Answer to Question #142540 in General Chemistry for s

Question

A sample of a common hydrocarbon is found to contain 40.2% carbon, 6.2% hydrogen and the rest is oxygen. What is its molecular formula if it has a molar mass of 179.0 g/mol?

Answer 1

Carbon - 40.2%

Hydrogen - 6.2%

Oxygen - 53.6%

Firstly we divide by their molecular masses

Carbon - 40.2/12 - 3.35

Hydrogen - 6.2/1 - 6.2

Oxygen - 53.6/16 - 3.35

Then we divide by the lowest ratio

Carbon - 3.35/3.35 - 1

Hydrogen - 6.2/3.35 - 1.85

Oxygen - 3.35/3.35 - 1

$\therefore$ The compound's empirical formula is $(CH_{1.85}O)_n$

Since the molecular mass of the compound is 179.0g/mol,

$(CH_{1.85}O)_n = 179\\ (12+ 1(1.85)n + 16 = 179\\ (29.85)n = 179\\ \therefore n = 6$

$\therefore$ The compound's molecular formula is $(CH_{1.85}O)_6$ = $C_6H_{11}O_6$