0.246 g of an organic compound containing 58.53% carbon and 4.06% hydrogen gave 22.4 mL of nitrogen at STP. What is the empirical formula of the compound?

Options

A. `C_(2)H_(7)N_(2)`
B. `C_(6)H_(5)NO_(2)`
C. `C_(2)H_(7)NO_(2)`
D. `C_(2)H_(2)NO_(2)`

Answer

B

Explanation

(i) Calculation of precentage composition : (i) Precentage of carbon = 58.53 (give).
(ii) Precentage of hydrogen = 4.06 (give).
(iii) Percentage of nitrogen .
` = (28 xx " vol. of" N_(2) "at STP" xx 100)/(22400 xx "Wt.of compound")`
`= (28 xx 22.4 xx 100)/(22400 xx 0.246) = 11.38%`
(iv) Precentage of oxygen
= 100 - (percentage of C + precentage of H + percentage of N) = 100 - (58.8 + 4.06 + 11.38) = 26.03
(ii) Calculation of empirical formual :

Hence the empirical formula of the compound is `C_(6)H_(5)NO_(2)`