CHEM1011 · Fundamentals Of Chemistry 1a
Bonding, Equations and Stoichiometry
This chapter is the quantitative engine of CHEM1011: it links how atoms bond (ionic transfer vs covalent sharing, with polar covalent in between) to how you count what reacts. You move from classifying a bond by electronegativity difference, to writing formulae from valence and polyatomic-ion charges, to balanced equations that conserve atoms and charge, and finally to the mole-ratio / limiting-reagent / percent-yield calculations that dominate the exam. Almost every numerical question in the early exam reduces to the same n = m/M → ratio → m = n×M machine, so mastering it here pays off across the whole paper.
What this chapter covers
- 011. Ionic vs covalent bonding: electron transfer vs shared pair; bonding pair vs lone pair
- 022. Polar covalent character from electronegativity difference (ΔEN)
- 033. Valence as a formula-prediction shortcut (C=4, N=3, O=2, H/Cl/Br/I=1)
- 044. Polyatomic ions: formula + charge you must memorise (NH₄⁺, NO₃⁻, CO₃²⁻, SO₄²⁻, PO₄³⁻, OH⁻)
- 055. Molecular vs empirical formula; empirical formula from percent composition
- 066. Balancing equations: conserve atoms AND net charge (coefficients only)
- 077. Stoichiometric mole ratios and mass↔mole conversion (n = m/M, m = n×M)
- 088. Limiting reagent and percent yield
Empirical formula from percent composition
- +1Assume a 100 g sample, so each percentage becomes a mass in grams: 52.2 g C, 13.0 g H, 34.8 g O.
- +1Convert each mass to moles with n = m/M: C = 52.2/12.01 = 4.35 mol; H = 13.0/1.008 = 12.9 mol; O = 34.8/16.00 = 2.18 mol.
- +1Divide every value by the smallest (2.18 mol): C = 4.35/2.18 = 2.00; H = 12.9/2.18 = 5.92 ≈ 6; O = 2.18/2.18 = 1.00.
- +1Read off the whole-number ratio C : H : O = 2 : 6 : 1, giving the empirical formula C₂H₆O.
Key terms
- Ionic bond
- An electrostatic attraction formed when a metal transfers one or more electrons to a non-metal, producing a cation (+) and an anion (−) — e.g. Na⁺ and Cl⁻ in NaCl.
- Covalent bond
- A bond in which two non-metal atoms share a pair of electrons. A shared pair sitting in a bond is a bonding pair; an unshared pair localised on one atom is a lone pair.
- Polar covalent bond
- A shared pair held unequally because the two atoms differ in electronegativity, giving the more electronegative atom a partial negative charge (δ−) and its partner a partial positive (δ+) — e.g. the O–H bonds in H₂O.
- Polyatomic ion
- A group of covalently bonded atoms carrying an overall charge that travels as a unit through reactions — e.g. NH₄⁺, NO₃⁻, CO₃²⁻, SO₄²⁻, PO₄³⁻, OH⁻. You must know both formula and charge.
- Empirical formula
- The simplest whole-number ratio of atoms in a compound (e.g. CH₂O), as opposed to the molecular formula, which is the actual atom count in one molecule (e.g. C₆H₁₂O₆).
- Limiting reagent
- The reactant that runs out first and therefore sets the maximum amount of product. Found by comparing (moles of each reactant ÷ its balanced-equation coefficient) and taking the smallest quotient.
Bonding, Equations and Stoichiometry FAQ
How do I tell whether a bond is ionic, polar covalent, or non-polar covalent?
Use the electronegativity difference (ΔEN) between the two atoms. Roughly: ΔEN below ~0.4 is non-polar covalent (electrons shared evenly); ~0.4 up to ~2 is polar covalent (shared but unevenly, giving δ+/δ− partial charges); and above ~2 is ionic (electrons effectively transferred). A quick sanity check is metal + non-metal → ionic, non-metal + non-metal → covalent.
How do I find the limiting reagent?
Balance the equation first — the coefficients are the mole ratio. Convert each reactant's mass to moles with n = m/M, then divide each reactant's moles by its own coefficient. The reactant with the smallest quotient is the limiting reagent; it caps the product. Never decide by mass alone, and don't compare raw moles unless the ratio happens to be 1:1.
What is the difference between empirical and molecular formula?
The empirical formula is the simplest whole-number ratio of elements (e.g. CH₂O); the molecular formula is the real number of atoms in one molecule (e.g. C₆H₁₂O₆ = glucose). Percent-composition or combustion data gives you the empirical formula; you need a separate molar-mass measurement to scale it up to the molecular formula (divide the molar mass by the empirical-formula mass to get the multiplier).
Why must I balance charge as well as atoms?
A balanced equation conserves both mass and charge. For molecular equations you only adjust coefficients until each element's atom count matches on both sides (balance C, then H, then O last for combustion). For ionic and acid-ionisation equations you must also make the net charge equal on each side — e.g. HCl(aq) → H⁺(aq) + Cl⁻(aq), where 0 = (+1) + (−1).
What is on the exam for this topic?
Expect to: classify a bond from ΔEN; write a formula from valence or by criss-crossing polyatomic-ion charges; derive an empirical formula from percent composition; balance an equation conserving atoms and charge; and run a full mass-to-mass calculation using n = m/M → mole ratio → m = n×M, identifying the limiting reagent and computing percent yield = (actual/theoretical) × 100%. Carry units and ~3 significant figures throughout. The datasheet gives constants and the periodic table, but polyatomic-ion charges are not provided — memorise them.
Exam move
Treat this chapter as one decision tree plus one calculation machine, and drill both until they are automatic. The decision tree handles the qualitative marks: bond type from ΔEN, formula from valence or criss-crossed polyatomic-ion charges, and balancing (coefficients only — atoms first, charge too for ionic/acid equations). The calculation machine handles every numerical question and is always the same three steps: n = m/M → scale by the mole ratio from the balanced equation → m = n×M. Burn in two non-negotiable habits: (1) find the limiting reagent by dividing each reactant's moles by its coefficient and taking the smallest — never by mass; and (2) base theoretical yield (and therefore percent yield) on the limiting reagent, not the reactant in excess. Memorise the common polyatomic ions cold (they are not on the datasheet), keep 3–4 sig figs through working and round only the final ratio/answer, and practise mixed problems (combustion, precipitation with a net ionic equation) so you can recognise which template applies under time pressure.
