Balearic Islands PAU General Sciences - independent foundations: Matter, materials, reactions and chemical reasoning

Structure, bonding, materials and inorganic chemistry

Atomic Structure and the Periodic Table

This independent foundation chapter develops how the quantum-mechanical model of the atom explains the layout of the periodic table and the trends you are asked to predict: electron configuration, ionisation energy, atomic and ionic radius, electronegativity, and the evidence available from mass spectra and successive ionisation energies.

Subatomic structure and isotopes

Atomic number Z = protons; mass number A = protons + neutrons; isotopes differ only in neutrons, ions only in electrons.
Relative atomic mass is the abundance-weighted mean of the isotope masses, not the mass number of any single isotope.
A mass spectrum gives m/z peaks with relative abundances; an identified intact singly charged molecular ion can support Mr. In a suitable small organic molecule, the M+1/M intensity ratio has an approximate contribution of 1.1 percentage points per carbon from carbon-13, but other isotopes and fragmentation must be considered.

Electron configuration

Fill by Aufbau order, obey the Pauli exclusion principle and Hund's rule (maximise unpaired spins within a subshell).
4s fills before 3d, but when transition metals ionise the 4s electrons are removed first.
Cr [Ar]3d5 4s1 and Cu [Ar]3d10 4s1 are the standard exceptions to expect.
Blocks are named for the subshell receiving the last electron.

Periodic trends

Across a period, nuclear charge rises while shielding is roughly constant, so effective nuclear charge increases: radius falls, ionisation energy and electronegativity rise.
Down a group, extra shells increase shielding and distance: radius rises, ionisation energy falls.
Successive ionisation energies jump sharply at shell boundaries; a large jump after the second value means two outer electrons, so Group 2.

Worked example

Rubidium is 72.2% 85Rb and 27.8% 87Rb. Ar = (85 × 0.722) + (87 × 0.278) = 61.37 + 24.19 = 85.56. Now an element with first ionisation energies 738, 1451 and 7733 kJ/mol: the third value is about five times the second, so two electrons are removed easily and the third comes from an inner shell - a Group 2 element.

Traps that cost marks

Quoting a periodic-table atomic mass as the mass number of a particular isotope, or forgetting to weight by abundance.
Defining first ionisation energy without "gaseous atoms" and "one mole".
Assuming Al has a higher first ionisation energy than Mg simply because it lies farther across the period; the measured Al value is lower, consistent with removing its higher-energy 3p electron.
Writing an electron configuration for a transition-metal ion without emptying 4s first.
Calling Zn a typical transition metal; its common ion is d10, so it lacks the characteristic variable oxidation states and colour.

Chemical Bonding and Molecular Structure

This independent foundation chapter develops how electrons are shared or transferred, how the resulting Lewis structure is converted into a three-dimensional shape by VSEPR theory, and how bonding and shape together explain polarity and physical properties.

Bond types

Ionic bonding transfers electrons to form a lattice; melting points are high and conduction requires mobile charge carriers, commonly in a melt or in aqueous solution when the compound dissolves. Lattice strength rises with ionic charge and falls with ion radius.
Covalent bonding shares electron pairs. Increasing bond order shortens the bond and increases bond enthalpy, so C≡C is shorter and stronger than C=C.
Metallic bonding uses delocalised electrons: malleable, ductile, good conductor. A dative bond (as in NH4+) becomes an ordinary covalent bond once formed.
Resonance delocalises electrons over equivalent positions, giving bond orders intermediate between the drawn structures, as in CO3^2− and benzene.

VSEPR shapes

Count electron domains around the central atom; double and triple bonds count as one domain, lone pairs count as one.
Repulsion order is lone pair–lone pair > lone pair–bond pair > bond pair–bond pair, so lone pairs compress adjacent angles.
2 domains linear (180°), 3 trigonal planar (120°), 4 tetrahedral (109.5°), 5 trigonal bipyramidal, 6 octahedral. In a trigonal bipyramid, lone pairs occupy equatorial positions.

Polarity

In the localized bond-dipole model, determine molecular polarity from the vector sum and actual charge distribution, not from the mere presence of polar bonds. CO2 and CCl4 cancel; H2O, NH3 and SO2 do not.
Electronegativity difference indicates bond polarity; molecular shape decides whether the dipoles add.

Worked example

SF4: sulfur has six valence electrons, four are used in S–F bonds leaving one lone pair, so there are 5 domains - trigonal bipyramidal electron geometry. The lone pair takes an equatorial site, giving a seesaw shape with F–S–F angles distorted from ideal trigonal-bipyramidal values. VSEPR predicts the seesaw arrangement qualitatively; exact angles require experimental or calculated structural data.

Traps that cost marks

Forgetting the lone pairs on the central atom when counting domains.
Expanding the octet on a period 2 atom; only period 3 and below can do this.
Confusing electron-pair geometry with molecular shape when a lone pair is present.
Assuming a molecule with polar bonds is automatically polar.
Describing ionic compounds as discrete molecules rather than a lattice.

States of Matter and Intermolecular Forces

This independent foundation chapter develops the gas laws, the assumptions and failures of the ideal-gas model, phase behaviour, and the ranking of melting and boiling points from the strength of intermolecular forces.

Gas behaviour

Ideal gas equation PV = nRT with P in Pa, V in m3, T in kelvin and R = 8.314 J mol−1 K−1.
Dalton's law: total pressure is the sum of partial pressures; each partial pressure is mole fraction × total pressure.
Kinetic theory assumes negligible molecular volume, no intermolecular forces, and elastic collisions. Real gases deviate most at high pressure and low temperature, when molecules are close together.
van der Waals equation (P + an²/V²)(V − nb) = nRT corrects for attraction (a) and molecular volume (b).

Intermolecular forces

London dispersion forces act between all particles and strengthen with more electrons and greater surface contact; they are the only force in non-polar molecules.
Dipole–dipole forces act between permanent dipoles. Hydrogen bonding is the strong special case requiring H directly bonded to N, O or F.
Ion–dipole forces operate in solutions of ionic compounds.
For otherwise comparable substances, stronger attractions often raise boiling point and enthalpy of vaporisation and lower vapour pressure. Melting also depends strongly on crystal packing and symmetry, so it is not ranked by attraction strength alone.

Phase behaviour

Vapour pressure rises with temperature; a liquid boils when its vapour pressure equals the external pressure.
A phase diagram shows the triple point and critical point; water's solid–liquid line slopes backwards, so ice melts under pressure.

Worked example

0.575 g of a volatile liquid gives 200 cm3 of vapour at 100 °C and 101 kPa. n = PV/RT = (101000 × 2.00 × 10−4)/(8.314 × 373) = 20.2/3101 = 6.51 × 10−3 mol. M = 0.575/6.51 × 10−3 = 88.3 g mol−1, consistent with a small ester such as C4H8O2.

Traps that cost marks

Substituting degrees Celsius or cubic centimetres directly into PV = nRT.
Claiming hydrogen bonding whenever hydrogen is present; it needs an N–H, O–H or F–H bond.
Forgetting that HF boils above HCl despite being lighter, because of hydrogen bonding.
Assuming ice is denser than water; the open hydrogen-bonded lattice makes it less dense.
Ignoring that dispersion forces increase with chain length and branching reduces them.

Inorganic Chemistry and Periodicity

This independent foundation chapter develops the descriptive chemistry that distinguishes groups and periods: s-block trends, halogen behaviour, the characteristic properties of transition metals, and the observations and equations you are expected to recall rather than derive.

Periodic trends

Group 2 reactivity with water increases down the group as ionisation energy falls. Hydroxide solubility increases down the group while sulfate solubility decreases, which is why BaSO4 is used as a precipitate.
Thermal stability of the usual Group 2 carbonates and nitrates generally increases down the group because the larger cation distorts the anion less.
Halogen oxidising power decreases down the group; halide reducing power increases. Hydrogen halides become less thermally stable and their aqueous solutions more acidic down the group.
Disproportionation, for example Cl2 + 2NaOH → NaCl + NaOCl + H2O.

Transition metals

Under the IUPAC Gold Book definition, a transition element has an atom with an incomplete d subshell or can form cations with one. Sc has an atomic 3d electron and qualifies under that definition; zinc is conventionally excluded because its atom and common ion retain a filled d subshell.
Characteristic properties: variable oxidation states (Mn from +2 to +7), catalytic activity, coloured complex ions and paramagnetism.
Ligands donate a lone pair to form coordinate bonds. Coordination numbers are commonly 6 (octahedral) and 4 (tetrahedral or square planar).
Ligand-field d–d transitions can produce colour where suitable d states are available. Simple aqueous Sc3+ and Zn2+ lack such transitions, but colour is not determined by d count alone: charge-transfer transitions can make other d0 or d10 complexes coloured.

Worked example

In [Fe(CN)6]3−, six cyanide ligands at −1 give −6; the overall charge is −3, so iron is +3. Fe3+ is d5. Cyanide is a strong-field ligand, so the configuration is low-spin t2g5 with one unpaired electron - the ion is paramagnetic and coloured.

Traps that cost marks

Reversing the hydroxide and sulfate solubility trends down Group 2.
Calling all d-block elements transition metals, including Zn without checking the atom-or-cation definition. Sc itself has an incomplete atomic d subshell.
Assuming every transition-metal ion is coloured.
Quoting a qualitative observation without a balanced equation.
Forgetting that halide reducing power increases, not decreases, down the group.

Physical chemistry, amounts, reactions and equilibria

Chemical Reactions and Stoichiometry

This independent foundation chapter develops the mole as the currency of chemistry: balancing equations, identifying the limiting reagent, calculating yields and atom economy, and handling concentrations, dilutions and gas volumes confidently.

Core relationships

n = m/M; n = cV with V in dm3; n = V/24.0 for a gas at room temperature and pressure, using the value given in the data provided.
One mole contains approximately 6.02 × 10^23 specified entities.
The balanced equation supplies the molar ratio; never use the ratio from an unbalanced equation.
For solutions, c in mol dm−3 = mol/dm3; convert cm3 to dm3 by dividing by 1000.

Limiting reagents

Convert every reactant to moles, divide each by its coefficient, and the smallest quotient is limiting.
Product amount is fixed by the limiting reagent; the excess reagent remains unreacted and can be calculated by subtraction.

Yield and atom economy

Percentage yield = actual mass or moles ÷ theoretical × 100; it measures how much of the expected product was obtained.
Atom economy = stoichiometrically weighted formula mass of the desired product ÷ total stoichiometrically weighted formula masses of the reactants × 100; it measures how much reactant mass ends up in the desired product, independent of how well the reaction runs.

Dilution and titration

Dilution: c1V1 = c2V2 because moles of solute are unchanged.
Titration: moles from cV, then use the equation ratio to reach the unknown.

Worked example

4.60 g of sodium (0.200 mol) reacts with 8.00 g of chlorine (0.113 mol) in 2Na + Cl2 → 2NaCl. The sodium needs 0.100 mol of chlorine, which is available, so sodium is limiting. n(NaCl) = 0.200 mol, so m = 0.200 × 58.5 = 11.7 g. The chlorine is in excess by 0.013 mol.

Traps that cost marks

Forgetting to balance before applying a mole ratio.
Quoting product mass from the excess reagent, which always overestimates the yield.
Adding masses or volumes of solutions instead of moles when deciding which is limiting.
Omitting water of crystallisation from a hydrated molar mass.
Treating atom economy and percentage yield as the same quantity.

Thermochemistry and Energetics

This independent foundation chapter develops enthalpy change: sign conventions, calorimetry calculations, Hess's law cycles, standard enthalpy of formation and combustion data, mean bond enthalpies, and the entropy and Gibbs free energy link to whether a reaction is feasible.

Enthalpy basics

ΔH is negative for exothermic reactions and positive for endothermic ones; standard pressure is 100 kPa; a temperature must be specified, commonly 298.15 K for tabulated thermochemical data. Standard state does not itself fix one universal temperature.
ΔHf° for an element in its standard state is zero by definition.
ΔH° = ΣΔHf°(products) − ΣΔHf°(reactants). For combustion data the sign convention reverses: ΔH = ΣΔHc(reactants) − ΣΔHc(products).

Calorimetry

q = mcΔT; for dilute aqueous solutions assume density 1.00 g cm−3 and c = 4.18 J g−1 K−1, so the mass is the total mass of solution.
ΔH per mole = −q/n, converted to kJ mol−1.

Hess's law and bond enthalpies

Any route between the same initial and final states gives the same ΔH; construct cycles, including Born–Haber cycles for ionic compounds.
ΔH = Σ bonds broken − Σ bonds formed. Mean bond enthalpies are averages over many compounds, so answers differ slightly from formation-data values.

Feasibility

ΔG = ΔH − TΔS; at specified constant temperature and pressure, negative actual reaction ΔG indicates a thermodynamically favorable forward change. It does not guarantee a measurable rate; ΔG° and actual ΔG are not interchangeable away from standard-state conditions. Entropy rises with more moles of gas, higher temperature and greater disorder; a reaction with positive ΔH can become feasible at high T if ΔS is positive.

Worked example

50.0 cm3 of 1.00 M HCl is mixed with 50.0 cm3 of 1.00 M NaOH and the temperature rises by 6.8 K. q = 100 g × 4.18 × 6.8 = 2842 J. n = 0.0500 mol of water formed. ΔH = −2.842/0.0500 = −56.8 kJ mol−1; the negative sign follows from the temperature rise.

Traps that cost marks

Omitting the minus sign, or using only one solution's mass in a neutralisation calculation.
Expecting the textbook value exactly; heat losses make the measured value less exothermic.
Getting the bond enthalpy sign the wrong way round.
Assigning a non-zero ΔHf° to an element.
Forgetting that ΔG, not ΔH alone, decides feasibility.

Chemical Kinetics and Equilibrium

This independent foundation chapter develops how reaction rates are measured and explained, how rate laws are deduced from experiment, and how equilibrium constants and Le Chatelier's principle predict the response of a system to changing conditions.

Rates and orders

Rate is the change in concentration per unit time, with units mol dm−3 s−1; for a reactant it is quoted as a positive value from the decrease.
Order is experimental and is not taken from the stoichiometric coefficients. Order with respect to a reactant is found by varying one concentration at a time.
Zero order: rate constant, so a concentration–time plot is linear. First order: constant half-life. For the simple integrated rate law −d[A]/dt = k[A]², 1/[A] against time is linear; an arbitrary overall second-order multireactant law needs its own analysis.

Collision theory and Arrhenius

A reaction occurs when collisions have energy at least equal to Ea and correct orientation. Increasing temperature often raises the fraction of collisions able to cross an activation barrier. Concentration and exposed surface area can affect encounter rates where the mechanism permits; their effects are not universally positive or explained by a higher energetic fraction.
k = Ae−Ea/RT; a plot of ln k against 1/T has gradient −Ea/R. Catalysts lower Ea.

Equilibrium

Equilibrium is dynamic: forward and reverse rates are equal and concentrations remain constant, but they are not equal to each other.
Kc = [products]^coefficients/[reactants]^coefficients, omitting pure solids and liquids.
K changes only with temperature. A catalyst supplies an alternative kinetic pathway and speeds attainment of equilibrium without changing the thermodynamic equilibrium constant or equilibrium composition.
Le Chatelier: raising concentration of a reactant drives the forward reaction; increasing pressure favours the side with fewer moles of gas; increasing temperature favours the endothermic direction.

Worked example

Doubling [A] while [B] is fixed quadruples the rate, so the reaction is second order in A; doubling [B] doubles the rate, so it is first order in B. Rate = k[A]²[B]. When rate = 2.0 × 10−3 mol dm−3 s−1 at [A] = [B] = 0.10 mol dm−3, k = 2.0 × 10−3/(0.0100 × 0.10) = 2.0 dm6 mol−2 s−1.

Traps that cost marks

Deriving the rate law from the balanced equation rather than from data.
Giving k units that do not match the overall order.
Claiming a catalyst increases yield or shifts equilibrium.
Adding an inert gas at constant volume and expecting a pressure shift; partial pressures of the reactants are unchanged.
Forgetting that K is unaffected by pressure and concentration changes.

Acids, Bases, and Salt Solutions

This independent foundation chapter develops Brønsted–Lowry theory, the distinction between strong and weak acids, pH and pKa calculations, buffer behaviour, titration curves and the hydrolysis that decides whether a salt solution is acidic, neutral or alkaline.

Theory

An acid donates a proton, a base accepts one; a conjugate pair differs by a single H+.
The first dissociation of H2SO4 and the usual aqueous dissociation of acids such as HCl, HNO3 and HClO4 are treated as strong in the relevant dilute-solution model; the second dissociation of H2SO4 requires a separate equilibrium calculation when significant; weak acids such as ethanoic acid (Ka ≈ 1.8 × 10−5) dissociate only partially.
Kw = [H+][OH−] = 1.0 × 10−14 at 298 K, so pH + pOH = 14 at that temperature.

Calculations

For a dilute fully dissociated monoprotic acid dominating over water autoionisation, [H+] ≈ c and pH ≈ −log c. Within that model, tenfold dilution raises pH by about one.
Weak acid: Ka = [H+]²/(c − [H+]), simplified to [H+]²/c when dissociation is under about 5%.
Buffer: pH = pKa + log([A−]/[HA]); at half neutralisation pH = pKa.
pKa = −log Ka; a smaller pKa means a stronger acid.

Salts and indicators

Strong acid + strong base gives a neutral salt; weak acid + strong base gives an alkaline solution because the anion hydrolyses; strong acid + weak base gives an acidic solution.
An indicator’s transition should lie within a sufficiently steep portion of the actual titration curve so its endpoint approximates equivalence to the required precision. Phenolphthalein and methyl orange are conventional examples, but suitability depends on the specific curve; merely naming acid/base classes is not enough.

Worked example

0.100 M ethanoic acid with Ka = 1.8 × 10−5: x²/0.100 = 1.8 × 10−5, so x = [H+] = 1.34 × 10−3 M and pH = −log(1.34 × 10−3) = 2.87. Only about 1.3% is dissociated, confirming weak behaviour and justifying the approximation.

Traps that cost marks

Using pH + pOH = 14 at temperatures other than 298 K.
Assuming the equivalence point of a weak acid–strong base titration is pH 7; it is above 7.
Inverting the buffer ratio, or forgetting that dilution hardly changes the pH of a buffer.
Ignoring the contribution of water autoionisation for very dilute or very weak acids.
Confusing strength (degree of dissociation) with concentration (amount per dm3).

Electrochemistry and Redox Reactions

This independent foundation chapter develops oxidation numbers, half-equation balancing in acidic and alkaline media, galvanic cell predictions from standard electrode potentials, and the quantitative link between charge, moles of electrons and mass deposited in electrolysis.

Oxidation numbers

A free element has oxidation number zero; a monatomic ion equals its charge; oxygen is usually −2 but −1 in peroxides; hydrogen is +1 except in hydrides; the sum equals the overall charge.
Oxidation is loss of electrons and an increase in oxidation number; reduction is gain of electrons and a decrease.
Disproportionation is simultaneous oxidation and reduction of the same element, as in Cl2 with cold dilute NaOH.

Balancing redox equations

Split into half-equations, balance atoms other than O and H, balance O with H2O, balance H with H+, and balance charge with electrons.
In alkaline solution, add OH− to both sides to neutralise H+, then cancel water.
Multiply each half-equation so the electron counts match before adding.

Cells and potentials

E°cell = E°cathode − E°anode; a positive standard cell potential corresponds to negative standard reaction Gibbs energy as written; actual composition and conditions can alter the actual cell potential.
E° is intensive: do not multiply it when you double a half-equation. ΔG° = −nFE° with F = 96 485 C mol−1.
Oxidation occurs at the anode. In a galvanic cell the anode is the negative electrode; in an electrolytic cell it is the positive one.

Worked example

Silver is deposited from AgNO3 solution using a current of 2.00 A for 30.0 min. Q = It = 2.00 × 1800 = 3600 C. n(e−) = 3600/96 485 = 0.0373 mol. Since Ag+ + e− → Ag, n(Ag) = 0.0373 mol and m = 0.0373 × 107.9 = 4.03 g.

Traps that cost marks

Reversing the sign conventions for galvanic and electrolytic cells.
Multiplying E° values when balancing a cell reaction.
Forgetting to balance charge with electrons before adding half-equations.
Assigning oxygen −2 in peroxides or hydrogen +1 in metal hydrides.
Mixing seconds and minutes in Q = It, or using the wrong number of electrons per ion.

Organic chemistry and macromolecules

Organic Chemistry and Functional Groups

This independent foundation chapter develops nomenclature, functional group recognition, structural, geometric and optical isomerism, reaction mechanisms and types, and the test-tube reactions that distinguish aldehydes, ketones, alkenes and alcohols.

Nomenclature and isomerism

Choose the longest chain containing the principal functional group, number to give the lowest locants, and apply the priority order: carboxylic acid > ester > amide > aldehyde > ketone > alcohol > amine > alkene > alkane.
Structural isomers differ in connectivity; positional isomers differ in where a group sits; for E/Z isomerism of an acyclic alkene, each alkene carbon needs two different substituents. A tetrahedral carbon with four different groups is a common source of chirality, but axial and other forms mean it is not universally required.

Reaction types

Alkanes undergo free-radical substitution with Cl2 or Br2 under UV light.
Alkenes undergo electrophilic addition; Markovnikov's rule places H on the carbon with more hydrogens; bromine water is decolourised.
Primary alcohols oxidise to aldehydes and then carboxylic acids with acidified dichromate; secondary alcohols give ketones; tertiary alcohols resist oxidation.
Haloalkanes undergo substitution (aqueous KOH) or elimination (hot ethanolic KOH).
Carbonyls: Appropriate aldehydes and ketones can form coloured 2,4-DNP derivatives under the specified test conditions. Tollens’ and Fehling’s are reducing tests rather than infallible labels: some other reducing species respond, and many aromatic aldehydes do not give the standard Fehling result. Interpret the complete supplied evidence and conditions.
Carboxylic acids with alcohols form esters under acid catalysis; esters hydrolyse back.

Worked example

A compound of formula C3H6O gives an orange precipitate with 2,4-DNP but no silver mirror with Tollens' reagent. Degree of unsaturation = (2×3 + 2 − 6)/2 = 1, consistent with one C=O. Positive 2,4-DNP confirms a carbonyl; the negative Tollens' test rules out an aldehyde, so the compound is the ketone propanone.

Traps that cost marks

Using Tollens' and Fehling's interchangeably, or expecting them to react with ketones.
Thinking tertiary alcohols oxidise to ketones.
Applying Markovnikov's rule backwards.
Calling any carbon with four bonds chiral; the four groups must differ.
Missing the C=C requirement that each carbon carries two different substituents for geometric isomerism.

Proteins, lipids and polymer structure

An organic functional group can appear in a small molecule or repeatedly along a macromolecule. Always identify the actual linkage and the model given. An amino acid can contain both amino and carboxyl functions. In the elementary peptide model, reaction of a carboxyl group with an amino group forms an amide linkage and eliminates water. Hydrolysis cleaves that linkage with water. A linear peptide containing n amino-acid residues has n minus one peptide links when its ends are not joined into a ring. This counting statement concerns the supplied linear structure, not every cyclic or branched biological assembly.

Protein primary structure describes amino-acid sequence. Secondary and tertiary organization depend on interactions and conformational constraints; denaturation can disturb higher-level structure without necessarily hydrolyzing the peptide backbone. Do not equate loss of enzyme activity with a change in DNA sequence or assume that all denaturation is reversible. If an exercise supplies an isoelectric point, a pH above that point commonly corresponds to a more negative average net charge, while a lower pH favors a more positive charge. Use the stated model rather than predicting a complete titration curve from one number.

A triacylglycerol contains glycerol joined to three fatty-acid residues through ester links. In an idealized saponification balance, base-promoted ester hydrolysis gives glycerol and carboxylate salts. It does not directly produce three neutral free acids while excess strong base remains. Unsaturation changes possible addition chemistry and chain packing. Compare otherwise similar structures: a cis double bond introduces a bend that can hinder close packing, but melting behavior also depends on chain length, composition and crystal structure.

For an alkene addition-polymerization model, the carbon-carbon double bond opens to join repeat units without eliminating water for every addition. Step-growth polymer formation may instead connect complementary functional groups; some such reactions eliminate a small molecule, often water or hydrogen chloride, while others do not. A diol and a dicarboxylic acid can form ester links in a simplified polyester model. Functionality, conversion, branching and crosslinking affect the resulting architecture, so step growth is not simply a synonym for one universal byproduct.

Thermoplastics can soften and flow under suitable heating because their chains are not permanently connected into one dense covalent network. A strongly crosslinked thermoset generally does not remelt into the original processing fluid; excessive heating can cause decomposition. Crosslinking restricts large-scale chain motion and may produce swelling rather than complete dissolution in an otherwise compatible solvent. These tendencies do not substitute for a particular material’s supplied data.

Distinguish mechanical size reduction from cleavage of polymer bonds. Fragmenting a plastic into smaller particles does not demonstrate chemical mineralization. Bio-based feedstock does not automatically imply biodegradability, and a compostability claim applies to specified conditions and tests. In a repeat-unit molar-mass calculation, state whether end groups, branching and the distribution of chain lengths are neglected before dividing a polymer molar mass by a repeat-unit mass.