Question

C7H6O3 + CH3COOH —> C9H8O4 + H2O
The above reaction between salicylic acid (C7H6O3) and acetic acid shows the synthesis of aspirin (C9H8O4). If the percent yield of the reaction is 33.3% then what mass of aspirin could be recovered from the reaction of 18.8 g of salicylic acid?

Answer 1

C7H6O3 + CH3COOH → C9H8O4 + H2O

The above reaction between salicylic acid (C7H6O3) and acetic acid shows the synthesis of aspirin (C9H8O4). If the percent yield of the reaction is 33.3% then what mass of aspirin could be recovered from the reaction of 18.8 g of salicylic acid?

Solution:

1. $n = \frac{m}{M}$;

$M(C_{7}H_{6}O_{3}) = 138\frac{g}{mole}$;

$n(C_{7}H_{6}O_{3}) = \frac{18.8g}{138\frac{g}{mole}} = 0.14$ mole;

2. $n(C_{9}H_{8}O_{4}) = n(C_{7}H_{6}O_{3})$. It is from the equation of reaction;

3. $m = n \times M$;

$M(C_{9}H_{8}O_{4}) = 180\frac{g}{mole}$;

$m(C_{9}H_{8}O_{4}) = 0.14$ mole $\times 180\frac{g}{mole} = 25.2$ g;

4. $\eta = \frac{m(\text{practical})}{m(\text{theoretical})} \times 100\%$;

$m(\text{practical}) = \frac{\eta \times m(\text{theoretical})}{100\%}$;

$m(C_{9}H_{8}O_{4}\text{practical}) = \frac{33.3\% \times 25.2g}{100\%} 8.4g$;

Answer: $m(C_{9}H_{8}O_{4}\text{practical})$ is 8.4g.

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