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Question
On analysis, a substance was found to contain:
C = 54.54%, H = 9.09%, O = 36.36%
The vapour density of the substance is 44, calculate its molecular formula.
Solution
Molecular weight of the compound = 88
Empirical formula weight of the compound = 2 × 12 + 4 × 1 + 16
= 24 + 4 +16
= 44
n = `"Molecular weight"/"Empirical formula weight"`
= `88/4`
= 2
∴ The molecular formula of the compound = C4H8O2
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