Question

Given 7.90 g of butanoic acid and excess ethanol, how many grams of ethyl butyrate would be synthesized, assuming a complete 100% yield?

Answer 1

Answer : 9.293 g

Explanation : The reaction is 

$C_{4}H_{8}O_{2} + C_{2}H_{5}OH ----\ \textgreater \ C_{6}H_{12}O_{2} + H_{2}O$

Here, if we assume the stoichiometric esterification ratio as 1:1 then,

no. of moles of  $C_{4}H_{8}O_{2}$ = 7.05 g/ 88 = 8.011 $X 10^{-2}$ moles

so here, no. of moles of ethyl butyl rate = no. of moles of butyric acid ;

$\frac{m (C_{4}H_{8}O_{2})}{M (C_{4}H_{8}O_{2})} = \frac{m (C_{6}H_{12}O_{8})}{M (C_{6}H_{12}O_{8})}$

$m(C_{6}H_{12}O_{8}))$ = (7.05 /88.11) X 116.16 = 9.29 g