Question

In a compound carbon \( =52.2 \% \), hydrogen \( =13 \% \), oxygen \( =34.8 \% \) are present and molecular mass of the compound is 92. Calculate molecular formula of the compound?

Answer 1

Emperical mass  = 2 * 12 + 6 * 1 + 16 

                               = 46

Molecular mass \( =92 \) \[ \begin{aligned} \text { Molecular Formula } & =\text { Emperical formula } \times \frac{\text { M.mass }}{E \cdot \text { mass }} \\ & =C_{2} H _{6} O \times \frac{92}{46} \\ & =\left(C_{2} H _{6} O \right) \times 2 \\ & = C _{4} H _{12} O _{2} \end{aligned} \]

Answer 2

Empirical formula = C₂H₆O

Empirical mass = 12 × 2 + 16 + 6 = 46

Molecular mass = 92

Therefore,

\(n = \frac{\text{Molecular mass}}{\text{Empirical mass}}\)

\(= \frac{92}{46}\)

\(= 2\)

Molecular formula = C₂H₆O × 2

= C₄H₁₂O₂