Chemistry • Year 11 • Module 2 • Lesson 3

Empirical and Molecular Formulas

Apply the four-step method and percentage composition calculations to worked scenarios, data tables, and real-compound contexts.

Apply · Data & Reasoning

1. Complete the worked solution — aspirin

Aspirin (acetylsalicylic acid) contains 60.00% carbon, 4.44% hydrogen, and 35.56% oxygen by mass. Its molar mass is 180.16 g mol⁻¹. The table below shows Steps 1–2 of the four-step method already completed. Complete Steps 3 and 4. (C = 12.011, H = 1.008, O = 15.999) 8 marks

Step	C	H	O
Step 1 — Assume 100 g: % → g	60.00 g	4.44 g	35.56 g
Step 2 — Convert to moles (n = m ÷ MM)	60.00 ÷ 12.011 = 4.996 mol	4.44 ÷ 1.008 = 4.405 mol	35.56 ÷ 15.999 = 2.223 mol
Step 3 — Divide by smallest (2.223); round to whole numbers
Empirical formula
Step 4 — MM(empirical formula) = ?
Multiplier n = MM(molecular) ÷ MM(empirical)
Molecular formula

Stuck? Revisit Worked Example 1 (empirical formula) and Worked Example 2 (molecular formula) in the lesson. The smallest mole value is 2.223.

2. Classify and complete — five compounds

The table below lists five compounds with their molecular formulas. Complete the empirical formula column, then tick whether the empirical and molecular formulas are the same or different. 10 marks (1 EF + 1 same/different each)

Compound	Molecular formula	Empirical formula	Same as MF?
Water	H₂O
Hydrogen peroxide	H₂O₂
Benzene	C₆H₆
Butane	C₄H₁₀
Phosphorus pentoxide	P₄O₁₀

2.1 For which of the five compounds above is the empirical formula identical to the molecular formula? Explain, in one sentence, what this tells you about the multiplier n for those compounds. 2 marks

Stuck? Revisit the Empirical vs Molecular Formula comparison table in the lesson. Divide each formula by the HCF of the subscripts.

3. Calculate percentage composition — ammonium nitrate

Ammonium nitrate, NH₄NO₃, is used as a fertiliser in Australian agriculture. Calculate the percentage by mass of each element. Show full working, including the molar mass calculation. Check that your percentages sum to 100 %. (N = 14.007, H = 1.008, O = 15.999) 5 marks

3.1 MM(NH₄NO₃) =

3.2 % N =

3.3 % H =

3.4 % O =

3.5 Check: = 100% ✓

Stuck? Revisit the “% composition” worked example in the lesson. NH₄NO₃ contains 2 N, 4 H, and 3 O atoms.

4. Find the missing element — ethanol combustion scenario

A student performs combustion analysis on an unknown organic compound. The results show 52.17% carbon and 13.04% hydrogen by mass. The percentage of the remaining element is not directly stated. 7 marks

4.1 Calculate the percentage of the unknown element and identify which element it most likely is. 2 marks

4.2 Use the four-step method to determine the empirical formula of the compound. Show all working. (C = 12.011, H = 1.008, O = 15.999) 4 marks

4.3 The compound’s molar mass is 46.07 g mol⁻¹. Calculate the multiplier n and state the molecular formula. 1 mark

Stuck? Percentages must sum to 100%. If C and H are given, the remainder is most likely oxygen for an organic compound. Revisit the Common Mistakes box in the lesson.

5. Contrast two compounds with the same empirical formula

Glucose (C₆H₁₂O₆, molar mass 180.16 g mol⁻¹) and fructose (C₆H₁₂O₆, molar mass 180.16 g mol⁻¹) are both found in Australian honey. 5 marks

5.1 State the empirical formula shared by glucose and fructose. Show that MM(empirical) and the multiplier n are the same for both compounds. 3 marks

5.2 The lesson states that glucose and fructose are “completely different substances” despite sharing the same molecular formula. Using only what this lesson teaches about empirical and molecular formulas, explain what this tells us about the completeness of the molecular formula as a description of a compound. 2 marks

Stuck? Revisit the callout box “Same empirical formula, different compounds” in the lesson.

Answers — Do not peek before attempting

Q1 — Aspirin worked solution

Step 3 — Divide by smallest (2.223):

C: 4.996 ÷ 2.223 = 2.248 • H: 4.405 ÷ 2.223 = 1.981 ≈ 2 • O: 2.223 ÷ 2.223 = 1.000

The ratio C : H : O = 2.248 : 2 : 1. Since 2.248 is not close to a whole number, multiply all values by 4: C = 8.99 ≈ 9, H = 7.92 ≈ 8, O = 4.00 = 4. All are now whole numbers.

Empirical formula: C₉H₈O₄

Step 4: MM(C₉H₈O₄) = 9(12.011) + 8(1.008) + 4(15.999) = 108.099 + 8.064 + 63.996 = 180.159 g mol⁻¹.

n = 180.16 ÷ 180.159 ≈ 1. The molecular formula is the same as the empirical formula: Molecular formula = C₉H₈O₄.

Marking note: Award marks for consistent application of the 4-step method. The key insight is recognising that 2.248 requires multiplication by 4 (not rounding to 2) to give whole-number subscripts. Accept C₉H₈O₄ for both empirical and molecular formula.

Q2 — Five compounds table

Water H₂O → EF = H₂O, Same. • Hydrogen peroxide H₂O₂ → EF = HO, Different. • Benzene C₆H₆ → EF = CH, Different. • Butane C₄H₁₀ → EF = C₂H₅, Different. • Phosphorus pentoxide P₄O₁₀ → EF = P₂O₅, Different.

Q2.1: Only water has the same empirical and molecular formula [1]. This tells us that the multiplier n = 1 for water — the empirical formula is already the simplest possible representation of the actual molecule [1].

Q3 — Ammonium nitrate % composition

MM(NH₄NO₃) = 2(14.007) + 4(1.008) + 3(15.999) = 28.014 + 4.032 + 47.997 = 80.043 g mol⁻¹ [1]

% N = 28.014 ÷ 80.043 × 100 = 35.00% [1]

% H = 4.032 ÷ 80.043 × 100 = 5.04% [1]

% O = 47.997 ÷ 80.043 × 100 = 59.96% [1]

Check: 35.00 + 5.04 + 59.96 = 100.00% ✓ [1]

Q4 — Ethanol combustion scenario

4.1: % unknown = 100 − 52.17 − 13.04 = 34.78%. For an organic compound (containing C and H), the third element is most likely oxygen [1 for calculation + 1 for identification].

4.2 (four-step method):

Step 1: C = 52.17 g, H = 13.04 g, O = 34.78 g [1]

Step 2: n(C) = 52.17 ÷ 12.011 = 4.344; n(H) = 13.04 ÷ 1.008 = 12.937; n(O) = 34.78 ÷ 15.999 = 2.174 [1]

Step 3: Divide by 2.174: C = 1.998 ≈ 2; H = 5.951 ≈ 6; O = 1.000 [1]. Empirical formula = C₂H₆O [1]

4.3: MM(C₂H₆O) = 2(12.011) + 6(1.008) + 15.999 = 24.022 + 6.048 + 15.999 = 46.069 g mol⁻¹. n = 46.07 ÷ 46.069 = 1.00 ≈ 1. Molecular formula = C₂H₆O (ethanol) [1].

Q5 — Glucose and fructose

5.1: C₆H₁₂O₆ → divide all subscripts by 6 (HCF) → empirical formula = CH₂O [1]. MM(CH₂O) = 12.011 + 2(1.008) + 15.999 = 30.026 g mol⁻¹. n = 180.16 ÷ 30.026 = 5.999 ≈ 6 for both glucose and fructose [1 for MM calculation + 1 for n].

5.2: The molecular formula identifies the number and type of atoms in a molecule, but does not uniquely identify a compound [1]. Glucose and fructose both have the molecular formula C₆H₁₂O₆ yet are completely different substances with different properties — this shows that knowing the molecular formula alone is not sufficient to identify a compound; additional information about the compound (such as physical properties or other experimental data) is needed [1].