Quantitive Chemistry 11

• Water of crystallisation


  Water that is chemically bonded into the crystal structure of a compound
• Degree of hydration


  The number of moles of water of water crystallisation chemically bonded to 1 mole of that compound
• CuSO4.5H2O is sometimes called copper (11) sulfate-5-water or copper (11) sulfate pentahyrdrate
• The 5H2O is part of the mass of the solid
• Determining the degree of hydration

  1. Heating to a constant mass

  2. Taking mass measurements before and after

  3. Titration
• Anhydrous substance


  Doesn't contain water of crystallisation
• Empirical Formula 


  This is the simplest whole number ratio of the atoms of each element in a compound. It is the formula that is determined from experimental mass
• Molecular Formula 


  This is the actual number of atoms of each element in a molecule. It is a simple multiple of the empirical formula
• Empirical and Molecular formula

  Mr can be used to determine the molecular formula from the empirical formula

  • cancel down the number of each type of atom to the lowest number when determining an empirical formula
• Example 1 - molecular and empirical formula 


  Empirical formula of a compound has be determined to be CH3 and the Mr of the compound is 30. Determine the molecular formula of the compound

  • The Mr of CH3 is 12 + 3 = 15
  • 30 divided by 15 = 2
  • the molecule formula is C2H6
• Example 2 - molecular and empirical formula 


  The empirical formula of a compound is CH2O and its Mr is 180. Determine the molecular formula of the compound

  • The Mr of CH2O is 12 + 2 + 16 =30
  • 180 divided by 30 = 6
  • Molecular formula is C6H12O6
• Example 3 - molecular and empirical formula

  The molecular formula of a compound is Na2S4O6. What is the empirical formula

  • Empirical formula is determined by simplifying the ratio by the simplest possible
  • This case divide by 2
  • NaS2O3
• Determining formulae from simple compounds

  Simple compounds are formed from two elements. Can work out the formula of a simple compound using valency values. Must be able to use information about % composition or mass to determine formula
• How to determine formulae from simple compounds

  • Can calculate number of moles by dividing the mass of each element by its Ar
  • Moles are converted to a simple ratio which is done by making the lowest mole value equal to 1 and reducing others in the same scale. Then divide them all by the lowest mole value
  • Most experiments involve heating to a constant mass
• Example 1 - determining formulae of simple compounds

  1.5g of magnesium combines with oxygen to give 2.5g of magnesium oxide. Find the formula of the oxide of magnesium

  • Mass of oxygen = 2.5g - 1.5g = 1.0g
  • element : magnesium Oxygen
  • mass : 1.5g 1.0g
  • Ar : 24 16
  • moles : 1.5/24 = 0.0625 1/16 = 0.0625
  • ratio : 1 1
  • Empirical formula : MgO
• Heating a solid to determine empirical formula


  1. Measure the mass of empty crucible and lid (mass=a)

  2. Measure the mass of lid and crucible containing some magnesium powder (mass=b)

  3. Heat contents of crucible with lid on a pipeclay triangle on a tripod over a Bunsen Burner

  4. Raise the lid a little now and again to let more air in

  5. Allow crucible and lid to cool and then measure mass

  6. Heat again and measure the mass again : repeat until mass is constant (mass=c)
• Results wouldn't be reliable without these measures in place
• Lack of reliability would be because not all the magnesium react or some product was lost in crucible
• Heating a solid to determine EF

  • The mass of magnesium is b-a, the mass of oxygen gained is c-b.
  • Data may be presented this way and you have to calculate masses needed to determine empirical formula
• Example 1 of heating solids to get EF

  An oxide of manganese contains 63.2% manganese. Determine the empirical formula of the oxide

  • 100 - 63.2 = 36.8 % oxygen
  • element: Manganese Oxygen
  • mass: 63.2 36.8
  • Ar: 55 16
  • moles: 63.2/55= 1.15 36.8/16= 2.3
  • ratio: /1.15. 1 2
  • EF: MnO2
• Example 2 of heating solids to get EF

  Sample of solid phosphorus was burned in excess oxygen. 0.775g of phosphorus reacted with 1.0g of oxygen.

  1. Determine empirical formula of oxide of phosphorus formed
  2. Given that the Mr of oxide of phosphorus is 284, determine the molecular formula

  • element : phosphorus oxygen
  • mass: 0.775g 1g
  • ar : 31 16
  • moles : 0.025 0.0625
  • ratio : 2 5
  • EF : P2O5
  • Mr of P2O5 is 62 + 80 = 142
  • 284/142 = 2, molecular formula is P4O10
• Heating a solid equipment

  • Crucible
  • crucible lid
  • pipe clay triangle
  • tripod
  • heat proof mat
  • heat e.g Bunsen burner
• Determining degree of hydration by heating to a constant mass 


  The method of determining empirical formula can also be applied to hydrated compounds

  • If hydrated compounds are heated, they lose water of crystallisation and their mass decreases as the anhydrous compound is formed
  • When the compound has been heated to a constant mass, the decrease in mass is the mass of the water lost
  • By using the mass of the anhydrous compound and the mass of the water lost, the degree of hydration can be determined
• Determining a degree of hydration equipment

  • Evaporating basin
  • hydrated compound
  • gauze
  • tripod
  • heat proof mat
  • heat e.g Bunsen burner
• Degree of hydration


  The number of water molecules (n) in a hydrated compound
• Degree of hydration


  • CuSO4.nH2O
• Finding the value of n

  1. Given: 4g of hydrated copper (II) sulfate produced 2.56g of anhydrous copper (II) sulfate

  2. Mass of water lost = 4g - 2.56g = 1.44g

  3. Moles of water = 1.44g / 18g/mol = 0.08 mol

  4. Moles of copper (II) sulfate = 2.56g / 160g/mol = 0.016 mol

  5. Ratio of moles of water to moles of copper (II) sulfate = 0.08 / 0.016 = 5

  6. Therefore, the formula is CuSO4.5H2O
• Compound


  Copper (II) Sulfate
• The empirical formula is CuSO4.5H2O
• Example 2 question 


  The following mass measurements were taken when a sample of hydrated aluminium nitrate, Al(NO3)3.nH20, was heated to a constant mass in an evaporating basin in a low temperature oven:

  • mass of evaporating basin = 54.13g
  • mass of basin + hydrated salt = 61.63g
  • mass of basin + contents after heating to a constant mass = 58.39g
  • Find the degree of hydration
• Example 2 answer

  • Mass of anhydrous salt = 58.39-54.13 = 4.26g
  • mass of water lost = 61.63-58.39 = 3.24g
  • compound: Aluminium Nitrate Water
  • formula: Al(NO3)3 H2O
  • mass: 4.26 3.24
  • mr: 213 18
  • moles : 0.02 0.18
  • ratio : 1 9
  • EF : Al(NO3)3.9H20
• Heating to a constant mass

  • When heated, some substances produce gases which are released to the atmosphere -these lose mass on heating
  • When heated, some substances react with gases in the air - these gain mass on heating
  • When a solid produces a gas on heating or reacts with a gas from the air, you should heat to a constant mass to ensure the reaction has gone to completion
  • Heating to a constant mass means that you heat the substance, allow it to cool and record the mass - repeated until mass no longer changes
• Process of heating to a constant mass

  Process involves measuring the mass of the empty container e.g crucible/evaporating basin. The substance is heated in a container and mass is measured on several different occasions, after cooling, until no further change
• Data involved in heating to a constant mass

  • Mass of crucible
  • mass of crucible and substance
  • mass of crucible and contents after heating for 2 minutes
  • mass of crucible and contents after heating for 4 minutes
  • mass of crucible and contents after heating for 6 minutes
• Heating constant mass

  • Heating to a constant mass has been achieved
  • substance gains mass on heating so it is reacting with gas in the air
  • gas in air that reacts is usually oxygen, though some substances may react with carbon dioxide and even nitrogen
  • mole calculations using theses masses can be carried out if Mr of substance is known
• Solution calculations 


  Slightly more complicated than solids because the number of moles depends on the volume of the solution in cm3 and on the concentration of the solution mol/dm3

  • A solution with a concentration of 1 mol/dm3 has 1 mole of solute dissolved in 1 dm3 of solution. 500cm3 of solution of concentration 1 mol/dm3, would have 0.5 moles of solute in that volume
• Calculating solution 


  When making a solution, certain mass of solid is dissolved in a certain volume of water. Number of moles of a solute in a solution can be calculated using:

  moles= volume x concentration / 1000
• Example 


  1.4g of KOH were dissolved completely in 100cm3 of water. Calculate the concentration of the solution formed in mol/dm3

  Answer

  • Mr of KOH = 39+16+1 = 56
  • moles = 1.4/56 = 0.025 mol
  • 0.025 x 1000 / 100 = 0.25 mol/dm3
• Determining moles of solute in a solution

  Same equation used to calculate number of moles of a solute in certain volume of solution

  moles = volume x concentration / 1000
• Example 


  45 cm3 of a 0.1 mol/dm3 solution of HCl were used in a titration. Calculate the number of moles of HCl used

  moles = 45 x 0.1. / 1000 = 0.0045 mol