Finally after a long wait, the final part (Part 4) of the long series, SPM Chemistry Form 4 notes on Chemical Formulae and Equations is here on Berry Berry Easy. In the first three parts, the relative atomic mass, relative formula mass, relative molecular mass, scales, properties, Standard Temperature Pressure (s.t.p.), empirical formula, molecular formula, covalent compounds, ions, ionic compounds and even prefixes were all discussed. All these are of utmost importance in understanding Chemistry. So all Form 4 students must understand them by now if you wish to understand other chapters. (For Form 5 students who have not mastered the aforementioned concepts, make sure you look back at previous posts and learn them before your SPM)
In this part, we’ll try to apply what you have learnt in the previous three parts with some examples. So get a pen and paper to follow through the steps. Good revision too if you have already understand it.
Form 4 – Terminology and Concepts: Chemical Formulae and Equations (Part 4)
1. Importance of chemical equation:
The types of reactants; the physical conditions; the quantity of reactants and products and stated in moles.
nA + nB –> pC + pD
2. Reactants are written in the left side of the reaction and products are written in the right side of the reaction.
- Example 1:
Word equation: Sodium hydroxide + sulphuric acid –> sodium sulphate + water
Chemical equation: NaOH + H2SO4 –> Na2SO4 + H2O
Balancing equation: 2NaOH + H2SO4 –> Na2SO4 + 2H2O
Complete chemical equation: 2NaOH + H2SO4 –> Na2SO4 + 2H2O
- Example 2:
Word equation: Aluminium + copper(II) oxide –> aluminium(III) oxide + copper
Chemical equation: Al + CuO –> Al2O3 + Cu
Balancing equation: 2Al + 3CuO –> Al2O3 + 3Cu
Complete chemical equation: 2Al + 3CuO –> Al2O3 + 3Cu
- Example 3:
Word equation: Nitrogen + hydrogen <–> ammonia
Chemical equation: N2 + H2 <–> NH3
Balancing equation: N2 + 3H2 <–> 2NH3
Complete chemical equation: N2 + 3H2 <–> 2NH3
3. Information obtainable from chemical equations.
- i) mass of reactants
- ii) volume of reacting gas
- iii) mass of products formed
- iv) volume of gas produced
2 cm3 of lead (II) nitrate solution is added to excess of potassium iodide solution.
How many molecules of potassium nitrate will be formed?
[Relative atomic mass: N, 14; O, 16; K, 39; I, 127; Pb, 207; Avogadro's constant: 6.02 x 1023 mol-1]
Step 1: Write a complete chemical equation.
- Pb(NO3)2(aq) + 2KI(aq) –> PbI2(s) + 2KNO3(aq)
- From the equation, 1 mole of Pb(NO3)2 reacts with 2 moles of KI formed 1 mole PbI2 of and 2 moles of KNO3.
Step 2: Convert to moles.
- No. of moles of Pb(NO3)2
= Mass of Pb(NO3)2 / Relative molecular mass
= 2 / [207 + 2(14 + 3 x 16)]
= 6.04 x 10-3 mol
Step 3: Ratio of moles.
- Number of moles of KNO3/ Number of moles of Pb(NO3)2
- Number of moles of KNO3
= (2 x 6.04 x 10-3) / 1
= 12.08 x 10-3 mol
Step 4: Convert to the number of molecules of potassium nitrate.
- Number of molecules of KNO3
= 12.08 x 10-3 x 6.02 x 1023
= 7.27 x 1021
Once you have completed all parts in this series, you would have mastered the most basic of knowledge in SPM Chemistry. So make sure all Berry Readers understand the basics before moving on to more advance topics.