calculations involving masses

• relative formula mass given relative atomic masses

  • relative formula mass (Mr) of compound - sum of all relative atomic masses of atoms in numbers shown
  • balanced chemical equation - sum of Mr of reactants = sum of Mr of products
• calculate formula from reacting masses

  • work out moles of each using moles = mass / molar mass
  • work out the ratio of moles
  • times the ratio so that you get the smallest whole numbers possible
  • find the formula by multiplying each element by their number in the ratio (remember to use little numbers not a big number at the front)

  this is empirical formula as it shows the simplest ratio
• empirical formula from the formula of molecule

  if you have a common multiple e.g. Fe2O4 , the empirical formula is the simplest whole number ratio, which would be FeO2

  if there is no common multiple, you already have the empirical formula
• molecular formula from empirical formula and relative molecular mass

  • find relative molecular mass of the empirical formula
  • divide relative molecular mass of compound by that of the empirical formula
  • multiply the number of each type of atom in the empirical formula by this number
  • e.g. if answer was 2 and the empirical formula was Fe2O3 then the molecular formula would be empirical formula x 2 = Fe4O6
• determine the empirical formula of magnesium oxide

  • weigh some pure magnesium - mass used
  • heat magnesium to burning in a crucible to form magnesium oxide, as the magnesium will react with the oxygen in the air
  • weigh the mass of the magnesium oxide - mass produced
  • then use the empirical formula
• law of conservation of mass - no atoms are lost or made during a chemical reaction so the mass of the products = mass of the reactants
• precipitate reaction in a closed system

  precipitate that forms is insoluble and is a solid, as all the reactants and products remain in the sealed reaction container then it is easy to show that the total mass is unchanged
• reaction in non-enclosed system

  the mass will change as some of the mass is lost when gas is given off
• moles = mass / Mr
• concentration (g dm3) = mass of solute (g) / volume (dm3)
• mass = conc x volume
• the number of atoms, molecules or ions in one mole of a given substance is the Avogadro constant - 6.02 x 1023 per mole
• the mass of one mole of particles is the ‘relative particle mass’ in grams
• the mass of a product formed is controlled by the mass of the reactant that is not in excess, the reaction continues until all the particles of the limiting reactant have been used up, the mass of product is directly proportional to the mass of the limiting reactant
• stoichiometry refers to the balancing numbers in front of compounds/elements in reaction equations
• balancing numbers in a symbol equation can be calculated from the masses of reactants and products

  • convert the masses in grams to amounts in moles (moles = mass/Mr)
  • convert the numbers of moles to simple whole number ratios
• for the reaction: Cu + O2 -> CuO (not balanced)

  127 g Cu react, 32g of oxygen react and 159g of CuO are formed

  work out the balanced equation using the masses given

  • moles : (moles = mass/Mr)
  • Cu: moles = 127 / 63.5 = 2
  • O2 : moles= 32 / (16 x 2) = 32/32 = 1
  • CuO moles = 159 / (16 + 63.5) = 2

  the ratio is 2:1:2 therefore the reaction is 2Cu + O2 -> 2CuO