C9: Chemical Calculations
Relative masses, moles and reacting masses
C9 Chemical Calculations
📋 Key Definitions
Relative atomic mass (Ar): The average mass of an atom of an element relative to 1/12th the mass of a carbon-12 atom. It is the number on the periodic table.
Relative formula mass (Mr): The sum of all the relative atomic masses of the atoms in a formula. Sometimes called relative molecular mass for molecules.
Mole: The amount of substance containing 6.02 × 10²³ particles (Avogadro's constant). One mole of any substance has a mass in grams equal to its relative formula mass.
Law of conservation of mass: In a chemical reaction, the total mass of the reactants equals the total mass of the products. No atoms are created or destroyed.
Key Formulae:
Number of moles = mass (g) ÷ Mr
Mass (g) = moles × Mr
Mr = mass (g) ÷ moles
📝 Relative Atomic Mass and Relative Formula Mass
The relative atomic mass (Ar) of each element is found on the periodic table. It accounts for the proportions of different isotopes of that element.
To calculate the relative formula mass (Mr), add up all the Ar values for every atom in the formula. Remember to multiply by the number of atoms of each type shown by subscripts in the formula.
Worked Example 1: Calculating Mr of H₂O
Find the Mr of water, H₂O.
Ar(H) = 1, Ar(O) = 16
Mr = (2 × 1) + 16 = 18
Worked Example 2: Calculating Mr of Ca(OH)₂
Find the Mr of calcium hydroxide, Ca(OH)₂.
Ar(Ca) = 40, Ar(O) = 16, Ar(H) = 1
Mr = 40 + (2 × 16) + (2 × 1) = 40 + 32 + 2 = 74
Note: The subscript 2 outside the brackets multiplies everything inside.
Worked Example 3: Calculating Mr of Al₂(SO₄)₃
Find the Mr of aluminium sulfate, Al₂(SO₄)₃.
Ar(Al) = 27, Ar(S) = 32, Ar(O) = 16
Mr = (2 × 27) + (3 × 32) + (3 × 4 × 16) = 54 + 96 + 192 = 342
⚖️ The Mole
Chemists measure substances in moles. One mole of any substance contains the same number of particles: 6.02 × 10²³ (Avogadro's constant).
The mass of one mole of a substance (in grams) is equal to its relative formula mass. For example, one mole of carbon-12 has a mass of exactly 12 g.
| Substance | Formula | Mr | Mass of 1 mole |
|---|---|---|---|
| Carbon | C | 12 | 12 g |
| Water | H₂O | 18 | 18 g |
| Sodium chloride | NaCl | 58.5 | 58.5 g |
| Carbon dioxide | CO₂ | 44 | 44 g |
| Calcium carbonate | CaCO₃ | 100 | 100 g |
Worked Example 4: Calculating moles from mass
How many moles are in 88 g of carbon dioxide, CO₂?
Mr(CO₂) = 12 + (2 × 16) = 44
Moles = mass ÷ Mr = 88 ÷ 44 = 2 mol
Worked Example 5: Calculating mass from moles
What is the mass of 0.5 moles of NaOH?
Mr(NaOH) = 23 + 16 + 1 = 40
Mass = moles × Mr = 0.5 × 40 = 20 g
⚖️ Conservation of Mass and Balanced Equations
Law of conservation of mass: The total mass of reactants equals the total mass of products. Atoms are rearranged but not created or destroyed in a chemical reaction.
This means the number of atoms of each element must be the same on both sides of a balanced symbol equation.
Worked Example 6: Balancing an equation
Balance the equation for the combustion of methane:
CH₄ + O₂ → CO₂ + H₂O
Step 1: Count atoms on each side.
Left: C=1, H=4, O=2. Right: C=1, H=2, O=3.
Step 2: Balance H by putting 2 before H₂O.
CH₄ + O₂ → CO₂ + 2H₂O
Step 3: Count O on right: 2 + 2 = 4. Put 2 before O₂.
CH₄ + 2O₂ → CO₂ + 2H₂O
Check: Left C=1, H=4, O=4. Right C=1, H=4, O=4. ✓
Worked Example 7: Using conservation of mass
When 12 g of carbon burns in oxygen, 44 g of carbon dioxide is produced. What mass of oxygen reacted?
Mass of reactants = mass of products
12 g + mass of O₂ = 44 g
Mass of O₂ = 44 − 12 = 32 g
Worked Example 8: Reacting masses from equations
In the reaction 2Mg + O₂ → 2MgO, what mass of magnesium oxide is produced from 4.8 g of magnesium?
Mr(Mg) = 24, Mr(MgO) = 24 + 16 = 40
Moles of Mg = 4.8 ÷ 24 = 0.2 mol
From the equation, 2 mol Mg → 2 mol MgO, so 1:1 ratio.
Moles of MgO = 0.2 mol
Mass of MgO = 0.2 × 40 = 8 g
📊 Understanding Balanced Symbol Equations
A balanced symbol equation shows the reactants and products with the correct number of atoms of each element on both sides. The large numbers in front of formulae (coefficients) tell you the ratio of moles of each substance that react.
State symbols: (s) = solid, (l) = liquid, (g) = gas, (aq) = dissolved in water (aqueous). Always include state symbols when asked.
| Word Equation | Balanced Symbol Equation | Mole Ratio |
|---|---|---|
| Hydrogen + oxygen → water | 2H₂(g) + O₂(g) → 2H₂O(l) | 2 : 1 : 2 |
| Iron + chlorine → iron(III) chloride | 2Fe(s) + 3Cl₂(g) → 2FeCl₃(s) | 2 : 3 : 2 |
| Calcium carbonate → calcium oxide + carbon dioxide | CaCO₃(s) → CaO(s) + CO₂(g) | 1 : 1 : 1 |
❓ Practice Questions
Q1: Foundation Calculate the relative formula mass of MgCl₂. (Ar: Mg = 24, Cl = 35.5)
Q2: Foundation How many moles are there in 20 g of NaOH? (Ar: Na = 23, O = 16, H = 1)
Q3: Higher In the reaction Fe₂O₃ + 3CO → 2Fe + 3CO₂, calculate the mass of iron produced from 160 g of Fe₂O₃. (Ar: Fe = 56, O = 16, C = 12)
Q4: Foundation Balance the equation: Al + O₂ → Al₂O₃
Q5: Higher 13 g of zinc reacts with excess hydrochloric acid. The equation is Zn + 2HCl → ZnCl₂ + H₂. Calculate the mass of zinc chloride produced. (Ar: Zn = 65, Cl = 35.5)
✅ Answers
- Mr(MgCl₂) = 24 + (2 × 35.5) = 24 + 71 = 95
- Mr(NaOH) = 23 + 16 + 1 = 40. Moles = 20 ÷ 40 = 0.5 mol
- Mr(Fe₂O₃) = (2 × 56) + (3 × 16) = 112 + 48 = 160. Moles of Fe₂O₃ = 160 ÷ 160 = 1 mol. From the equation, 1 mol Fe₂O₃ → 2 mol Fe. Moles of Fe = 2 mol. Mass of Fe = 2 × 56 = 112 g
- 4Al + 3O₂ → 2Al₂O₃. Check: Left: Al=4, O=6. Right: Al=4, O=6. ✓
- Moles of Zn = 13 ÷ 65 = 0.2 mol. From the equation, 1 mol Zn → 1 mol ZnCl₂, so moles of ZnCl₂ = 0.2 mol. Mr(ZnCl₂) = 65 + (2 × 35.5) = 136. Mass of ZnCl₂ = 0.2 × 136 = 27.2 g
🎯 Exam Tips
- Always show your working in calculation questions — you may get marks for method even if the final answer is wrong
- When calculating Mr, watch out for brackets — the subscript outside a bracket multiplies everything inside it
- Use the correct units: g for mass, mol for moles
- In reacting mass calculations, always find the moles first, then use the mole ratio from the balanced equation
- Check your balanced equations by counting atoms on each side before doing any calculations
- If a gas is produced in a reaction in an open container, the mass appears to decrease — this does not break the law of conservation of mass
🔬 Required Practical
Required Practical Activity
There is no specific required practical for C9 in Combined Science. However, you should be able to use conservation of mass to calculate reacting masses from balanced equations — practise the method of finding moles first, then using the mole ratio.
🔢 Maths Skills
Mathematical Skills
Moles = mass / Mr: First calculate Mr, then divide the given mass by Mr to find moles.
Reacting mass calculations: 1) Write the balanced equation. 2) Calculate moles of the known substance. 3) Use the mole ratio to find moles of the unknown. 4) Multiply by Mr to find the mass.
Example: What mass of MgO forms from 6 g Mg? Mg + O₂ → MgO. Moles of Mg = 6/24 = 0.25. Ratio 1:1 so moles of MgO = 0.25. Mass = 0.25 × 40 = 10 g.
⚠️ Common Misconceptions
Watch Out!
1 mole of any substance has a mass of 1 gram. Wrong: 1 mole = 1 gram Correct: 1 mole has a mass in grams equal to the substance's Mr — 1 mole of H₂O is 18 g, 1 mole of CO₂ is 44 g
Mass is not conserved when a gas is produced in a reaction. Wrong: mass is not conserved when gas is produced Correct: mass is always conserved — if a gas escapes from an open container the total mass appears to decrease, but the mass of gas still counts
✍️ 6-Mark Question
Extended Answer
6 marks: Explain how to calculate the mass of product from a balanced symbol equation.
First, write the balanced symbol equation and identify the mole ratio between the known reactant and the desired product. Then calculate the Mr of the known substance and use it to find the number of moles: moles = mass ÷ Mr. Next, use the mole ratio from the balanced equation to calculate the moles of the product. Finally, multiply the moles of product by its Mr to find the mass: mass = moles × Mr. Always show working and include units.
Mark scheme: 1 mark for balanced equation with mole ratio; 1 mark for calculating Mr; 1 mark for moles = mass/Mr; 1 mark for using mole ratio; 1 mark for mass = moles × Mr; 1 mark for clear explanation of the sequence.
📊 AO3: Analyse & Evaluate
Analysis and Evaluation
In the reaction 2Fe₂O₃ + 3C → 4Fe + 3CO₂, a student uses 32 g of Fe₂O₃ and 6 g of carbon. Mr(Fe₂O₃) = 160, Ar(C) = 12.
Question: Which reactant is in excess? Show your working and explain what this means.
Answer: Moles of Fe₂O₃ = 32/160 = 0.2 mol. Moles of C = 6/12 = 0.5 mol. Ratio requires 2 mol Fe₂O₃ : 3 mol C, so for 0.2 mol Fe₂O₃ need 0.3 mol C. Only 0.3 mol C is needed but 0.5 mol is available, so carbon is in excess. Fe₂O₃ is the limiting reactant — it will run out first and determine how much iron is produced.
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