1. A.P. Ch. 3 Review Work Stoichiometry

2. Atomic Mass Average of isotope masses based on their abundance Ex. Carbon has atomic mass of 12.01 amu 12 C has mass of 12.00 amu and 98.89% 13 C 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

3. Mole A mole is equal to 6.02x10 23 of those things Ex. 1 mole of Carbon contains 6.02x10 23 atoms 1 mole of H 2 O contains 6.02x10 23 molecules All of the atomic masses on the periodic table are equivalent to 1 mole of those elements in grams Like a dozen represents a certain number of an object, a mole represents a certain amount of particles

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

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

6. Empirical/Molecular Formulas Molecular formula: shows how many of each element are present in a compound Glucose is C 6 H 12 O 6 Empirical formula: shows smallest whole number ratio Glucose is CH 2 O Polio Virus Empirical Formula

7. Determining Chemical Formulas Empirical Formula: Starting with % composition of each element 1.Change % to mass (10% of C is 10 grams C) 2.Convert each mass to moles using Per. Table 3.Divide each elements # of moles by smallest # 4.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  CH 4

8. Molecular Formula: Using emp. Form., molar mass 1.Take molar mass of emp. form 2.Divide molar mass by emp. form mass 3.Multiply Empirical formula coefficients by that factor Mol. FormMol. Form Mass Emp. FormEmp. Form Mass = ____X________90 g____ CH 2 O ? 30 g C3H6O3C3H6O3 =

9. Chemical Equations CH 4 (g) + O 2 (g)  CO 2 (g) + H 2 O (g) Reactants Products ***Note: Remember back to Dalton’s 4 th 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 1.Start with atoms appearing least # of times per side 2.Put coefficients in front of entire molecule to balance 3.If you have to use a half # to balance an atom, double the entire equation at the end 22

10. Stoichiometric Conversions Typically involve using a balanced chemical equation to change from a reactant amount to a product amount Steps (Usually): 1.Convert any given amounts to moles 2.Determining limiting reactant (if any) 3.Use limiting reactant moles to convert to moles of product using mole/mole ratio 4.Convert moles product to desired units

11. 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

12. Example Using 2 H 2 + O 2  2 H 2 O how many grams of H 2 O can be made with 5.00 grams of H 2 and 32.0 grams of O 2 ?