1. empirical formula
Combustion analysis of a hydrocarbon showed that g of the compound contained g of carbon. Determine the empirical formula of the compound.

2. empirical formula of methane
elements in compound
carbon
hydrogen
mass in grams
(or in %)
(74.9%)
(25.1%)
number of moles
47.98/12.01
= 3.99 mole
(74.9/12.01 = 6.24 moles)
16.02/1.01
= moles
(25.1/1.01 = 25.1 moles)
ratio of moles
4 (6.25)
16 (25)
most simple ratio of moles
4/4 = 1
16/4 = 4
empirical formula
CH4

3. molecular formula
Calculate the molecular formula of a compound with a molecular mass of 84g and an empirical formula of CH2.

4. 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

5. 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?

6. AGAIN DO NOT ROUND UP OR DOWN BUT LOWEST WHOLE NUMBER RATIO
molecular formula
carbon
hydrogen
mass of element
93.71
6.29
number of moles
93.71/12 = 7.8
6.29/ 1 = 6.29
DO NOT ROUND UP OR DOWN
most simple ratio
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 5 4
empirical formula:
C5H4
ratio molecular formula /empirical formula:
128g/ 64g = 2
molecular formula = 2 x C5H4 =
C10H8

7. 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.

8. Empirical formula hydrated salt
compound
MgSO4
H2O
mass (in g or %)
4.58
4.86
molar mass
120.08
18.01
number of moles
4.58/120 = 0.039
4.86/18 = 0.27
most simple molar ratio
0.039/0.039 = 1
0.27/0.039 = 6.93
empirical formula
MgSO4.7H2O