IAL AS Chemistry – Topic 1: Formulae, Equations and Amount of Substance
Edexcel International AS Chemistry – Revision Notes, Exam Practice & Formula Sheet
1. Key Definitions
- Atom: Smallest unit of an element retaining its chemical properties.
- Element: Substance made of only one type of atom.
- Ion: Charged particle formed by loss or gain of electrons.
- Molecule: Two or more atoms covalently bonded.
- Compound: Two or more different elements chemically bonded.
- Empirical formula: Simplest whole-number ratio of atoms.
- Molecular formula: Actual number of atoms in a molecule.
2. The Mole and Avogadro Constant
The mole (mol) is the unit for amount of substance.
Avogadro constant:
6.02 × 1023 mol−1
One mole of any substance contains 6.02 × 1023 particles.
3. Chemical Equations
Balanced Equations
- Mass and charge must be conserved
- Correct stoichiometric coefficients required
State Symbols
- (s) solid
- (l) liquid
- (g) gas
- (aq) aqueous
Ionic Equations
- Only reacting ions shown
- Spectator ions removed
- Charges must balance
4. Relative Masses
- Relative atomic mass (Ar): Based on carbon-12 scale
- Relative molecular mass (Mr): Sum of Ar values
- Relative formula mass: Used for giant ionic/covalent structures
- Molar mass: Mass per mole (g mol−1)
Parts Per Million (ppm)
Used for very low concentrations, especially atmospheric gases.
5. Concentration of Solutions
mol dm−3:
Concentration = moles ÷ volume (dm3)
g dm−3:
Concentration = mass (g) ÷ volume (dm3)
Titration calculations are not required.
6. Empirical and Molecular Formula
- Convert mass to moles
- Divide by smallest number
- Convert to whole-number ratio
- Use molar mass to find molecular formula
7. Reacting Masses
- Balance the equation
- Convert given quantity to moles
- Apply mole ratio
- Convert back to mass or volume
8. Volumes of Gases
Molar volume at RTP: 24.0 dm3 mol−1
Ideal Gas Equation
pV = nRT
- p = pressure (Pa)
- V = volume (m3)
- n = moles
- R = 8.31 J mol−1 K−1
- T = temperature (K)
9. Percentage Yield & Atom Economy
Percentage yield:
(actual yield ÷ theoretical yield) × 100
Atom economy:
(molar mass of desired product ÷ total molar mass of all products) × 100
10. Exam-Style Questions (Worked)
Question 1
Calculate the number of moles in 9.0 g of water (H2O).
Answer:
Mr(H2O) = (2 × 1) + 16 = 18
Moles = 9.0 ÷ 18 = 0.50 mol
Question 2
What volume will 0.25 mol of CO2 occupy at RTP?
Answer:
Volume = moles × 24.0
Volume = 0.25 × 24.0 = 6.0 dm3
Question 3
A compound contains 40% carbon, 6.7% hydrogen and 53.3% oxygen. Find the empirical formula.
Answer:
C: 40 ÷ 12 = 3.33
H: 6.7 ÷ 1 = 6.7
O: 53.3 ÷ 16 = 3.33
Divide by smallest (3.33):
C : 1, H : 2, O : 1
Empirical formula = CH2O
11. Quick Formula Sheet
- moles = mass ÷ molar mass
- concentration = moles ÷ volume
- volume of gas = moles × 24.0
- pV = nRT
- % yield = actual ÷ theoretical × 100
- atom economy = desired ÷ total × 100
12. MCQ Practice
1. How many particles are in 1 mol of a substance?
A. 6.02 × 1022
B. 6.02 × 1023
C. 12.0
D. 24.0
Answer: B
2. Which unit is correct for molar mass?
A. g dm−3
B. mol dm−3
C. g mol−1
D. kg mol−1
Answer: C
3. What volume does 1 mol of gas occupy at RTP?
A. 22.4 dm3
B. 18.0 dm3
C. 24.0 dm3
D. 1.0 dm3
Answer: C
End of Topic 1 – IAL AS Chemistry
