Module 2: Introduction to Quantitative Chemistry — Worked Solutions (Preliminary Chemistry)
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Worked examples for Preliminary Chemistry Module 2: Introduction to Quantitative Chemistry. Each shows where the marks are awarded, the key idea, and the full solution explained by your choice of tutor — Stella, Ella or Cassie.
In short: full step-by-step worked solutions for Preliminary Chemistry — Module 2: Introduction to Quantitative Chemistry. Every question is worked through with the method and reasoning shown, so you can check how to get the answer, not just the final result.
How to use these
Try each question first, then check your working. Use the tutor tabs to read the full solution in the style that suits you: Stella is direct and challenging, Ella is warm and explains the why, and Cassie is concise and analytical.
Example 1 — Empirical formula from percentage composition
Question
A compound is found by analysis to contain 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Determine the empirical formula of the compound. ($M(\text{C}) = 12.01$, $M(\text{H}) = 1.008$, $M(\text{O}) = 16.00\ \text{g mol}^{-1}$.)
Solution
Assume 100 g, so the percentages become grams. Convert each to moles with $n = m / M$.
- C: $40.0 / 12.01 = 3.331\ \text{mol}$
- H: $6.7 / 1.008 = 6.647\ \text{mol}$
- O: $53.3 / 16.00 = 3.331\ \text{mol}$
Divide through by the smallest (3.331):
- C: $1.00$, H: $2.00$, O: $1.00$
The ratio is $1:2:1$, so the empirical formula is $\text{CH}_2\text{O}$.
Don't round mid-calculation — carry the decimals through and only round the final ratio to whole numbers.
Empirical formula is the simplest whole-number ratio of atoms, and atoms are counted in moles, not grams — so our job is to get to moles.
The neat trick is to assume we have exactly 100 g of the compound. Then "40.0%" simply means 40.0 g, and so on. Now convert each mass to moles using $n = m / M$:
- C: $40.0 / 12.01 = 3.331\ \text{mol}$
- H: $6.7 / 1.008 = 6.647\ \text{mol}$
- O: $53.3 / 16.00 = 3.331\ \text{mol}$
To find the ratio, we divide every value by the smallest one (here 3.331), which scales the smallest to 1:
- C: $1.00$, H: $2.00$, O: $1.00$
That gives a clean $1:2:1$, so the empirical formula is $\text{CH}_2\text{O}$. The reason we divide by the smallest is that it turns the moles into a relative ratio we can read as atom counts.
Assume 100 g → % becomes grams. $n = m/M$:
- C: $40.0/12.01 = 3.331$
- H: $6.7/1.008 = 6.647$
- O: $53.3/16.00 = 3.331$
Divide by smallest ($3.331$):
- C $1.00$ : H $2.00$ : O $1.00$
Empirical formula: $\text{CH}_2\text{O}$.
Where the marks go
- 1 mark: Assumes a 100 g sample and converts each percentage to a mass
- 1 mark: Correctly calculates moles of each element using $n = m/M$
- 1 mark: Divides through by the smallest mole value to find the ratio
- 1 mark: States the correct empirical formula $\text{CH}_2\text{O}$
Key idea
Empirical formula = simplest whole-number mole ratio: assume 100 g, convert masses to moles with $n = m/M$, then divide by the smallest.
Example 2 — Percentage composition
Question
Calculate the percentage by mass of nitrogen in ammonium nitrate, $\text{NH}_4\text{NO}_3$. ($M(\text{N}) = 14.01$, $M(\text{H}) = 1.008$, $M(\text{O}) = 16.00\ \text{g mol}^{-1}$.)
Solution
Find the molar mass, then the mass contributed by nitrogen.
$M(\text{NH}_4\text{NO}_3) = 2(14.01) + 4(1.008) + 3(16.00) = 28.02 + 4.032 + 48.00 = 80.05\ \text{g mol}^{-1}$.
Nitrogen appears twice: mass of N $= 2 \times 14.01 = 28.02\ \text{g}$.
$\%\,\text{N} = \dfrac{28.02}{80.05} \times 100 = 35.0\%$.
Count all the nitrogen atoms — there are two in this formula, one from the ammonium and one from the nitrate. Miss one and you halve your answer.
Percentage composition asks: of the total mass of the compound, what fraction is nitrogen? So we need the mass of nitrogen and the mass of the whole formula unit.
First the molar mass. Counting every atom in $\text{NH}_4\text{NO}_3$: two N, four H, three O.
$M = 2(14.01) + 4(1.008) + 3(16.00) = 28.02 + 4.032 + 48.00 = 80.05\ \text{g mol}^{-1}$.
The nitrogen contribution is the two N atoms: $2 \times 14.01 = 28.02\ \text{g}$.
Now the fraction, as a percentage:
$\%\,\text{N} = \dfrac{28.02}{80.05} \times 100 = 35.0\%$.
The key thing is to include both nitrogen atoms — it's easy to spot the N at the front and forget the one inside the nitrate group.
Molar mass of $\text{NH}_4\text{NO}_3$:
- $2(14.01) + 4(1.008) + 3(16.00) = 80.05\ \text{g mol}^{-1}$
Mass of N (2 atoms): $2 \times 14.01 = 28.02\ \text{g}$.
$\%\,\text{N} = \dfrac{28.02}{80.05} \times 100 = 35.0\%$.
Where the marks go
- 1 mark: Correct molar mass of $\text{NH}_4\text{NO}_3$ ($80.05\ \text{g mol}^{-1}$)
- 1 mark: Correct total mass of nitrogen ($2 \times 14.01 = 28.02\ \text{g}$)
- 1 mark: Correct percentage $35.0\%$ to appropriate significant figures
Key idea
Percentage by mass $= \dfrac{\text{mass of element in one mole}}{\text{molar mass of compound}} \times 100$ — count every atom of the element in the formula.
Frequently asked questions
Step-by-step solutions to Introduction to Quantitative Chemistry questions in Preliminary Chemistry, with the full method shown for each — so you can follow the reasoning, not just the final answer.
Attempt each question yourself first, then compare your method — not just your answer — against the worked solution. The aim is to learn the approach so you can handle unfamiliar questions, rather than memorising answers.
Yes — they follow the NESA syllabus for Preliminary Chemistry, using the methods and notation expected in exams and assessments.