Relative Atomic Mass (Ar)
Relative Atomic Mass: The weighted mean mass of an atom of an element compared with 1/12 of the mass of K,atom of C-12.
Relative Molecular Mass (Mr)
Relative Molecular Mass: The weighted mean mass of a molecule of an element compared with 1/12 of the mass of an atom of C-12.
To Calculate the relative molecular mass, Mr, of a molecule you add up all of the individual relative atomic masses for the elements contained within the molecule.
Na=23 + O=16 + H=1
Mole: the amount of substance containing as many particles as there are carbon atoms in exactly 12g of C-12.
Molar Mass: The mass, in g, per mole of a substance. Units are gmol-1
Molar masses are calculated using the following equation:
Amount (n) = mass / M
Calculate the amount, in moles, of carbon in 11g of carbon dioxide (CO2)
n=mass/m n=11/m C=12 and O=16 m= 12 + (16x2) = 44
Calculate the mass, in g, in 6.0mol of SiO2
Amount (n) = mass/M
Si= 28.1 and O=16 M= 28.1 + (16x2) = 60.1
mass= nxM mass= 6.0 x 60.1
The Avogardo Constant
The number of particles in one mole of any substance is called the Avogadro Constant. It is represented using the symbol NA and has the value 6.02x10^23mol-1.
Simply for every One Mole of any substance there are 6.02x10^23 atoms
The Formula is: Atoms/Ions = amount of moles X Avogadro Constant
= n X NA
EG Calculate the total number of atoms of chlorine in 35.5g of chlorine atoms
n = Mass/M n= 35.5/35.5= 1mol
Atom = n x NA = 1 x 6.02 x 10^23
= 6.02 x 10^23
Empirical Formula: The simplest whole number ratio of atoms of each element present in a compound.
EG A chloride of Iron contains 65.5% chlorine. What is the empirical formula?
element present: Iron Chlorine
% by Mass (g): 34.5 65.5
Relative atomic mass (Ar): 55.8 35.5
Moles (mass/M): 34.5/55.8= 0.618 65.5/35.5= 1.845
Ratio (moles/0.618) 1 : 3
Empirical Formula = FeCl3
Water of Crystallisation
Hydrated: When water of crystallisation is present in a crystal compound.
Anhydrous: When all the waters of crystallisation have been removed from a compound.
Water of Crystallisation: The water present in a compound giving the compound a crystalline appearance.
EG A sample of 7.38g of MgSO4 crystals ost 3.78g of water on heating.Calculate x in the formula MgSO4 . xH2O
Mass 3.60 3.78
Mr 24.3+32.1+(16x4)= 120.4 (1x2)+16= 18.0
Moles (mass/Mr) 0.0299 0.21
Ratio 1 : 7
therefore x = 7
Molecular Formula: the actual number of atoms of each element in a molecule.