Chemical Reactions, Stoichiometry and the Mole Concept
Equations and the mole
The mole links mass, particle number and gas volume: n = m/M, N = n × 6.022 × 10²³, and at STP one mole of gas occupies 22.4 L (24.5 L at 25 °C and 100 kPa). Balance equations by inspection, then use mole ratios from coefficients; coefficients give the ratio of moles, never the ratio of masses.
Limiting reagent and yield
Identify the limiting reagent by dividing moles of each reactant by its coefficient; the smallest quotient limits. Theoretical yield comes from the limiting reagent; percentage yield = actual/theoretical × 100. Percentage atom economy = (molar mass of desired product / total molar mass of reactants) × 100. These are different measures: yield concerns efficiency of the process, atom economy concerns design.
Solutions
c = n/V with V in litres. Dilution obeys c₁V₁ = c₂V₂. Titration calculations combine this with a balanced equation.
Traps
- Using masses directly in mole ratios.
- Forgetting to convert cm³ to dm³ (divide by 1000).
- Using the excess reagent to find theoretical yield.
- Confusing empirical and molecular formula; molecular formula is a whole-number multiple of empirical.
- Ignoring significant figures when reporting.
Rule of thumb
Write a roadmap: mass → moles → mole ratio → moles → mass. Every arrow needs a conversion factor.
Worked example
4.0 g of calcium carbonate (M = 100 g/mol) reacts with excess HCl. n = 0.040 mol. From CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂, n(CO₂) = 0.040 mol, so V = 0.040 × 24.5 = 0.98 L at 25 °C.
Practical link
Record burette readings to two decimal places, and repeat titrations until concordant results agree within 0.10 mL.