Chemistry • Year 11 • Module 2 • Lesson 3
Empirical and Molecular Formulas
Lock in the core vocabulary, the four-step method, and the relationship between empirical and molecular formulas before tackling harder questions.
1. Term–definition match
The definitions below are shuffled. In the right-hand column write the matching term from this list: empirical formula, molecular formula, percentage composition, multiplier (n), combustion analysis, whole-number ratio. 6 marks (1 each)
| # | Definition | Matching term |
|---|---|---|
| 1.1 | The simplest whole-number ratio of atoms of each element in a compound; it cannot be simplified further. | |
| 1.2 | The actual number of atoms of each element in one molecule; always a whole-number multiple of the empirical formula. | |
| 1.3 | The mass of each element expressed as a percentage of the total molar mass of the compound. | |
| 1.4 | The integer that converts the empirical formula into the molecular formula: MF = n × EF. | |
| 1.5 | An experimental technique that burns a compound in excess O2 and measures the masses of CO2 and H2O produced to find the percentages of carbon and hydrogen. | |
| 1.6 | Dividing all atom counts (or mole values) by the highest common factor to obtain the smallest set of integers. |
2. True or false — with correction
Circle T or F for each statement. If the statement is false, write the corrected version on the line below it. 12 marks (1 T/F + 1 correction each)
2.1 The empirical formula always shows the actual number of atoms in one molecule. T / F
2.2 Glucose (C6H12O6) and acetic acid (C2H4O2) share the same empirical formula. T / F
2.3 Percentage composition data alone is sufficient to determine the molecular formula of a compound. T / F
2.4 In Step 1 of the four-step method, you assume a 100 g sample so that the percentage value becomes a mass in grams directly. T / F
2.5 The multiplier n = MM(molecular) ÷ MM(empirical) must always be a whole number. T / F
2.6 If dividing mole values by the smallest gives a ratio of 1.5 : 1 : 0.5, the empirical formula subscripts are obtained by multiplying all values by 3. T / F
3. Fill-in-the-blank paragraph
Use the word bank to complete the passage. Each word is used once. 8 marks (1 per blank)
Word bank:
empirical · molecular · molar mass · moles · multiplier · percentage composition · ratio · smallest
The ___________ formula gives the simplest whole-number ratio of atoms in a compound, while the ___________ formula gives the actual number of atoms per molecule. To find the empirical formula from experimental data, a chemist first calculates the ___________ of each element in the compound. Assuming a 100 g sample converts each percentage directly into a mass, which is then divided by the element's ___________ to find the number of ___________. Dividing each mole value by the ___________ mole value produces the whole-number ___________ that defines the empirical formula. To convert the empirical formula to the molecular formula, the ___________ n = MM(molecular) ÷ MM(empirical) is applied.
4. Function recall
Answer each question in 1–2 sentences using precise terms from the lesson. 8 marks (2 each)
4.1 Why is it mathematically valid to assume a 100 g sample in Step 1 of the empirical formula method?
4.2 Why must the multiplier n always be a whole number?
4.3 What additional piece of information (beyond percentage composition) is needed to determine the molecular formula of a compound?
4.4 After dividing all mole values by the smallest, a student obtains C : H : O = 1 : 1.33 : 0.5. What integer should they multiply all values by to obtain whole-number subscripts?
5. Sequence and label the four-step method
The four steps of the empirical/molecular formula method are listed out of order below. Write the correct step number (1–4) beside each step, then write a one-sentence explanation of why that step is performed. 8 marks (1 number + 1 reason each)
| Step # | Action | Why this step is done |
|---|---|---|
| Divide all mole values by the smallest mole value to get the whole-number ratio. | ||
| If the molar mass is given, calculate n = MM(molecular) ÷ MM(empirical) and multiply all subscripts by n. | ||
| Assume a 100 g sample — treat each percentage directly as a mass in grams. | ||
| Convert each elemental mass to moles using n = m ÷ MM. |
Q1 — Term–definition match
1.1 empirical formula • 1.2 molecular formula • 1.3 percentage composition • 1.4 multiplier (n) • 1.5 combustion analysis • 1.6 whole-number ratio.
Q2 — True / false with correction
2.1 False. The empirical formula shows the simplest whole-number ratio of atoms, not the actual number per molecule. The molecular formula shows the actual atom count.
2.2 True. Both C6H12O6 and C2H4O2 reduce to CH2O (divide glucose by 6, acetic acid by 2).
2.3 False. Percentage composition gives the empirical formula only. The molecular molar mass is also needed to calculate the multiplier n and obtain the molecular formula.
2.4 True.
2.5 True. A molecule contains a discrete whole number of atoms; fractions of an atom do not exist.
2.6 False. To convert 1.5 : 1 : 0.5 to whole numbers, multiply all by 2 (giving 3 : 2 : 1), not by 3.
Q3 — Cloze paragraph
In order: empirical / molecular / percentage composition / molar mass / moles / smallest / ratio / multiplier.
Q4.1 — Why assume 100 g?
Assuming 100 g converts each percentage value directly into a mass in grams without any additional calculation (e.g. 40% carbon becomes 40 g carbon). This works because the empirical formula depends only on the ratio of moles, which is independent of the total sample mass.
Q4.2 — Why must n be a whole number?
Molecules contain discrete, whole numbers of atoms — you cannot have half an atom. The multiplier n represents how many empirical formula “units” make up one molecule, so it must be a positive integer. Values like 5.999 are rounded to 6 due to rounding in atomic masses.
Q4.3 — Additional information needed
The molar mass (molecular mass) of the compound is needed. It allows the calculation of n = MM(molecular) ÷ MM(empirical formula), which then converts the empirical formula to the molecular formula.
Q4.4 — Correct multiplier for 1 : 1.33 : 0.5
Multiply all values by 6: 1×6 = 6, 1.33×6 ≈ 8, 0.5×6 = 3 → C6H8O3. (Note: 1.33 ≈ 4/3, so ×3 gives 4/3×3 = 4 and 0.5×3 = 1.5 — still not whole. Try ×6: 1.33×6 ≈ 8, 0.5×6 = 3 — whole numbers. Accept ×6.)
Q5 — Four-step method sequence and reasons
| Correct Step # | Action | Why |
|---|---|---|
| 3 | Divide all mole values by the smallest | Scaling to the smallest value gives the simplest integer ratio — the empirical formula. |
| 4 | Scale to molecular formula using n | Empirical formula alone cannot tell us the actual atom count; n bridges the two formulas. |
| 1 | Assume a 100 g sample | Converts percentages to gram masses directly, simplifying all subsequent division. |
| 2 | Convert grams to moles using n = m ÷ MM | Atoms have different masses, so a mole ratio (not a mass ratio) is needed for the empirical formula. |