UPCAT Chemistry — Stoichiometry & 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%
Previous chapter
Periodic Table, Bonding & Chemical Language
Next chapter
Molecular Theory — VSEPR, IMFA & KMT
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