Summary Complete Formulae, Equations and Amounts of Substance Revision Notes (A Level Edexcel)

FORMULAE, EQUATIONS AND AMOUNTS OF SUBSTANCE


1. know that the mole (mol) is the unit for amount of a substance
 Moles measures the amount of a substance. The unit is the mol
 mol = mass ÷ Mr
 1 mole of a sample contains 6.02 x 1023 atoms.
 The mole is the number of atoms present in exactly 12.0g of carbon-12
(numerically equal to Avogadro’s number)
o 1 mol of 23Na atoms = 23g
o 1 mol Cl2 molecules = 71g


2. be able to use the Avogadro constant, L, (6.02 x 1023 mol-1) in calculations
 Avogadro’s constant, L, has the unit mol-1
 L = 6.02 x 1023
o How many atoms are there in 2 mol Na?
 Number of atoms in 2 mol Na = 2 mol x 6.02 x 10 23 mol-1
 1.204 x 1024 atoms of Na
o How many protons are there in 1 mol Ca?
 Calcium’s atomic number is 20. In one mol Ca, there are 20 mol
protons.
 20 mol protons = 20 x 6.02 x 1023 protons.
 There are 1.204 x 1025 protons in 20 mol Ca.
o What is the mass of a nucleon (proton OR neutron)?
 In one atom of hydrogen, there is one proton, one electron (with
negligible mass) and no neutrons. The Mr of H is 1.0 g mol-1
 One atom = 1 ÷ (6.02 x 1023) mol
 1 ÷ (6.02 x 1023) mol = mass ÷ 1.0 g mol-1
 Mass = 1 ÷ (6.02 x 1023) mol x 1.0 g mol-1
 Mass of a nucleon = 1.66 x 10-24g


3. know that the molar mass of a substance is the mass per mole of the
substance in g mol-1
 The molar mass is the mass of a substance per mole of the substance
 It is measured in g mol-1
 It has the same numerical value as the relative formula mass (M r)
 The relative formula mass = the sum of the relative atomic masses (atoms or
ions) in a substance. There are no units, as it is the mass of a compound relative
to the mass of 1/12 the mass of one atom of 12C. It is a ratio.
 It can be worked out using: molar mass = mass ÷ mol
o The molar mass of Al2S3 is 150.3 g mol-1



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, 4. know what is meant by the terms ‘empirical formula’ and ‘molecular formula’
 Empirical formula = the simplest whole number ratio of atoms of each element in
a compound
 Molecular formula = the actual number of atoms of each element in a molecule
o Molecular formula – ‘in a molecule’
o Empirical formula – ‘in a compound’


5. be able to use experimental data to calculate
i. empirical formula
 We need to find the number of moles of each atom present in the compound
 Experiments can tell us the mass of each element present in a compound
 Using the equation n = m ÷ Mr, we can work out the number of moles of atoms of
each element present
 Once we know the number of moles of atoms of each element present, we can
find the simplest whole number ratio between them
 Finally, we express this information as a chemical formula
o A compound is 52.9% aluminium and 47.1% of oxygen by mass
o Assuming that we have in total 1 g of the substance, we can guess the
masses of the different elements in the compound. The total mass used
doesn’t matter, as long as the ratio of the masses of each element remains
the same.
 Al – 0.529g
 O – 0.471g
o We can work out the molar mass (g/mol) of each element by looking at its
Mr
 Al – 27.0 g/mol
 O – 16.0 g/mol
o We can work out the number of moles with mole = mass ÷ molar mass
 Al – 0.01959… mol
 O – 0.0294… mol
o We can work out the ratio, by dividing both numbers by the smallest value
of moles (in this case, divide by 0.01959)
 Al –1
 O – 1.5
o However, this is not a whole-number ratio. The empirical formula of the
compound is Al2O3


ii. molecular formulae including the use of pV = nRT for gases and volatile
liquids
 The first way to work out the molecular formula is by using the formula mass of
the empirical formula and the formula mass of the molecular formula.


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