A.P. Ch. 3 Review Work Stoichiometry

Atomic Mass Average of isotope masses based on their abundance Ex. Carbon has atomic mass of 12.01 amu 12C has mass of 12.00 amu and 98.89% 13C has mass of 13.003 amu and 1.11% (0.9889 x 12.00) + (0.0111 x 13.003) = 12.01 amu Diamond (pure Carbon) in Kimberlite

Calculate the Atomic Mass Copper has two isotopes: 65Cu is 30.91%, 63Cu is 69.09%

Mole Like a dozen represents a certain number of an object, a mole represents a certain amount of particles A mole is equal to 6.02x1023 representative particles Ex. 1 mole of carbon contains 6.02x1023 atoms 1 mole of H2O contains 6.02x1023 molecules All of the atomic masses on the periodic table are equivalent to 1 mole of those elements in grams

Molar Mass The mass of one mole of any substance For compounds, add all of the atomic masses of every element together Example: CH4 12.01 g/mol + 4(1.008 g/mol) = 16.04 g/mol

Percent Composition The % of each element in a compound Take total masses of each element, divide by molar mass of compound, multiply by 100

Empirical/Molecular Formulas Molecular formula: shows how many of each element are present in a compound Glucose is C6H12O6 Empirical formula: shows smallest whole number ratio Glucose is CH2O

Determining Chemical Formulas Empirical Formula: Starting with % composition of each element Change % to mass (10% of C is 10 grams C) Convert each mass to moles using molar mass Divide each elements # of moles by smallest # If you get whole numbers you are done, if not multiply each by a factor to get whole numbers 74.8% C, 25.2% H = 74.8g C, 25.2g H = 6.23 mol C, 25.0 mol H 6.23/6.23 = 1 C, 25.0/6.23 = 4 H  CH4

Determine Empirical Formula 87.4% Nitrogen, 12.6% Hydrogen

Molecular Formula: Using emp. Form., molar mass Take molar mass of emp. form Divide molar mass by emp. form mass Multiply Empirical formula coefficients by that factor Mol. Form Mol. Form Mass Emp. Form Emp. Form Mass = C3H6O3 ____X____ ____90 g____ CH2O ? = 30 g

Calculate Molecular Formula Empirical formula: NH2 Approximate molar mass: 32 g/mole

Chemical Equations CH4 (g) + O2 (g)  CO2 (g) + H2O (g) 2 2 Reactants Products ***Note: Remember back to Dalton’s 4th theory, that reactions are simply the rearranging of atoms *** And, according to the conservation of mass, the masses and elements must be the same before and after Balancing Equations: to ensure equal atomic quantities Start with atoms appearing least # of times per side Put coefficients in front of entire molecule to balance If you have to use a half # to balance an atom, double the entire equation at the end

Half-Number Example C2H6 + O2  CO2 + H2O

Stoichiometric Conversions Typically involve using a balanced chemical equation to change from a reactant amount to a product amount Steps (Usually): Convert any given amounts to moles Determining limiting reactant (if any) Use limiting reactant moles to convert to moles of product using mole/mole ratio Convert moles product to desired units

Limiting/Excess Reagents Limiting reagent: reactant that runs out first, must use to determine how much product can theoretically be made Excess reagent: left-over reactant Can determine which is which by comparing given moles to stoichiometric ratios in equation Percent Yield: experimental mass x 100 theoretical mass

Example Using 2 H2 + O2  2 H2O how many grams of H2O can be made with 5.00 grams of H2 and 32.0 grams of O2? If the experimental mass of H2O is 32.5 grams, what is the percent yield?