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UPCAT ChemistryStoichiometry & Chemical ReactionsRevision Notes

Revision notes for UPCAT Chemistry Stoichiometry & Chemical Reactions — designed for time-pressed reviewers. These notes skip the basics and focus on what University of the Philippines consistently tests, so you spend your revision hours on the content most likely to appear on exam day.

Exam context

The University of the Philippines College Admission Test is conducted by University of the Philippines and is scheduled for Mid-2026 (announced by UP Admissions). The Chemistry subtest is marked as "Core" in the official pattern, and Stoichiometry & Chemical Reactions appears in position 4th of 7 in the UPCAT Chemistry review rotation. Passing mark: UPG ≤ 2.2 typical. Recent UPCAT 2026 papers have drawn roughly 20 questions from this subject.

Stoichiometry & Chemical Reactions - Revision notes

Stoichiometry is the quantitative study of chemical reactions and formulas. It allows us to predict how much reactant we need and how much product we can make in chemical reactions. This chapter combines the concepts of chemical reactions with mathematical calculations to solve real-world chemistry problems. Understanding stoichiometry is crucial for UPCAT and other college entrance exams as it forms the foundation for advanced chemistry topics.

Sections

Formulas

Example

Zn + I₂ → ZnI₂ (zinc iodide formation)

Formula

Synthesis: A + B → AB

Variables

A and B are reactants, AB is the product

Application

Formation of compounds from elements

Example

2Al₂O₃ → 4Al + 3O₂ (aluminum oxide decomposition)

Formula

Decomposition: AB → A + B

Variables

AB is the reactant, A and B are products

Application

Breaking down compounds using heat or electricity

Example

C₆H₆ + O₂ → CO₂ + H₂O (benzene combustion)

Formula

Combustion: CₓHᵧ + O₂ → CO₂ + H₂O

Variables

CₓHᵧ is hydrocarbon fuel, O₂ is oxygen

Application

Burning of fuels for energy

Exam Tips

  • Practice identifying reaction types by looking at the reactants and products
  • Remember that coefficients apply to the entire formula, not just the first element
  • Use the activity series to predict single replacement reactions
  • Always check if your balanced equation follows the law of conservation of mass

Key Points

  • Chemical reactions follow predictable patterns that help us classify and understand them
  • Synthesis reactions combine simple substances to form complex compounds (A + B → AB)
  • Decomposition reactions break complex compounds into simpler substances (AB → A + B)
  • Single replacement reactions involve one element replacing another in a compound (AB + X → AX + B)
  • Double displacement reactions involve exchange of ions between compounds (AB + XY → AY + XB)
  • Combustion reactions involve burning with oxygen to produce CO₂ and H₂O
  • Acid-base reactions produce salt and water when acids react with bases

Definitions

Term

Reactants

Definition

Substances at the beginning of a chemical reaction that undergo chemical change

Importance

Essential for identifying what materials are needed for a reaction

Term

Products

Definition

Substances formed at the end of a chemical reaction

Importance

Shows what you get from a chemical reaction

Term

Coefficient

Definition

Number in front of a chemical formula indicating the number of molecules or formula units

Importance

Critical for balancing equations and stoichiometric calculations

Section Title

Types of Chemical Reactions

Common Mistakes

  • Confusing reactants and products in chemical equations
  • Changing subscripts instead of coefficients when balancing equations
  • Not recognizing reaction patterns for classification
  • Forgetting to balance combustion reactions properly

Formulas

Example

5g H₂O ÷ 18g/mol = 0.28 mol H₂O

Formula

Number of moles = mass (g) / molar mass (g/mol)

Variables

mass in grams, molar mass from periodic table

Application

Converting between mass and moles

Example

0.28 mol H₂O × 6.022×10²³ = 1.67×10²³ molecules

Formula

Number of entities = moles × 6.022 × 10²³

Variables

moles of substance, Avogadro's number

Application

Finding number of atoms, molecules, or ions

Example

H in H₂O: (2.0g / 18.0g) × 100 = 11.1%

Formula

Percent composition = (mass of element / total mass) × 100

Variables

mass of specific element, total compound mass

Application

Finding percentage of each element in a compound

Exam Tips

  • Always write down the given information and what you're solving for
  • Use dimensional analysis to check your units
  • Round atomic masses to appropriate significant figures
  • Practice converting between grams, moles, and number of particles

Key Points

  • A mole is the SI unit for amount of substance, equal to 6.022 × 10²³ entities (Avogadro's number)
  • Molar mass is the mass of one mole of a substance, expressed in g/mol
  • Molecular mass is the sum of atomic masses in a compound, expressed in amu
  • The mole concept allows us to count atoms and molecules by weighing them
  • Avogadro's number connects the microscopic and macroscopic worlds
  • Molar mass can be calculated by adding atomic masses from the periodic table

Definitions

Term

Mole

Definition

Amount of substance containing 6.022 × 10²³ entities

Importance

Fundamental unit for quantitative chemistry calculations

Term

Avogadro's Number

Definition

6.022 × 10²³ entities per mole

Importance

Connects atomic scale to measurable quantities

Term

Molar Mass

Definition

Mass of one mole of a substance in grams per mole

Importance

Conversion factor between mass and moles

Section Title

The Mole Concept and Molar Mass

Common Mistakes

  • Confusing molecular mass (amu) with molar mass (g/mol)
  • Forgetting to multiply by the number of atoms when calculating molar mass
  • Using wrong units in mole calculations
  • Not properly accounting for coefficients in chemical formulas

Formulas

Example

0.5 mol NaCl in 2L solution = 0.25 M NaCl

Formula

Molarity (M) = moles of solute / liters of solution

Variables

moles of dissolved substance, total volume in liters

Application

Preparing solutions of known concentration

Example

0.1 mol sugar in 0.5 kg water = 0.2 m solution

Formula

Molality (m) = moles of solute / kg of solvent

Variables

moles of dissolved substance, mass of solvent in kg

Application

Temperature-independent concentration measure

Example

2M × 0.1L = 0.5M × 0.4L (dilution check)

Formula

M₁V₁ = M₂V₂

Variables

Initial and final molarity and volume

Application

Dilution calculations in laboratory

Exam Tips

  • Always identify what type of concentration is being asked for
  • Remember that molarity changes with temperature but molality doesn't
  • Use the dilution equation for mixing solutions problems
  • Practice unit conversions between mL and L

Key Points

  • Molarity measures concentration as moles of solute per litre of solution
  • Molality measures concentration as moles of solute per kilogram of solvent
  • Dilution involves adding solvent to decrease concentration
  • The dilution equation M₁V₁ = M₂V₂ is essential for dilution calculations
  • Concentration calculations are crucial in laboratory work and industry

Definitions

Term

Molarity

Definition

Moles of solute per litre of solution

Importance

Most common concentration unit in chemistry

Term

Solute

Definition

Substance being dissolved in a solution

Importance

Key component for concentration calculations

Term

Solvent

Definition

Substance doing the dissolving, usually water

Importance

Determines the solution properties

Section Title

Solutions and Molarity

Common Mistakes

  • Confusing molarity with molality
  • Using volume of solvent instead of total solution volume
  • Forgetting to convert mL to L in molarity calculations
  • Not accounting for the change in total volume when mixing solutions

Formulas

Example

If theoretical = 10g, actual = 8g, then % yield = 80%

Formula

Percent yield = (actual yield / theoretical yield) × 100

Variables

actual yield from experiment, theoretical yield from calculation

Application

Measuring reaction efficiency

Example

2H₂ + O₂ → 2H₂O gives 2:1:2 mole ratio

Formula

Mole ratio = coefficients from balanced equation

Variables

coefficients of reactants and products

Application

Converting between different substances in reaction

Exam Tips

  • Always start with a balanced chemical equation
  • Use dimensional analysis with mole ratios as conversion factors
  • Identify the limiting reactant by calculating how much product each reactant can make
  • Show all steps clearly for partial credit on calculations

Key Points

  • Stoichiometry uses balanced equations to calculate quantities in reactions
  • Mole ratios from balanced equations are conversion factors
  • Mass-to-mass calculations require converting through moles
  • Limiting reactant determines how much product can be formed
  • Excess reactant is left over after the reaction is complete
  • Theoretical yield is calculated from stoichiometry
  • Actual yield is measured experimentally and is usually less than theoretical
  • Percent yield compares actual to theoretical yield

Definitions

Term

Stoichiometry

Definition

Quantitative study of reactants and products in chemical reactions

Importance

Essential for predicting reaction outcomes

Term

Limiting Reactant

Definition

Reactant that is completely consumed and limits product formation

Importance

Determines maximum amount of product possible

Term

Theoretical Yield

Definition

Maximum amount of product that can be formed from given reactants

Importance

Baseline for comparing actual experimental results

Section Title

Stoichiometric Calculations

Common Mistakes

  • Not balancing the equation before doing stoichiometric calculations
  • Using mass ratios instead of mole ratios
  • Assuming all reactants are limiting without checking
  • Confusing theoretical and actual yield

Formulas

Example

If empirical = CH₂O and molecular mass = 180, then n = 6, so molecular = C₆H₁₂O₆

Formula

Molecular formula = (Empirical formula) × n

Variables

n = molecular mass / empirical formula mass

Application

Finding actual molecular formula from empirical formula

Exam Tips

  • Convert percentages to grams (assume 100g sample)
  • Convert grams to moles for each element
  • Find the simplest ratio by dividing by the smallest number of moles
  • If ratios aren't whole numbers, multiply by appropriate factor

Key Points

  • Empirical formula shows the simplest whole number ratio of elements
  • Molecular formula shows the actual number of atoms in a molecule
  • Molecular formula is a whole number multiple of empirical formula
  • Percent composition data can be used to find empirical formulas
  • Molecular mass is needed to convert empirical to molecular formula
  • Combustion analysis is used for compounds containing C, H, and O

Definitions

Term

Empirical Formula

Definition

Simplest whole number ratio of elements in a compound

Importance

Shows basic composition without molecular structure

Term

Molecular Formula

Definition

Actual number of atoms of each element in a molecule

Importance

Shows true molecular composition

Term

Formula Mass

Definition

Sum of atomic weights in empirical formula

Importance

Used to determine molecular formula from empirical formula

Section Title

Empirical and Molecular Formulas

Common Mistakes

  • Not reducing to simplest whole number ratios for empirical formula
  • Confusing empirical and molecular formulas
  • Rounding too early in calculations
  • Not using molecular mass to find the multiplier n

Connections

  • Stoichiometry connects to atomic structure through molar mass calculations
  • Chemical reactions relate to thermodynamics through energy changes
  • Solution chemistry connects to electrochemistry and acid-base reactions
  • Empirical formulas relate to molecular geometry and bonding
  • Percent yield connects to chemical kinetics and reaction mechanisms
  • Limiting reactants relate to industrial processes and cost optimization

Exam Strategy

For UPCAT and college entrance exams, focus on mastering unit conversions and dimensional analysis. Practice identifying limiting reactants and calculating percent yields. Memorize key formulas for molarity and percent composition. Work through complete stoichiometry problems step-by-step, showing all work for partial credit. Pay special attention to balancing equations as this is the foundation for all stoichiometric calculations. Review common polyatomic ions and their formulas for faster problem solving.

Quick Review Questions

What is the molar mass of Ca(OH)₂?

Ca = 40.1 + 2(O = 16.0) + 2(H = 1.0) = 40.1 + 32.0 + 2.0 = 74.1 g/mol

How many molecules are in 2.5 moles of H₂O?

2.5 mol × 6.022 × 10²³ molecules/mol = 1.51 × 10²⁴ molecules

What is the molarity of a solution containing 0.5 mol NaCl in 250 mL of solution?

M = 0.5 mol ÷ 0.25 L = 2.0 M NaCl

In the reaction 2H₂ + O₂ → 2H₂O, what is the mole ratio of H₂ to H₂O?

From the balanced equation, 2 moles H₂ produce 2 moles H₂O, giving a 1:1 ratio

If the theoretical yield is 15g and actual yield is 12g, what is the percent yield?

% yield = (12g ÷ 15g) × 100% = 80%

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