Chemical Reactions and Stoichiometry

FE Exam Weight: Chemistry contributes 5–8 questions (~6% of the 110-question FE Other Disciplines exam). Most are quick plug-and-chug problems. The exam is open to the searchable NCEES FE Reference Handbook, so the constants (Avogadro's number, R, molar volume) and equation forms are provided — your job is to FIND the right relation and APPLY it without unit errors.

Atomic Structure and the Periodic Table

An atom has a nucleus of protons (charge +1, define the element via atomic number Z) and neutrons (neutral), surrounded by electrons (charge −1). A neutral atom has electrons = protons. Isotopes share Z but differ in neutron count, so atomic weights on the table are weighted averages of isotope masses.

The periodic table organizes elements by increasing Z. Key trends moving left→right across a period: atomic radius decreases, ionization energy and electronegativity increase. Moving top→bottom down a group: radius increases, ionization energy decreases. Valence electrons (outermost shell) determine bonding behavior — group 1 (alkali metals) lose one electron, group 17 (halogens) gain one, group 18 (noble gases) are inert.

Chemical Bonding

Electronegativity difference (ΔEN) predicts bond character: ΔEN > ~1.7 → ionic; 0.4–1.7 → polar covalent; < 0.4 → nonpolar covalent. Polar molecules (asymmetric charge, like H₂O) dissolve ionic and polar solutes ("like dissolves like").

** Electrons fill orbitals (s, p, d, f) by increasing energy; the outermost shell holds the valence electrons that drive reactivity. Atoms tend to gain, lose, or share electrons to reach a stable octet (eight valence electrons), which explains why sodium (one valence electron) readily forms Na⁺ and chlorine (seven) forms Cl⁻. 01 g/mol. You will compute molecular weights constantly, so reading them off the periodic table quickly is essential.

Solutions and Concentration

Most FE chemistry happens in aqueous solution, so concentration units matter:

A dilution preserves moles of solute: M₁V₁ = M₂V₂. To make 500 mL of 0.10 M HCl from 2.0 M stock: V₁ = M₂V₂/M₁ = (0.10 × 500)/2.0 = 25 mL of stock diluted to 500 mL. Mixing up molarity (per liter of solution) and molality (per kg of solvent) is a frequent trap.

The Mole Concept

Convert with: moles = mass (g) ÷ molar mass (g/mol), and particles = moles × 6.022 × 10²³.

Balancing Equations

Mass is conserved — each element's atom count must match on both sides. Balance Fe₂O₃ + CO → Fe + CO₂: set Fe to 2 (→ 2Fe), then balance C and O. Result: Fe₂O₃ + 3CO → 2Fe + 3CO₂ (Fe 2=2, C 3=3, O 6=6 ✓).

Worked Stoichiometry Example

How many grams of CO₂ form when 100 g of propane (C₃H₈, MW = 44.1) burns completely?

Balanced: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

1. Moles C₃H₈ = 100 ÷ 44.1 = 2.268 mol
2. Mole ratio = 3 mol CO₂ per 1 mol C₃H₈
3. Moles CO₂ = 2.268 × 3 = 6.804 mol
4. Mass CO₂ = 6.804 × 44.0 = ≈ 299 g

The Ideal Gas Law

$PV = nRT$

Variables: P = absolute pressure, V = volume, n = moles, T = absolute temperature in kelvin (a near-universal trap — never use °C), R = universal gas constant. Pick R to match units:

Example: Volume of 2.0 mol of an ideal gas at 300 K and 2.0 atm: V = nRT/P = (2.0)(0.08206)(300)/2.0 = 24.6 L. Combined-gas shortcut for a fixed amount of gas: P₁V₁/T₁ = P₂V₂/T₂.

Limiting Reagent, Yield, and Equilibrium

The limiting reagent runs out first. Compute how much product each reactant alone could make; the smallest controls theoretical yield. Then percent yield = (actual ÷ theoretical) × 100%.

For a reversible reaction aA + bB ⇌ cC + dD, the equilibrium constant is K = ([C]ᶜ[D]ᵈ)/([A]ᵃ[B]ᵇ). K ≫ 1 favors products; K ≪ 1 favors reactants. Le Chatelier's principle: a disturbed equilibrium shifts to oppose the change — adding reactant or removing product shifts right; increasing pressure shifts toward fewer gas moles; raising temperature shifts an exothermic reaction left (toward reactants).