Practice: Valproic acid, used to treat seizures, is composed of carbon, hydrogen and oxygen. A 0.165 g sample is combusted to produce 0.166 grams of water and 0.403 grams of carbon dioxide. If the molar mass is 144 g/mol, what is the molecular formula?
Under Combustion Analysis, the empirical formula of a compound is determined through a combustion reaction.
Combustion Analysis
Concept #1: Combustion Analysis
Example #1: A 0.2500 g sample contains carbon, hydrogen and oxygen and undergoes complete combustion to produce 0.3664 g of CO2 and 0.1500 g of H2O. What is the empirical formula of the compound?
Practice: When hydrocarbons are burned in a limited amount of air, both CO and CO2 form. When 0.460 g of a hydrocarbon was burned in air, 0.477 g of CO, 0.749 g CO2, and 0.460 g of H2O were formed. What is the empirical formula of the compound?
Concept #2: Combustion of Non-Hydrocarbons
Example #2: A solvating agent (MW = 147.0 g/mol) that contains C, H, and Cl is used for spectroscopic processes. Determine its molecular formula when a 0.250 g sample creates 0.451 g CO2 and 0.0617 g of H2O upon combustion.
Practice: If a compound made of calcium, carbon, nitrogen undergoes combustion in oxygen to create 2.389 g CaO, 1.876 g CO2, and 3.921 g NO2, determine its empirical formula.
Practice: The combustion of 4.16 grams of a compound which contains only C,H,O and F yields 7.7 g CO2 and 2.52 g H2O. Another sample of the compound with a mass of 3.63 g is found to contain 0.58 g F. What is the empirical formula of the compound?