1. Moles, Formulas, Reactions & Stoichiometry
Lecture 3 Topics Brown, chapter 3
1. Moles
Molecular mass (aka molecular weight)
2. Molar converstions
Percent composition
Empirical formulas
3. Stoichiometry: balancing chemical equations
4. Patterns of Chemical Reactivity
Combination
Decomposition
Combustion
Exchange
5. Stoichiometry & Conversions
Limiting reactants
Theoretical & percent yield

2. Conversions: moles to molecules to atoms
Percent composition is molar.
Empirical formulae are molar.
Molecular formulae are molar.
The general process of advancing scientific knowledge by making experimental observations and by formulating hypotheses, theories, and laws.
It’s a systematic problems solving process AND it’s hands-on….. Experiments must be done, data generated, conclusions made.
This method is “iterative”; it requires looping back and starting over if needed. [Why do you think they call it REsearch?] Often years, decades or more of experiments are required to prove a theory.
While it’s possible to prove a hypothesis wrong, it’s actually NOT possible to absolutely prove a hypothesis correct as the outcome may have had a cause that the scientist hasn’t considered.

3. Interconversion of moles, mass, atoms
Chemistry students are often frustrated by the subject’s lack of canned equations. Problem solving depends much more on strategy & process: logic. Use these ‘devices’ to develop problem-solving strategies.
mass (grams)
moles
atoms / molecules
molar mass (MW)
g/mole
Avogadro’s number
molecules/mole
The Conversion Flower can be used to convert: moles <--> grams
moles <--> molecules
moles A <--> moles B
moles <--> liter p

4. Conversions with moles
How many grams is 0.55 mol Cu?
STRATEGY: moles > grams
atomic mass
0.55 mol g = 35 g
1 mol
How many moles contain 6.96x1024 atoms of Na?
STRATEGY: atoms > moles
the mole
6.96x1024 Na atoms mole = 11.6 mol
6.02x1023 atoms
How many molecules in 25 g of NaOH?
STATEGY: g > moles > molecules
molar mass the mole
25 g mole x1023 molecules = 3.8x1023 g
40.00 g mole p

5. Percent composition
Percent composition: what percent of a molecule’s mass is contributed by each element?
Calculated using atomic and molar masses.
Percentage composition of C12H22O11?
FW/MW = 342 amu
% C = (12)(12.0 amu) x 100% = 42.1%
342 amu
Calculate mass of each element.
Calculate MW (sum masses in 1.)
Calculate each element’s mass as % of the MW. = (part/whole)(100)
% should sum to 100!
% H = (22)(1.0 amu) x 100% = 6.4%
342 amu
% O = (11)(16.0 amu) x 100% = 51.5%
342 amu
H2O
(2.02/18.02)(100) = % H (16.00/18.02)(100) = 88.8% O
C6Cl5OH
(72.07/266.32)(100) = % C (177.25/266.32)(100) = % Cl
(16.00/266.32)(100) = 6.01% O
(1.01/266.32)(100) = 0.38% H
p.88

6. Empirical formula from chemical analysis
Empirical formulae can be calculated from % composition data obtained via simple chemical analysis.
Steps: 1. Begin by assuming you have 100 g of the substance & calculate g each element. 2. Convert g to moles. Using atomic mass. 3. Calculate mole ratios: divide all by the smallest number of moles. 4. Convert to formula ratios. You may need to multiply by small whole numbers to get
to whole number ratios.
A pure hydrocarbon (molecule made wholly of C & H) is 85.7% C. Calculate the empirical formula of the compound.
C 85.7 g 1 mol = 7.14 mol /7.14 = > 1 so CH2 is empirical formula
12.01 g
H 14.3 g mol = 14.2 mol /7.14 = > 2
1.01 g
What’s an empirical formula? While a molecular formula tells the reader what type and what number of atoms make up a single molecule, the empirical formula is a simplied version of the molecular formula & tells the reader ONLY the type & relative ratios of atoms in a molecule. Empirical formulas do not define molecules.
An Fe/O compound is 69.9% Fe.
Fe g/55.85 = 1.25 mol/1.25 = 1 x 2 = 2
O g/16.00 = 1.88 mol/1.25 = x 2 = 3
So the formula is Fe2O3
p.95

7. Calculate molecular from empirical formula
A molecular formula can be calculated from an empirical formula ONLY if the MW is known.
1. Calculate empirical formula (previous slide), & then empirical weight. 2. Calculate the ratio of MW/EW.
3. Multiply empirical formula subscripts by that ratio.
Vitamin C (ascorbic acid, MW 176 g/mol) has a percent composition of:
40.19% C
4.58% H
54.50% O
40.19 g/12.01 g/mol = mol/3.35 = 1.00 x 3 = 3
4.58 g/1.01 g/mol = mol/3.35 = 1.37 x 3 = 4 EF= C3H4O3
54.50 g/15.99 g/mol = 3.41 mol/3.35 = x 3 = 3 EW = 88.1 g/mol
MW = 176 g/mol = So multiply the empirical subscripts by 2 to
EW g/mol get the molecular formula -> C6H8O6
Mesitylene, a minor component of crude oil, has an empirical formula of C3H4. Its molar mass is 121 amu. What’s its molecular formula?
empirical mass: (3)(12.011) + (4)( ) = amu
121 amu/ amu = 3 multiple --> C9H12
p.95