Empirical Formula and Composition: HSC Chemistry Practice Paper
Test your empirical formula and percentage composition skills with this original 40-mark timed HSC Chemistry practice paper. Includes complete worked solutions and a detailed marking guide.
This original tutorgum HSC Chemistry practice paper focuses on percentage composition and empirical formula calculations. It tests your ability to convert mass data into mole ratios, work with non-integer ratios, calculate percentage composition, and use an independently supplied molar mass to determine a molecular formula.
The paper is designed for focused revision rather than as a full-course examination. All questions are original tutorgum practice questions. This is not an official, endorsed, past, or predicted NESA paper.
01Exam instructions
| Assessment detail | Conditions |
|---|---|
| Suggested reading time | 5 minutes |
| Suggested working time | 50 minutes |
| Total marks | 40 |
| Questions | 6 |
| Equipment | Scientific calculator, writing equipment, and periodic table |
These are suggested conditions for this practice paper, not official examination rules.
Assessed topics
- Calculating percentage composition by mass from a chemical formula.
- Deriving empirical formulae from percentage composition.
- Deriving empirical formulae from measured masses of constituent elements.
- Converting non-integer mole ratios into the smallest whole-number ratio.
- Using an independently supplied molar mass to determine a molecular formula.
Excluded content
This paper does not assess reaction stoichiometry, concentration calculations, limiting reagents, chemical equilibrium, bonding theory, or organic reaction pathways.
Data and assumptions
Use the following relative atomic masses throughout the paper.
| Element | Relative atomic mass, \(A_r\) |
|---|---|
| H | 1.0 |
| C | 12.0 |
| N | 14.0 |
| O | 16.0 |
| Mg | 24.3 |
| Fe | 55.8 |
| Cu | 63.5 |
- All percentage compositions are percentages by mass.
- Assume all samples are pure compounds.
- Use the supplied relative atomic masses rather than alternative rounded values.
- Show your working, including mole calculations and ratios.
- Unless otherwise stated, give numerical answers to 3 significant figures.
- Empirical formulae must use the smallest possible whole-number subscripts.
All questions are compulsory.
02Questions
Question 1 (6 marks)
An unknown compound contains 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass.
(a) Assume you have a 100.0 g sample of the compound. Calculate the amount, in moles, of each element present. (3 marks)
(b) Determine the empirical formula of the compound. Show how you obtain the simplest whole-number mole ratio. (3 marks)
Question 2 (6 marks)
A sample of copper oxide contains 6.35 g of copper and 0.80 g of oxygen.
(a) Calculate the amount, in moles, of copper and oxygen in the sample. (2 marks)
(b) Determine the simplest whole-number mole ratio of copper to oxygen. (2 marks)
(c) Write the empirical formula of the copper oxide. Explain why the mass ratio of copper to oxygen cannot be used directly as the formula subscripts. (2 marks)
Question 3 (6 marks)
An iron oxide contains 69.9% iron and 30.1% oxygen by mass.
(a) Assume you have a 100.0 g sample. Calculate the amount, in moles, of iron and oxygen present. (2 marks)
(b) Determine the simplest mole ratio of iron to oxygen. (2 marks)
(c) Convert your ratio into the smallest whole-number ratio and state the empirical formula of the iron oxide. Show your working. (2 marks)
Question 4 (7 marks)
A 4.30 g sample of an unknown compound contains 2.40 g of carbon, 0.50 g of hydrogen, and 1.40 g of nitrogen.
(a) Calculate the amount, in moles, of each element in the sample. (3 marks)
(b) Determine the simplest whole-number mole ratio of carbon, hydrogen, and nitrogen. (2 marks)
(c) Write the empirical formula of the compound. (1 mark)
(d) Calculate the percentage composition by mass of nitrogen in the compound. (1 mark)
Question 5 (7 marks)
Magnesium nitrate has the chemical formula \(\ce{Mg(NO3)2}\).
(a) Calculate the molar mass of magnesium nitrate. (2 marks)
(b) Calculate the percentage composition by mass of each element in magnesium nitrate:
- Magnesium
- Nitrogen
- Oxygen
Give each answer to 3 significant figures. (3 marks)
(c) Calculate the mass of oxygen present in a 29.66 g sample of pure magnesium nitrate. (2 marks)
Question 6 (8 marks)
An unknown hydrocarbon contains 85.7% carbon and 14.3% hydrogen by mass.
An independently measured molar mass for this compound is 56.0 g mol\(^{-1}\).
(a) Determine the empirical formula of the hydrocarbon. Show your mole calculations and simplest whole-number ratio. (4 marks)
(b) Calculate the empirical formula mass of the compound. (1 mark)
(c) Use the independently supplied molar mass to determine the number of empirical formula units in one molecule. Hence, determine the molecular formula of the hydrocarbon. (3 marks)
03Worked solutions and marking guide
Solution 1
(a) Calculate the amount of each element (3 marks)
The percentage composition is given by mass.
Using a 100.0 g sample means that each percentage can be treated directly as a mass in grams.
The amount of each element is calculated using:
\[ n=\frac{m}{M} \]
where:
- \(n\) is the amount in mol,
- \(m\) is the mass in g,
- \(M\) is the molar mass in g mol\(^{-1}\).
For carbon:
\[ n(\ce{C})=\frac{40.0}{12.0} \]
\[ =3.33\ \mathrm{mol} \]
For hydrogen:
\[ n(\ce{H})=\frac{6.7}{1.0} \]
\[ =6.7\ \mathrm{mol} \]
For oxygen:
\[ n(\ce{O})=\frac{53.3}{16.0} \]
\[ =3.33\ \mathrm{mol} \]
(b) Determine the empirical formula (3 marks)
The next step is to divide every mole amount by the smallest mole amount.
The smallest value is approximately 3.33 mol.
\[ \begin{aligned} \ce{C}:&\quad \frac{3.33}{3.33}=1\\ \ce{H}:&\quad \frac{6.7}{3.33}=2.01\\ \ce{O}:&\quad \frac{3.33}{3.33}=1 \end{aligned} \]
The simplest whole-number ratio is:
\[ \ce{C:H:O}=1:2:1 \]
Therefore, the empirical formula is:
\[ \boxed{\ce{CH2O}} \]
Marking guide
| Part | Required working | Marks |
|---|---|---|
| (a) | Correct amount of carbon | 1 |
| (a) | Correct amount of hydrogen | 1 |
| (a) | Correct amount of oxygen | 1 |
| (b) | Correct division by the smallest mole amount | 1 |
| (b) | Correct whole-number ratio, 1:2:1 | 1 |
| (b) | Correct empirical formula, \(\ce{CH2O}\) | 1 |
| Total | 6 |
Solution 2
(a) Calculate the amount of each element (2 marks)
Use:
\[ n=\frac{m}{M} \]
For copper:
\[ n(\ce{Cu})=\frac{6.35}{63.5} \]
\[ =0.100\ \mathrm{mol} \]
For oxygen:
\[ n(\ce{O})=\frac{0.80}{16.0} \]
\[ =0.0500\ \mathrm{mol} \]
(b) Determine the simplest mole ratio (2 marks)
Divide both mole amounts by the smallest value, 0.0500.
\[ \begin{aligned} \ce{Cu}:&\quad \frac{0.100}{0.0500}=2\\ \ce{O}:&\quad \frac{0.0500}{0.0500}=1 \end{aligned} \]
Therefore:
\[ \ce{Cu:O}=2:1 \]
(c) Write the empirical formula (2 marks)
The empirical formula uses the simplest whole-number mole ratio as its subscripts.
There are two copper atoms for every one oxygen atom.
Therefore:
\[ \boxed{\ce{Cu2O}} \]
The mass ratio cannot be used directly because empirical formula subscripts represent relative numbers of atoms, not their relative masses. Different elements have different molar masses, so their masses must first be converted into moles.
Marking guide
| Part | Required working | Marks |
|---|---|---|
| (a) | Correct amount of copper | 1 |
| (a) | Correct amount of oxygen | 1 |
| (b) | Correct division by the smallest mole amount | 1 |
| (b) | Correct ratio, 2:1 | 1 |
| (c) | Correct empirical formula, \(\ce{Cu2O}\) | 1 |
| (c) | Explains that subscripts represent atom ratios rather than mass ratios | 1 |
| Total | 6 |
Solution 3
(a) Calculate the amount of each element (2 marks)
Assume a 100.0 g sample.
This gives:
- Iron: 69.9 g
- Oxygen: 30.1 g
Calculate the number of moles of each element.
For iron:
\[ n(\ce{Fe})=\frac{69.9}{55.8} \]
\[ =1.25\ \mathrm{mol} \]
For oxygen:
\[ n(\ce{O})=\frac{30.1}{16.0} \]
\[ =1.88\ \mathrm{mol} \]
(b) Determine the simplest mole ratio (2 marks)
Divide both values by the smallest mole amount, approximately 1.25.
\[ \begin{aligned} \ce{Fe}:&\quad \frac{1.25}{1.25}=1\\ \ce{O}:&\quad \frac{1.88}{1.25}=1.50 \end{aligned} \]
The mole ratio is approximately:
\[ \ce{Fe:O}=1:1.50 \]
(c) Convert to a whole-number ratio and determine the empirical formula (2 marks)
This is an important step.
You cannot simply round 1.50 to 2. Doing so would change the relative proportions of the elements.
Instead, multiply both sides of the ratio by 2.
\[ 1:1.50 \]
\[ (1\times2):(1.50\times2) \]
\[ =2:3 \]
The smallest whole-number ratio is therefore:
\[ \ce{Fe:O}=2:3 \]
The empirical formula is:
\[ \boxed{\ce{Fe2O3}} \]
Marking guide
| Part | Required working | Marks |
|---|---|---|
| (a) | Correct amount of iron | 1 |
| (a) | Correct amount of oxygen | 1 |
| (b) | Correct division by the smallest mole amount | 1 |
| (b) | Correct ratio, approximately 1:1.50 | 1 |
| (c) | Correct conversion to 2:3 | 1 |
| (c) | Correct empirical formula, \(\ce{Fe2O3}\) | 1 |
| Total | 6 |
Solution 4
(a) Calculate the amount of each element (3 marks)
The sample contains:
- 2.40 g carbon
- 0.50 g hydrogen
- 1.40 g nitrogen
Calculate the number of moles of each element.
For carbon:
\[ n(\ce{C})=\frac{2.40}{12.0} \]
\[ =0.200\ \mathrm{mol} \]
For hydrogen:
\[ n(\ce{H})=\frac{0.50}{1.0} \]
\[ =0.50\ \mathrm{mol} \]
For nitrogen:
\[ n(\ce{N})=\frac{1.40}{14.0} \]
\[ =0.100\ \mathrm{mol} \]
(b) Determine the simplest whole-number mole ratio (2 marks)
Divide each mole amount by the smallest value, 0.100.
\[ \begin{aligned} \ce{C}:&\quad \frac{0.200}{0.100}=2\\ \ce{H}:&\quad \frac{0.50}{0.100}=5\\ \ce{N}:&\quad \frac{0.100}{0.100}=1 \end{aligned} \]
Therefore:
\[ \ce{C:H:N}=2:5:1 \]
(c) Write the empirical formula (1 mark)
Use the mole ratio as the formula subscripts.
\[ \boxed{\ce{C2H5N}} \]
(d) Calculate the percentage composition of nitrogen (1 mark)
Percentage composition is calculated using:
\[ \%\text{ by mass} = \frac{\text{mass of element}} {\text{total mass of compound}} \times100 \]
For nitrogen:
\[ \%\ce{N}=\frac{1.40}{4.30}\times100 \]
\[ \boxed{32.6\%} \]
Marking guide
| Part | Required working | Marks |
|---|---|---|
| (a) | Correct amount of carbon | 1 |
| (a) | Correct amount of hydrogen | 1 |
| (a) | Correct amount of nitrogen | 1 |
| (b) | Correct division by the smallest mole amount | 1 |
| (b) | Correct ratio, 2:5:1 | 1 |
| (c) | Correct empirical formula, \(\ce{C2H5N}\) | 1 |
| (d) | Correct percentage calculation and answer, 32.6% | 1 |
| Total | 7 |
Solution 5
(a) Calculate the molar mass of magnesium nitrate (2 marks)
The chemical formula is:
\[ \ce{Mg(NO3)2} \]
First, count the atoms of each element.
- 1 magnesium atom
- 2 nitrogen atoms
- 6 oxygen atoms
The molar mass is the sum of the molar masses contributed by all atoms in one formula unit.
\[ \begin{aligned} M&=24.3+2(14.0)+6(16.0)\\ &=24.3+28.0+96.0\\ &=\boxed{148.3\ \mathrm{g\,mol^{-1}}} \end{aligned} \]
(b) Calculate percentage composition (3 marks)
Use the following relationship:
\[ \%\text{ element} = \frac{\text{mass contribution of element}} {\text{total molar mass}} \times100 \]
For magnesium:
\[ \%\ce{Mg}=\frac{24.3}{148.3}\times100 \]
\[ \boxed{16.4\%} \]
For nitrogen, remember that there are two nitrogen atoms in each formula unit.
\[ \%\ce{N}=\frac{2(14.0)}{148.3}\times100 \]
\[ \boxed{18.9\%} \]
For oxygen, there are six oxygen atoms.
\[ \%\ce{O}=\frac{6(16.0)}{148.3}\times100 \]
\[ \boxed{64.7\%} \]
Check that the percentages add to approximately 100%.
\[ 16.4+18.9+64.7=100.0\% \]
This is a useful way to check your calculation.
(c) Calculate the mass of oxygen in a 29.66 g sample (2 marks)
The percentage of oxygen is 64.7%.
Convert this percentage into a decimal fraction:
\[ \frac{64.7}{100}=0.647 \]
Multiply by the total sample mass.
\[ m(\ce{O})=29.66\times0.647 \]
\[ \boxed{19.2\ \mathrm{g}} \]
Using the unrounded oxygen mass fraction gives the same answer to 3 significant figures.
Marking guide
| Part | Required working | Marks |
|---|---|---|
| (a) | Correct atom counts in the formula | 1 |
| (a) | Correct molar mass, 148.3 g mol\(^{-1}\) | 1 |
| (b) | Correct magnesium percentage, 16.4% | 1 |
| (b) | Correct nitrogen percentage, 18.9% | 1 |
| (b) | Correct oxygen percentage, 64.7% | 1 |
| (c) | Correct use of oxygen mass fraction | 1 |
| (c) | Correct mass, 19.2 g | 1 |
| Total | 7 |
Solution 6
(a) Determine the empirical formula (4 marks)
The hydrocarbon contains:
- 85.7% carbon
- 14.3% hydrogen
Assume a 100.0 g sample.
This means the sample contains:
- 85.7 g carbon
- 14.3 g hydrogen
Calculate the number of moles of each element.
For carbon:
\[ n(\ce{C})=\frac{85.7}{12.0} \]
\[ =7.14\ \mathrm{mol} \]
For hydrogen:
\[ n(\ce{H})=\frac{14.3}{1.0} \]
\[ =14.3\ \mathrm{mol} \]
Next, divide both amounts by the smallest mole amount, approximately 7.14.
\[ \begin{aligned} \ce{C}:&\quad \frac{7.14}{7.14}=1\\ \ce{H}:&\quad \frac{14.3}{7.14}=2.00 \end{aligned} \]
The simplest whole-number ratio is:
\[ \ce{C:H}=1:2 \]
Therefore, the empirical formula is:
\[ \boxed{\ce{CH2}} \]
(b) Calculate the empirical formula mass (1 mark)
The empirical formula is \(\ce{CH2}\).
Its formula mass is the sum of the relative atomic masses of its atoms.
\[ M_{\text{empirical}}=12.0+2(1.0) \]
\[ \boxed{14.0\ \mathrm{g\,mol^{-1}}} \]
(c) Determine the molecular formula (3 marks)
The independently supplied molar mass is:
\[ M_{\text{molecular}}=56.0\ \mathrm{g\,mol^{-1}} \]
The molecular formula must be a whole-number multiple of the empirical formula.
Calculate the multiplier:
\[ \text{Multiplier} = \frac{\text{molecular molar mass}} {\text{empirical formula mass}} \]
Substitute:
\[ =\frac{56.0}{14.0} \]
\[ =4 \]
This means the molecular formula contains four empirical formula units.
The empirical formula is \(\ce{CH2}\), so multiply every subscript by 4.
\[ (\ce{CH2})_4 \]
Therefore:
\[ \boxed{\ce{C4H8}} \]
The molecular formula is \(\ce{C4H8}\).
Notice that the independently supplied molar mass was not needed to calculate the empirical formula. It was used afterwards to determine how many empirical formula units make up one molecule.
Marking guide
| Part | Required working | Marks |
|---|---|---|
| (a) | Correct amount of carbon | 1 |
| (a) | Correct amount of hydrogen | 1 |
| (a) | Correct simplest mole ratio, 1:2 | 1 |
| (a) | Correct empirical formula, \(\ce{CH2}\) | 1 |
| (b) | Correct empirical formula mass, 14.0 g mol\(^{-1}\) | 1 |
| (c) | Correct multiplier calculation, 4 | 1 |
| (c) | Correct application of multiplier to carbon subscript | 1 |
| (c) | Correct molecular formula, \(\ce{C4H8}\) | 1 |
| Total | 8 |
End of paper
Total available: 40 marks.
The key skill throughout this assessment is keeping mass and mole ratios separate. Chemical formula subscripts describe relative numbers of atoms, so always convert the available mass data into moles before deriving an empirical formula. For percentage composition, work in the opposite direction: calculate each element’s contribution to the total formula mass, then express it as a percentage.