Undergraduate
  1. General chemistry  1.1. Introduction to Chemistry  1.2. Atomic Structure  1.2.1. Subatomic particles  1.2.2. Atomic Model  1.2.3. Quantum numbers  1.2.4. Electron Configuration  1.2.5. Periodic trends  1.2.6. Isotopes and their applications  1.3. Chemical bond  1.3.1. Ionic bond  1.3.2. Covalent bonds  1.3.3. Metal Bond  1.3.4. Hybridization  1.3.5. Molecular geometry and VSEPR theory  1.3.6. Bond polarity and dipole moment  1.4. Stoichiometry  1.4.1. Mole concept  1.4.2. Empirical and molecular formula  1.4.3. Limiting reactant and excess reactant  1.4.4. Percent Yield and Purity  1.4.5. Dilution calculation  1.5. States of matter  1.5.1. Gas Laws  1.5.2. Matter and Intermolecular Forces  1.5.3. Solid and Crystal Structures  1.5.4. Phase diagram  1.5.5. Plasma and supercritical fluid  1.6. Chemical reactions  1.6.1. Types of Chemical Reactions  1.6.2. Balancing Chemical Equations  1.6.3. Thermochemical reactions  1.6.4. Reaction energy profiles  1.7. Solutions and Mixtures  1.7.1. Concentration units  1.7.2. Syndrome properties  1.7.3. Solubility and Solubility Rules  1.7.4. Henry's Law and Raoult's Law  1.7.5. Partition coefficient  1.8. Chemical equilibrium  1.8.1. Law of mass action  1.8.2. Le Chatelier's principle  1.8.3. Reaction quotient  1.8.4. To use this online calculator for Equilibrium Constant, enter Equilibrium Constant (Eq.) and hit the calculate button. Here is how the Equilibrium Constant calculation can be explained with given input values -> 0.05 = 0.05/0.  1.8.5. Acid-base balance  1.9. Acids and bases  1.9.1. Arrhenius, Brønsted–Lowry, and Lewis definitions  1.9.2. pH and pOH  1.9.3. Acid-base titration  1.9.4. Buffer Solution  1.9.5. Acid-base indicator  1.9.6. Polyprotic Acids and Bases  1.10. Electrochemistry  1.10.1. redox reactions  1.10.2. Galvanic and Electrolytic Cells  1.10.3. Nernst equation  1.10.4. Electrolysis  1.10.5. Faraday's laws of electrolysis  1.11. Thermodynamics  1.11.1. Laws of Thermodynamics  1.11.2. Enthalpy, entropy and Gibbs free energy  1.11.3. Spontaneity of reactions  1.11.4. Thermodynamic cycles (Hess's law, Carnot cycle)  1.12. Kinetics  1.12.1. Rate law and reaction order  1.12.2. Activation Energy and Catalyst  1.12.3. Collision theory  1.12.4. Reaction mechanism  1.12.5. Transition state theory
  2. Organic chemistry  2.1. Structure and relationships  2.1.1. Hybridisation in Carbon Compounds  2.1.2. Resonance and aromatization  2.1.3. Molecular orbitals  2.1.4. Inductive and mesomeric effects  2.2. Hydrocarbons  2.2.1. Hydrocarbons  2.2.2. Alkene  2.2.3. Alkynes  2.2.4. Aromatic compounds  2.2.5. Cycloalkanes and Conformation  2.3. Functional Group  2.3.1. Alcohols and phenols  2.3.2. Ethers and epoxides  2.3.3. Aldehyde and ketone  2.3.4. Carboxylic acids and derivatives  2.3.5. Amin  2.3.6. Esters and amides  2.3.7. Thiols and sulfides  2.4. Stereoscopic  2.4.1. Isomerism in Stereochemistry  2.4.2. Chirality and optical activity  2.4.3. Structural analysis  2.4.4. Diastereomers and Enantiomers  2.5.1. Substitution reactions  2.5.2. Elimination reactions  2.5.3. Addition reactions  2.5.4. Radical reactions  2.5.5. Rearrangement reactions  2.5.6. Pericyclic Reactions  2.6. Spectroscopy and structural analysis  2.6.1. Infrared Spectroscopy  2.6.2. Nuclear magnetic resonance spectroscopy  2.6.3. Mass Spectrometry  2.6.4. UV-visible spectroscopy  2.6.5. X-ray crystallography  2.7. Polymer chemistry  2.7.1. Addition and Condensation Polymers  2.7.2. Polymerization mechanism  2.7.3. Biodegradable Polymer  2.7.4. Conducting and smart polymers
  3. Inorganic chemistry  3.1. Coordination chemistry  3.1.1. Ligands and coordination compounds  3.1.2. Crystal field theory  3.1.3. Molecular Orbital Theory in Coordination Compounds  3.1.4. Jahn–Taylor distortion  3.2. Main Group Chemistry  3.2.1. Alkali and alkaline earth metals  3.2.2. Halogens and Noble Gases  3.2.3. Transition metals and their complexes  3.2.4. Lanthanides and Actinides  3.3. Solid state chemistry  3.3.1. Crystal lattices and unit cells  3.3.2. Defects in solids  3.3.3. Electrical and magnetic properties  3.3.4. Superconductivity  3.4. Bio-inorganic chemistry  3.4.1. Metal ions in biology  3.4.2. Enzymatic reactions with metals  3.4.3. Metal poisoning and detoxification  3.4.4. Metalloproteins and Metalloenzymes
  4. Physical chemistry  4.1. Quantum Chemistry  4.1.1. Wave–particle duality  4.1.2. Schrödinger Equation  4.1.3. Quantum Numbers and Orbitals  4.1.4. Atomic and molecular spectroscopy  4.2. Statistical mechanics  4.2.1. Maxwell–Boltzmann distribution  4.2.2. Partitioning functions  4.2.3. Bose–Einstein and Fermi–Dirac statistics  4.3. Chemical thermodynamics  4.3.1. Gibbs Free Energy  4.3.2. Phase transition  4.3.3. Thermodynamic efficiency  4.4. Surface chemistry  4.4.1. Absorption and Catalysis  4.4.2. Colloids and Emulsions
  5. Analytical Chemistry  5.1. Classical methods  5.1.1. Gravimetric Analysis  5.1.2. titrimeter  5.2. Instrumental methods  5.2.1. Chromatography  5.2.2. Spectroscopy  5.2.3. Electroanalytical methods  5.2.4. X-ray diffraction
  6. Industrial Chemistry  6.1. Petrochemistry  6.2. Chemical engineering principles  6.3. Green Chemistry  6.4. Pharmaceuticals and Drug Synthesis
  7. Biochemistry  7.1. Carbohydrates  7.2. Proteins and enzymes  7.3. Lipids and membranes  7.4. Nucleic acids  7.5. Metabolism and Bioenergetics  7.6. Enzyme kinetics and mechanism

Undergraduate → General chemistry → Stoichiometry ↓


Empirical and molecular formula


Chemistry allows us to understand the structure of the various substances that make up our world. Two key concepts that help us understand the structure of chemical compounds are the empirical formula and the molecular formula. In this article, we will explore these concepts extensively, understand the differences between them, and practice determining these formulas through examples.

Empirical formula

The empirical formula of a compound is the simplest whole-number ratio of the atoms of each element in the compound. It provides the relative number of atoms of each element, but it does not tell the exact number of atoms present in the molecular entity. The empirical formula is based solely on ratios, and thus, it does not give information about isomers, structures, or the actual number of atoms in the compound.

Steps to determine the empirical formula

To determine the empirical formula of a compound, follow these general steps:

  1. Start with mass: Find the mass of each element in grams. If you are given percentages, take them as the mass of a 100-gram sample.
  2. Convert mass to moles: Convert each mass to moles by dividing the mass of each element by its atomic mass (found on the periodic table).
  3. Find the simplest ratio: Divide the number of each mole by the smallest number of moles calculated.
  4. Adjust for whole numbers: If necessary, multiply the ratio by a whole number to get a whole number for each element.

Example: Determining the empirical formula

Suppose we have a compound sample that contains 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass.

  1. Convert mass percent to grams: C = 40.0, g, H = 6.7, g, O = 53.3, g
  2. Convert grams to moles: 
    Carbon: 40.0, g div 12.01, g/mol = 3.33, mol
Hydrogen: 6.7, g div 1.008, g/mol = 6.65, mol
Oxygen: 53.3, g div 16.00, g/mol = 3.33, mol
  3. Find the simplest ratio by dividing by the smallest number of moles: 
    Carbon: 3.33, mol div 3.33, mol = 1
Hydrogen: 6.65, mol div 3.33, mol = 2
Oxygen: 3.33, mol div 3.33, mol = 1
  4. Draw up the empirical formula: CH 2 O

Molecular formula

The molecular formula of a compound shows the exact number of atoms of each element in the molecule. Unlike the empirical formula, the molecular formula is not always the simplest ratio. The empirical formula can be thought of as a basic unit that can be multiplied to match the actual number of atoms in the molecule.

Steps to determine molecular formula

You can determine the molecular formula as follows:

  1. Determine the empirical formula: First, determine the empirical formula using the steps above.
  2. Find the empirical formula mass: Calculate the molar mass of the empirical formula.
  3. Divide the molar mass: Divide the molar mass of the compound by the empirical formula mass to get the ratio.
  4. Multiply to get the molecular formula: Multiply the entire empirical formula by this ratio to get the molecular formula.

Example: Determination of molecular formula

The empirical formula of a compound is CH 2 O, and its molar mass is about 180.18 g/mol.

  1. Calculate the empirical formula mass: 
    C: 12.01, g/mol + H: (2 times 1.008, g/mol = 2.016, g/mol) + O: 16.00, g/mol = 30.03, g/mol
  2. Find the ratio: 180.18, g/mol div 30.03, g/mol ≈ 6
  3. Multiply the empirical formula by the ratio to get the molecular formula: 
                    C( 1 × 6)H( 2 × 6)O( 1 × 6) = C 6 H 12 O 6

Comparison of empirical and molecular formulas

The difference between empirical and molecular formulas lies in their representation of the composition of the compound. The empirical formula provides the simplest ratios, while the molecular formula gives the exact count of each type of atom in the compound. Let's look at this through a graphic representation.

CH 2 O C 6 H 12 O 6 ×6

Practice problems

To understand these concepts further, let's practice some problems:

Problem 1

The empirical formula of a substance with a molar mass of 60 g/mol is CH 4 Find the molecular formula.

  1. Determine the empirical formula mass. 
                    C: 12.01, g/mol + H: (4 times 1.008, g/mol = 4.032, g/mol) = 16.042, g/mol
  2. Calculate the ratio: 60, g/mol div 16.042, g/mol ≈ 3.74 (round off to the nearest whole number, which is 4)
  3. Determine the molecular formula: C 4 H 16

Problem 2

A compound contains 29.1% nitrogen and 70.9% oxygen by mass, and its molar mass is 92 g/mol. Determine the empirical and molecular formulas.

  1. Convert mass percent to grams: N = 29.1, g, O = 70.9, g
  2. Convert grams to moles: 
    Nitrogen: 29.1, g div 14.01, g/mol = 2.08, mol
Oxygen: 70.9, g div 16.00, g/mol = 4.43, mol
  3. Find the simplest ratio by dividing by the smallest number of moles: 
    Nitrogen: 2.08, mol div 2.08, mol = 1
Oxygen: 4.43, mol div 2.08, mol ≈ 2.13
    Since 2.13 is not a whole number, multiply both by 2 to get whole numbers: 
    Nitrogen: 1 times 2 = 2
Oxygen: 2.13 times 2 = 4.26 ≈ 4
  4. Empirical Formula: N 2 O 4
  5. Calculate the empirical formula mass: 
    N: (2 times 14.01, g/mol = 28.02, g/mol) + O: (4 times 16.00, g/mol = 64.00, g/mol) = 92.02, g/mol
  6. Determine the ratio: 92, g/mol div 92.02, g/mol = 1
  7. Molecular Formula: N 2 O 4

Conclusion

Understanding the concepts of empirical and molecular formulas is essential to determine the structure of chemical compounds. The empirical formula provides a simplified representation of the compound, while the molecular formula reveals the exact number of atoms. Mastering these concepts involves practicing with different compounds, analyzing their structure, and calculating the correct formulas.


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Empirical and molecular formula

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