only)

• 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.

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

4.3.2.3 · Using moles to balance equations (HT only)

• 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.

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

4.3.2.4 · Limiting reactants (HT only)

• 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.

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

4.3.2.5 · Concentration of solutions

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

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

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

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

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.
• Higher tier: AQA restricts "calculate the theoretical mass of a product from a given mass of reactant and the balanced equation for the reaction" to Higher tier; calculating percentage yield from an actual and a theoretical yield is examined on both tiers.
• Exam calculations should show the relationship used, substitution with units and a suitably rounded result.

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

4.3.3.2 · Atom economy (chemistry only)

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

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

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

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

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

• 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.

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

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

• 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.

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