Learning Objectives
7 objectivesBy the end of this note, you should be able to:
- Calculate empirical and molecular formulae from experimental composition data.
- Use balanced equations to convert between masses of reactants and products.
- Use molar volume and the ideal gas equation to calculate gas volumes.
- Calculate percentage yield and percentage atom economy in chemical processes.
- Determine a formula or confirm an equation using experimental data.
- Carry out Core Practical 1 to measure the molar volume of a gas.
- Link ionic and full equations to observations in displacement, acid and precipitation reactions.
Empirical and Molecular Formulae
The empirical formula is the simplest whole-number ratio of atoms of each element in a compound, found by converting masses or percentages into moles.
The molecular formula gives the actual number of atoms of each element in one molecule, and is always a whole-number multiple of the empirical formula.
The procedure for empirical formulae has four steps:
- Divide each element’s mass (or percentage) by its relative atomic mass to obtain moles.
- Divide every mole value by the smallest mole value.
- Round to the nearest whole-number ratio.
- Write the formula using these whole numbers as subscripts.
To convert from empirical to molecular formula, divide the relative molecular mass of the compound by the empirical formula mass to obtain a multiplier. Multiply every subscript in the empirical formula by this multiplier to get the molecular formula.
Key Equations
Empirical-to-molecular multiplier:
$$n=\frac{{M}_{r}(\text{molecular})}{{M}_{r}(\text{empirical})}$$
Variables: $n$ is the whole-number multiplier; $M_r$ values are dimensionless. SI unit: dimensionless.
Worked Example: Finding the Molecular Formula of a Hydrocarbon
A hydrocarbon contains 85.7% carbon and 14.3% hydrogen by mass. Its relative molecular mass is 56.0. Determine its molecular formula.
Step 1 — Convert percentages to moles.
For carbon:
$${n}_{C}=\frac{85.7}{12.0}$$
$${n}_{C}=7.142\text{ mol}$$
For hydrogen:
$${n}_{H}=\frac{14.3}{1.0}$$
$${n}_{H}=14.30\text{ mol}$$
Step 2 — Divide by the smallest value.
$$\frac{7.142}{7.142}=1.000$$
$$\frac{14.30}{7.142}=2.002$$
The empirical formula is CH₂.
Step 3 — Find the multiplier.
$${M}_{r}(\text{empirical})=12.0+2(1.0)=14.0$$
$$n=\frac{56.0}{14.0}$$
$$n=4$$
Step 4 — Multiply.
The molecular formula is C₄H₈.
The molecular formula contains four times the atoms of the empirical formula, consistent with butene (or another C₄H₈ alkene isomer).
MisconceptionStudents often stop at the empirical formula when the question asks for the molecular formula. Always check whether $M_r$ is provided, because that signals a molecular formula is required.
Exam TipRead the question wording carefully and confirm which formula is asked for.
Reacting Masses from Equations
Balanced chemical equations allow a known mass of one reactant or product to be converted into the mass of any other species in the equation, using the mole ratio from the coefficients.
The procedure has three steps:
- Calculate moles of the known substance using $n = m/M_r$.
- Use the stoichiometric ratio in the balanced equation to find moles of the required substance.
- Convert moles back to mass using $m = n M_r$.
Key Equations
Moles from mass:
$$n=\frac{m}{{M}_{r}}$$
Variables: $n$ = amount of substance in mol; $m$ = mass in g; $M_r$ = relative molecular mass (dimensionless). SI unit of $n$: mol.
Worked Example: Mass of Calcium Oxide from Calcium Carbonate
Calcium carbonate decomposes on heating: CaCO₃(s) → CaO(s) + CO₂(g). Calculate the mass of calcium oxide produced from 25.0 g of calcium carbonate.
Step 1 — Calculate moles of CaCO₃.
Equation used
$$n=\frac{m}{{M}_{r}}$$
Given
$$m=25.0\text{ g}$$
$${M}_{r}({\text{CaCO}}_{3})=40.1+12.0+3(16.0)=100.1$$
Substitution:
$$n=\frac{25.0}{100.1}$$
$$n=0.2498\text{ mol}$$
Step 2 — Apply the 1:1 mole ratio.
$$n(\text{CaO})=0.2498\text{ mol}$$
Step 3 — Convert moles to mass.
Equation used
$$m=n\times {M}_{r}$$
$${M}_{r}(\text{CaO})=40.1+16.0=56.1$$
$$m=0.2498\times 56.1$$
$$m=14.0\text{ g}$$
Approximately 14.0 g of calcium oxide forms, illustrating that the mass decreases because gaseous CO₂ is released into the atmosphere.
Gas Volumes Using Molar Volume
At room temperature and pressure, one mole of any gas occupies 24.0 dm³ (24,000 cm³), allowing direct conversion between moles and gas volumes for any species in an equation.
This relationship comes from Avogadro’s law: equal volumes of any gases under the same conditions contain equal numbers of molecules. The identity of the gas does not affect its molar volume because the gas particles are far apart compared with their size.
Key Equations
Moles from gas volume at room temperature and pressure (RTP):
$$n=\frac{V}{24.0}$$
Variables: $n$ = moles in mol; $V$ = volume in dm³; 24.0 = molar volume at RTP in dm³ mol⁻¹. SI unit of $n$: mol.
If the volume is given in cm³, divide by 24,000 instead.
Worked Example: Volume of Hydrogen from a Reactive Metal
0.486 G of magnesium reacts completely with excess dilute hydrochloric acid at RTP: Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g). Calculate the volume of hydrogen produced.
Step 1 — Calculate moles of Mg.
Equation used
$$n=\frac{m}{{M}_{r}}$$
$$n=\frac{0.486}{24.3}$$
$$n=0.0200\text{ mol}$$
Step 2 — Apply the 1:1 mole ratio.
$$n({\text{H}}_{2})=0.0200\text{ mol}$$
Step 3 — Convert moles to volume at RTP.
Equation used
$$V=n\times 24.0$$
$$V=0.0200\times 24.0$$
$$V=0.480{\text{ dm}}^{3}$$
480 Cm³ of hydrogen is produced, a typical scale for a school gas-collection experiment over water.
Ideal Gas Equation pV = nRT
The ideal gas equation pV = nRT is used when conditions differ from RTP, or when a gas is produced from a volatile liquid, because volume depends strongly on both temperature and pressure.
Pressure must be in pascals (Pa), volume in cubic metres (m³) and temperature in kelvin (K). Convert °C to K by adding 273. Convert kPa to Pa by multiplying by 1000. Convert dm³ to m³ by dividing by 1000, and cm³ to m³ by dividing by 1,000,000.
Key Equations
Ideal gas equation:
$$pV=nRT$$
Variables: $p$ = pressure in Pa; $V$ = volume in m³; $n$ = moles in mol; $R = 8.31$ J K⁻¹ mol⁻¹ (gas constant); $T$ = temperature in K. SI unit of $n$: mol.
Worked Example: Mass of a Volatile Liquid from Vapour Volume
A volatile liquid produces 245 cm³ of vapour at 100 °C and 100 kPa. The mass of the liquid is 0.586 g. Calculate the relative molecular mass.
Step 1 — Convert all units to SI.
Pressure conversion:
$$p=100\times 1000$$
$$p=100,000\text{ Pa}$$
Volume conversion:
$$V=\frac{245}{1,000,000}$$
$$V=2.45\times {10}^{-4}{\text{ m}}^{3}$$
Temperature conversion:
$$T=100+273$$
$$T=373\text{ K}$$
Step 2 — Calculate moles using the ideal gas equation.
Equation used
$$n=\frac{pV}{RT}$$
$$n=\frac{100,000\times 2.45\times {10}^{-4}}{8.31\times 373}$$
$$n=7.90\times {10}^{-3}\text{ mol}$$
Step 3 — Calculate $M_r$.
Equation used
$${M}_{r}=\frac{m}{n}$$
$${M}_{r}=\frac{0.586}{7.90\times {10}^{-3}}$$
$${M}_{r}=74.2$$
An $M_r$ of 74.2 is consistent with butan-1-ol (C₄H₁₀O, $M_r$ = 74.0), a typical volatile liquid in this experiment.
Examiner InsightMarks are routinely lost when temperature is left in °C or pressure in kPa. Always rewrite all three values in SI units before substituting.
Exam TipShow the unit conversions on separate lines before the substitution into pV = nRT.
Percentage Yield and Atom Economy
Percentage yield measures how much of the theoretical product is actually obtained, while percentage atom economy measures how efficiently the atoms of reactants end up in the desired product.
Yield is typically less than 100% because of incomplete reactions, side reactions, losses during purification (filtration, transfer, recrystallisation) and reversible reactions that do not go to completion. Atom economy depends only on the balanced equation, so a reaction can have high yield but low atom economy if many atoms become unwanted by-products.
A high atom economy is essential for green chemistry, because it minimises waste and reduces raw material costs in industrial processes.
Key Equations
Percentage yield:
$$\%\text{ yield}=\frac{\text{actual mass of product}}{\text{theoretical mass of product}}\times 100$$
Variables: masses in g. SI unit: %.
Percentage atom economy:
$$\%\text{ atom economy}=\frac{{M}_{r}\text{ of desired product}}{\text{sum of }{M}_{r}\text{ of all products}}\times 100$$
Variables: $M_r$ values dimensionless, weighted by stoichiometric coefficients. SI unit: %.
Worked Example: Yield and Atom Economy in Ester Production
6.00 G of ethanol reacts with excess ethanoic acid to produce ethyl ethanoate and water: CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O. The actual mass of ester obtained is 8.80 g. Calculate the percentage yield and the atom economy.
Step 1 — Calculate moles of ethanol.
$$n=\frac{6.00}{46.0}$$
$$n=0.1304\text{ mol}$$
Step 2 — Theoretical moles and mass of ester.
The mole ratio is 1:1.
$$n(\text{ester})=0.1304\text{ mol}$$
$${M}_{r}({\text{CH}}_{3}{\text{COOC}}_{2}{\text{H}}_{5})=88.0$$
$$m(\text{theoretical})=0.1304\times 88.0$$
$$m(\text{theoretical})=11.48\text{ g}$$
Step 3 — Percentage yield.
$$\%\text{ yield}=\frac{8.80}{11.48}\times 100$$
$$\%\text{ yield}=76.7\%$$
Step 4 — Atom economy.
$$\%\text{ atom economy}=\frac{88.0}{88.0+18.0}\times 100$$
$$\%\text{ atom economy}=83.0\%$$
The yield is moderate because esterification is reversible and does not reach completion. The atom economy is relatively high because water is the only by-product.
MisconceptionAtom economy is calculated from the balanced equation, not from experimental masses. It does not depend on yield.
Exam TipUse $M_r$ values weighted by stoichiometric coefficients, not actual masses.
Determining Formulae by Experiment
A formula or equation can be confirmed experimentally by measuring the masses or volumes of substances reacting and using the resulting mole ratios to deduce the stoichiometry.
Common experimental designs include heating a metal in oxygen and recording mass gain to deduce the metal oxide formula, decomposing a hydrated salt to find the number of water molecules of crystallisation, and reacting a metal with acid while measuring hydrogen volume to confirm a balanced equation.
Evaluation of data is part of this skill. Identify systematic errors such as incomplete reaction, loss of solid as smoke, gas escaping before collection, or absorption of moisture from the air. Identify random errors in mass or volume readings. Compare the calculated mole ratio with whole-number values and judge whether deviations are within experimental uncertainty.
Examiner InsightWhen asked to evaluate data, name the specific error and state its direction of effect on the calculated formula. Generic answers about “human error” are not credited.
Exam TipLink each error to whether the calculated value would be too high or too low.
Worked Example: Water of Crystallisation in Hydrated Magnesium Sulfate
4.92 G of hydrated magnesium sulfate, MgSO₄·xH₂O, is heated until all water is driven off. The mass of anhydrous MgSO₄ remaining is 2.40 g. Determine the value of x.
Step 1 — Mass of water lost.
$$m({\text{H}}_{2}\text{O})=4.92-2.40$$
$$m({\text{H}}_{2}\text{O})=2.52\text{ g}$$
Step 2 — Moles of MgSO₄ and water.
$$n({\text{MgSO}}_{4})=\frac{2.40}{120.4}$$
$$n({\text{MgSO}}_{4})=0.01994\text{ mol}$$
$$n({\text{H}}_{2}\text{O})=\frac{2.52}{18.0}$$
$$n({\text{H}}_{2}\text{O})=0.1400\text{ mol}$$
Step 3 — Mole ratio.
$$\frac{0.1400}{0.01994}=7.02$$
Step 4 — Round to whole number.
$$x=7$$
The salt is MgSO₄·7H₂O (Epsom salt). The slight excess above 7.00 may indicate residual moisture in the crucible; heating to constant mass minimises this error.
Core Practical 1: Molar Volume of a Gas
The molar volume of a gas can be measured by reacting a known mass of a reactive solid with excess acid and collecting the gas evolved over water in an inverted measuring cylinder or gas syringe.
To determine the molar volume of hydrogen gas at room temperature and pressure by reacting a known mass of magnesium with excess dilute hydrochloric acid.
- Weigh accurately a small piece of magnesium ribbon (approximately 0.04 g) using a balance reading to two or three decimal places.
- Fill a measuring cylinder completely with water and invert it over a water trough, ensuring no air is trapped.
- Place excess dilute hydrochloric acid (1.0 mol dm⁻³) in the conical flask, enough to react fully with the magnesium.
- Drop the magnesium into the acid and immediately fit the bung with the delivery tube positioned under the inverted cylinder.
- Allow the reaction to complete and collect all the hydrogen produced.
- When effervescence stops, record the final volume of hydrogen at room temperature and pressure.
- Record the laboratory temperature and atmospheric pressure.
- Independent variable (IV): mass of magnesium used, measured to ±0.001 g.
- Dependent variable (DV): volume of hydrogen collected, measured in cm³ from the inverted cylinder.
- Control variables (CV): acid in excess (ensures all Mg reacts), same temperature and pressure (recorded), same purity of magnesium (clean ribbon, oxide layer removed by emery paper).
The mole ratio Mg:H₂ is 1:1, so moles of H₂ equal moles of Mg. Dividing the measured volume by moles of H₂ gives molar volume close to 24,000 cm³ mol⁻¹ at RTP, confirming Avogadro’s prediction.
Hydrogen partly dissolves in water and some gas may escape before the bung is fitted; this causes the measured volume to be too low and molar volume calculated to be lower than 24.0 dm³ mol⁻¹. Adding the magnesium quickly and using cold water minimise these losses.
SafetyHydrogen is highly flammable; keep all flames away. Dilute HCl is an irritant; wear eye protection and avoid skin contact.
Worked Example: Calculating Molar Volume from Practical Data
0.0486 G of magnesium fully reacts with excess dilute HCl. The volume of hydrogen collected at RTP is 48.5 cm³. Calculate the molar volume.
Step 1 — Moles of Mg.
$$n(\text{Mg})=\frac{0.0486}{24.3}$$
$$n(\text{Mg})=2.00\times {10}^{-3}\text{ mol}$$
Step 2 — Moles of H₂ (1:1 ratio).
$$n({\text{H}}_{2})=2.00\times {10}^{-3}\text{ mol}$$
Step 3 — Molar volume.
$${V}_{m}=\frac{V}{n}$$
$${V}_{m}=\frac{48.5}{2.00\times {10}^{-3}}$$
$${V}_{m}=24,250{\text{ cm}}^{3}{\text{ mol}}^{-1}$$
$${V}_{m}=24.3{\text{ dm}}^{3}{\text{ mol}}^{-1}$$
The measured molar volume is very close to the theoretical value of 24.0 dm³ mol⁻¹ at RTP, confirming the reliability of the experiment.
Ionic Equations and Test-Tube Reactions
An ionic equation shows only the species that change during a reaction, with state symbols indicating whether each substance is solid, liquid, gas or aqueous, and links directly to the visible observations in test-tube chemistry.
To convert a full equation into an ionic equation, write all aqueous strong electrolytes as separate ions. Cancel out spectator ions that appear unchanged on both sides. Keep solids, liquids, gases and weak electrolytes as full formulae. Check that both atoms and charges balance.
Three reaction types are examined in this section:
- Displacement reactions occur when a more reactive metal or halogen displaces a less reactive one from solution.
- Acid reactions include neutralisation by alkali, reaction with reactive metal, reaction with carbonate, and reaction with metal oxide.
- Precipitation reactions occur when two aqueous solutions mix to form an insoluble solid.
| Reaction | Full equation | Ionic equation | Observation |
|---|---|---|---|
| Displacement (metal) | Zn(s) + CuSO₄(aq) → ZnSO₄(aq) + Cu(s) | Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s) | blue solution fades; pink-brown solid coats zinc |
| Displacement (halogen) | Cl₂(aq) + 2KBr(aq) → 2KCl(aq) + Br₂(aq) | Cl₂(aq) + 2Br⁻(aq) → 2Cl⁻(aq) + Br₂(aq) | colourless to orange solution |
| Acid + alkali | HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l) | H⁺(aq) + OH⁻(aq) → H₂O(l) | temperature rises; no visible colour change |
| Acid + metal | 2HCl(aq) + Mg(s) → MgCl₂(aq) + H₂(g) | 2H⁺(aq) + Mg(s) → Mg²⁺(aq) + H₂(g) | effervescence; metal disappears |
| Acid + carbonate | 2HCl(aq) + CaCO₃(s) → CaCl₂(aq) + H₂O(l) + CO₂(g) | 2H⁺(aq) + CaCO₃(s) → Ca²⁺(aq) + H₂O(l) + CO₂(g) | effervescence; solid dissolves; gas turns limewater milky |
| Acid + metal oxide | 2HCl(aq) + CuO(s) → CuCl₂(aq) + H₂O(l) | 2H⁺(aq) + CuO(s) → Cu²⁺(aq) + H₂O(l) | black solid dissolves; blue solution forms |
| Precipitation | AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq) | Ag⁺(aq) + Cl⁻(aq) → AgCl(s) | white precipitate forms |
| Precipitation | Pb(NO₃)₂(aq) + 2KI(aq) → PbI₂(s) + 2KNO₃(aq) | Pb²⁺(aq) + 2I⁻(aq) → PbI₂(s) | bright yellow precipitate forms |
MisconceptionSpectator ions are present but not removed from existence; they remain dissolved in solution. They are simply omitted from the ionic equation because they are unchanged.
Exam TipAlways include state symbols in ionic equations, especially (s) for precipitates and (g) for gases.
QUICK RECAP
Key Points
- Empirical formula: simplest whole-number ratio of atoms from moles.
- Molecular formula = empirical formula × ($M_r$ / empirical mass).
- Moles from mass: $n = m / M_r$.
- Moles from gas at RTP: $n = V / 24.0$ (V in dm³).
- Ideal gas: $pV = nRT$; use Pa, m³, K.
- Convert °C to K by adding 273.
- 24.0 dm³ mol⁻¹ at RTP for any gas.
- % yield = (actual mass / theoretical mass) × 100.
- % atom economy = $M_r$ (desired) / $M_r$ (all products) × 100.
- Atom economy is calculated only from the balanced equation.
- Yield < 100% due to incomplete reaction, side reactions, transfer losses.
- Core Practical 1: Mg + excess HCl → measure H₂ volume to find $V_m$.
- Sources of error: H₂ dissolves in water, gas escapes before bung fitted.
- Ionic equations omit spectator ions; always include state symbols.
- Displacement: more reactive species replaces less reactive in solution.
- Precipitation: aqueous ions combine to form an insoluble solid.
- Acid + carbonate gives CO₂ gas turning limewater milky.
- Acid + metal gives H₂ effervescence.
- Acid + alkali gives water; H⁺(aq) + OH⁻(aq) → H₂O(l).
- Mole ratios from balanced equations are essential at every step.
CAN I…? PROGRESS CHECK
Self-Assessment
- Can I calculate empirical and molecular formulae from percentage composition data?
- Can I convert between mass, moles and gas volume using $n = m/M_r$ and $n = V/24$?
- Can I apply $pV = nRT$ with correct SI unit conversions?
- Can I calculate the percentage yield of a reaction from experimental and theoretical masses?
- Can I calculate the percentage atom economy of a reaction from a balanced equation?
- Can I distinguish between yield and atom economy and explain why both matter industrially?
- Can I describe Core Practical 1 including apparatus, variables, and sources of error?
- Can I calculate the molar volume of a gas from experimental data?
- Can I write balanced ionic equations with state symbols for displacement reactions?
- Can I write balanced ionic equations with state symbols for typical acid reactions?
- Can I write balanced ionic equations with state symbols for precipitation reactions?
- Can I match observations such as effervescence, colour changes and precipitates to ionic equations?