$\begin{array}{1 1}(a)\;C_4O_3&(b)\;C_{12}O_9\\(c)\;C_6 O_2&(d)\;CO\end{array}$

The empirical formula $=C_4O_3$

Empirical formula mass =$(4\times 12)+(3\times 16)=96$

Molecular mass =290

$n=\large\frac{Mol.mass}{Emp.mass}=\large\frac{290}{96}$

$\Rightarrow 3$(approx)

Molecular formula=$n\times Empirical\;formula$

$\qquad\qquad\qquad\;=3\times C_4O_3$

$\qquad\qquad\qquad\;= C_{12} O_9$

Hence (b) is the correct option.

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