Stoichiometry and Chemical Calculations
What this domain tests
The stoichiometry items test the mole as the central currency of chemistry: converting mass, volume and concentration into moles, applying the balanced equation, identifying the limiting reagent and reporting the answer with sensible significant figures.
Core rules
- n = m/M; n = cV (V in dm³); at STP one mole of ideal gas occupies 22.7 dm³ at 1 bar (22.4 dm³ at 1 atm).
- Balance the equation first; the coefficients give the mole ratio, never the mass ratio.
- Limiting reagent method: convert every reactant to moles, divide each by its coefficient, and the smallest quotient is limiting.
- Percent yield = (actual/theoretical) × 100, always calculated from the limiting reagent.
- Empirical formula from mass percentages: divide by atomic mass, then by the smallest quotient; the molecular formula is a whole-number multiple found from the molar mass.
- Dilution: c₁V₁ = c₂V₂. In titration, moles of titrant = moles of analyte × (coefficient ratio).
Worked example
5.00 g of N₂ reacts with 5.00 g of H₂ to form ammonia. Which is limiting, and what mass of NH₃ forms? N₂: 5.00/28.02 = 0.1785 mol; ÷1 = 0.1785. H₂: 5.00/2.02 = 2.475 mol; ÷3 = 0.825. Nitrogen gives the smaller quotient, so N₂ is limiting. Moles NH₃ = 2 × 0.1785 = 0.3570 mol. Mass = 0.3570 × 17.03 = 6.08 g. Hydrogen is present in large excess (1.94 mol left over).
Traps that cost marks
- Choosing the reactant with the smallest mass rather than the smallest mole-per-coefficient quotient.
- Forgetting hydrate water in the molar mass (CuSO₄·5H₂O is 249.7, not 159.6 g mol⁻¹).
- Reporting more significant figures than the data allow - three in the example above, not six.
- Forgetting that only the limiting reagent limits the yield, and that percent yield is measured against it.