Amount of Substance (Moles) - Unit 5 - Year 10 Chemistry Revision 2024

• Mole (Avogadro's Number)


  6.022 x 10^23 particles of a substance (atoms, molecules or ions)
• Aᵣ (Relative Atomic Mass)


  the average mass of an atom compared to 1/12 of the mass of an atom of carbon-12
• Mᵣ (Relative Molecular/Formula Mass)


  the mass, in grams, of one mole of a compound

  (found by adding the Aᵣ of all the atoms in the compound)
• Number of Moles Formula


  n = m / Mᵣ

  number of moles = mass given (in g) / molar mass
• Reacting Mass Calculations


  1. Write out the balanced equation

  2. Note the molar ratio

  3. Write the known mass(es) under the equation

  4. Calculate the Mᵣ of the relevant substances

  5. Calculate the number of moles of the substance with a known mass

  6. Use the mole ratio to calculate the number of moles of the substance with an unknown mass

  7. Calculate the required mass of the substance
• What mass of H₂ is liberated when 1.15g Na reacts with excess water?

  2Na + 2H₂O → 2NaOH + H₂

  
  

  Na : H₂

  2 : 1

  
  

  Mᵣ(Na) = 23

  Mᵣ(H₂) = 2

  
  

  n = m / Mᵣ

  n(Na) = 1.15/23 = 0.05 moles

  ∴ n(H₂) = 0.05 / 2 = 0.025 moles

  
  

  m = n x Mᵣ

  m(H₂) = 0.025 x 2 = 0.05g
• Empirical Formula


  the simplest whole number ratio of atoms of each element in a compound

  
• Molecular Formula


  the actual number of atoms of each element in a compound
• Calculating Empirical Formula


  1. Write out the elements present

  2. Write out the % by mass of each element

  3. Calculate number of moles present in 100g of the compound

  4. Divide each by the smallest number of moles to produce the ratio of moles of each of the elements

  5. Write empirical formula
• A compound contains 40% carbon, 6.7% hydrogen and the rest oxygen. Calculate its empirical formula.


  In 100g, there is...

  40g C

  6.7g H

  53.3 O

  
  

  Aᵣ(C) = 12

  Aᵣ(H) = 1

  Aᵣ(O) = 16

  
  

  n(C) = 40/12 = 3.3 (2sf)

  n(H) = 6.7/1 = 6.7 (2sf)

  n(O) = 53.3/16 = 3.3 (2sf)

  
  

  C : H : O

  3.3 : 6.7 : 3.3

  1 : 2 : 1

  
  

  Empirical Formula = CH₂O
• Calculating Molecular Formula


  1. Calculate Mᵣ of the empirical formula

  2. Divide Mᵣ of molecule (given in question) by Mᵣ of empirical formula to find scaling factor

  3. Multiply each part of the empirical formula by scaling factor
• The Mᵣ of a compound with empirical formula CH₂O is 60. Calculate the molecular formula.


  Mᵣ(CH₂O) = 12 + (2 x 1) + 16 = 30

  scale factor = 60/30 = 2

  
  

  CH₂O x 2 = C₂H₄O₂
• Molar Volume


  the volume occupied by 1 mole of any gas at room temperature and pressure (RTP)

  
  

  1 mole of gas = 24000cm³ (24dm³)
• Number of Mole Formulas (Gas)


  n = volume (in cm³) / 24000
• 1 dm³ to cm³

  1000 cm³ (1 litre)
• 1 m³ to cm³

  1 000 000 cm³
• 10g of calcium carbonate is reacted with excess hydrochloric acid. What volume of carbon dioxide is formed at RTP?


  CaCO₃ (s) + 2HCl (aq) → CaCl₂ (aq) + H₂O (l) + CO₂ (g)

  
  

  CaCO₃ : CO₂

  1 : 1

  
  

  Mᵣ(CaCO₃) = 40 + 12 + (3 x 16) = 100

  Mᵣ(CO₂) = 12 + (2 x 16) = 44

  
  

  n = m / Mᵣ

  n(CaCO₃) = 10/100 = 0.1 moles

  ∴ n(CO₂) = 0.1 moles

  
  

  V = n x 24000

  V(CO₂) = 0.1 x 24000 = 2400cm³ (2.4dm³)