Question

etermining a Chemical Formula Submit for Grading Current score: 1/4 pts Sample A 4.287 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 11.09 grams of CO2 and 2.270 grams of H20 are produced In a separate experiment, the molar mass is found to be 136.2 g/mol. Determine the empirical formula and the molecular formula of the organic compound. Enter the elements in the order C, H, O empirical formula - molecular formula = Check & Submit Answer Show Approach

Answer 1

A 4.287 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 11.09 grams of CO, and 2.270 grams of H O are produced. In a separate experiment, the molar mass is found to be 136.2 g/mol. Determine the empirical formula and the molecular formula of the organic compound. Enter the elements in the order C,H,O empirical formula - CAHO molecular formula - C H202 moles of CO2 = (11.09 g)/(44.01 g/mol) = 0.252 mol moles of carbon = 0.252 mol weight of carbon = (0.252 mol) x (12.01 g/mol) = 3.02 g moles of H2O = (2.27 g)/(18.02 g/mol) = 0.126 mol moles of Hydrogen = 2 x 0.126 mol = 0.252 mol weight of Hydrogen = (0.252 mol) (1.01 g/mol) = 0.254 g weight of oxygen = 4.287 g - (3.02 g +0.254 g) = 1.01 g moles of oxygen = (1.01 g)/(16 g/mol) = 0.063 mol Simplest mol ration C = (0.252 mol) / (0.063 mol) = 4 H = (0.252 mol) / (0.063 mol) = 4 O = (0.063 mol) / (0.063 mol) = 1 Empirical formula is: C4H40 Empirical mass = 68.08 g/mol molecular mass = 136.2 g/mol Molecular formula: CgH802

When 3.649 grams of a hydrocarbon, CH, were burned in a combustion analysis apparatus, 11.45 grams of CO, and 4.687 grams of H,0 were produced. In a separate experiment, the molar mass of the compound was found to be 56.11 g/mol. Determine the empirical formula and the molecular formula of the hydrocarbon Enter the elements in the order presented in the question. empirical formula - CH2 molecular formula - CAH moles of CO2 = (11.45 g)/(44.01 g/mol) = 0.26 mol moles of carbon = 0.26 mol moles of H20 = (4.687 g)/(18.02 g/mol) = 0.260 mol moles of Hydrogen = 2 x 0.260 mol = 0,52 mol Simplest mol ration C = (0.26 mol) / (0.26 mol) = 1 H = (0.52 mol) / (0.26 mol) = 2 Empirical formula is: CH2 Empirical mass = 14.03 g/mol molecular mass = 56.11 g/mol ratio = (56.11/ 14.03) = 4 Molecular formula: C4H8