AP Chemistry
Stoichiometry
In the thermite reaction, a
mixture of powdered aluminum
and powdered iron(III) oxide
react to yield iron and aluminum
oxide. The reaction burns hot
enough to be useful in underwater welding.
2 Al + Fe2O3  2 Fe + Al2O3 + energy
Chemical Equations
In a reaction:
atoms are rearranged
mass
AND
energy
are conserved
charge
Balancing Chemical Equations
law of
conservation
of mass
=
same # of atoms
of each type on each
side of equation
Hint: Start with most complicated substances first
and leave simplest substances for last.
Solid lithum reacts w/oxygen
to form solid lithium oxide.
4 Li(s)
+
O2(g)
Li+
O2–
2 Li2O(s)
+
Aqueous aluminum sulfate reacts w/aqueous barium
chloride to form a white precipitate of barium sulfate.
The other compound remains in solution.
Ba2+ Cl–
Al3+ SO42–
Al2(SO4)3 (aq) + 3 BaCl2 (aq)
3 BaSO4 (s) + 2 AlCl3 (aq)
Methane gas (CH4) reacts with
oxygen to form carbon dioxide
gas and water vapor.
Furnaces burn primarily methane.
CH4(g) + 2 O2(g)
CaC2(s) +
H2O(l)
CO2(g) + 2 H2O(g)
C2H2(g) +
CaO(s)
3 CaSi2 + 2 SbI3
6 Si + 2 Sb + 3 CaI2
2 Al + 3
6 CH3OH
2 Al(CH3O)3 + 3 H2
2 C2H2(g) + 5 O2(g)
C3H8 + 5 O2
2 C4H10 + 13 O2
2 CO2(g) + 2 H2O(l)
4
3 CO2 + 4 H2O
8
4 CO2 +10
5 H2O
Complete combustion of a hydrocarbon,
or of a compound containing C, H, and O
(e.g., methanol, CH3OH) yields CO2 and H2O.
Another pattern of reactivity:
alkali
metal
hydrogen
+ water
+
metal
hydroxide
gas
e.g., 2 Rb(s) + 2 H2O(l)  2 RbOH(aq) + H2(g)
Two (of the several) Types of Reactions
combination (synthesis): simpler substances combine
to form more complex substances
-- form:
A + B  AB
AB + C  ABC
sodium + chlorine gas
2 Na + Cl2
A + B + C  ABC
sodium chloride
2 NaCl
decomposition: complex substances are broken
down into simpler ones
-- form:
AB  A + B
ABC  AB + C
lithium chlorate
Li+ ClO3–
2 LiClO3
water
ABC  A + B + C
lithium chloride + oxygen
Li+
Cl–
2 LiCl + 3 O2
hydrogen gas + oxygen gas
2 H2O
2 H2 + O2
formula weight: the mass of all of the atoms in
a chemical formula (unit is amu)
-- If the substance is a molecular substance
(e.g., C3H8), then the term molecular weight
is also used.
molar mass: the mass of one
mole of a substance
(unit is usually grams)
-- recall that 1 mole of any = 6.02 x 1023 particles
substance
of that substance
Find the molar mass and formula weight of
ammonium phosphate.
NH4+
PO43–
(NH4)3PO4 N:
3 (14.0 g) = 42.0 g
H: 12 (1.0 g)
= 12.0 g
P:
1 (31.0 g) = 31.0 g
O:
4 (16.0 g) = 64.0 g
m.m. =
149.0 g
f.w. = 149.0 amu
percentage composition: the mass % of each
element in a compound
equation:
g element
x 100
% of element =
molar mass of compound
Find the percentage of oxygen, by mass, in
calcium nitrate.
Ca(NO3)2
6(16.0)
%O=
= 58.5% O
40.1  2(14.0)  6(16.0)
Empirical Formula and Molecular Formula
lowest-terms
formula
shows the true number and
type of atoms in a m’cule
Compound
Molecular
Formula
Empirical
Formula
glucose
C6H12O6
CH2O
propane
C3H8
C3H8
butane
C4H10
C2H5
naphthalene
C10H8
C5H4
sucrose
C12H22O11
C12H22O11
octane
C8H18
C4H9
Finding an Empirical Formula from Experimental Data
1. Find # of g of each element.
2. Convert each g to mol.
3. Divide each “# of mol” by the smallest “# of mol.”
4. Use ratio to find formula.
“What’s your flavor
of ice cream?”
A ruthenium/sulfur
compound is 67.7% Ru.
Find its empirical formula.
 1 mol Ru 
  0.670 mol Ru  0.670  1
67.7 g Ru 
 101.1 g Ru 
 1 mol S 
  1.006 mol S  0.670  1.5
32.3 g S 
 32.1 g S 
RuS1.5
Ru2S3
A sample of a compound contains 4.63 g lead, 1.25 g
nitrogen, and 2.87 g oxygen. Name the compound.
 1 mol Pb 
  0.0223 mol Pb  0.0223  1
4.63 g Pb 
 207.2 g Pb 
 1 mol N 
  0.0893 mol N  0.0223  4
1.25 g N 
 14.0 g N 
 1 mol O 
  0.1794 mol O  0.0223  8
2.87 g O 
 16.0 g O 
?
PbN4O8
?
Pb(NO2)4
Pb? 4 NO2–
lead(IV) nitrite