Answer:
Empirical formula= C4H20
Molecular formula = C8H4O2
Step-by-step explanation:
It is given that 81.7 mg produced 222 mg of CO2 and 45.4 mg of H2O.
Note:Molar mass of C = 12g/mol
Molar mass of CO2 = 44g/mol
Therefore mass of C in 222mg of CO2= 12/44*(0.222)
=0.0605g
Note: Molar mass of H=1g/mol
Molar mass of H2O=18g/mil
Therefore, mass of H in 45.4mg of H2O= 1/18*(0.0454)
=0.0025g
Therefore, Mass of O = 0.0817-0.0025-0.0605
=0.0187g
Divide through by the atomic mass:
C=0.0605/12=0.00504
H=0.002/1=0.002
O=0.0187/16=0.00117
Divide through by the smallest number
C=0.00504/0.00117 = 4.3 ; Approx 4
H=0.002/0.00117=1.7 ; Approx 2
O=0.00117/0.00117=1
The empirical formula= C4H20
To calculate Molecular formula, (C4H20)n = 162
((12*4)+(1*2)+16)n=162
66n=162
n=162/66=2.45 ; Approx 2
Molecular formula = C8H4O2