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Molecular weight of $CH_3CH \text{=} CH_2 = 42$

Molecular weight of $Cl_2 = 71$

Molecular weight of $Cl CH_3CH \text{=} CH_2 = 76.5$

% Atom economy = $ \frac{ \text{Molecular weight of desired product} } {\text{Sum of the molecular weight of desired product}} \times 100$

$= \frac{76.5}{42 +71} \times 100$

$\therefore$ **% Atom economy = 67.7**

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