Question

6.00 g of a certain Compound X, known to be made of carbon, hydrogen and perhaps oxygen, and to have a molecular molar mass of 78. g/mol, is burned completely in excess oxygen, and the mass of the products carefully measured: product carbon dioxide mass 20.31 g 4.16 g water Use this information to find the molecular formula of X. X 5 ?

Answer 1

|44.01 g of CO2 contains 12.01 g of C. so mass of C present in Mass of Hydrocarbon Mass of CO2 6.00 20.31 * (12.01 ÷44.01) g 20.31 g g of CO2 20.31 Mass of H2O 4.160 g 5.542 g Molar Mass of CO2 44.01 g/mol 18.015 g of H2O contains 2*1.008 g of H 18.015 g/mol 4.16 * (2* 1.008 ÷18.015) g so mass of H present in g of H2O Molar mas of H2O 4.16 0.466 g Mass of O = Mass of Hydrocarbon - Mass of C -Mass of H 5.542 g 0.466 g 6 g -0.01 g ; so can be neglected O mass Now we get 5.5424 g 0.47 g 0.000 g н Notes Atom н Mass in g 5.542 0.466 Molar Mass in g/mol 1.008 Moles of atom = Mass in g / Molar Mass (g/mol) 12.01 5.542 12.01 0.466 1.008 Convert Mass to moles Divide mass with molar mass 0.461 0.462 Divide the all mole values with small value to get simplest Simplest ratio 0.461 0.461 0.462 0.461 ratio 1.000 1.001 Nearest whole number 1.0000 1.0 Multiply with to get nearest whole number Whole number so we get н So emperical formula Emperical formula mass *12.01 + *1.008 CH g mol-1 13.02 Molar mass = g/mol 78 where n= Molar mass / emperical formula mass Emperical formula * n Molecular Formula = CH 6. 78.00 13.02 Molecular Formula = C6H6