Answer to Question #261360 in General Chemistry for Dan

Question

How do you get from 103.94-100% to 50.9-x?

Answer 1

Kindly note:

This question is seeking clarification of the question;

" If 50.9g of aspirin ($C_9H_80_4)$ are produced from 78.9g of $C_7H_60_3$

What is the percentage yield from the reaction below?;

$C_7H_60_3{(s)}+C_4H_60_{3}{(s)}$

$\to\>C_9H_80_4{(s)}+HC_2H_30_2{(aq)}$ "

Solution

Taking $C=12,H=1,0=16$

Molar weight of $(C_7H_60_3)=(12×7)+(1×6)+(16×3)$

$=138g$

Molar weight of $(C_9H_80_4)=(12×9)+(1×8)+(16×4)$

$=180g$

Theoretically, $138g$ of $(C_7H_60_3)$ yield $180g$ of Aspirin

Proportionally $79.8g$ of $(C_7H_60_3)$

should yield;

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

(104.09g is the Theoretical value)

But the actual yield by $79.8g$ of $C_2H_60_3$ is $50.9g$ of Aspirin which is smaller than expected $104.09g$

Meaning :

Either, some of materials

1) were wasted

2) They failed to react or

3) The products were not

captured

$\therefore$ Percentage yield

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

$=\frac{50.9}{104.09}×100\%$

48.9$\%$