An aromatic compound contains `69.4%` carbon and `5.8%` hydrogen. A sample of `0.303g` of this compound was analysed for nitrogen by Kjeldahl's method. The ammonia evolved was absorbed in 50 ml of 0.05 M sulphuric acid. The excess of acid required 25 ml of 0.1 M sodium hydroxide for neutralization. Determine the molecular formula of the compound if its molecular weight is 121. Draw two possible structures for this compound.

Answer

Calculation of `%` nitrogen
`50ml "of" % nitrogen
(`because` Normality of `H_(2)SO_(4))=2xx` molarity)
Excess of acid requires 25 ml of `0.1M` or `0.1 N NaOH`
(`because` Normality of NaOH=molarity of NaOH)
25 ml of 0.1N NaOH-=25 ml of `0.1 N H_(2)SO_(4)`
`therefore` vol. of `0.1 NH_(2)SO_(4)` used for the neutralisation of `NH_(3)=50-25=25ml`
NOw we know that, `%` of nitrogen `=(1.4xx"Normaility of acid"xx"Voll. of acid")rarr`
`=(1.4xx0.1xx25)/(0.303)=11.55%`; Hence `%` of oxygen `=100-(69.4+5.8+11.55)=13.25`