Transcript Notes 3 student

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
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 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
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
Polio Virus
Empirical
Formula
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  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
C3H6O3
____X____
____90 g____
CH2O
=
30? g
Chemical Equations
CH4 (g) + 2 O2 (g) 
Reactants
CO2 (g) + 2 H2O (g)
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
Stoichiometric Conversions
•
•
1.
2.
3.
4.
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?