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 amu12C 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

Mole

• A mole is equal to 6.02x1023 of those things

• 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

• Like a dozen represents a certain number of an object, a mole represents a certain amount of particles

Molar MassMolar 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

Polio Virus Empirical Formula

Determining Chemical FormulasEmpirical Formula:

Starting with % composition of each element1. Change % to mass (10% of C is 10 grams C)2. Convert each mass to moles using Per. Table3. 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 CH4

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. Form Mol. Form Mass

Emp. Form Emp. Form Mass

=

____X____ ____90 g____

CH2O ? 30 g

C3H6O3

=

Chemical Equations CH4 (g) + O2 (g) CO2 (g) + H2O (g)

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

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

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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

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?