calculations
the mole
the unit for amount of substance (mol)
relative formula mass
The symbol for the relative formula mass is Mr and it refers to the total mass of the substance
To calculate the Mr of a substance, you have to add up the relative atomic masses of all the atoms present in the formula
The number of atoms, molecules or ions in a mole (1 mol) of a given substance is the Avogadro constant. The value of the Avogadro constant is 6.02 x 1023 per mole
calculations linking moles, Ar and Mr
One mole of any element is equal to the relative atomic mass of that element in grams or for a compound the relative formula mass in grams
To find the mass of one mole of a compound, we add up the relative atomic masses
mass = moles x Mr
reacting masses
Chemical equations can be used to calculate the moles or masses of reactants and products
Then, the ratio between the substances is identified using the balanced chemical equation
Once the moles have been determined they can then be converted into grams using the relative atomic or relative formula masses
percentage yield
it is calculated by actual yield / theoretical yield
The percentage yield is a good way of measuring how successful a chemical process is
The theoretical yield is the amount of product that would be obtained under perfect practical and chemical conditions, calculated from reacting masses
The actual yield is the recorded amount of product obtained
In practice, you never get 100% yield in a chemical process for several reasons
Yield is the term used to describe the amount of product you get from a reaction
water of crystalisation example question
a sample of copper (II) sulfate, CuSO4.xH20, weighing 6.25g was heated gently and reweighed on cooling. the new mass of the alt was 4.00g. calculate the value of x and hence find the salt formula.
divide the mass in grams by the Mr for both the water and the copper sulfate, then divide them by the smallest result to got the ratio of copper sulfate molecules to water molecules. the ratio for the water is x
MgO combustion practical
aim: To determine the empirical formula of magnesium oxide by combustion of magnesium
Measure mass of crucible with lid. Add sample of magnesium into crucible and measure mass Strongly heat the crucible for several minutes. Lift the lid frequently to allow air into the crucible for the magnesium to fully oxidise without letting magnesium oxide smoke escape. Continue heating until the mass of crucible remains constant, Measure the mass of crucible and contents
empirical formula
mass of metaL
Subtract mass of crucible from magnesium and the mass of the empty crucible
Mass of oxygen
Subtract mass of the magnesium used from the mass of magnesium oxide
divide each of the two masses by the relative atomic masses of the elements and simplify the ratio. Represent the ratio into the form ‘MxOy‘ E.g, MgO
CuO reduction practical
aim: To determine the formula of copper (II)oxide by reduction with methane
Place metal oxide into a horizontal boiling tube and measure the mass. Support the tube in a horizontal position and pass over a steady stream of methane and burn off the excess gas. Heat the copper (II)oxide strongly until metal oxide changes colour (all O2 removed) Measure mass of the tube remaining metal powder
empirical formula
mass of oxygen
Subtract mass of the remaining metal powder from the mass of metal oxide
mass of metal
Measure mass of the remaining metal powder
Divide each of the two masses by the relative atomic masses of elements and simplify the ratio
definitions
empirical formula
molecular formula
Empirical formula: gives the simplest whole number ratio of atoms of each element in the compound
The molecular formula is the formula that shows the number and type of each atom in a molecule
calculating empirical and molecular formula
molecular formula
empirical formula
Divide the relative formula mass of the molecular formula by the relative formula mass of the empirical formula. Multiply the number of each element present in the empirical formula by this number to find the molecular formula
calculated from knowledge of the ratio of masses of each element in the compound - grams/RAMs