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JEE Advanced levelSingle correct optionDifficulty 3 of 3mc147

A compound contains 40% C, 6.67% H and 53.33% O by mass. 0.90 g of its vapour occupies 224 mL at STP (22.4 L/mol). Its molecular formula is ( $C = 12$ , $H = 1$ , $O = 16$ )

A $C X_{3} H X_{6} O X_{3}$
B $CH X_{2} O$
C $C X_{2} H X_{4} O X_{2}$
D $C X_{6} H X_{12} O X_{6}$

Answer: A

Mole ratio C:H:O = 40/12 : 6.67/1 : 53.33/16 = 3.33 : 6.67 : 3.33 = 1:2:1, empirical formula

CH X_{2} O

(30). Molar mass = 0.90/(0.224/22.4) = 90. n = 90/30 = 3, so

C X_{3} H X_{6} O X_{3}

.

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