Calculations Involving Masses

1.41 Describe the limitations of particular representations and models, to include dot and cross, ball and stick models and two- and three-dimensional representations
  • Main limitation is that it applies really well only to the small class of solids composed of Group 1 and 2 elements with highly electronegative elements such as the halogens
  • In covalent molecular, the dot-cross diagrams don’t express the relative attraction of shared electrons due to electronegativity (learn about this at A-level more)
1.42 Describe most metals as shiny solids which have high melting points, high density and are good conductors of electricity whereas most non-metals have low boiling points and are poor conductors of electricity
1.43 Calculate relative formula mass given relative atomic masses
  • Relative formula mass (Mr) of a compound: sum of the relative atomic masses of the atoms in the numbers shown in the formula
  • In a balanced chemical equation:

sum of Mr of reactants in quantities shown = sum of Mr of products in quantities shown

1.44 Calculate the formulae of simple compounds from reacting masses and understand that these are empirical formulae
  • divide mass by formula mass to find moles
  • divide by the smaller amount of moles to get a ratio
  • use the ratio to form the simplified empirical formula (e.g. if 2 Fe and 3 O then it is Fe2O3)
1.45 Deduce: the empirical formula of a compound from the formula of its molecule, and the molecular formula of a compound from its empirical formula and its relative molecular mass
  • 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
  • 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
    • If answer was 2 and the empirical formula was Fe2O3 then the molecular formula would be empirical formula x 2 = Fe4O6
1.46 Describe an experiment to determine the empirical formula of a simple compound such as magnesium oxide
  • Heat magnesium to burning in a crucible to form magnesium oxide, as the magnesium will react with the oxygen in the air
  • Empirical formula of a compound gives the lowest whole number ratio of atoms of each element present in the compound.
  • Empirical formula of magnesium oxide is determined by reacting magnesium metal with oxygen from the air to produce the magnesium oxide.
  • Known quantities: 
mass of magnesium used mass of magnesium oxide produced
  • Required calculations: 
mass oxygen = mass magnesium oxide – mass magnesium moles magnesium = mass magnesium ÷ molar mass magnesium 
moles oxygen = mass oxygen ÷ molar mass oxygen

Lowest whole number ratio of moles of magnesium to moles of oxygen is determined. 
Usually requires both the moles of magnesium and the moles of oxygen to be divided by whichever is the lowest number.

1.47 Explain the law of conservation of mass applied to: a closed system including a precipitation reaction in a closed flask and a non-enclosed system including a reaction in an open flask that takes in or gives out a gas
  • 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
    • Therefore, chemical reactions can be represented by symbol equations, which are balanced in terms of the numbers of atoms of each element involved on both sides of the equation.
  • With a precipitation reaction – 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
  • Does not hold for a reaction in an open flask that takes in or gives out a gas, since mass will change from what it was at the start of the reaction
1.48 Calculate masses of reactants and products from balanced equations, given the mass of one substance
  • Find moles of that one substance: moles = mass / molar mass
  • Use balancing numbers to find the moles of desired reactant or product
  • Mass = moles x molar mass to find mass

1.49 Calculate the concentration of solutions in g dm^-3
  • Concentration of a solution can be measured in mass per given volume of solution e.g. grams per dm^3 (g/dm^3)
  • To calculate mass of solute in a given volume of a known concentration do mass = conc x vol i.e. g = g/dm^3 x dm^3 (think about the units!!!)
1.50 (HT only) Recall that one mole of particles of a substance is defined as: the Avogadro constant number of particles (6.02 x 10^23 atoms, molecules, formulae or ions) of that substance and a mass of ‘relative particle mass’ g
  • The number of atoms, molecules or ions in a mole of a given substance is the Avogadro constant: 6.02 x 10^23 per mole.
1.51 (HT only) Calculate the number of: moles of particles of a substance in a given mass of that substance and vice versa, particles of a substance in a given number of moles of that substance and vice versa and particles of a substance in a given mass of that substance and vice versa
  • Chemical amounts are measured in moles. The symbol for the unit mole is mol.
  • The mass of one mole of a substance in grams is numerically equal to its relative formula mass.
  • For example, the Ar of Iron is 56, so one mole of iron weighs 56g.
  • The Mr of nitrogen gas (N2) is 28 (2*14), so one mole is 28g.
  • One mole of a substance contains the same number of the stated particles, atoms, molecules or ions as one mole of any other substance
  • You can convert between moles and grams by using this triangle:
    • g how many moles are there in 42g of carbon?
      • Moles = Mass / Mr = 42/12 = 3.5 moles
    • The number of particles, atoms, molecules or ions in a mole of a given substance is the Avogadro constant: 6.02 x 10^23 per mole.
1.52 (HT only) Explain why, in a reaction, the mass of product formed is controlled by the mass of the reactant which is not in excess
  • In a chemical reaction with 2 reactants you will often use one in excess to ensure that all of the other reactant is used
    • The reactant that is used up / not in excess is called the limiting reactant since it limits the amount of products
  • If a limiting reactant is used, there will be less product since there are less reactants, this means a smaller mass of product or a decrease in moles of product – think about balancing the moles: less moles on left means less moles on right
1.53 (HT only) Deduce the stoichiometry of a reaction from the masses of the reactants and products
  • Balancing numbers in a symbol equation can be calculated from the masses of reactants and products by converting the masses in grams to amounts in moles and converting the numbers of moles to simple whole number ratios (use above formula as well)