Amounts of substance

• Relative atomic mass


  weighted average mass of all the isotopes/ 1/12 mass of one atom of carbon -12
• Relative atomic mass calculation

  (mass x abundance of each isotopes) / total abundance
• the mole


  the amount of substance that contains as many particles as there in exactly 12g of carbon-12
• Avogadro constant


  6.022 x 10^23
• a mole of any substance contains the same number of particles as a mole of any other substance. the mass of one mole of a substance is its Mr in grams. This is its molar mass ( g mol-1)
• number of moles = mass/Mr
• mass = number of moles x Mr
• number of moles = volume ( dm3) x concentration
• dm3 to cm3= x1000
• ideal gas equation


  pV = nRT
• P, pressure= Pa

  V, volume= m3

  n, number of moles

  r= gas constant

  t= temperature in kelvin= +273
• volume


  v= nRT/ p
• pressure


  nRT/ v
• moles


  n= pV/ RT
• density


  p= m/v
• 1cm3 of water has a mass of 1g, the density of the water is 1 g cm-3
• empirical formula


  the simplest whole number ratio of atoms of each element present in a compound
• molecular formula


  actual number of atoms of each element present in a molecule
• How to calculate empirical formula

  1. If Sulfur has a mass of 3.2 g and Oxygen has a mass of 3.2g
  2. their Ar's are 32.1 and 16
  3. divide mass by Mr to get moles
  4. get the simplest ratio
  5. and ef will be SO2
• To calculate molecular formula

  • calculate empirical formula
  • Calculate relative mass of empirical formula
  • use Mr to calculate how many times bigger the molecular formula is compared to the empirical formula
• known moles = unknown moles
• standard solution


  known concentration of one solution
• methyl orange colour


  in acid-red

  in base- yellow

  end point- orange
• phenolphthalein


  acid- colourless

  base- pink

  end point- colourless
• How to make a standard solution

  • record the mass of a weighing boat and solid
  • tip solid into a beaker and record the mass of the empty weighing boat
  • determine actual mass of solid used by calculating difference between the two masses
  • dissolve solid in deionised water
  • using a funnel, transfer to a volumetric flask
  • make up to line, sitting proud on meniscus
  • stopper the flask and invert 3 times
• to find the concentration of a solution of a base


  • Fill a burette with a standard solution of the acid
  • pipette a known volume of base into conical flask
  • add the solution of the acid from the burette until the indicator just changes colour, known as end point
  • record volume of acid added: record burette readings to nearest 0.05 cm3
  • perform a rough titration, the accurate ones until titres are concordant ( within 0.10 cm3 of each other)
  • calculate mean titre
• percentage error= uncertainty/ value x100%
• Back titrations


  used to analyse substances that are not soluble in water but do react with acids

  • a known mass of solid is reacted with an excess of acid
  • resulting solution is titrated with a standard solution of base to determine the amount of acid left
• two indigestion tablets, containing Mg(OH)2, were dissolved in 25.0cm3 of 1.00moldm-3 of HCl. the resulting solution was titrated with 0.750moldm-3 aq NaOH. 17.4cm3 of aq NaOH was required to neutralise. determine mass, in mg, of Mg(OH)2 in each tablet


  n(HCl)= 1.00 x 2.5x10-2 = 2.50 x 10-2

  n(NaOH) = 0.750 X 17.4x10-3= 0.01305

  n(HCL) left over = n(NaOH) = 0.01305

  Mg(OH)2, n(HCL) reacted= n(HCL) original - n(Hcl) left over

  2.5x10-2 - 0.01305= 0.01195 reacts with Mg(OH)2

  0.01195/2= 5.975x10-3

  mass= 58.3 x 5.975x10-3= 0.348g

  0.348/2= 0.174g = 174mg
• hydrated salt

  contains water
• water of crystallisation

  the water molecules which form an essential part of the crystalline structure of a hydrated salt
• to calculate the number of moles of water of crystallisation

  • know the masses of the hydrated and anhydrous salts
  • mass of water = mass of hydrated salt- mass of anhydrous salt
  • work out number of moles of anhydrous salt and of water
  • find simplest whole number ratio
• heating to constant mass

  remove H2O
• percentage yield

  actual yield of product/ theoretical yield of product x 100
• theoretical yield

  number of moles expected if reactants were completely converted into products
• reasons for low percentage yield

  • incomplete reaction/ reversible reaction
  • side reactions
  • loss of product during transfer or purification steps
  • impurities in the reactants
• atom economy

  Mr of desired product/sum of Mr of all products x 100
• addition reactions

  have an atom economy of 100%
• Substitution and elimination reactions

  less than 100% atom economy