\(C_{2}H_{4}\)
\(C_{3}H_{4}\)
\(C_{3}H_{6}\)
\(C_{4}H_{8}\)
Correct answer is C
Given the molecular formula, \(CH_{2}\).
The empirical formula will be \((CH_{2})n\)= 42
\((12+(2\times1))n =42\)
⇒ \(14n =42\)
n = 3, hence the empirical formula is \(C_{3}H_{6}\)
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