Salam Academy
WCH11 · Module

Formulae, equations and amounts of substance

3 subtopics≈ 135 min in total
  1. Ideal gases and lattices

    An ideal gas has particles of negligible volume, no intermolecular forces and perfectly elastic collisions. It obeys pV=nRTpV=nRT with pp in Pa, VV in m³, TT in K and R=8.31 J K−1mol−1R=8.31\ \text{J K}^{-1}\text{mol}^{-1}. Real gases deviate at high pressure (particle volume matters) and low temperature (intermolecular forces matter); they behave most ideally at high temperature and low pressure. Solids can be giant ionic, giant molecular (covalent), simple molecular or metallic lattices.

    ≈ 40 min
  2. The mole and equations

    Relative masses are measured against one-twelfth of a carbon-12 atom. One mole contains 6.02×10236.02\times10^{23} particles (the Avogadro constant, LL), and has a mass in grams equal to MrM_r: n=mMn=\frac{m}{M}. Empirical formulae come from composition by mass (or combustion data); the molecular formula is a whole-number multiple matching MrM_r. Equations balance atoms and charge; ionic equations omit spectator ions. Water of crystallisation is found from mass lost on heating. Atom economy =Mr of desired product∑Mr of all products×100=\frac{M_r\text{ of desired product}}{\sum M_r\text{ of all products}}\times100.

    ≈ 45 min
  3. Reacting quantities and titrations

    Every calculation runs through moles: mass (n=mMn=\frac{m}{M}), solution (n=cVn=cV, VV in dm³), gas (n=V24.0n=\frac{V}{24.0} dm³ at r.t.p.). Use the balanced equation's mole ratio to link substances. The limiting reagent is used up first and sets the yield. Titrations find unknown concentrations. In a back titration, an excess of reagent is added and the leftover is titrated; subtracting gives the amount that reacted. Percentage yield =actualtheoretical×100=\frac{\text{actual}}{\text{theoretical}}\times100.

    ≈ 50 min