Spring Final Exam
Stoichiometry Review
Molar Mass
the mass of one mole of a substance
PbO2
HNO3
Pb:
1 (207.2 g) = 207.2 g
O:
2 (16.0 g)
= 32.0 g
H:
1 (1.0 g)
= 1.0 g
N:
1 (14.0 g)
= 14.0 g
O:
3 (16.0 g)
= 48.0 g
239.3 g
63.0 g
percentage composition: the mass % of each
element in a compound
g element
x 100
% of element =
molar mass of compound
Find % composition.
PbO2
(NH4)3PO4
(see calcs above)
207.2 g Pb : 239.2 g = 86.6% Pb
32.0 g O : 239.2 g = 13.4% O
42.0 g N
12.0 g H
31.2 g P
64.0 g O
: 149.0 g
: 149.0 g
: 149.0 g
: 149.0 g
=
=
=
=
28.2% N
8.1% H
20.8% P
43.0% O
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.
A compound is 45.5% yttrium and 54.5% chlorine.
Find its empirical formula.
 1 mol Y 
  0.512 mol Y  0.512  1
45.5 g Y 
 88.9 g Y 
 1 mol Cl 
  1.535 mol Cl  0.512  3
54.5 g Cl 
 35.5 g Cl 
YCl3
To find molecular formula…
A. Find empirical formula.
B. Find molar mass of
empirical formula.
C. Find x = MM molecular
MM empirical
D. Multiply all parts of
empirical formula by x.
(How many empiricals “fit into” the molecular?)
A carbon/hydrogen compound is 7.7% H and has a
molar mass of 78 g. Find its molecular formula.
 1 mol H 
  7.7 mol H  7.69  1
7.7 g H 
 1.0 g H 
 1 mol C 
  7.69 mol C  7.69  1
92.3 g C 
 12.0 g C 
emp. form.  CH
mmemp = 13 g
78 g
=6
13 g
C6H6
A compound has 26.33 g nitrogen, 60.20 g oxygen,
and molar mass 92 g. Find molecular formula.
 1 mol N 

26.33 g N 
 14.0 g N 
 1.881mol N  1.881  1
 1 mol O 
  3.763 mol O  1.881  2
60.20 g O 
 16.0 g O 
NO2
mmemp = 46 g
92 g
=2
46 g
N2O4
1 mol = molar mass (in g)
Mass
(g)
Volume
(L or dm3)
Island Diagram:
1 mol = 22.4 L
MOLE
(mol)
1 mol = 22.4 dm3
Particle
(at. or
m’c)
1 mol = 6.02 x 1023
particles
a. Diagram has four islands.
b. “Mass Island” for elements or compounds
c. “Particle Island” for atoms or molecules
d. “Volume Island”: for gases only
1 mol @ STP = 22.4 L = 22.4 dm3
What mass is 1.29 mol Iron (II) nitrate ?
Fe2+ NO31–
1.29 mol
(
179.8 g
1 mol
)
Fe(NO3)2
= 232 g
How many molecules is 415 L sulfur dioxide at STP?
SO2
415 L
(
1 mol
22.4 L
)(
)
6.02 x 1023 m’c
1 mol
= 1.12 x 1025 m’c
What mass is 6.29 x 1024 m’cules aluminum sulfate ?
Al3+
SO42–
Al2(SO4)3
342.3 g
1 mol
6.29 x 1024 m’c
6.02 x 1023 m’c
1 mol
(
)(
)
= 3580 g
At STP, how many g is 87.3 L of nitrogen gas?
N2
87.3 L
(
1 mol
22.4 L
)(
28.0 g
1 mol
)=
109 g
•During a chem. rxn.; atoms are rearranged
(NOT created or destroyed!)
•Chemical equations must be balanced to
show the relative amounts of all
substances.
•Balanced means: each side of the
equations has the same # of atoms of each
element.
CH4 + O2 —> H2O + CO2
Unbalanced
CH4 + 2O2 —> 2H2O + CO2 Balanced
Cellulose reacts with oxygen gas to form
carbon dioxide gas & liquid water.
C6H10O5 + 6O2  6CO2 + 5H2O
6
10
17
C
H
O
6
10
17
Balanced!!!
Nitroglycerin decomposes to form nitrogen gas,
oxygen gas, carbon dioxide gas & water vapor
2 C3H5(NO3)3  3N2 + O2 + 6CO2 + 5H2O
6
10
6
18
C
H
N
O
6
10
6
19
Not
Balanced!
• When balancing equations, you may
change coefficients as much as you need
to, but you may never change
subscripts because you can’t change what
substances are involved.
2 H2 (g) + O2 (g)  2 H2O (l)
1. Atom Inventory:
Reactants
Products
4 2
H
2 4
2
O
1 2
2 Add coefficients to balance
3. Double check to make sure it is all
BALANCED!
• Balancing Chemical Equations practice