Analysis of 0.89 g of an organic compound containing C, H, N and O gave 1.32 g of CO2, 0.63 g of H2O and 224 mL of N2 at STP (molar volume 22.4 L mol^-1). The molar mass of the compound is 89 g mol^-1 (C = 12, H = 1, N = 14, O = 16). Its molecular formula is:

JEE Advanced levelSingle correct optionDifficulty 3 of 3mc030

Analysis of $0.89 g$ of an organic compound containing C, H, N and O gave $1.32 g$ of $CO X_{2}$ , $0.63 g$ of $H X_{2} O$ and $224 mL$ of $N X_{2}$ at STP (molar volume $22.4 L mo l^{- 1}$ ). The molar mass of the compound is $89 g mo l^{- 1}$ (C = 12, H = 1, N = 14, O = 16). Its molecular formula is:

= 1.32 \times 12/44 = 0.36

g; H

= 0.63 \times 2/18 = 0.07

g; N

= 0.224/22.4 \times 28 = 0.28

g; O by difference

= 0.18

g. Moles: C 0.03, H 0.07, N 0.02, O 0.01, ratio

3 : 7 : 1 : 2

. Empirical formula mass

= 89

, equal to the molar mass, so the formula is

C X_{3} H X_{7} NO X_{2}