DO NOW:
1. Define Stoichiometry
2. If 2 moles of nitrogen gas reacts to form ammonia, how
many moles of hydrogen is required? Grams? How
many moles of ammonia is produced? Grams?
Do Now: Compare and contrast
1. An element and a compound
2. A substance and a mixture
3. Homogeneous and heterogeneous mixtures
4. List 5 methods of separating mixtures or
decomposing compounds
Elements and compounds
Compare
Contrast
 Both stable forms of matter
 Elements are the smallest stable
 Represented by symbols
(elements) or formulas
(compounds)
 Both have fixed composition
form of matter that can exist by
itself.
 Cannot be broken down into
stable components
 Cannot be broken down by
ordinary physical or chemical
means
 Only one type of atom
 Displayed on periodic table
 Compounds are 2 or more types of
atoms chemically bonded
 Properties significantly (usually)
different from component
elements
 Can be broken down into stable
component atoms
Substances and Mixtures
Compare
Contrast
 Both forms of matter
 Substances have fixed
 Both composed of atoms and
molecules
composition
 Can be represented by
symbols or formulas
 Mixtures are physically put
together
 Not bonded
 Each substance retains
physical and chemical
properties (most)
 No fixed composition
(varies)
Homogeneous and Heterogeneous
Compare
Contrast
 Both forms of matter
 Homogeneous
 Uniform composition (and
properties) for given sample
(different samples have
differing composition and
properties)
 Appears to be in single phase
 Heterogeneous
 Not uniform throughout
 Separate, distinct regions
visible
 Varying properties and
composition WITHIN same
sample
 Both physical combinations
of component substances
 Both NOT chemically
combined
 Components of both retain
(most) physical and chemical
properties
Separation Methods
Mixtures
Compounds
 Filtration
 solid-liquid phase heterogenous
only
 Distillation
 liquid-liquid, solid liquid
homogenous.
 Based on differences in boiling point
(volatitily)
 Chromatography
 Paper, liquid, gas
 Separation due to differences in
polarity / mass / size /
attraction/bonding between mobile
and immobile media
 Mechanical
 Sieve/Mesh
 Magnets
 Physical
 Electrolysis (electric current)
 2H2O  2H2
+ O2
 Heat
 2Ag2O  4Ag
+ O2
 Reaction of compound with
more active metal/non-metal
 Ca + CuCl2  Cu + CaCl2
 Light
 2H2O2  2H2O + O2
3.1: ATOMIC MASS
A. C-12, the Relative Standard
1. C-12 is assigned a mass of exactly 12 atomic mass units (amu)
2. Masses of all elements are determined in comparison to the
carbon -12 atom (12C) the most common isotope of carbon
3. Comparisons are made using a mass spectrometer
B. Atomic Mass (Average atomic mass, atomic weight)
1. Atomic masses are the average of the naturally occurring
isotopes of an element
2. Atomic mass does not represent the mass of any actual atom
3. Atomic mass can be used to "weigh out" large numbers of
atoms.
Average atomic mass = Σ (% of each isotope)(atomic mass of each isotope)
100
Mass Spectoscopy
 Used to identify the varying masses of isotopes in a
naturally occurring sample of any given element (also
compounds--later)
 How it works
 Vaporized atoms are bombarded with electrons to form
positive ions
 Positive ions are accelerated using an electric field
 Ions are deflected in the magnetic field according to their
mass and charge (always positive)


Lighter/more highly charged particles are deflected MORE
Comparisons of % deflected at each point are made
 Detected and converted to a readable output (mass spectra)
• Plots charge/mass ratio (x-axis) vs. relative abundance (y-
axis)
 Peaks represents isotope with that mass and relative
percent (consistent with relative abundance)
 Used to solve for average atomic mass
Average Atomic Mass =
Σ (% of each isotope)(atomic mass of each isotope)
100
Example
 What element is this
likely to be? Analyze
 # isotopes?

4
 Relative abudance?

88 most abudant
 Average atomic
mass closest to?

<88
 Element id?

Strontium
Example
• What element is
this likely to be?
Analyze
– # isotopes?
• 10
– Relative
abudance?
• 118 and 121 most
abudant
– Average atomic
mass closest to?
• Between 118-121
– Element id?
• Tin
 Note: Not to scale on either axis
 Identify diatomic element. Analyze
 # isotope?

3
 Relative abundance

16 most abudant
 Most likely to be?

oxygen
Why so many peaks?
-O2 can form from any
combination of the
isotope 16-16 (32), 16-17
(33), 16-18 (34), 17-17
(34), 17-18 (35), 18-18
(36)
 Is the above a complete/true mass spectrum of water?
Why or why not?
 What other peaks/relative abundance would you expect?


2, 3 (Extremely tiny)
24, 23, 22, 21, 20, 19 (quite small)
3.2: The MOLE
A. Avogadro's number
1. 6.022 x 1023 units = 1 mole
2. Named in honor of Avogadro (he did NOT discover it)
B. Measuring moles
1. An element's atomic mass expressed in grams contains 1
mole of atoms of that element
a. 12.01 grams of carbon is 1 mole of carbon (any random sample)
b. 12 grams of carbon-12 is 1 mole of carbon-12 (specific isotope)
3.3: MOLAR MASS
A. Molar Mass (Gram molecular weight)
1.
The mass in grams of one mole of a compound
a. Mass of _______________ particles
2. The sum of the masses of the component atoms in a compound
a. Molar mass of ethane (C2H6):
Mass of 2 moles of C = 2(12.01 g)
Mass of 6 moles of H = 6(1.008 g)
30.07 g
3. Examples:
a. Determine the molar mass of calcium carbonate.
b. A certain sample contains 4.50 moles of calcium carbonate.
Determine the mass of the carbonate ion present.
3.4: PERCENT COMPOSITION
A. Calculating any percentage
1. "The part, divided by the whole, multiplied by 100"
B. Percentage Composition
1. Mass of each element expressed as a percent of the total mass
of the compound.
2. Calculate the percent of each element in the total mass of the
compound
(#atoms of the element)(atomic mass of element) x 100
(molar mass of the compound)
C. Examples:
1. Determine percent composition of Carvone (C10H14O).
2. Determine percent composition of penicillin
(C14H20N2SO4).
3.5: Determining the Formula of A
Compound
A. Empirical Formula: simplest whole number ratio of
atoms of each element in a compound.
1.
2.
Entirely different and unrelated compounds, with
entirely different molecular formulas may have the
same empirical formula.
Example: Benzene, C6H6; Ethene, C2H2; 1,3,5,7cyclooctatetraene, C8H8 are very different compound
but each have a 1:1 ratio of C to H atoms in their
molecular formula, so all have an empirical formula of
CH.
B. Determining the empirical formula:
1. Determine the percentage of each element in your
compound
2. Treat % as grams, and convert grams of each element to
moles of each element
3. Find the smallest whole number ratio of atoms. Find
the smallest number of moles calculated and divide ALL
the results of the calculations, by that number.
NOTE: avoid rounding up or down too much at this stage,
and be lenient with significant figures.
4. If the ratio is not all whole number, it should include a
recognizable decimal (.25, .333, .500). Multiply each by
an integer (4, 3, 2 as appropriate) so that all elements are
in whole number ratio
Example: Determine empirical formula of compound
composed of 71.65% Cl, 24.27% C and 4.07% H.
C. Molecular formula: tells us exactly how many atoms
of each element are present in the compound, rather
than just the simplest whole number ratio.
1.
2.
3.
Simple multiple of the empirical formula
Indicate ACTUAL formula
May or may not indicate information about the actual
structure of category of compound (more later)
B. Determining the molecular formula
1. Find the empirical formula mass (of previously
established empirical formula)
2. Divide the known (given) molecular mass by the
empirical formula mass, deriving a whole number, n
3. Multiply the empirical formula by n to derive the
molecular formula
Example: The molar mass of the sample (previous slide)
is known to be 98.96 g/mol
3.6: CHEMICAL EQUATIONS
A. Chemical reactions
1. Reactants are listed on the left hand side
2. Products are listed on the right hand side
3. Atoms are neither created nor destroyed (conservation
of mass)
a. All atoms present in the reactants must be accounted for
among the products, in the same number
b. No new atoms may appear in the products that were not
present in the reactants
B. The Meaning of a Chemical Reaction
1. Physical States
a. Solid - (s)
b. Liquid - (l)
c. Gas - (g)
d. Dissolved in water (aqueous solution) - (aq)
2. Relative numbers of reactants and products
a. Coefficients give atomic/molecular/mole ratios
WHY?
multiplying ALL coefficients by the same value (avogadro’s
number) does not affect the relationship or ratio.
3.7 Balancing Chemical Equations
A. Determine what reaction is occurring
1. It is sometimes helpful to write this in word form:
Hydrogen + oxygen  water
B. Write the unbalanced equation
1. Focus on writing correct atomic and compound formulas
H2 + O2  H2O
C. Balance the equation by inspection
1. It is often helpful to work systematically from left to right
2H2 + O2  2H2O
D. Include phase information
2H2 (g) + O2 (g)  2H2O (l)
3.8 Stoichiometric Calculations:
Amounts of Reactants and Products
A. Balance the chemical equation
B. Convert value given (generally mass, volume,
particles, density or other data) of reactant or product
to moles
C. Compare moles of the known to moles of the desired
substance
1. A ratio derived from the coefficients in the balanced
equation
D. Convert from moles back to required value (mass,
volume, particles, density or other)
3.9 Calculations Involving a
Limiting Reactant
A. Concept of limiting reactant (limiting reagent):
" I want to make chocolate chip cookies. I look around my
kitchen (I have a BIG kitchen!) and find 40 lbs. of
butter, two lbs. of salt, 1 gallon of vanilla extract, 80 lbs.
of chocolate chips, 200 lbs. of flour, 150 lbs. of sugar, 150
lbs. of brown sugar, ten lbs. of baking soda and TWO
eggs. It should be clear that it is the number of eggs
that will determine the number of cookies that I can
make."
1. The limiting reactant controls the amount of
product that can form
B. Solving limiting reactant problems
1. Convert grams of reactants to moles
2. Use stoichiometric ratios to determine the limiting reactant
3. Solve as before, beginning the stoichiometric calculation with
the grams of the limiting reactant
C. Calculating Percent Yield
1. Actual (Experimental) yield - what you got by actually
performing the reaction
2. Theoretical yield - what stoichiometric calculation says the
reaction SHOULD have produced
Actual Yield
Theoretical Yield
x 100% = percent yield