Moles, Formulas, Reactions & Stoichiometry
Lecture 3 Topics
Brown, chapter 3
1. Moles
Molecular mass (aka molecular weight)
3.4
3.3
2. Molar converstions
• Percent composition
• Empirical formulas
3.3
3.5
3. Stoichiometry: balancing chemical equations
3.1
4. Patterns of Chemical Reactivity
• Combination
• Decomposition
• Combustion
• Exchange
3.2
5. Stoichiometry & Conversions
• Limiting reactants
• Theoretical & percent yield
3.6
3.7
Conversions: moles to molecules to atoms
Percent composition is molar.
Empirical formulae are molar.
Molecular formulae are molar.
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)
atoms /
molecules
moles
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.93 - 94
Conversions with moles
How many grams is 0.55 mol Cu? STRATEGY: moles ------------> grams
atomic mass
0.55 mol
63.55 g = 35 g
1 mol
How many moles contain 6.96x1024 atoms of Na?
STRATEGY: atoms -------------------> moles
the mole
6.96x1024 Na atoms
1 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
1 mole
40.00 g
6.02x1023 molecules = 3.8x1023 g
1 mole
p.93 - 94
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?
% C = (12)(12.0 amu) x 100% = 42.1%
342 amu
1. Calculate mass of each element.
2. Calculate MW (sum masses in 1.)
3. Calculate each element’s mass as
% of the MW.
= (part/whole)(100)
4. % 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) =
(16.00/18.02)(100) =
C6Cl5OH
FW/MW = 342 amu
11.2% H
88.8% O
(72.07/266.32)(100) =
(177.25/266.32)(100) =
(16.00/266.32)(100) =
(1.01/266.32)(100) =
27.06% C
66.56% Cl
6.01% O
0.38% H
p.88
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
H
85.7 g 1 mol = 7.14 mol/7.14 =
12.01 g
14.3 g 1 mol = 14.2 mol/7.14 =
1.01 g
1.00 -> 1 so CH2 is empirical formula
1.99 -> 2
An Fe/O compound is 69.9% Fe.
Fe
O
69.9 g/55.85 = 1.25 mol/1.25 = 1
30.1 g/16.00 = 1.88 mol/1.25 = 1.50
x2=2
x2=3
So the formula is Fe2O3
p.95
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 40.19 g/12.01 g/mol = 3.35 mol/3.35 = 1.00
x3=3
4.58 mol/3.35 = 1.37
x 3 = 4 EF= C3H4O3
4.58% H 4.58 g/1.01 g/mol =
x 3 = 3 EW = 88.1 g/mol
54.50% O 54.50 g/15.99 g/mol = 3.41 mol/3.35 = 1.02
MW = 176 g/mol = 2
EW
88.1 g/mol
So multiply the empirical subscripts by 2 to
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)(1.00794) = 40.065 amu
121 amu/40.065 amu = 3 multiple --> C9H12
p.95