empirical formula
Combustion analysis of a hydrocarbon
showed that 64.00 g of the compound
contained 47.98 g of carbon.
Determine the empirical formula of the
compound.
empirical formula of methane
elements in
compound
mass in grams
(or in %)
number of moles
ratio of moles
most simple ratio
of moles
empirical formula
carbon
hydrogen
47.98 (74.9%)
16.02
47.98/12.01
= 3.99 mole
(74.9/12.01 = 6.24
moles)
4 (6.25)
4/4 = 1
(25.1%)
16.02/1.01
= 16.02 moles
(25.1/1.01 = 25.1
moles)
16 (25)
16/4 = 4
CH4
molecular formula
Calculate the molecular formula of a
compound with a molecular mass of
84g and an empirical formula of CH2.
molecular formula
• mass empirical formula = 14g
• molar mass of formula = 84g
• whole number ratio (n) of molar mass/empirical
mass = 84 g/14 g = 6
• molecular formula = empirical formula x
whole number ratio: (CH2 ) x 6 = C6H12
molecular formula
Naphthalene, best known as
‘mothballs’, is composed of carbon
(93.71%) and hydrogen (6.29%). If
the molar mass of the compound is
128g, what is the molecular formula
of naphthalene?
molecular formula
mass of element
number of moles
most simple ratio
carbon
hydrogen
93.71
6.29
93.71/12 = 7.8
6.29/ 1 = 6.29
DO NOT ROUND UP OR DOWN
7.8/ 6.29 = 1.25
6.29/6.29 = 1
AGAIN DO NOT ROUND UP OR DOWN BUT LOWEST
WHOLE NUMBER RATIO
lowest whole number
ratio = multiply by 4
empirical formula:
ratio molecular formula
/empirical formula:
molecular formula = 2 x
C 5H 4 =
5
4
C 5H 4
128g/ 64g = 2
C10H8
Empirical formula hydrated salt
When 9.44g hydrated magnesium
sulphate, MgSO4.nH2O, was heated until
there was no further mass decrease,
4.58g of anhydrous magnesium was left
behind. Determine the empirical
formula.
Empirical formula hydrated salt
compound
mass (in g or %)
molar mass
number of moles
most simple molar
ratio
empirical formula
MgSO4
4.58
120.08
4.58/120 = 0.039
0.039/0.039 = 1
H2O
4.86
18.01
4.86/18 = 0.27
0.27/0.039 = 6.93
MgSO4.7H2O