Answer:
a) CH2
b) C18H36
c) 80%
Explanation:
CO2 = 44.009 g/mol --> 78.44 g = 1.782 mol of CO2
H2O - 18.015 g/mol --> 32.12 g = 1.782 mol of H2O
a) 1.782 mol of CO2 contain 1.782 mol of C, 1.782 mol of H2O contain 1.782 x 2 mol of H
Thus, Emperical formulae of Hydrocarbon = CH2
b) CH2 = 14.027 g
252.48g / 14.027 g = 18
Molecular formulae of Hydrocarbon = C18H36
c) 1.782 mol of CH2 was combusted
1.782 x (14.027 g) = 24.996 g
% of sample equaling HydroCarbon = 24.996 g/31.25 g = 79.98 % = 80 %