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Science · Honors Chemistry

Chapter 1: Quantitative Foundations

Empirical and Molecular Formulas

A ratio is not a count.

Lesson
3
Time
About 25 minutes
0 of 10 done
Part 1 of 9Something to Notice
Practice
Using mathematics and computational thinking
Crosscutting concept
Scale, proportion, and quantity
Core idea
PS1.A: Structure and Properties of Matter

Step 1: Something to Notice

Watch first. The explanation comes later.

Watch the idea first — 39 seconds. Then read on, and try it yourself in the next step.

Three substances with completely different properties give identical percentage composition to three figures.

What further measurement separates them?

Step 2: Find Out

Isolate the supplied amounts and note the ratio in which they actually react.

Step 1 — Predict

Two compounds have the same percentage composition. Are they the same substance?

Choose what you think will happen. You cannot see the experiment until you do — guessing first is what makes it worth watching.

Step 3: So Here Is Why

Now the explanation, after you have seen it happen.

Atoms, not grams

Convert masses to moles first: a formula is a ratio of atoms, not of masses.

The empirical formula is the simplest ratio

A ratio

Divide by the smallest, then clear fractions.

Empirical formula The simplest whole-number ratio of atoms in a compound.

Molar mass selects the multiple

A count

CH₂O, C₂H₄O₂ and C₆H₁₂O₆ share one empirical formula.

Molecular formula The actual number of each atom in a molecule.

Combustion analysis supplies the composition

The route

Carbon as CO₂ and hydrogen as H₂O, both weighed.

Oxygen is found by difference

Least reliable

So it carries both uncertainties and is the least reliable figure.

A ratio is not a count

The empirical formula gives the simplest whole-number ratio of atoms. Determining the molecular formula additionally requires the molar mass, because many molecules share a ratio.

Combustion analysis works backwards

The carbon dioxide and water produced give the carbon and hydrogen in the original; oxygen is found by difference. It is stoichiometry run in reverse to identify an unknown.

Step 4: A Common Mistake

Lots of people think

The empirical formula tells you what the molecule is.

Step 5: Why Isomers Defeat Formulas Entirely

The same idea somewhere new.

Ethanol and dimethyl ether share the molecular formula C₂H₆O and are a drinkable liquid boiling at 78 °C and a gas boiling at −24 °C. Molecular formula fixes composition and says nothing about connectivity, so it cannot distinguish them; a structural formula is needed. As molecules grow the problem grows with them, since C₆H₁₄ has five isomers and C₁₀H₂₂ has seventy-five. The general lesson is that each level of formula answers a narrower question than the last — empirical gives a ratio, molecular gives a count, structural gives an arrangement — and knowing which one a piece of data can support is most of the skill.

Connectivity

Step 6: Think It Through

Practice makes it stick.

Three Substances, One Composition

Problem 1 of 2

A gas, vinegar and a sugar share a percentage composition. What separates them?

Why Oxygen Last

Problem 2 of 2

In combustion analysis, why is oxygen found by difference?

Ratio Then Count

1 of 5

What is the first step from percentage composition?

2 of 5

What does an empirical formula give?

3 of 5

What extra measurement gives the molecular formula?

4 of 5

What does combustion analysis capture?

5 of 5

What does a molecular formula fail to show?

Step 7: Quick Check

Show what you know.

Question 1 of 1

Does an empirical formula identify a compound?

Step 8: Explain It in Writing

Claim, evidence, then reasoning.

The question

Explain how empirical and molecular formulas are determined and why both steps are needed.

Fill in all three boxes. The reasoning box is the one that matters most.

What You Found Out

  • Convert masses to moles first, because a formula is a ratio of atoms.
  • The empirical formula is the simplest whole-number ratio.
  • Molar mass selects which multiple of it the molecule actually is.
  • Combustion analysis gives carbon and hydrogen directly and oxygen by difference.