Answer:
Explanation:
9.708 mg of CO₂ will contain 12 x 9.708 / 44 = 2.64 g of C .
3.969 mg of H₂O will contain 2 x 3.969 / 18 = .441 g of H .
mg of O in given liquid B = 3.795 - ( 2.64 + .441 ) = .714 mg
ratio of mg of C , H , O in the compound = 2.64 : .441 : .714
ratio of no of atoms of C , H , O in the compound
= 2.64 / 12 : .441 /1 : .714 / 16
= .22 : .44 : .0446
= .22 / .22 : .44 / .22 : .044 / .22
= 1 : 2 : .2
1 x 5 : 2 x 5 : .2 x 5
= 5 : 10 : 1
empirical formula of the compound = C₅H₁₀O
Volume of 89.8 mL at 1 .00 atm at 200⁰C
volume of gas at 1 atm and 0⁰C = 89.8 x 273 / 473 mL
= 51.83 mL
51.83 mL weighs .205 g
22400 mL will weigh .205 x 22400 / 51.83 g
= 88.6 g
So molecular weight = 88.6
Let molecular formula be (C₅H₁₀O)ₙ
molecular weight = n ( 5 x 12 + 10 + 16 )
= 86 n
86 n = 88.6
n = 1 approx
So molecular formula is same as empirical formula
C₅H₁₀O is molecular formula .
