Question

Combustion analysis of 0.800 grams of an unknown hydrocarbon yields 2.613 g CO2 and 0.778 g H2O. What is the percent composition of the hydrocarbon?

Answer 1

let in compound number of moles of C and H be x and y respectively

Number of moles of CO2 = mass of CO2 / molar mass CO2
= 2.613/44
= 0.059386

Number of moles of H2O = mass of H2O / molar mass H2O
= 0.778/18
= 0.043222

Since 1 mol of CO2 has 1 mol of C
Number of moles of C in CO2= 0.059386
Number of moles of H = 2*0.043222 = 0.086444
mass of C = moles of C * molar mass of C
= 0.059386 mol * 12 g/mol
= 0.713 g

mass of H = moles of H * molar mass of H
= 0.086444mol * 1 g/mol
= 0.0864 g

% of C = mass of C * 100 / total mass
= 0.713*100/0.800
= 89.1 %
% of H = 100 - % of C
= 100 - 89.1
= 10.9 %