Question

if 69.9 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)

Explanation:

Step1: Calculate molar masses

The molar mass of $C_7H_6O_3$:
$M_{C_7H_6O_3}=(7\times12.01)+(6\times1.01)+(3\times16.00)=138.12\ g/mol$
The molar mass of $C_9H_8O_4$:
$M_{C_9H_8O_4}=(9\times12.01)+(8\times1.01)+(4\times16.00)=180.17\ g/mol$

Step2: Calculate moles of $C_7H_6O_3$

$n_{C_7H_6O_3}=\frac{m_{C_7H_6O_3}}{M_{C_7H_6O_3}}=\frac{79.8\ g}{138.12\ g/mol}=0.578\ mol$

Step3: Determine moles of $C_9H_8O_4$ from stoichiometry

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 full equation, though the by - product is not needed here), the mole ratio of $C_7H_6O_3$ to $C_9H_8O_4$ is 1:1. So, the theoretical number of moles of $C_9H_8O_4$, $n_{C_9H_8O_4}^{theo}=0.578\ mol$

Step4: Calculate theoretical mass of $C_9H_8O_4$

$m_{C_9H_8O_4}^{theo}=n_{C_9H_8O_4}^{theo}\times M_{C_9H_8O_4}=0.578\ mol\times180.17\ g/mol = 104.14\ g$

Step5: Calculate percent yield

Percent yield $=\frac{m_{C_9H_8O_4}^{actual}}{m_{C_9H_8O_4}^{theo}}\times100\%=\frac{69.9\ g}{104.14\ g}\times100\% = 67.1\%$

Answer:

67.1%