AP CHEMISTRY NOTES Ch 3 Stoichiometry

3.1 Counting by Weighing Find the average mass = total mass of substance / number of substance

3.2 Atomic Masses Find the average atomic mass = total % of amu of isotopes Average Atomic mass Ex: Average atomic mass of carbon 98.89%x12amu+1.11%x13amu=12.01amu (Given atomic mass)

3.3 The mole Abbreviated mol Avogadro’s number Conversion factor 6.022 x 1023 Conversion factor Ex. Carbon 6.022 x 1023 atoms)(12 amu/atom) = 12g 6.022 X 1023 amu = 1g

Molecule Mole Atoms 1 mole= 6.022x1023 particles 1 molecule= ____ atoms Atoms

3.4 Molar Mass Calculating molar mass Molar mass and numbers of molecules

Mass Atoms Mole Particle 1 mole= 6.022x1023 particles 1 mole= molar mass Mass Mole Particle 1 molecule= ____ atoms Atoms

3.5 Percent Composition of Compounds Mass percent 1 (from the formula) = Mass of an element in a compound mass of 1 mol of compound Mass percent of C, H, O in C2H5OH

3.5 Percent Composition of Compounds Try Ex 3.9 Mass percent of C, H, O in C2H5OH Mass of C in 1 mol C2H5OH x 100% = Mass of 1 mol C2H5OH Mass of H in 1 mol C2H5OH x 100% = Mass of O in 1 mol C2H5OH x 100% = 52.14% 13.13% 34.73% Add up 100.00%

3.6 Determining the Formula of a Compound From an experiment to find mass of C, H, O

Unknown Compound (0.1156g) with C, H, N CO2 : 0.1638g (all C converted to CO2) H2O : 0.1676g (all H converted to H2O Mass of C in a compound = 0.1638g CO2 x 12.01g C = 44.01 g CO2 Find mass % of C 0.04470g C x 100% = 0.1156g compound 0.04470g C 38.67 %

Mass % of H CO2 : 0.1638g (all C converted to CO2) H2O : 0.1676g (all H converted to H2O Mass of H in a compound = 0.1676 g H2O x 2.016g H = 18.02 g H2O Find mass % of H 0.01875g H x 100% = 0.1156g compound 0.01875g H 16.22 %

Mass % of N Find mass % of N 100.00% - (% of C + % of H) 100.00% - (38.67% + 16.22%) = Assume there is a 100.00 g compound Mass of carbon Mass of hydrogen Mass of nitrogen 45.11 %N 38.67 g 16.22 g 45.11 g

Moles of C, H, N 38.67 g C x 1 mol C = 12.01 g C 16.22 g H x 1 mol H = 45.11 g N x 1 mol N = 14.01 g N 3.220 mol C 16.09 mol H 3.219 mol N

Smallest whole number ratio of atoms divide by the smallest mol C: 3.220mol = 3.220mol H: 16.09 mol H = N: 3.219 mol N = 1.000 = 1 4.997 = 5 1.000 = 1

Determine the formula CH5N as an empirical formula C2H10N2, C3H15N3 might be a molecular formula Represented by (CH5N)n

Empirical Formula Determination Since mass percentage gives the number of grams of a particular element per 100 grams of compound, base the calculation on 100 grams of compound. Each percent will then represent the mass in grams of that element. Determine the number of moles of each element present in 100 grams of compound using the atomic masses of the elements present.

Empirical Formula Determination Divide each value of the number of moles by the smallest of the values. If each resulting number is a whole number (after appropriate rounding), these numbers represent the subscripts of the elements in the empirical formula. If the numbers obtained in the previous step are not whole numbers, multiply each number by an integer so that the results are all whole numbers.

Molecular Formula Determination-1 Obtain the empirical formula. Compute the mass corresponding to the empirical formula. Calculate the ratio Molecular mass Empirical formula mass The integer from the previous step represents the number of empirical formula units in one molecule. When the empirical formula subscripts are multiplied by the integer, the molecular formula results. Molecular formula = (empirical formula) x molar mass                      empirical formula mass

Molecular Formula Determination-2 Using the mass percentages and the molar mass, determine the mass of each element present in one mole of compound. Determine the number of moles of each element present in one mole of compound. The integers from the previous step represent the subscripts in the molecular formula.

Pg 119 problems 70 72 74 76 78 80

3.7 Chemical equation Chemical equation Reactants Products States The atoms have been reorganized Bonds have been broken New ones have been formed Reactants Products States (s), (l), (g), (aq)

3.8 Balancing Chemical Equations 4 _ NH3(g)+_O2(g)_NO(g)+_H2O(g) 5 4 6