Combustion Analysis A 0.1156g sample composed of C, H, N is burned to produce 0.1638g CO2 and 0.1676g H2O Assume all C in the sample is converted to CO2 Assume all H is converted to H2O To solve: All C CO2, all HH2O (N is remainder) Convert g CO2 to moles CO2 then find moles C Convert g H2O to moles H2O, then find moles H -gN = 1 1

Limiting Reactants Reactant that is “used up” in a chemical reaction Excess reactant – reactant that is left over Given amounts of more than one reactant 2 ways to solve: Find the moles of each reactant, compare to moles in the reaction Solve the problem twice (with each reactant), the reactant that produces the smallest amount is limiting 2 2

Why Use Excess Reactant? Can cause the reaction to go faster Use the least expensive as the excess Can shift the reaction towards the products Equilibrium reaction – reversible reaction 3 3

Yield Theoretical yield = calculated amount of product Use stoichiometry Experimental yield = actual yield Amount obtained in a lab Would be given in a problem “obtained”, “recovered”, etc. Percent yield = experimental/theoretical x 100 Normally less than 100% Incomplete reaction Impure reagents Experimental error Side reactions 4 4