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Empirical and molecular formulas from percent composition A compound has a percent composition of 39.97%C, 13.41% H and the rest N. Determine the empirical formula The empirical formula gives the smallest whole number of ratio of atoms, the mole ratio can also be solved o Assume 100g of sample, convert percent composition to mass of each element o Convert mass to moles of each element o Compare the moles of each element to find ratio o Use a coefficient if needed to reach a whole number ratio o Report empirical formula To calculate: %mass = mass of element/ molar mass of compound all x100% Theoretical yield: the greatest possible amount of product that can be made in a chemical reaction based on the amount of the limiting reactant o All of the limiting reactants will be completely Percent yield: the percentage of the theoretical yield of a chemical reaction that is usually produced o The actual amount produced needs to be provided %yield = actual/theoretical x 100% Limiting reactant or reagent: the reactant that limits the amount of product that can be reproduces Excess reactant: the reactant that does not limit the amount of product that can be produced, some will be left over Example of finding molecular formula: A severely poisons compound is 21.46% K. 26.34% O and 52.20% Os and has a molar mass of 364.42 g/mol what is its molecular formula 21.46% K(1 mol K/ 39.098) = .5489 mol K/ .2744 = 2.001 Mol K 26.34 g O (1 mol O/ 15.999g O) = 1.646 mol O / .2744 = 5.998 mol O 52.20 g Os (1 mol Os/ 190.23 g Os) = .2744 mol Os/ .2744 = 1 Mol Os Answer: K2OsO6 Combustion analysis o Mass percent is determined by combustion analysis o The compound is weighted, heated in O2, all its bonds break down and form oxides, which are weighed o C -> CO2, H-> H2O, O is determined indirectly

Reactants = reagents/reactants Products are to the right Physical states ▪ (s) solid ▪ (l) liquid ▪ (g) gas ▪ (aq) aqueous Sometimes all reactants are completely consumed: sometimes one or more is in excess o o

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