3. Amount of substance

• Conversions of mass


  1 tonne = 1000kg

  1 tonne = 1,000,000 g

  1 kg = 1000g

  1 g = 1000 mg
• conversions of volume

  1 m3 = 1000 dm3

  1 m3 = 1,000,000 cm3

  1 dm3 = 1000 cm3
• what is a mole?


  the amount of substance which contains as many elementary particles as there are atoms in exactly 12g of carbon 12
• converting to moles


  1 mole = 6.02 x10 23 (Avogadro’s constant) particles

  mole = number of particles / 6.02 x10 23
• moles, mass and mr

  moles = mass/mr

  mass = moles x mr

  mr = mass/moles
• diatomic molecules

  Ar will be multiplied by 2

  oxygen = 16x2 = 32

  hydrogen = 1x2 = 2
• Empirical formula


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

  e.g. Benzene C6H6 = CH

  Heptene C7H14 = CH2
• Calculating the empirical formulae from percentage composition

  Fe (44%) and Cl (56%)

  Percentage/mr

  44/55.8=0.789 mol

  56/35.5=1.577 mol

  Divide by smallest value

  1.577/0.789=1.999

  therefore ratio = 1:2

  FeCl2
• empirical formulae

  K (26.6%) Cr (35.4%) O (38%)


  26.6/39.1 =0.68 mol

  35.4/52 =0.68 mol

  38/16 =2.375 mol

  ratio = 1:1:3.5 (must be a whole number)

  ratio= 2:2:7

  K2Cr2O7
• work out the empirical formulae for chloride of cobalt which contains 54.7 % chlorine

  10–54.7=45.3

  Cl - 1.54

  Co- 0.77

  ratio= 2:1

  CoCl2
• a hydrocarbon of molar mass 142 g/mol has an empirical formulae of C5H11. What is the molecular formula?

  n = m/mr

  n = 142/60+11 = 2 mol

  therefore = C10H22
• a hydrocarbon contains 0.72g of carbon and 0.12g of hydrogen the molar mass is 42 find the molecular formula

  nC= 0.72/12=0.06

  nH= 0.12/1=0.12

  C:H = 1:2 = CH2

  42/12+2 = 3 mol

  C3H6
• Working out how many moles of each substance are used up or produced using stoichiometric numbers


  4X + Y = 2Z

  4 moles of X reacted with 1 mole of Y produces 2 moles of Z

  2 moles of X reacted with 0.5 moles of Y produces 1 mole of Z
• reacting gases


  n=v(dm3)/24

  n=v(cm3)/24000
• ideal gas equation


  pV=nRT

  pascals

  volume (m3)

  moles

  gas constant (8.314 Jmol-1k-1)

  temp (kelvin)
• Reacting solutions 


  concentration: moldm-3

  N=cV (dm3)

  N = cV/1000

  c = nx1000/v

  V = nx1000/c
• Calculating uncertainty


  Number of readings x uncertainty equipment / quantity measured x 100
• calculating percentage yield


  actual yield / theoretical yield x 100
• calculating atom economy


  sum of molar masses of desired product / sum of molar masses of all products x 100
• 2 ways to prepare a standard solution


  Dissolving a solid in distilled water.

  Diluting a known concentration of an existing solution with distilled water
• preparation of a standard solution (dissolving a solid)


  Measure a known mass, by difference, using a mass balance.

  Dissolve in a small amount of water in a flask/beaker and then transfer to a 250cm3 volumetric flask (use a funnel)

  Rinse the original flask with water and transfer washings to the volumetric flask.

  Carefully make up to the line with water (bottom of the meniscus on the line) using a dropping pipette.

  • Stopper + Invert to mix.
• preparation of a standard solution (diluting a known concentation)


  Using pipette, transfer the original solution to a new volumetric flask

  • Carefully fill the volumetric flask to the line, with more distilled water.

  • Stopper.

  • Invert to mix.

  In both cases, the number of moles does not change when you dilute with wate