An aromatic compound contains 69.4% C and 5.8% H. A sample of 0.303g of this compound was analysed for nitrogen by Kjeldahl method. The ammonia evolved was absorbed in 50mL of 0.05M H_(2)SO_(4). The excess acid required 25mL of 0.1M NaOH for neutralisation. Determine the molecular formula of the compound if its molecular weight is 121. Draw two possible structures for this compound.

An aromatic compound contains 69.4% C and 5.8% H. A sample of 0.303g o

1.	An aromatic compound contains 69.4% C and 5.8% H. A sample of 0.303g of this compound was analysed for nitrogen by Kjeldahl method. The ammonia evolved was absorbed in 50mL of 0.05M H_(2)SO_(4). The excess acid required 25mL of 0.1M NaOH for neutralisation. Determine the molecular formula of the compound if its molecular weight is 121. Draw two possible structures for this compound.
Answer» Solution :Let us first calculate the percentage of N in the compound. As all the N is converted to `NH_(3)`, applying POAC for N atoms Moles of N in the compound= moles of N in `NH_(3)` `=1 XX` moles of `NH_(3)` = EQ. of `NH_(3)` = m.e. of `NH_(3)//1000` = m.e. of `H_(2)SO_(4)//1000` `=("total m.e. of " H_(2)SO_(4)-" m.e. of excess" H_(2)SO_(4))/(1000)` `=("total m.e. of " H_(2)SO_(4)- "m.e. of " NaOH)/(1000)` `=(50 xx 0.1 -25 xx 0.1)/(1000)` =0.0025 `therefore` wt. of `N= 0.0025 xx 14= 0.035` and % of `N= (0.035)/(0.303) xx 100= 11.55%` `therefore` % of O `= 100- (69.4+ 5.8 + 11.55)` = 13.25% `therefore` moles of `C: H: N: O= (69.4)/(12): (5.8)/(1): (11.55)/(14): (13.25)/(16)` `=5.8: 5.8: 0.825: 0.825` `=7: 7:1:1` `therefore` empirical formula of the aromatic compound is `C_(7)H_(7)NO` (121) As the molecular weight is also 121, molecular formula is `C_(7)H_(7)NO` Since the compound is aromatic, it may be written as `C_(6)H_(5)CH=NOH`(benzaldoxime) with the following isomeric STRUCTURES: `{:(C_(6)H_(5)-C-H),("\|\|"),(""N-OH):} and {:(C_(6)H_(5)-C-H),("\|\|"),(""HO-N):}`