Question

1. aspirin, ( c_{9}h_{8}o_{4} ), is produced by the reaction of salicylic acid, ( c_{7}h_{6}o_{3} ), and acetic anhydride, ( c_{4}h_{6}o_{3} ) according to the following balanced equation:

( c_{7}h_{6}o_{3}(s)+c_{4}h_{6}o_{3}(l)\to c_{9}h_{8}o_{4}(s)+c_{2}h_{4}o_{2}(l) )
if 2.04 g of aspirin, ( c_{9}h_{8}o_{4} ), is produced from the reaction of 3.03 g ( c_{7}h_{6}o_{3} ) with excess ( c_{4}h_{6}o_{3} ), what is the percentage yield?
(molar mass ( c_{7}h_{6}o_{3}=138.1g/mol ), ( c_{9}h_{8}o_{4}=180.2g/mol ))
a. 28.8 %
b. 29.0 %
c. 50.9 %
d. 51.6 %
e. 67.3 %

Explanation:

Step1: Calculate the moles of salicylic acid ($C_7H_6O_3$)

Use the formula $n=\frac{m}{M}$, where $n$ is the number of moles, $m$ is the mass, and $M$ is the molar mass.
For $C_7H_6O_3$, $m = 3.03\ g$ and $M=138.1\ g/mol$.
$n_{C_7H_6O_3}=\frac{3.03\ g}{138.1\ g/mol}\approx0.0219\ mol$

Step2: Determine the moles of aspirin ($C_9H_8O_4$) produced theoretically

From the balanced equation $C_7H_6O_3(s)+C_4H_6O_3(l)\to C_9H_8O_4(s)+C_2H_4O_2(l)$, the mole ratio of $C_7H_6O_3$ to $C_9H_8O_4$ is $1:1$.
So, $n_{C_9H_8O_4}^{theoretical}=n_{C_7H_6O_3} = 0.0219\ mol$

Step3: Calculate the mass of aspirin produced theoretically

Use $m = n\times M$. For $C_9H_8O_4$, $n = 0.0219\ mol$ and $M = 180.2\ g/mol$
$m_{C_9H_8O_4}^{theoretical}=0.0219\ mol\times180.2\ g/mol\approx3.95\ g$

Step4: Calculate the percentage yield

Use the formula $\text{Percentage yield}=\frac{m_{actual}}{m_{theoretical}}\times100\%$
Here, $m_{actual} = 2.04\ g$ and $m_{theoretical}\approx3.95\ g$
$\text{Percentage yield}=\frac{2.04\ g}{3.95\ g}\times100\%\approx51.6\%$

Answer:

D. 51.6%