Practice Molecular Formula Questions with Solutions

What You Actually Need to Know About Molecular Formulas

Molecular formulas tell you exactly which atoms are in a molecule and how many of each you have. That's it. Nothing fancy. If you can't write one from scratch or determine one from data, your chemistry foundation is cracked.

This guide gives you real practice problems with real solutions. No theory dumps. No motivational garbage. Just the work.

The Core Concept (Keep It Simple)

A molecular formula shows:

The elements present in the compound
The number of atoms of each element

Example: C₆H₁₂O₆ means 6 carbon atoms, 12 hydrogen atoms, and 6 oxygen atoms bonded together.

Compare this to an empirical formula, which shows the simplest whole-number ratio of atoms. C₆H₁₂O₆ simplifies to CH₂O — same ratio, but not the actual molecule.

How to Find a Molecular Formula: The Method That Actually Works

You need two things:

The empirical formula of the compound
The molar mass (molecular weight) from experimental data

Then you follow this formula:

n = Molecular mass ÷ Empirical formula mass

Multiply each subscript in the empirical formula by n. That's your molecular formula.

Practice Questions with Full Solutions

Problem 1: From Empirical Formula to Molecular Formula

Question: A compound has an empirical formula of CH₂O and a molar mass of 180 g/mol. Find its molecular formula.

Solution:

Step 1: Find the empirical formula mass.

C = 12.01, H = 1.01, O = 16.00

CH₂O = 12.01 + (2 × 1.01) + 16.00 = 30.03 g/mol

Step 2: Calculate n.

n = 180 ÷ 30.03 = 6

Step 3: Multiply the empirical formula by 6.

C(1×6)H(2×6)O(1×6) = C₆H₁₂O₆

That's glucose. Done.

Problem 2: Finding Empirical Formula from Percent Composition

Question: A compound is 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass. Find its empirical formula.

Solution:

Assume you have 100 g of the compound. That gives you:

40.0 g C
6.7 g H
53.3 g O

Convert to moles:

C: 40.0 g ÷ 12.01 g/mol = 3.33 mol
H: 6.7 g ÷ 1.01 g/mol = 6.63 mol
O: 53.3 g ÷ 16.00 g/mol = 3.33 mol

Divide by the smallest number (3.33):

C: 3.33 ÷ 3.33 = 1
H: 6.63 ÷ 3.33 = 2
O: 3.33 ÷ 3.33 = 1

Empirical formula = CH₂O

Problem 3: Combustion Analysis Problem

Question: Combustion of 2.50 g of a hydrocarbon produces 7.67 g of CO₂ and 3.14 g of H₂O. Find the empirical formula.

Solution:

Step 1: Find carbon from CO₂.

7.67 g CO₂ × (12.01 g C ÷ 44.01 g CO₂) = 2.10 g C

Step 2: Find hydrogen from H₂O.

3.14 g H₂O × (2.02 g H ÷ 18.02 g H₂O) = 0.352 g H

Step 3: Check for other elements. It's a hydrocarbon, so only C and H are present. Oxygen comes from the air.

Step 4: Convert to moles.

C: 2.10 g ÷ 12.01 g/mol = 0.175 mol
H: 0.352 g ÷ 1.01 g/mol = 0.348 mol

Step 5: Divide by smallest value.

C: 0.175 ÷ 0.175 = 1
H: 0.348 ÷ 0.175 = 2

Empirical formula = CH₂

Problem 4: Finding Molecular Formula with Combustion Data

Question: A compound containing only C, H, and O has a molar mass of 116 g/mol. Combustion of 0.150 g produces 0.396 g of CO₂ and 0.162 g of H₂O. Find the molecular formula.

Solution:

Step 1: Find mass of C.

0.396 g CO₂ × (12.01 ÷ 44.01) = 0.108 g C

Step 2: Find mass of H.

0.162 g H₂O × (2.02 ÷ 18.02) = 0.0182 g H

Step 3: Find mass of O.

Mass of sample = 0.150 g

Mass of O = 0.150 - 0.108 - 0.0182 = 0.0238 g O

Step 4: Convert to moles.

C: 0.108 ÷ 12.01 = 0.00900 mol
H: 0.0182 ÷ 1.01 = 0.0180 mol
O: 0.0238 ÷ 16.00 = 0.00149 mol

Step 5: Divide by smallest.

C: 0.00900 ÷ 0.00149 = 6.04 ≈ 6
H: 0.0180 ÷ 0.00149 = 12.1 ≈ 12
O: 0.00149 ÷ 0.00149 = 1

Empirical formula = C₆H₁₂O

Step 6: Find empirical formula mass.

C₆H₁₂O = (6 × 12.01) + (12 × 1.01) + 16.00 = 100.14 g/mol

Step 7: Calculate n.

n = 116 ÷ 100.14 = 1.16 ≈ 1

Molecular formula = C₆H₁₂O

Common Mistakes That Will Cost You Points

Forgetting to divide by the smallest number of moles when finding the empirical formula. Always do this step.
Using atomic masses wrong. Memorize the common ones or keep a reliable table. Approximations fail on closer problems.
Not checking your work. Multiply your empirical formula by n and see if it matches the given molecular mass. It should.
Confusing empirical and molecular formulas. The empirical formula is the simplified ratio. The molecular formula is the actual molecule.
Rounding too early. Keep extra significant figures in calculations. Round only at the end.

Quick Reference: Molecular vs Empirical Formulas

Compound	Empirical Formula	Molecular Formula
Glucose	CH₂O	C₆H₁₂O₆
Benzene	CH	C₆H₆
Hydrogen peroxide	HO	H₂O₂
Formaldehyde	CH₂O	CH₂O
Acetic acid	CH₂O	C₂H₄O₂

Tools and Resources for Practice

You don't need expensive textbooks. Here's what actually works:

Khan Academy — Free videos on formula determination. Solid basics.
ChemCollective virtual labs — Practice problems with immediate feedback.
Textbook end-of-chapter problems — Old editions are cheap. Work every single one.
Past exam papers — Your school or AP materials. These are gold.

Getting Started: Your Action Plan

If you're struggling with molecular formulas, here's what to do:

Master the mole concept first. You can't solve formula problems without understanding moles. This is non-negotiable.
Memorize common atomic masses. C = 12, H = 1, O = 16, N = 14. You'll use these constantly.
Practice empirical formula calculation until you can do it without thinking. Percent composition → moles → divide by smallest → empirical formula.
Learn the n calculation backwards and forwards. Molecular mass ÷ empirical mass = n.
Work through 20+ practice problems. Reading solutions isn't the same as solving them yourself.

The Bottom Line

Molecular formula problems follow a predictable pattern. Calculate moles, find ratios, determine n, multiply. That's the entire process.

The only way to get faster is practice. Stop watching videos and start solving problems.