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Stoichiometric relationships

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Chapter 1: Quantitative chemistry

1.1 The mole concept and Avogadro’s constant

• The amount of substance (n) is measured in moles (mol).

• The mole concept applies to all species: atoms, molecules, ions, electrons, formula units.

• 1 mol contains the same number of chemical species as there are atoms in exactly 12 g of the isotope carbon-12 .

• 1 mol of any substance contains 6.02  1023 species.

• 6.02  1023 mol–1 is called Avogadro’s constant (L). It has units as it is the number of particles per mole.

• The relative atomic mass (Ar) of an element is the average mass of an atom according to relative abundances of its isotopes, on a scale where the mass of one atom of is 12 exactly. It has no units.

• The relative molecular mass (Mr) is the sum of the relative atomic masses of the atoms in the molecular formula.

• The relative formula mass of an ionic compound is the sum of the relative atomic masses of the ions in the formula.

• The molar mass (M) is the relative mass expressed in g and has units of g mol–1. • Number of mol = mass / molar mass

• Number of particles = number of mol  Avogadro’s constant

• n = m / M

• N = nL

1.2 Formulas

• The molecular formula shows the number of atoms of each element present in a molecule.

• The empirical formula gives the ratio of the atoms of different elements in a compound. It is the molecular formula expressed as its simplest ratio. The molecular formula is a whole-number multiple of the empirical formula.

• The empirical formula of a compound containing the elements X, Y and Z can be determined by completing the following table:   Mass/ g or % of X Mass/ g or % of Y Mass/ g or % of Z Mass/ g mX mY mZ n/mol = mX/MX = mX/MX = mX/MX Simples ratio (divide by smallest amount in previous row) 1.3 Chemical equations

• The coefficients in a chemical equation describe the relative amounts of reactants and products. For a reaction pX + qY rZ where p, q and r are coefficients: nX/nY = p/q or nZ/nZ = p/r etc

• State symbols indicate the state of a substance: (s) solid, (l) liquid, (g) gas and (aq) aqueous solution or dissolved in water. 1.4 Mass and gaseous volume relationships in chemical reactions

• The limiting reactant determines the theoretical yield of product. The other reagents are in excess.

• The theoretical yield is the mass or amount of product produced according to the chemical equation assuming 100% reaction of the limiting reagent.

• Percentage yield = (experimental yield/ theoretical yield)  100%

• The kelvin is the SI unit of temperature: T (K) = T (°C) + 273

• Units of volume: 1 dm3 = 1 litre = 1  10–3 m3 = 1  103 cm3 = 1000 ml   For a fixed mass of Ideal Gas at constant T: P = k1/V (k1 constant) For a fixed mass of Ideal Gas at constant P: P = k2T The combined gas law For a fixed mass: P1V1/T1 = P2V2/T2 The ideal gas equation PV = nRT When SI units are used R has the value 8.31 J K–1 mol–1. T must be in K.

• Temperature (in K) is a measure of the average kinetic energy of the particles. Particles have minimum kinetic energy at absolute zero (0 K).

• As kinetic energy = ½mv2 and all gases have the same kinetic energy at the same temperature, particles with smaller mass move faster.

• Avogadro’s hypothesis states that equal volumes of different gases contain equal numbers of particles at the same temperature and pressure.

• Number of mol = volume/molar volume n = V/Vmol • The coefficients in a chemical equation describe the relative amounts of reactants and products. Gay–Lussac’s law: For a reaction pX (g) + qY (g)  rZ (g) where p, q and r are coefficients: volX / volY = p / q or volX / volZ = p / r etc.

• Molar volume, Vm, of any gas at STP = 22.4 dm3.

• STP for gases is standard temperature (0°C or 273 K) and pressure (1 atmosphere or 100 kPa). 1.5 Solutions

• Density = mass / volume

• ρ = m/v •

A solution is a homogeneous mixture of a liquid (the solvent) with another substance (the solute). The solute can be solid, liquid or gas but the solvent is generally a liquid.

• Concentration is the amount of solute in a known volume of solution. It can be expressed either in g dm–3 or mol dm–3. Concentration in mol dm−3 is often represented by square brackets around the substance: [solute] (mol dm−3) = nsolute (mol) / Vsolution (dm3) nsolute = [solute]  Vsolution (dm3) nsolute = [solute]  Vsolution (cm3) / 1000

• Titration is a chemical technique in which one solution is used to analyse another solution to find its concentration or amount.