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Q.
59 g of an amide obtained from a carboxylic acid, RCOOH, upon heating with alkali liberated 17 g of ammonia. The acid is
Organic Chemistry – Some Basic Principles and Techniques
Solution:
$RCONH_2 + NaOH \rightarrow RCOONa + NH_3$
As 1 mole of NH3 is given by one mole of amide.
Thus mol. mass of $RCONH_2$ is
A + 12 + 16 + 14 + 2 = 59 or A + 44 = 59
A = 59 - 44 = 15
Hence alkyl group with mol. mass A = 15 is $CH_3$
Thus, acid is $CH_3COOH$ .