Slide presentation
Deducing Molecular Formula
General Chemistry • Chemical Compounds
Topic 1 · Actual formula
Deducing Molecular Formula
The empirical formula gives the simplest atom ratio. The molecular formula gives the actual number of atoms in one molecule, found by combining the empirical formula with molar mass.
Learning target
Find the multiplier from molar mass, then multiply every empirical formula subscript to get the molecular formula.
Same ratio, larger actual molecule.
Why it matters
Same empirical formula can represent different molecules.
Empirical formula data tells the simplest ratio, but chemists often need the actual molecular formula to identify the compound, compare structures, and calculate reaction quantities accurately.
Compound identity
CH2O could represent different compounds. Molar mass helps identify the actual molecule.
Structure reasoning
The molecular formula shows how many atoms are available for bonding and structure.
Quantitative chemistry
Molecular formulas connect composition, molar mass, and stoichiometry.
Core concept
The molecular formula is a whole-number multiple of the empirical formula.
The empirical formula cannot be multiplied by a fraction. The multiplier must be a whole number because molecular formulas count actual atoms.
Empirical formula
Start with the simplest atom ratio.
Formula mass
Add atomic masses in the empirical formula.
Multiplier
Divide molar mass by empirical formula mass.
Molecular formula
Multiply every subscript by the multiplier.
Example: If the empirical formula is CH2O and \(n=6\), the molecular formula is C6H12O6.
Vocabulary
Separate formula mass, molar mass, and multiplier.
The calculation works only when each quantity is interpreted correctly. The empirical formula mass belongs to the simplest formula unit, while molar mass belongs to the actual molecule.
| Term | Meaning | Unit | Example |
|---|---|---|---|
| Empirical formula | Simplest whole-number atom ratio | formula | CH2O |
| Molecular formula | Actual atom count in one molecule | formula | C6H12O6 |
| Empirical formula mass | Mass of one empirical formula unit | g/mol | CH2O ≈ 30.03 g/mol |
| Molar mass | Mass of one mole of actual molecules | g/mol | Glucose ≈ 180.18 g/mol |
| Multiplier | Whole-number factor from empirical to molecular formula | unitless | \(180.18/30.03=6\) |
Main relationship
Molar mass tells how many empirical units fit inside the molecule.
First calculate the empirical formula mass. Then compare the actual molar mass to that empirical formula mass.
The multiplier \(n\) should be a whole number after reasonable rounding. Then multiply every empirical subscript by \(n\).
If \(n=1\)
The empirical formula and molecular formula are the same.
If \(n>1\)
The molecular formula has the same ratio but larger subscripts.
Interactive simulation
Change the empirical formula and molar mass.
Build an empirical formula using C, H, and O subscripts. Then adjust the molar mass to see when the multiplier becomes a whole number.
Formula multiplier tester
Calculated molecular formula
C₆H₁₂O₆
Empirical mass: 30.03 g/mol; multiplier: 6
Static fallback model
For empirical formula CH2O and molar mass 180.2 g/mol, \(n=180.2/30.03\approx6\). The molecular formula is C6H12O6.
The longer molar mass bar represents several empirical formula units inside one molecule.
Dynamic relationship
The multiplier increases as molar mass increases.
For a fixed empirical formula, the ratio of molar mass to empirical formula mass determines the multiplier. The graph highlights whole-number multiplier targets.
The plotted point uses the interactive formula and molar mass from the previous slide.
Worked example
Find the molecular formula from CH2O and molar mass 180.18 g/mol.
The empirical formula gives the ratio C:H:O = 1:2:1. The molar mass tells how many times that ratio appears in the actual molecule.
Calculate empirical formula mass
CH2O: \(12.01 + 2(1.008) + 16.00 = 30.03\ \text{g/mol}\).
Find the multiplier
\(n=\frac{180.18}{30.03}=6.00\)
Multiply every subscript
CH2O × 6 gives C6H12O6.
Final answer: The molecular formula is C6H12O6.
Common mistake
Do not multiply only one subscript.
The multiplier applies to the entire empirical formula. Every subscript must be multiplied so the atom ratio stays the same.
Incorrect reasoning
“If CH2O has multiplier 6, make only carbon larger: C6H2O.”
This changes the ratio from 1:2:1 to 6:2:1.
Correct reasoning
Multiply all subscripts: C becomes 6, H becomes 12, and O becomes 6. The ratio remains 1:2:1.
Practice check
Deduce the molecular formula.
Question: A compound has empirical formula NO2 and molar mass 92.02 g/mol. What is the molecular formula?
Show answer
Empirical formula mass
NO2: \(14.01 + 2(16.00)=46.01\ \text{g/mol}\).
Multiplier
\(n=\frac{92.02}{46.01}=2\)
Molecular formula
Multiply NO2 by 2 to get N2O4.
Reasonableness check
The molecular formula mass of N2O4 is twice the empirical formula mass of NO2, matching the given molar mass.
Apply the topic
Use molecular formulas after empirical formulas are known.
A common problem path is percent composition → empirical formula → empirical formula mass → molecular formula. The molar mass supplies the final scale factor.
Open the calculator
Practice using empirical formula mass and molar mass to determine molecular formulas.
Try related questions
Check your ability to find the multiplier and apply it to every subscript.
Composition
Find simplest ratio.
Empirical formula
Calculate formula mass.
Molar mass
Find multiplier.
Molecular formula
Actual atom count.
Final summary
The molecular formula scales the empirical formula to the actual molecule.
Empirical formula
Shows the simplest whole-number atom ratio.
Molecular formula
Shows the actual number of atoms in one molecule.
Multiplier
\(n=\frac{\text{molar mass}}{\text{empirical formula mass}}\), rounded to a whole number.
Subscripts
Multiply every empirical formula subscript by \(n\).
Key idea: Empirical formula gives proportion; molar mass tells how many of that proportion appear in the actual molecular formula.