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hscscience Chem · Y11
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Module 2 · L3 of 20 ~35 min ⚡ +50 XP in Learn · +25 to complete

Empirical & Molecular Formulas

In the 1800s, chemists could burn a compound and weigh the products, but they had no way to know if glucose (C₆H₁₂O₆) and acetic acid (CH₂O, scaled up) were the same substance or different ones. The empirical formula brought order to that chaos.

Today's hook, Glucose (C₆H₁₂O₆) and formaldehyde (CH₂O) share the same atom ratio but are completely different substances. So what does percentage composition tell us, and what doesn't it?
0/5QUESTS
Start here

Warm up and orient yourself

Recall earlier work, make a prediction, then scan the formulas you are about to use.

Warm up first

Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.

Worksheets

Practise this lesson

Four printable worksheets that build from the foundations up to exam-style questions, start at whatever level suits you.

Build Foundations & guided practice Apply Application practice Master Mastery challenge Exam Exam-style questions
01
Recall, your gut answer first
+5 XP warm-up

Glucose (C₆H₁₂O₆) and formaldehyde (CH₂O) share the same ratio of atoms but are completely different substances. If a chemist measured the percentage by mass of each element in an unknown compound, what would that tell them, and what wouldn't it tell them?

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02
Formula reference · this lesson
core formulas

$$\%\ \text{composition} = \dfrac{\text{mass of element}}{MM_{\text{compound}}} \times 100$$

$n = \%\ \div\ MM_{\text{element}}$  (treat % as grams in a 100 g sample)

whole-number ratio = divide each n by smallest n

molecular formula = empirical formula × n  where $n = MM_{\text{molecular}} \div MM_{\text{empirical}}$

Key insight: the empirical formula is the simplest whole-number ratio; the molecular formula is the actual count, always a whole-number multiple of the empirical formula.

What you'll master, and the words for it

03
What you'll master
Know

Key facts

  • Definition of empirical formula
  • Definition of molecular formula
  • That molecular formula = empirical × n
Understand

Concepts

  • Why two compounds can share an empirical formula
  • What percentage composition tells us
  • How experimental data leads to a formula
Can do

Skills

  • Calculate % composition from a molecular formula
  • Derive empirical formula from % composition data
  • Find molecular formula given empirical formula + MM
04
Key terms
Empirical formula
The simplest whole-number ratio of atoms of each element in a compound (e.g. CH₂O for glucose).
Molecular formula
The actual number of atoms of each element in one molecule (e.g. C₆H₁₂O₆ for glucose).
Whole-number ratio
Dividing all atom counts by the highest common factor to get the smallest integer set.
Multiplier (n)
The integer that converts empirical formula to molecular formula: molecular formula = n × empirical formula.
Combustion analysis
An experimental technique that burns a compound in excess O₂ and measures CO₂ and H₂O produced to find C and H percentages.
Percentage composition
The mass of each element expressed as a percentage of the total molar mass.

Empirical formula vs molecular formula

05
Empirical vs molecular formula
core concept · +3 XP at end
Beyond the syllabus. Empirical formula is the Module 2 endpoint. The syllabus asks you to determine percentage composition and empirical formula from experimental data, and that is what will be examined. Molecular formula is the step after it, supporting rather than separately prescribed, so it will not be tested in its own right. Learn it anyway: it is one line, the multiplier n = M(molecular) ÷ M(empirical), and it is the check that tells you your empirical formula was right. Put your practice into getting the ratio out of the percentages; the multiplication at the end is the easy half.

A chemical formula describes the composition of a substance. There are two types you need to know for this course, and the distinction between them matters in both theory and experiment.

Empirical formula, CH₂O
The simplest whole-number ratio of atoms in the compound. Cannot be simplified further. Derived from experimental data.
Molecular formula, C₆H₁₂O₆
The actual number of each atom in one molecule. Always a whole-number multiple of the empirical formula.
Same empirical formula, different compounds: Glucose (C₆H₁₂O₆), fructose (C₆H₁₂O₆), and acetic acid (C₂H₄O₂) all share the empirical formula CH₂O. This is why you need the molar mass to determine the molecular formula, empirical data alone isn't enough.

Percentage composition

Percentage composition tells you what fraction (by mass) each element contributes to a compound. You can calculate it from the molecular formula, or derive the molecular formula from it.

% of element X = (number of X atoms × M(X)) ÷ MM(compound) × 100

Example, % C in CO₂: MM(CO₂) = 44.009 g mol⁻¹
% C = (1 × 12.011) ÷ 44.009 × 100 = 27.29%
% O = (2 × 15.999) ÷ 44.009 × 100 = 72.71%
Check: 27.29 + 72.71 = 100% ✓

The empirical formula is the simplest whole-number ratio of atoms (e.g. CH₂O); the molecular formula is the actual count per molecule (e.g. C₆H₁₂O₆). Multiple compounds can share an empirical formula, molar mass is required to distinguish them. Percentage composition = (n × Melement) ÷ MMcompound × 100; all percentages must sum to 100%.

Pause, copy the highlighted definition into your book before moving on.

Did you get this? True or false: glucose (C₆H₁₂O₆) and formaldehyde (CH₂O) have the same empirical formula but different molecular formulas.

Quick check: A compound is reported as 40% C and 6.7% H. What must you do next?

The four-step method, percentages to formula

06
Method: from % data to molecular formula
core concept

We just saw that empirical and molecular formulas differ, and that percentage composition links them. That raises a question: given only % composition data, how do you systematically derive the empirical or molecular formula? This card answers it → with a four-step method that works for every HSC problem.

This is a four-step method. Every empirical/molecular formula problem in the HSC follows this sequence. Memorise the steps and the logic behind each one.

Four steps from percentage composition to an empirical and molecular formula, with a worked example.
When you get a ratio like 1 : 1.5: This is not a whole number, multiply everything by 2 to get 2 : 3. Common non-integer results and their fixes: ×0.5 → multiply by 2 | ×0.33 → multiply by 3 | ×0.25 → multiply by 4.

Four-step method: (1) assume 100 g so % = grams; (2) convert each element's mass to moles using n = m ÷ MM; (3) divide all by the smallest mole value to get the empirical formula; (4) if MM is given, find the multiplier n = MM(molecular) ÷ MM(empirical), must be a whole number.

Add the highlighted method to your notes before the check below.

Fill the blanks: drag each token into the matching blank.

100 g moles smallest MM

Step 1: assume a ___ sample. Step 2: convert each element to ___. Step 3: divide all by the ___ value. Step 4: scale using ___ if given.

Did you get this? True or false: the multiplier n that converts an empirical formula to a molecular formula must always be a whole number.

Odd one out three of these compounds share the empirical formula CH₂O. Which one does not?

Two worked examples

Worked examples · reveal as you go

Worked example 1 · empirical formula from % composition +5 XP on full reveal

A compound contains 40.00% carbon, 6.72% hydrogen, and 53.28% oxygen by mass. Determine its empirical formula. (C = 12.011, H = 1.008, O = 15.999)

1
Assume 100 g sample: $\mathrm{C}: 40.00\ \mathrm{g}$ · $\mathrm{H}: 6.72\ \mathrm{g}$ · $\mathrm{O}: 53.28\ \mathrm{g}$
% becomes g directly, cleanest entry into the calculation
2
$n(\mathrm{C}) = 40.00 \div 12.011 = 3.330\ \mathrm{mol}$
$n(\mathrm{H}) = 6.72 \div 1.008 = 6.667\ \mathrm{mol}$
$n(\mathrm{O}) = 53.28 \div 15.999 = 3.330\ \mathrm{mol}$
Convert to moles, atoms have different masses so equal masses ≠ equal atoms
3
Divide by smallest ($3.330$):
$\mathrm{C}: 1.000$ · $\mathrm{H}: 2.002 \approx 2$ · $\mathrm{O}: 1.000$
Dividing scales so the minimum becomes 1; round sensibly
4
Ratio $\mathrm{C} : \mathrm{H} : \mathrm{O} = 1 : 2 : 1$
Already whole numbers, no multiplication needed
5
Empirical formula: $\mathrm{CH_2O}$ ✓
Final answer, simplest whole-number ratio
Step 1 of 5
Worked example 2 · molecular formula from empirical + MM +5 XP on full reveal

A compound has the empirical formula CH₂O and a molar mass of 180.16 g mol⁻¹. Determine its molecular formula.

1
$\mathrm{MM(CH_2O)} = 12.011 + 2(1.008) + 15.999 = 30.026\ \mathrm{g}\ \mathrm{mol}^{-1}$
Calculate MM of the empirical unit first
2
$n = MM_{\text{molecular}} \div MM_{\text{empirical}} = 180.16 \div 30.026 = 5.999 \approx 6$
Multiplier n must be a whole number, atoms can't be fractional
3
Multiply each subscript by 6: $\mathrm{C}: 1 \times 6 = 6$ · $\mathrm{H}: 2 \times 6 = 12$ · $\mathrm{O}: 1 \times 6 = 6$
Every atom gets the same factor, ratio stays, scale changes
4
Molecular formula: $\mathrm{C_6H_{12}O_6}$ (glucose) ✓
Final answer, actual atom count per molecule
Step 1 of 4
Sort the steps+7 XP

A compound is 40.00% C, 6.72% H, 53.28% O. Put the empirical-formula steps in order.

  • Divide each by the smallest mole value (3.330): C = 1.000, H = 2.002, O = 1.000.
  • Assume a 100 g sample so each percentage becomes grams: 40.00 g C, 6.72 g H, 53.28 g O.
  • Write the empirical formula from the whole-number ratio: CH₂O.
  • Convert each mass to moles using n = m ÷ MM: n(C) = 3.330, n(H) = 6.667, n(O) = 3.330.

The three traps, then quick-fire drill

Common errors · the 3 traps that cost marks

1

Not rounding to a whole number after dividing

After dividing by the smallest mole value, you might get 1.999 or 3.002. These are whole numbers affected by rounding in the molar masses, write them as 2 and 3 respectively. Only multiply by an integer (×2, ×3, ×4) if the result is genuinely non-integer, like 1.5 or 1.33.

Fix: Round values within 0.05 of a whole number. For results like 1.5, 1.33, or 1.25, multiply through instead.

2

Confusing empirical and molecular formula in the answer

If the question asks for the empirical formula, give the simplified ratio. If it asks for the molecular formula, give the actual formula with the multiplier applied. Writing CH₂O when the question wants C₆H₁₂O₆ (or vice versa) will not score marks even if your method was correct.

Fix: Underline the word "empirical" or "molecular" in the question before you start, answer exactly what was asked.

3

Forgetting to check percentages sum to 100%

If a question gives you only two of three elements' percentages, you must calculate the third by subtraction (100 − given percentages). If you ignore this, you'll miss an entire element from the formula.

Fix: Before starting, add all given percentages. If they don't total 100%, the remainder is another element, usually oxygen.

Quick-fire practice · 5 reps +2 XP per reveal

1

A compound contains 52.17% C, 13.04% H, and 34.78% O by mass. Find the empirical formula.

2

If the compound in Q1 has molar mass 46.07 g mol⁻¹, what is its molecular formula?

3

Calculate the % composition by mass for water (H₂O). (H = 1.008, O = 15.999)

4

A compound is 43.64% P and 56.36% O. Its molar mass is 283.9 g mol⁻¹. Find the molecular formula.

5

A compound has the empirical formula CH₃. Its molar mass is 30.07 g mol⁻¹. Find the molecular formula.

Revisit your thinking

12
Revisit your thinking

Earlier you were asked: If a chemist measured the percentage by mass of each element in an unknown compound, what would that tell them, and what wouldn't it tell them?

Percentage composition gives you the empirical formula, the simplest whole-number ratio of atoms. But it cannot tell you the molecular formula (the actual number of atoms per molecule) unless you also know the molar mass. Glucose and acetic acid have the same empirical formula (CH₂O) but are completely different substances.

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Interactive Tool, Empirical Formula Worker Open fullscreen ↗

Try this: keep the first compound (40.00% C, 6.71% H, 53.29% O), press Show atom share and compare the two bars for hydrogen. Then drag the sample mass off 100 g and watch which column changes. Type the empirical formula and check it, then open step 4 and try each molar mass.

Use the Empirical Formula Worker with the first compound (40.00% C, 6.71% H, 53.29% O) and press Show atom share. Hydrogen is 6.7% of the mass. What share of the atoms is hydrogen?
Do this next

Practise independently

Attempt at least one calculation or explanation in your own words before checking the model answers.

01
Multiple choice
+2 XP per correct · +5 bonus if perfect

Pick your answer, then rate your confidence. That tells the system what to drill next.

02
Short answer
UnderstandLower-order3 marks

Q1. Define the term empirical formula and explain why two compounds with the same empirical formula are not necessarily the same substance. Use an example in your answer.

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ApplyMid-order5 marks

Q2. A compound is found to contain 43.64% phosphorus and 56.36% oxygen by mass. Its molar mass is 283.9 g mol⁻¹. Determine the molecular formula of the compound. Show all working. (P = 30.974, O = 15.999)

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ApplyLower-order4 marks

Q3. Calculate the percentage by mass of each element in ammonium sulfate, (NH₄)₂SO₄. Show all working, including the molar mass calculation. Check that your percentages sum to 100%. (N = 14.007, H = 1.008, S = 32.06, O = 15.999)

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EvaluateHigher-order5 marks

Q4. A student analyses an unknown compound and reports the following: "The compound contains 52.14% C, 13.13% H, and 34.73% O. I calculated the empirical formula as CH₃O." Evaluate this claim. Show full working to verify or correct the empirical formula, and explain what additional information would be needed to determine the molecular formula. (C = 12.011, H = 1.008, O = 15.999)

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CreateHigher-order6 marks

Q5. An organic chemist isolates a new compound from a plant extract. Combustion analysis gives 54.55% C and 9.09% H; the remainder is assumed to be oxygen. Spectroscopic data suggest the molar mass is approximately 88 g mol⁻¹. (a) Determine the empirical formula. (b) Determine the molecular formula. (c) Propose a structural formula (name or draw) consistent with the molecular formula and suggest a possible functional group. (C = 12.011, H = 1.008, O = 15.999)

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📖 Comprehensive answers (click to reveal)

Multiple choice, drill bank

Empirical vs molecular formula. Empirical = simplest ratio; molecular = actual count per molecule.

Reducing a molecular formula to its empirical formula. C₄H₈O₂: divide all subscripts by 4 (HCF = 4) → CH₂O.

Empirical formula from percentage composition. n(C) = 75.0÷12.011 = 6.244; n(H) = 25.0÷1.008 = 24.802. Ratio H:C ≈ 4. Empirical formula: CH₄.

Molecular formula from the empirical formula and MM. MM(CH₃) = 15.035. n = 30.07÷15.035 = 2. Molecular formula = C₂H₆.

Clearing fractional subscripts. Multiply all by 2: C = 2, H = 3, O = 1 → C₂H₃O.

Short answer model answers

Q1 (3 marks): The empirical formula shows the simplest whole-number ratio of atoms of each element in a compound [1]. Two compounds can have the same empirical formula but different molecular formulas because the molecular formula is a whole-number multiple of the empirical formula, multiple different multiples are possible [1]. For example, glucose (C₆H₁₂O₆) and acetic acid (C₂H₄O₂) both have the empirical formula CH₂O, but they are completely different substances [1].

Q2 (5 marks):

n(P) = 43.64÷30.974 = 1.409; n(O) = 56.36÷15.999 = 3.523
Divide by 1.409: P = 1.000, O = 2.500 → ×2: P = 2, O = 5 → EF = P₂O₅
MM(P₂O₅) = 2(30.974) + 5(15.999) = 141.943 g mol⁻¹
n = 283.9÷141.943 = 1.999 ≈ 2
Molecular formula: P₄O₁₀

Q3 (4 marks):

MM((NH₄)₂SO₄) = 2(14.007) + 8(1.008) + 32.06 + 4(15.999) = 28.014 + 8.064 + 32.06 + 63.996 = 132.134 g mol⁻¹
% N = 28.014÷132.134 × 100 = 21.20%
% H = 8.064÷132.134 × 100 = 6.10%
% S = 32.06÷132.134 × 100 = 24.26%
% O = 63.996÷132.134 × 100 = 48.43%
Check: 21.20 + 6.10 + 24.26 + 48.43 = 99.99% ≈ 100% ✓

Q4 (5 marks): n(C) = 52.14÷12.011 = 4.341; n(H) = 13.13÷1.008 = 13.025; n(O) = 34.73÷15.999 = 2.171. Divide by 2.171 → C: 2.00, H: 6.00, O: 1.00. Empirical formula = C₂H₆O. The student's claim of CH₃O is incorrect. To find the molecular formula, the molar mass would be needed; MM(C₂H₆O) = 46.07 g mol⁻¹.

Q5 (6 marks): (a) % O = 100 − 54.55 − 9.09 = 36.36%. n(C) = 54.55÷12.011 = 4.542; n(H) = 9.09÷1.008 = 9.018; n(O) = 36.36÷15.999 = 2.273. Divide by 2.273: C = 2, H = 4, O = 1 → EF = C₂H₄O. (b) MM(C₂H₄O) = 44.053 g mol⁻¹. n = 88÷44.053 ≈ 2. Molecular formula = C₄H₈O₂. (c) C₄H₈O₂ could be an ester (e.g. methyl propanoate or ethyl acetate) or a carboxylic acid (e.g. butanoic acid). Functional group: ester (–COO–) or carboxylic acid (–COOH).

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