Advanced Empirical Formulae

Combustion 0.202 g of a sample of organic compound containing elements carbon, hydrogen and oxygen to produce carbon dioxide 0.361 g and 0.147 g of water. If the relative molecular mass is 148, determine the molecular formula.

ANSWER:

mass of carbon in carbon dioxide = 12 x 0.361 / 44 = 0.0984 g
mass of hydrogen in water = 2 x 0.147 / 18 = 0.0163 g
mass  of oxygen in compound = 0.202 - (0.0984 + 0.0163) = 0.0873 g

	Carbon, C	Hydrogen , H	Oxygen , O
Mass (g)	0.0984	0.0163	0.0873
Number of mole (mol)	0.0082	0.0163	0.0055
Mole ratio	0.0082/0.055 = 1.5	0.0163/0.0055 = 3	0.0055/0.0055 = 1
Simplest ratio	3	6	2
Empirical formula	C3H6O2

Molecular formula =
(C₃H₆O₂)n = 148
3(12) + 6(1) + 2(16) = 148
74n = 148
n = 2

molecular formula = C₆H₁₂O₄