Topic 3
7 subtopics · Cambridge IGCSE Chemistry 0620
Key Idea
The mole is the unit for amount of substance. 1 mole = 6.02 x 10^23 particles (Avogadro's number). Molar mass = Mr in g/mol.
Explanation
Key formulae: - n = m / Mr (moles = mass / molar mass) - m = n x Mr - Mr = m / n Relative formula mass (Mr): sum of all relative atomic masses in the formula. Example: Mr of H2O = (2x1) + 16 = 18 g/mol Example: Mr of CaCO3 = 40 + 12 + (3x16) = 100 g/mol
Worked Examples
Practice Questions
What is the relative formula mass (Mr) of aluminium sulfate, Al2(SO4)3? (Ar: Al=27, S=32, O=16)
Key Idea
Use a balanced equation and mole ratios to calculate the mass of reactants or products in a reaction.
Explanation
Steps for reacting mass calculations: 1. Write a balanced chemical equation. 2. Write the mole ratio from the equation. 3. Convert given mass to moles: n = m / Mr. 4. Use mole ratio to find moles of the required substance. 5. Convert moles back to mass: m = n x Mr.
Worked Examples
Practice Questions
CuO(s) + H2(g) to Cu(s) + H2O(l). Calculate the mass of copper produced when 8.0 g of CuO reacts completely. (Ar: Cu=64, O=16) [3 marks]
Key Idea
The limiting reagent is the reactant that is completely used up first, determining the maximum amount of product that can be formed.
Explanation
Steps to identify the limiting reagent: 1. Convert masses of BOTH reactants to moles. 2. Divide each by its coefficient in the balanced equation. 3. The smaller value identifies the limiting reagent. 4. Use the limiting reagent to calculate the product. The excess reagent is the one that is NOT fully used up. Some of it remains at the end of the reaction.
Worked Examples
Practice Questions
N2(g) + 3H2(g) to 2NH3(g). 28.0 g of N2 and 6.0 g of H2 are reacted. (Ar: N=14, H=1) a) Identify the limiting reagent. Show full working. [3 marks] b) Calculate the mass of NH3 that can be produced. [2 marks]
Key Idea
At room temperature and pressure (r.t.p.), 1 mole of any gas occupies 24 dm3 (24,000 cm3). V = n x 24.
Explanation
Key formulae: - V (dm3) = n x 24 - n = V / 24 - Convert cm3 to dm3: divide by 1000. This applies to ANY gas at r.t.p. regardless of its identity — because equal volumes of gases at the same temperature and pressure contain equal numbers of molecules (Avogadro's Law).
Worked Examples
Practice Questions
CaCO3(s) + 2HCl(aq) to CaCl2(aq) + H2O(l) + CO2(g). Calculate the volume of CO2 produced at r.t.p. when 5.0 g of CaCO3 reacts completely. (Mr: CaCO3 = 100) [3 marks]
Key Idea
Concentration = moles / volume (in dm3). Used in titration calculations to find unknown concentrations.
Explanation
Key formulae: - c = n / V (mol/dm3) - n = c x V - V = n / c - 1000 cm3 = 1 dm3 (always convert cm3 to dm3 by dividing by 1000) Titration calculation steps: 1. Calculate moles of the known solution: n = c x V. 2. Use mole ratio from balanced equation to find moles of unknown. 3. Calculate concentration of unknown: c = n / V.
Worked Examples
Practice Questions
20.0 cm3 of H2SO4 is neutralised by 25.0 cm3 of 0.100 mol/dm3 NaOH. Equation: H2SO4 + 2NaOH to Na2SO4 + 2H2O. Calculate the concentration of the H2SO4. [3 marks]
Key Idea
Empirical formula = simplest whole number ratio of atoms. Molecular formula = actual number of atoms. Molecular formula = n x empirical formula.
Explanation
Steps to find empirical formula from percentage composition: 1. Treat % as grams (assume 100 g of compound). 2. Divide each mass by its Ar to get moles of each element. 3. Divide all values by the smallest to get the simplest ratio. 4. Round to whole numbers if very close. To find molecular formula: 1. Find Mr of the empirical formula. 2. Ratio = given Mr / empirical Mr. 3. Molecular formula = empirical formula x ratio.
Worked Examples
Practice Questions
A compound contains 52.2% C, 13.0% H, and 34.8% O by mass. Its Mr = 46. a) Find the empirical formula. Show your working. [3 marks] b) Find the molecular formula. [2 marks]
Key Idea
% yield = (actual yield / theoretical yield) x 100. % purity = (mass of pure substance / total mass) x 100.
Explanation
PERCENTAGE YIELD: % yield = (actual yield / theoretical yield) x 100 Actual yield is less than theoretical yield because of: reversible reactions, side reactions, loss during transfer. PERCENTAGE COMPOSITION: % by mass of element = (Ar x number of atoms / Mr) x 100 Example: % Fe in Fe2O3 = (2x56 / 160) x 100 = 70% PERCENTAGE PURITY: % purity = (mass of pure substance / total mass of sample) x 100
Worked Examples
Practice Questions
A student reacts 6.5 g of zinc with excess sulfuric acid. The theoretical yield of ZnSO4 is 16.1 g, but only 12.9 g is obtained. Calculate the percentage yield. [2 marks]