Module 2.1.3- Amount of substance

• Empirical formula


  The simplest whole number ratio of atoms of each element present in a compound
• Avogrados constant

  6.02x10^23

  
• Mass equation linking moles and Mr

  Mass= Molar mass (Mr) x Moles

  
• Calculation of empirical formula from mass/percentage


  1. Find the number of moles (n = m/M)2. Simplify (divide by smallest)

  
• Water of crystallisation


  Water molecules that form an essential part of the crystalline structure of a compound
• Formula of a hydrated salt


  Using data given in the table:

  Calculate moles of anhydrous salt

  Calculate moles of water

  Find smallest whole number ratio

  
• Factors that may affect experimental data in determining the formula of a hydrated salt


  -that not all water has been lost: heat to constant mass until no more mass change-further decomposition of the salt (difficult to distinguish the colour change)
• Converting from cm^3 -> dm^3 -> m^3

  Divide by 1000 each time (left to right)Multiply by 100 each time (right to left)

  
• Converting from mol dm^3 -> g dm^3


  multiply by Mr (right to left)
• Concentration formula


  moles/volume (dm^3)

  
• Molar gas equation (RTP)


  moles = volume (dm^3)/ 24dm^3

  
• Ideal gas equation


  PV = nRT

  
• Converting from Celsius to Kelvin


  C + 273 = K

  
• Stoichiometry


  the relationship between the relative quantities of substances taking part in a reaction or forming a compound, typically a ratio of whole integers.
• Percentage yield formula


  (actual yield/theoretical yield) x 100

  
• Reasons why theoretical yield isn't reached

  reaction may not have gone to completion

  side reactions

  purification of product may have resulted in the loss of some products
• How to work out limiting reagent


  1. Make sure equation is balanced2. Work out moles3. Use ratios to see how many moles react to get how many moles of product4. Which ever one is the smallest (when amount of moles is the same) is the limiting reagent.
• Atom economy


  (Molecular mass of desired products/ Molecular mass of all products) x 100
• Benefits of a high atom economy


  more sustainable as full use of raw materials less waste going to environment less money separating waste from product