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Chemistry 3% exam weight

Stoichiometry (Stoikiometri / Perhitungan Kimia)

Part of the UTBK/SNPMTN (Indonesia) study roadmap. Chemistry topic chemis-007 of Chemistry.

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Topic 7 — Stoichiometry (Stoikiometri / Perhitungan Kimia)

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Stoichiometry is the quantitative study of reactants and products in balanced chemical reactions. Every calculation starts from the mole (mol), the SI unit for amount of substance.

QuantityFormulaVariables
Moles from massn = m / Mrm = mass (g), Mr = molar mass (g/mol)
Moles from gas at STPn = V / 22.4V = gas volume (L) at 0 °C, 1 atm
Moles from particlesn = N / N_AN = particle count, N_A = 6.022 × 10²³
MolarityM = n / VV = solution volume (L)
Percent yield% = (actual / theoretical) × 100%both in grams
  • Always balance the equation first before applying mole ratios.
  • STP means 0 °C and 1 atm; the molar volume is 22.4 L/mol.
  • Identify the limiting reactant by comparing the mole-to-coefficient ratio, not by mass alone.

🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

The Mole as the Central Converter

Stoichiometric problems all funnel through the mole. Convert mass to moles with n = m / Mr, gases with n = V / 22.4 L/mol at STP, and particles with n = N / N_A. Once two species share the unit “mol”, their stoichiometric coefficients in the balanced equation act as a conversion ratio.

Empirical vs. Molecular Formula

From a percent composition by mass, divide each element’s percentage by its atomic mass to get a mole ratio. Round to whole numbers (multiply through if needed) to produce the empirical formula. Divide the given molar mass by the empirical-formula mass to find the integer n that converts empirical → molecular formula: (empirical formula)ₙ.

Limiting Reactant and Yield

Compare n / coefficient for every reactant. The smallest value runs out first and determines the theoretical yield. The percent yield compares the actual mass recovered against this theoretical ceiling: % yield = (actual / theoretical) × 100%. Yields above 100% signal measurement error or impure product.

Solution Concentration

TermDefinitionWhen to use
Molarity (M)mol solute / L solutionTitrations, dilution problems
Molality (m)mol solute / kg solventColligative properties, temperature-stable
Mass percent(g solute / g solution) × 100%Mixing without volume data
DilutionM₁V₁ = M₂V₂Adding solvent to a stock solution

Typical UTBK Question Pattern

Expect a multi-step problem combining two conversions — for example, “How many mL of 0.5 M HCl are needed to neutralise 2.0 g of CaCO₃?” — which links mass → moles → mole ratio → moles of acid → volume via molarity.

🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Edge Cases and Mechanics

  • Non-STP gases: at 25 °C and 1 atm the molar volume is ≈ 24.5 L/mol; failing to convert the given conditions before dividing is a frequent deduction.
  • Density-based molality: when given solution density (g/mL), compute solvent mass as (total mass − solute mass). Mass of solution = density × volume in mL.
  • Mixed units: convert mL → L and mg → g before plugging into n = m / Mr or M = n / V.
  • Gas-collection over water: subtract water-vapour pressure from total pressure before applying PV = nRT, since the collected gas is mixed with water vapour.

Worked Example

Reaction: 2 H₂(g) + O₂(g) → 2 H₂O(l). If 4.0 g H₂ reacts with 32.0 g O₂, find the limiting reactant and theoretical yield of H₂O.

  1. n(H₂) = 4.0 / 2.0 = 2.0 mol; n / coefficient = 2.0 / 2 = 1.0.
  2. n(O₂) = 32.0 / 32.0 = 1.0 mol; n / coefficient = 1.0 / 1 = 1.0.
  3. The ratios tie, so both reactants are fully consumed. Theoretical H₂O = 2.0 mol × 18 g/mol = 36 g.

Common Mistakes

  • Selecting the reactant with smaller mass as limiting — always use n / coefficient.
  • Forgetting to balance; mole ratios from an unbalanced equation are wrong by construction.
  • Computing % yield using the excess reactant mass — yield is anchored to the limiting reactant.
  • Reporting molarity when the question asks for molality (or vice versa) when density is supplied.

Practice Prompts

  1. A 25.0 mL solution of 0.150 M NaOH is diluted to 250 mL. Calculate the new molarity and the mass of NaOH originally dissolved.
  2. Combustion of 1.20 g of an unknown hydrocarbon (CₓHᵧ) produces 3.52 g CO₂ and 1.44 g H₂O; the molar mass is 84 g/mol. Determine the empirical and molecular formulas.

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