Answer to Question #266161 in General Chemistry for grizz

Question

If 65.7 g of aspirin (C₉H₈O₄) are produced from 79.8 g of C₇H₆O₃, what is the percent yield from the reaction below?

C₇H₆O₃ (s) + C₄H₆O₃ (l) → C₉H₈O₄ (s) + HC₂H₃O₂ (aq).

Answer 1

Taking $C=12,H=1, and \>O=16$

Molar weight of $C_7H_6O_3=(12×7)+(1×6)+(16×3)$

$\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>=138g$

Molar weight of $C_9H_8O_4=(12×9)+(1×8)+(16×4)$

$=180g$

Theoretically $138g$ of $C_7H_6O_3$ yields $180g$ of Aspirin

$\therefore79.8g$ of $C_7H_6O_3$ should yield

$\frac{79.8}{138}×180=104.09g$ of Aspirin

Actual yield $=65.7g$

$\therefore\%$ Yield $=\frac{Actual\> Yield}{Theoretical \>Yield}×100\%$

$=\frac{65.7}{104.09}×100\%=63.118\%$