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Question
Solve the following problem:
A compound on analysis gave the following percentage composition by mass: H = 9.09; O = 36.36; C = 54.55. Mol mass of compound is 88 g mol−1. Find its molecular formula.
Solution
Given: Percentage of H, O, C = 9.09%, 36.36%, 54.55% respectively.
Molar mass of the compound = 88 g mol–1
To find: The molecular formula of the compound
Calculation:
Moles of C = `"% of C"/"Atomic mass of C"=54.55/12` = 4.546 mol
Moles of H = `"% of H"/"Atomic mass of H"=9.09/1.0` = 9.090 mol
Moles of O = `"% of O"/"Atomic mass of O"=36.36/16` = 2.272 mol
Hence, the ratio of number of moles of C:H:O is
`4.546/2.272` = 2, `9.090/2.272` = 4 and `2.272/2.272` = 1
Hence, empirical formula is C2H4O.
Empirical formula mass = 24 + 4 + 16 = 44 g mol–1 Hence,
r = `"Molar mass"/"Empirical formula mass"=(88 "g mol"^-1)/(44 "g mol"^-1)` = 2
∴ Molecular formula = r × empirical formula
Molecular formula = 2 × C2H4O = C4H8O2
Molecular formula of the compound = C4H8O2
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