A primary alkanol has a molecular mass of 60. What is the structural formula of the compound? [C= 12.0, H= 1.0, O= 16.0)
Alkanols have the formula \(C_{n}H_{2n+1}OH\).
⇒ \( 12n + (1\times (2n+1)) + 16.0 + 1.0 = 60\)
= (14n +18 = 60\)
\(n = \frac{60-18}{14} = \frac{42}{14} = 3\)
Hence, the alkanol is gotten by putting n=3,
=\(C_{3}H_{7}OH\)
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