topic c3- quantitative chemistry

• relative formula mass

  • Mr
• e.g. find the relative atomic mass of MgCl2

  Ar of Mg=24 Ar of Cl= 35.5

  24+35.5+35.5= 95 Mr= 95
• percentage mass of an element in a compound=

  Ar x number of atoms of that element/ Mr of the compound x100
• e.g. find the % mass of sodium in sodium carbonate Na2CO3

  23 + 12 (16 x 3) = 106 %mass= 23 x 2 / 106 x 100= 43%

  Mr of Na2C03= 106
• the mole

  • mole= amount of substance
  • mass= no moles/ Mr
• e.g. how many moles are there in 66g of CO2?

  Mr of CO2= 44

  No. moles= 66/ 44= 1.5 mol
• e.g. what mass of carbon is there 4 moles of carbon dioxide?

  no. moles x mass

  4 x 12 = 48 g
• e.g. calculate the mass of 0.2 mol of potassium bromide

  Mr = 39 + 80 = 119

  mass = 119 x 0.2 = 23.8g or 24g
• avagadro's constant

  6.023 x 10 ^23
• avagadro's constant

  6.023x10^23
• relative formula mass (Mr)

  sum of all the relative atomic masses (Ar) of the atoms in the molecular formula
• percentage mass of an element in a compound

  Ar x number of atoms of that element/

  Mr of the compound

  x100
• number of moles

  mass/ Mr
• conservation of mass

  no atoms are created or destroyed in a chemical reaction, so the total masses of reactants and products are the same
• conservation of mass

  mass is always conserved
• e.g. show that mass is conserved in this reaction

  mass equal on both sides:

  2Li + F2 -> 2LiF

  14 + 38 -> 52

  52 52
• e.g. show that mass is conserved in this reaction

  both sides have an equal mass- have to write this underneath working to get the mark:

  H2SO4 + 2NaOH -> Na2SO4 + 2H2O

  98 + 80 142 + 36

  178 178

  because:

  2+32+(16x4)= 98 23+23+16+16+2= 80 98+80=178

  23+23+32+(16x4)= 142 4+16+16= 36 142+26+178
• e.g. 84g of N2 reacts completely with 18g of H2 to produce 102g of NH3- the mole and equations

  divide mass by Mr to calculate moles

  divide by the smallest number of moles (3)

  Mr of N2=28 H2=2 NH3=17

  N2=84/28=3 3/3=1

  H2=18/2=9 9/3=3

  NH3=102/17=6 6/3=2

  N2 + 2H2 -> 2NH3
• limiting reactants- e.g. calculate the mass of potassium chloride produced when 23.8g of potassium bromide reacts in an excess of chlorine

  Mr of KBr=119 Mr of KCl=74.5

  Cl2 + 2KBr -> 2KCl

  119 74.5

  no. moles = 23.8 / 119 = 0.200

  mass= no.moles x Mr

  74.5 x 0.200= 14.9g of potassium chloride
• concentration of solutions

  • concentration (g/dm^3)= mass of solute (g)/ volume of solvent (dm^3)
  • mass (g)= conc. (g/dm^3) x volume (dm^3)
• balanced equations

  • a symbol equation with the same number of atoms of each element on both sides
  • 2Mg + O2 -> 2MgO
  • they tell you how many moles of each substance take part in the reaction
• limiting reactants 


  • a reactant that gets completely used up in a reaction, so limits the amount of products formed
  • all the other reactants are in excess