Question

A 4.043g sample of a compound containing only carbon, hydrogen, and oxygen is burned in an excess of dioxygen, producing 9.191g CO2 and 3.762g H2O. What percent by mass of oxygen us contained in the original sample? (please try to show steps)

Answer 1

Number of moles of CO2 = mass of CO2 / molar mass CO2

= 9.191/44

= 0.2089

Number of moles of H2O = mass of H2O / molar mass H2O

= 3.762/18

= 0.209

Since 1 mol of CO2 has 1 mol of C

Number of moles of C in CO2= 0.2089

Since 1 mol of H2O has 2 mol of H

Number of moles of H = 2*0.209 = 0.418

Molar mass of O = 16 g/mol

mass O = total mass - mass of C and H

= 4.043 - 0.2089*12 - 0.418*1

= 1.1184

mass % of O = mass of O * 100 / mass of compound

= 1.1184*100 / 4.043

= 27.7 %

Answer: 27.7 %