2.1 Amount of chemical change

A balanced chemical equation conserves every element and charge. Its coefficients show the mole ratio of reactants and products.

Write the products and state symbols, balance atoms with coefficients rather than changing formula subscripts, then check every element and the charge.

For Al + O₂ → Al₂O₃, preserve the formulas and choose coefficients 4Al + 3O₂ → 2Al₂O₃. Changing O₂ to O or Al₂O₃ to another subscript would change the substances rather than balance them. Add state symbols from chemical evidence or stated conditions, not from atom counting.

Representative question

[Maximum number: 2]

Write an equation for the reaction, including all state symbols.

$\mathrm{CaCO}_{3}(\mathrm{~s})+2 \mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{CaCl}_{2}(\mathrm{aq})+\mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})$

correct products AND state symbols

correct balancing

Marking guidance:

Do not accept $\mathrm{H}_{2} \mathrm{CO}_{3}(\mathrm{aq})$ for M1.

Accept ionic equation.

$known amount → moles → coefficient ratio → required amount$

Read the required-to-known ratio from the balanced equation. For gases at fixed temperature and pressure, the same coefficient ratio applies to volumes; always finish with the requested unit.

For N₂ + 3H₂ → 2NH₃, 0.50 mol N₂ corresponds to 1.00 mol NH₃ when H₂ is sufficient. Keep the conversion chain visible—given unit to moles, coefficient ratio, then requested unit—so molar mass, concentration or gas volume is used at the correct end.

Given or requested representation	Amount bridge	Condition/unit checkpoint
mass	n = m/M	use a consistent mass unit with molar mass
solution	n = cV	convert volume to dm³ when c is mol dm⁻³
gas at stated T/P	n = V/Vₘ, or use the given gas relation	do not mix molar volumes from different conditions
particles	n = N/Nₐ	specify atoms, molecules, ions or formula units

Convert the known representation to moles, apply the balanced-equation coefficient ratio, then convert once to the requested representation.

Representative question

[Maximum number: 3]

Calculate the volume of $2.00 \mathrm{~mol} \mathrm{dm}^{-3}$ sulfuric acid required to react completely with 10.0 g of thallium (I) hydroxide.

amount $\mathrm{TIOH}\left\langle\left\langle=\frac{10.0}{221.39}\right\rangle\right\rangle=0.0452$
amount $\mathrm{H}_{2} \mathrm{SO}_{4}=\ll 0.0452 \times 1 / 2>=0.0226$
volume of $\mathrm{H}_{2} \mathrm{SO}_{4} \ll=\frac{0.0226}{2.00} \gg$
$=0.0113 \mathrm{dm}^{3} / 11.3 \mathrm{~cm}^{3}$

Marking guidance:

Award [3] for correct final answer.

The limiting reactant is used up first and determines the maximum, or theoretical, yield. An excess reactant remains after the reaction is complete.

Convert each reactant to moles, divide by its balanced-equation coefficient, and identify the smallest normalized amount as limiting. Use that reactant's ratio to calculate product.

For 2H₂ + O₂ → 2H₂O with 3.0 mol H₂ and 2.0 mol O₂, compare n/coefficient: 1.5 for H₂ and 2.0 for O₂, so H₂ limits and forms 3.0 mol H₂O. A smaller starting mass is not necessarily limiting; the decision depends on moles relative to coefficients.

Representative question

[Maximum number: 2]

Deduce which reactant is limiting. Use sections 1, 4 and 7 of the data booklet.

$\begin{array}{ll} n\left(\mathrm{CaCO}_{3}\right) & <=\frac{3.162 \mathrm{~g}}{100.09 \mathrm{~g} \mathrm{~mol}^{-1}}>= \text { AND } & 0.0316<\mathrm{mol}> n(\mathrm{HCl})<=4.00 \mathrm{~mol} \mathrm{dm} \mathrm{~m}^{-3} \times 20.0 \times 10^{-3} \mathrm{dm}^{3}> =0.0800<\mathrm{mol}> \end{array}$

$\mathrm{CaCO}_{3}$ is limiting

Do not award M2 without working/answer seen for M1

$percentage yield = (experimental yield / theoretical yield) × 100\%$

Find theoretical yield from the limiting reactant and stoichiometric ratio before comparing it with the measured experimental yield. Do not divide the product mass by a starting mass directly.

If stoichiometry predicts 10.0 g but 8.20 g is isolated, percentage yield is 82.0%. A value above 100% signals wet or impure product, measurement error or an incorrect theoretical yield; it is not evidence that the reaction created extra conserved matter.

Representative question

[Maximum number: 2]

1.72 g of methyl methanoate is produced from 2.83 g of methanoic acid and excess of the other reagent. Determine the percentage yield.

ALTERNATIVE 1

expected yield $<=2.83 \times \frac{60.06}{46.03}>=3.69$ «g» $\) percentage yield « \(=100 \times \frac{1.72}{3.69} »=46.6<\% \gg$

ALTERNATIVE 2

«amount of methanoic acid used $=\frac{2.83}{46.03}=» 0.0615$ «mol»

«expected amount of methyl methanoate $=0.0615 \mathrm{~mol}$ »

«actual amount of methyl methanoate $=\frac{1.72}{60.06}=0.0286 \mathrm{~mol}$ »

percentage yield $\mu=\frac{0.0286}{0.0615} \times 100 »=46.5 \%$

Marking guidance:

Award [2] for correct final answer.

Award [0] for 60.8\% (simple ratio of starting and final masses).

$atom economy = (Mr of desired product / total Mr of reactants) × 100\%$

Atom economy measures how much of the reactant mass is represented by the desired product. Low atom economy means more reactant mass becomes by-products and waste.

Apply stoichiometric coefficients to every formula mass before forming the ratio. Atom economy is fixed by the chosen equation and desired product, whereas percentage yield measures experimental recovery; a reaction can have high atom economy but poor yield, or the reverse.

Representative question

[Maximum number: 2]

Calculate the atom economy for the synthesis of ethyl ethanoate by the reaction below. Use sections 1 and 7 of the data booklet.

2 \mathrm{C}_{2} \mathrm{H}_{6}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})+\frac{3}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}_{2}(\mathrm{l})+2 \mathrm{HCl}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})

$88.12 \mathrm{~g} \mathrm{~mol}^{-1}$

$\times 100$
$2 \times 30.08 \mathrm{gmol}^{-1}+70.90 \mathrm{gmol}^{-1}+3 / 2 \times 32.00 \mathrm{gmol}^{-1} /(88.12 / 179.06)$

49.21\%

Marking guidance:

Award [2] for correct final answer.

Retrieve the route: balance the equation, convert through the coefficient mole ratio, identify the limiting reactant, calculate theoretical and percentage yield, then assess atom economy.

Check formula subscripts, state symbols, ratio direction, units, the smallest normalized reactant amount, the theoretical-yield denominator, and the total reactant mass used for atom economy.

ConceptIB Chemistry HL

2.1 Amount of chemical change

Balanced Chemical Equations

Balancing with State Symbols

Assessment in practice

Mole Ratios in Equations

Reacting-Quantity Calculations

Assessment in practice

Limiting and Excess Reactants

Finding the Limiting Reactant

Assessment in practice

Theoretical and Percentage Yield

Calculating Percentage Yield

Assessment in practice

Atom Economy

Calculating Atom Economy

Assessment in practice

Amount of Chemical Change Summary

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