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Ch.3 - Chemical ReactionsWorksheetSee all chapters
All Chapters
Ch.1 - Intro to General Chemistry
Ch.2 - Atoms & Elements
Ch.3 - Chemical Reactions
BONUS: Lab Techniques and Procedures
BONUS: Mathematical Operations and Functions
Ch.4 - Chemical Quantities & Aqueous Reactions
Ch.5 - Gases
Ch.6 - Thermochemistry
Ch.7 - Quantum Mechanics
Ch.8 - Periodic Properties of the Elements
Ch.9 - Bonding & Molecular Structure
Ch.10 - Molecular Shapes & Valence Bond Theory
Ch.11 - Liquids, Solids & Intermolecular Forces
Ch.12 - Solutions
Ch.13 - Chemical Kinetics
Ch.14 - Chemical Equilibrium
Ch.15 - Acid and Base Equilibrium
Ch.16 - Aqueous Equilibrium
Ch. 17 - Chemical Thermodynamics
Ch.18 - Electrochemistry
Ch.19 - Nuclear Chemistry
Ch.20 - Organic Chemistry
Ch.22 - Chemistry of the Nonmetals
Ch.23 - Transition Metals and Coordination Compounds
Empirical Formula
Molecular Formula
Combustion Analysis
Combustion Apparatus
Polyatomic Ions
Naming Ionic Compounds
Writing Ionic Compounds
Naming Ionic Hydrates
Naming Acids
Naming Molecular Compounds
Balancing Chemical Equations
Limiting Reagent
Percent Yield
Mass Percent
Functional Groups in Chemistry
Additional Guides
Theoretical Yield (IGNORE)
Balancing Chemical Equations Worksheet (IGNORE)
Limiting Reactant (IGNORE)

The molecular formula gives the actual number of atoms.

Molecular Formula

Concept #1: Molecular Formula

Example #1: After a workout session, lactic acid (M = 90.08 g/mol) forms in muscle tissue and is responsible for muscle soreness. Elemental analysis shows that this compound contains 40% C, 6.7% H and 53.3% O. Determine the molecular formula.

Practice: What is the molecular formula for the following compound? 

Empirical Formula: NPCl                    Molar Mass: 347.64 g/mol

Practice: Cortisol (MW = 362.47 g/mol), a known steroid hormone, is found to contain 69.6% carbon, 8.34% hydrogen, and 22.1% oxygen by mass. What is its molecular formula?

Practice: Elemental analysis of a pure compound indicated that the compound had 72.2% C, 8.50% H and the remainder as O. If 0.250 moles of the compound weighs 41.55 g, what is the molecular formula of the compound?