Moles & Chemical Calculations

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CognitoMoles — calculations and reacting masses

CognitoMoles calculations with equations

Key facts & methods

The mole concept

1 mole of a substance contains 6.02 × 10²³ particles (Avogadro's number).
Molar mass (Mᵣ): the mass of 1 mole in grams. Numerically equal to the relative formula mass (Mᵣ).
Examples: Mᵣ of H₂O = 2+16 = 18. So 1 mole of H₂O = 18 g. Mᵣ of NaCl = 23+35.5 = 58.5.
Key equation: moles = mass ÷ molar mass. Or: n = m ÷ Mᵣ.
Rearranged: mass = moles × Mᵣ. Molar mass = mass ÷ moles.

Using equations to calculate reacting masses

The ratio of moles in a balanced equation = ratio of reacting amounts.
Method: (1) Write balanced equation. (2) Find moles of given substance. (3) Use ratio from equation. (4) Calculate mass of required substance.
E.g. CaCO₃ → CaO + CO₂. If 50 g CaCO₃: moles = 50/100 = 0.5. Moles CO₂ = 0.5 (1:1 ratio). Mass CO₂ = 0.5 × 44 = 22 g.
Mᵣ of CaCO₃ = 40+12+(3×16) = 100. Mᵣ of CO₂ = 12+(2×16) = 44.

Limiting reactants and excess (Higher)

The limiting reactant is the one that is completely used up — it determines the maximum yield.
The other reactant is in excess — some remains after the reaction.
To find limiting reactant: calculate moles of each reactant, compare to the ratio required by the equation.
The product yield is calculated from the moles of the limiting reactant.
Exam tip: always start by calculating moles = mass ÷ Mᵣ for each substance.

Exam questions — 4 questions · 10 marks · AQA 8462 style

3 marksCalculate molesAQA 8462 P1 style 🔢 Calculator

(a) Calculate the number of moles in 36 g of water (H₂O). (b) What mass of carbon dioxide (CO₂) contains 0.25 moles? (3 marks)

Hint: (a) Calculate Mᵣ, then n = m/Mᵣ. (b) Rearrange to mass = n × Mᵣ.

+30 XP

4 marksReacting massesAQA 8462 P2 style 🔢 Calculator

Magnesium reacts with oxygen: 2Mg + O₂ → 2MgO. Calculate the mass of magnesium oxide produced when 12 g of magnesium reacts completely with excess oxygen. (Mᵣ: Mg = 24, O = 16) (4 marks)

Hint: Step 1: moles of Mg = mass/Mᵣ. Step 2: use ratio from balanced equation. Step 3: mass of MgO = moles × Mᵣ(MgO).

+40 XP

4 marksLimiting reactant (Higher)AQA 8462 Higher style 🔢 Calculator

N₂ + 3H₂ → 2NH₃. 28 g of nitrogen and 9 g of hydrogen are mixed. (a) Calculate the moles of each. (b) Identify the limiting reactant. (c) Calculate the mass of ammonia produced. (Mᵣ: N₂=28, H₂=2, NH₃=17) (4 marks)

Hint: Calculate moles of each. Check if the ratio is satisfied: does the equation need 3× as many moles of H₂ as N₂? Whichever runs out first is limiting.

+40 XP

1 markMolar massAQA 8462 style

What is the molar mass of calcium carbonate CaCO₃? (Aᵣ: Ca=40, C=12, O=16)

A 68 g/mol

B 84 g/mol

C 100 g/mol

D 116 g/mol

1 markMoles calculationAQA 8462 style

How many moles are in 44 g of CO₂? (Mᵣ of CO₂ = 44)

A 0.5 mol

B 1 mol

C 2 mol

D 88 mol

Moles & Chemical Calculations

The mole concept

Using equations to calculate reacting masses

Limiting reactants and excess (Higher)

Module complete! 🎉