The Mole Concept

Amount of substance (mol)

Amount of substance (mol)

n=m (g)/Mr

Amount= mass (in grams) ÷ relative Molecular mass

mass (g)

n × Mr

Amount × relative molecular mass

relative molecular mass

m ÷ n

Mass (g) ÷ amount (mol)

Amount of gas

V(dm³) ÷ 24

OR V (cm³) ÷ 24000

Depends on unit of volume

Volume of gas (dm³)

V (dm³) = n ×24

Amount of substance in solution

n=( V×C ) ÷ 1000

Volume (cm³) × Concentration ÷ 1000

If volume is in dm³, do not ÷1000

Neutralisation reaction

Ca×Va×b = Cb×Vb×a

a=acid

b=base

C=concentration

V=volume

Percentage yield

(Actual yield ÷ theoretical yield ) ×100

Percentage purity

(Mass of pure substance ÷ mass of contaminated substance) ×100