A 2.912 g sample of a compounds containing only C, H, and O was completely oxidized in a reaction that yielded 3.123 g of water and 7.691 g of carbon dioxide. Determine the empirical formula and molecular formula of the compound if it has a molar mass of 100.1 g/mol.

Explanation:

18 gram of water contains 2 g of hydrogen

3.123 gram of water will contain 2 x 3.123 / 18 = .347 g of hydrogen .

44 gram of carbon dioxide contains 12 g of carbon

7.691 gram of carbon dioxide will contain 12 x 7.691 / 44 = 2.1 g of carbon .

So the sample will contain 2.912 - ( .347 + 2.1 ) g of oxygen .

= .465 g of oxygen .

moles of Carbon = 2.1 / 12 = .175

moles of hydrogen = .347 / 1 = .347

moles of oxygen = .465 / 16 = .029

Ratio of moles of carbon , oxygen and hydrogen ( C,O,H )

= 0.175 : 0.029 : 0.347

= .175/ .029 : 1 : .347 / .029

= 6 : 1 : 12

So empirical formula = C₆H₁₂O

Let the molecular formula be

molecular weight = n ( 6 x 12 + 12x 1 + 16)

= 100 n

Given 100 n = 100.1

n = 1

Molecular formula = C₆H₁₂O.

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