Unit 6b:
Stoichiometry
• Review Book: Topic 6
• Textbook: Ch. 7 & 9
I. Chemical Formulas
­ atoms are soooooo small...
LOTs of them in any given sample of a
substance
A. Mole (mol) = 6.02 x 1023 things
1 dozen eggs = 12 eggs
1 mol of atoms = 6.02 x 1023 atoms
­ subscripts represent moles of
atoms/ions
Ex:
If a mole of moles were digging a
mole of holes, what would you see?
a mole of molasses!!!!
NaCl:
1 mol of Na atoms
1 mol of Cl atoms
H2O
2 mol of H atoms
1 mol of O atoms
II. Math & Formulas
A. Molar mass/Gram­formula mass (GFM)
­ mass of 1 mole of a substance
add atomic masses
of all elements in
formula
­ calculate:
­ unit: g
ex 1: calculate the GFM of Ba(NO3)2
G:
S:
S:
add atomic masses on P.T.
# of
mass
atoms x
137 = 137
x
1
Ba
= 28
14
x
2
N
= 96
16
x
6
O
261 g
C:
II. Math & Formulas
B. Mole­Mass calculations
*Table T* Mole Calculations
# of moles = given mass (g)
GFM
calculate twice
II. Math & Formulas
ex 2: What is the mass of 3 moles of
Ba(NO3)2 (GFM = 261 g)?
G:
S:
S:
C:
Table T
# of moles = given mass (g)
GFM
3
1
=
X
261
X
=
783 g
more than 1 mol
...greater than GFM
less than 1 mol
....smaller than GFM
II. Math & Formulas
II. Math & Formulas
ex 3: How many moles are in 300. g of
Ba(NO3)2 (GFM = 261 g)?
C. Mole­Particle calculations
conversion factor:
G:
S:
S:
Table T
# of moles = given mass (g)
GFM
1 mol = 6.02 x 1023 particles
equation similar to one on Table T:
= given particles
(6.02 x 1023)
# moles
X
1
=
300.
261
X
=
1.14 mol
same as ex 2
C:
II. Math & Formulas
II. Math & Formulas
ex 4: How many molecules are in 2
moles of CO2?
G:
ex 5: How many moles are in
3.01 x 1023 atoms of Na?
G:
S:
S:
S:
S:
C:
Table T
# of moles = given particles
(6.02 x 1023)
2
1
=
X
= 1.20 x 1024 molecules
X
(6.02 x 1023)
greater than 1 mol
...more than 6.02 x 1023 particles
C:
Table T
# of moles = given particles
(6.02 x 1023)
X
1
=
X
= 0.500 mol
(3.01 x 1023)
(6.02 x 1023)
same as ex 4
less than 1 mol
...less than 6.02 x 1023
II. Math & Formulas
II. Math & Formulas
D. % Composition
ex 6:
1. Molecular & ionic compounds
*Table T*
% by mass
=
x 100
part
whole
What is the percent by mass of
sodium in NaCl? (GFM = 58 g)
G:
S:
Table T
S:
%
=
% Na
=
23
58
% Na
=
40. %
C:
part
x 100
whole
x 100
29 is 50% of 58, so 23 will be < 50%
II. Math & Formulas
II. Math & Formulas
2. Hydrates
2. Hydrates
ex 7:
ex 8:
% water in CuSO4 5 H2O?
(GFM = 250. g)
G:
A 10.0 g sample of hydrate
was heated until the mass
was 7.24 g.
% H2O in hydrate?
G:
S:
like % comp.
S:
% H­2O
% H­2O
= part
x 100
whole
* find mass of H2­O (difference)
like % comp.
S:
% H­2O =
= (5 x 18) x 100
250.
% H­2O =
C:
S:
% H­2O =
C:
III. Mole Ratios
x 100
% H­2O = (10.0 ­ 7.24) x 100
10.0
36.0 %
5 x 18 = 90, < 50% of 250
27.6 %
10.0 ­ 7.24 = 2.76, < 50% of 10
B. Problems
ex 9:
A. Balanced equations are like recipes
Given balanced eq:
4NH3 + 5O2
*Coefficients represent MOLE amounts
ex:
part
whole
2H2 + O2
2H2O
4NO + 6H2O
if 12 mol of O2 consumed ­ how
many mol of NO are produced?
G:
mix 2 mol H2 and 1 mol O2,
get 2 mol H2O
S:
set up mole proportion
S:
eq:
actual:
5 = 4
12
X
5X = 48
5
5
mole ratio of H2 to H2O is 2:2 or 1:1
X = 9.6 mol of NO
C:
IV. Mass­Mass Problems
ex 10:
Given balanced eq:
4NH3 + 5O2
4NO + 6H2O
how many grams of H2O are
produced by completely
reacting 34g of NH3?
G:
S:
set up mass proportion
S:
eq:
actual:
4(17g) =
34g
68 X
68
=
6(18g)
X
3672
68
X = 54 g H2O
C:
34 g of NH3 is 2 mol...we should
make 3 mol of H2O from that
(3 x 18g) = 54g
what you do to one coefficient,
you must do to the other to keep
ratio the same
5 to 12 (a little more than doubled)
4 to 9.6 (a little more than doubled)