1 Principles of chemistry

A solid keeps its shape because its particles cannot move from place to place. A liquid flows because close particles can move past one another. A gas fills its container because widely separated particles move freely in all directions.

State symbols record the before-and-after state: evaporation is (l) → (g), melting is (s) → (l), condensation is (g) → (l), and sublimation of iodine is (s) → (g).

Gas particles move quickly but not straight from one end of a tube to the other. Their random directions and collisions with air particles and the tube walls make the visible result take time.

In salt water, salt is the solute, water is the solvent and salt water is the solution. If added salt remains undissolved after thorough stirring at a fixed temperature, the solution above it is saturated.

Worked example: 17.6 g of dry salt was dissolved using 50.0 g of water. Solubility = (17.6 ÷ 50.0) × 100 = 35.2 g per 100 g of water. The numerical scaling factor is 2 because 100 g is twice 50.0 g.

If solubility falls from 90 to 30 g per 100 g water on cooling, 60 g crystallises from 100 g water. From 50 g water, the mass is (60 ÷ 100) × 50 = 30 g.

Control the temperature throughout saturation, measure the correct water volume, add enough solid, stir thoroughly, filter before sampling and evaporate gently. Reheat, cool and reweigh until the mass is constant if checking that all water has been removed.

In a particle diagram, identical particles made from one atom type represent an element. Identical particles containing different atom types bonded together represent a compound. More than one kind of particle in the same box represents a mixture.

Measure the melting point of a solid or boiling point of a liquid and compare both the value and whether the transition is sharp. Adding an impurity commonly lowers and broadens a solid's melting point, so the range is as important as a single recorded temperature.

For sea water, simple distillation collects pure water while salt remains. Filtration cannot remove dissolved salt. To obtain dry salt instead, crystallise it; heating every solution to dryness can decompose the solid or produce poor crystals.

A spot that travels farther is usually more soluble in the mobile solvent, relative to its attraction to the stationary paper. The dye travelling nearest the solvent front therefore has the greatest movement under those conditions.

Worked example: the spot moves 9.7 cm from the start line and the solvent front moves 12.0 cm. $R_f = 9.7/12.0 = 0.808\ldots = 0.81$ to two significant figures. $R_f$ has no unit because it is a ratio of two distances in the same unit.

Compare an unknown's $R_f$ with reference values obtained under the same conditions. A matching value supports an identification; different solvents, papers or temperatures can change the value.

For a fair comparison, use the same solvent, type of paper and start-line arrangement. If a sample does not move, repeat with a different suitable solvent because the component may be insoluble in the first solvent.

A molecule is a discrete group of two or more atoms held together by covalent bonds. Its formula shows both the elements present and the number of atoms, so one molecule of methanol, $\ce{CH3OH}$, contains six atoms in total.

A neutral atom has equal numbers of protons and electrons, so their positive and negative charges cancel. Neutrons add mass but no charge; electrons contribute very little to the total mass.

For $\ce{^{205}_{81}Tl}$: protons = 81, neutrons = $205-81=124$, and a neutral atom has 81 electrons. Another thallium isotope must still have 81 protons but can have a different number of neutrons.

For isotopes 24, 25 and 26 with abundances 79.2%, 10.0% and 10.8%: $A_r=[(24\times79.2)+(25\times10.0)+(26\times10.8)]/100=24.316$, which is 24.3 to one decimal place.

The answer must lie between the lightest and heaviest isotope masses and should be closest to the mass of the most abundant isotope. Keep the unrounded value until the final requested precision.

An element is located by the intersection of its group and period. Magnesium is in Group 2, Period 3; boron is in Group 3, Period 2. Use the element's atomic number to identify it before reading its coordinates.

Start with the atomic number, fill each inner shell before the next, and confirm that the shell totals add back to the atomic number. The number of occupied shells should match the period.

Copper conducts electricity and copper oxide is basic, supporting classification as a metal. Sulfur and chlorine form acidic oxides, supporting classification as non-metals. Silicon dioxide is acidic because it reacts with basic calcium oxide.

Locate the element first, then compare its position with the stepped divide. Period 3 begins with metallic sodium, magnesium and aluminium and continues into non-metals such as sulfur and chlorine.

The relationship works in reverse for the first 20 elements. Group 5, Period 3 means three occupied shells and five outer electrons, so the configuration is 2,8,5.

Chemical reactions involve outer-shell electrons. Atoms in one group therefore tend to lose, gain or share the same number of electrons and form similar types of ions or bonds. Group 1 atoms each have one outer electron and tend to lose that one electron in reactions.

A full outer shell is a stable arrangement, so a noble-gas atom has little tendency to gain, lose or share electrons. Helium has a full first shell of 2; neon, argon and the later noble gases have a full outer shell of 8.

Write correct formulae first; count each element; change the coefficient before a whole formula; recount; then add state symbols from the information given: (s), (l), (g) or (aq). For an unfamiliar reaction, use the supplied names, formulae and conditions rather than inventing products.

For $\ce{Mg(NO3)2.6H2O}$, count brackets and water separately: $24 + 2(14) + 6(16) + 6[2(1)+16] = 256$. A coefficient before a formula is not part of one formula unit and must not be included in its $M_r$.

The mole lets chemical amounts be compared using the coefficients in a balanced equation. For $\ce{N2 + 3H2 -> 2NH3}$, the amount ratio is 1 mol : 3 mol : 2 mol.

For $2.00\,\mathrm{g}$ of $\ce{CuSO4}$ with $M_r=159.5$, $n=2.00/159.5=0.0125\,\mathrm{mol}$. Check that division makes sense: the mass is much less than one molar mass, so the amount must be less than 1 mol.

For $\ce{CH4 + 2O2 -> CO2 + 2H2O}$, $32\,\mathrm{g}$ of methane is $32/16=2$ mol. The 1:2 ratio requires 4 mol of oxygen, so its mass is $4\times32=128\,\mathrm{g}$.

Make sure actual and theoretical yields refer to the same product and use the same unit. If the theoretical yield is $29.9\,\mathrm{g}$ and the actual yield is $23.92\,\mathrm{g}$, the yield is $23.92/29.9\times100=80.0\%$.

If $12.5\,\mathrm{g}$ of $\ce{CuSO4.xH2O}$ leaves $8.0\,\mathrm{g}$ of $\ce{CuSO4}$, water lost is $4.5\,\mathrm{g}=0.25$ mol and salt is $8.0/159.5\approx0.050$ mol. The ratio 1:5 gives $\ce{CuSO4.5H2O}$.

The molecular formula is a whole-number multiple of the empirical formula. Therefore both formulae have the same elemental ratio, but only the molecular formula states the actual atom count in a molecule.

For $25.0\,\mathrm{cm^3}$ of $0.0500\,\mathrm{mol\,dm^{-3}}$ solution, $V=25.0/1000=0.0250\,\mathrm{dm^3}$, so $n=0.0500\times0.0250=0.00125\,\mathrm{mol}$.

For reacting gases, first convert the known quantity to moles, use the balanced-equation coefficient ratio, then multiply the required gas moles by the molar gas volume. For $60\,\mathrm{cm^3}$ of gas at rtp, $n=60/24000=0.0025\,\mathrm{mol}$.

Subtract container masses to obtain sample masses. In combustion, oxygen mass is oxide minus metal; in reduction, oxygen mass is oxide minus metal remaining. Divide metal mass by the metal's $A_r$ and oxygen mass by 16, then simplify the mole ratio.

In calcium chloride, one calcium atom loses two outer electrons and two chlorine atoms each gain one. This forms $\ce{Ca^{2+}}$ and two $\ce{Cl^-}$ ions, each with a full outer shell.

Read a Roman numeral as the positive charge on a variable-charge metal ion: iron(II) is $\ce{Fe^{2+}}$ and iron(III) is $\ce{Fe^{3+}}$. Keep a polyatomic ion together as one charged unit.

Write both ion charges, find the lowest common total charge, and convert that into ion counts. Use brackets when more than one polyatomic ion is required; omit a subscript 1.

For $\ce{Na2O}$, draw two $[\ce{Na}]^+$ ions and one $[\ce{O}]^{2-}$ ion. The oxide outer shell has eight electrons: six originally from oxygen and one transferred from each sodium atom.

Electron transfer forms the positive and negative ions; the attraction between their opposite charges is the bond. In an ionic solid, each ion is attracted to oppositely charged neighbours throughout the lattice.

Strong electrostatic attractions act between oppositely charged ions throughout the lattice. A large amount of thermal energy is needed to overcome enough of these attractions for the ions to move apart, so melting and boiling points are high.

In molten or aqueous ionic compounds, positive and negative ions move in opposite directions and carry charge through the liquid. Melting or dissolving changes mobility; it does not create electrons that conduct.

Each atom contributes to or uses the shared pair, so the electrons count in the outer shell of both bonded atoms. One shared pair is a single covalent bond; two or three shared pairs form double or triple bonds.

The shared electrons lie between the nuclei and are attracted to both. This two-nucleus attraction holds the atoms together; saying only that electrons are shared describes the arrangement but not the electrostatic mechanism.

Use the same accounting for the required range: $\ce{H2}$, $\ce{O2}$, $\ce{N2}$, halogens and hydrogen halides; $\ce{H2O}$, $\ce{NH3}$ and $\ce{CO2}$; and up-to-two-carbon molecules such as methane, ethane, ethene and halogen derivatives. For $\ce{CO2}$, carbon shares two pairs with each oxygen; each oxygen also has two lone pairs.

Melting or boiling separates molecules by overcoming intermolecular forces. Because these forces are weak, little thermal energy is needed, so simple molecular substances are often gases, liquids or low-melting solids.

Larger, heavier molecules usually have more electrons and stronger intermolecular attractions. More thermal energy is therefore needed to separate the molecules, giving a higher melting or boiling point.

To melt or boil the substance, many strong covalent bonds throughout the network must be broken. This requires a large amount of thermal energy, so giant covalent substances are solids with high melting and boiling points.

Diamond and graphite have high melting points because many strong covalent bonds in their giant structures must be broken. $\ce{C60}$ has a much lower melting point because melting overcomes weak forces between its molecules, not the covalent bonds within each cage.

Their electrons are held in covalent bonds or localized around atoms, and their molecules are neutral. Giant covalent compounds such as silicon dioxide also lack mobile ions or delocalised electrons, so charge cannot flow through the structure.

Repeat the pattern beyond one pair of ions so the diagram clearly shows a lattice rather than a molecule. Place the electron symbols in the spaces throughout the array, not attached to one particular ion.

Outer electrons are no longer associated with one metal atom; they move throughout the structure. Their negative charge attracts the positive ions in every direction and holds the giant metallic lattice together.

The charge carriers are electrons, not moving metal ions. When a force shifts one layer, the electron attraction is not tied to fixed pairs of atoms, so bonding can continue across the rearranged layers.

Their molecules are neutral, and their electrons are held in covalent bonds or localized around atoms. Applying a potential difference therefore provides no mobile ions or delocalised electrons to carry charge between electrodes.

Electrolysis therefore requires an ionic substance to be molten or dissolved. Cations move toward the negative electrode and anions toward the positive electrode, carrying charge through the electrolyte.

Opposite charges attract: cations move toward the negative cathode and anions move toward the positive anode. The electrode name stays the same throughout this electrolytic-cell context.

For a molten compound, only its ions compete: the cation forms the element at the cathode and the anion forms the element at the anode. In an aqueous solution, water also supplies hydrogen and hydroxide ions; sodium is not deposited because hydrogen is reduced more readily, while copper is deposited from $\ce{Cu^{2+}}$.

At the cathode, put electrons on the left because cations gain them. At the anode, put electrons on the right because anions or hydroxide ions lose them. Recount atoms, then confirm that total charge is equal on both sides.

When investigating current against gas volume, keep electrolyte, concentration, electrode area, separation and time constant. Plot volume against current, identify anomalies from the pattern and use a best-fit line; at fixed time, greater current produces more gas.