Ch. 9 Notes (Stoichiometry)
Stoichiometry –
The study of quantities of materials that
are consumed and produced in chemical
reactions.
Origin (Greek):
stoicheion (meaning "element") and
metron (meaning "measure").
So basically we are measuring the elements.
Three new concepts

Mole Ratio
Using a balanced equation to convert moles
of one formula to another.


Limiting Reactant
VERY
Also called
IMPORTANT.
Limiting Reagent.
Predictions
A chemicalor
reaction
“theoretical
stops yields”
if you run
are out of
determined
one of the by
reactants.
the Limiting Reactant.
Percent Yield
Comparing actual yield to theoretical yield.
Previous concepts
Ch.3
Unit-conversion/factor-label method
Ch. 6 Formula writing
Ch. 7 Mole concept
Ch. 8 Balancing equations
Ch. 3 Review
Unit-conversions/Factor-Label Method.
Conversion factor
1 L = 1000 mL
45 mL x
1 L
1000 mL
= 0.045 L
Ch. 6 Review Formula Writing
Ionic Compounds
Calcium chloride
Ca2+ +
2 Cl-1
Sodium carbonate
2 Na+
+
CO32Aluminum nitrate
Al3+
+
3 NO3-1

 Na2CO3
 Al(NO3)3
Molecular Compounds
Sulfur dioxide
CaCl2
SO2
Ch. 7 Review: Mole Concept
1 mole = Molar mass
1 mole = 22.4 L or 22.4 dm3 for gases @STP
1 mole = 6.02  1023 rep. particles
Use Avogadro’s number when you see key words:
atoms, molecules, formula units.
Review of Molar Masses
SO2
32.06 g 16.00 g
x
2
x
1
32.06 g
+ 32.00 g = 64.06 g SO2
1 mole SO2 = 64.06 g SO2
Note: dm3 = L
Ch. 8 Review Balancing equations
N2
+
3 H2

2 NH3
The coefficients can represent:
moles
molecules
volume of gas in L @STP
Solving a Stoichiometry Problem
Always start with a balanced equation.
2. Convert what you are given (grams, liters,
particles) to moles.
3. Use the mole ratio to convert moles of given
to moles of what you want.
4. Convert to desired product in terms of mass,
volume, or number of particles.
1.
DAY 44 Class Activity (Chapter 9 Mole Ratio) Worksheet
1. How many moles of H2 would be needed if you
wanted to make 12 moles of NH3 assuming you have
unlimited amounts of N2?
Start with a balanced chemical equation:
N2
Given:
12 mol NH3
X
+
3
2
3 H2 → 2 NH3
mol H2
Want:
=
18
mol H2
Problem 1: Suppose you wanted to make NI3
1 N2
3 I2
+
2 NI3

Balance the equation 1st.
 How many moles of nitrogen do you have?
Given:
14.0 gg N
N22
X
1
28.02
1 mole = molar mass
N2
mol N2
Want:
=
0.500 mol N2
7
N
Do the Calculation:
14.01 g
x ÷ 28.02
2
14.0
= 0.499643
=
0.500
Nitrogen
28.02 g
14.01
Problem 1: Suppose you wanted to make NI3
1 N2
+
3 I2

2 NI3
 How many moles of iodine will react with given N2?
Given:
3.01 x 1023 molec. N2 X
1
mol N2
6.02 x 1023
23 formulas
You
1
Now
mole
can’t
you
= 6.02
change
can change
x 10
formulas
molecules
until
you’re
and doa the
mole.
moley moley.
mol I2
X
Want:
=
1.50
Do the Calculation:
3.01E23 ÷ 6.02E23 x 3 = 1.50
mol I2
Problem 1: Suppose you wanted to make NI3
1 N2
+
3 I2

2 NI3
 Predict how many moles of NI3 are produced with N2.
Given:
44.8 L N2
X
1 mol N2
22.4
mol NI3
X
Now
you
can change
formulas
You
1
mole
can’t
= 22.4
change
L formulas
until
and doa the
moley moley.
you’re
mole.
Do the Calculation:
44.8 ÷ 22.4 x 2 = 4
Want:
=
4.00
mol NI3
Problem 1: Suppose you wanted to make NI3
N2
+
3 I2
2 NI3

 Convert moles of NI3 to grams. (Theoretical yield)
Given:
0.04 mol NI3
X
Want:
394.71 g NI3
1
1 mole = molar mass NI3
NI3
=
7
N
126.90 g
Nitrogen
14.01 g
14.01
Do
the
Calculation:
x
3
x
1
380.70 g = 394.71 g
14.010.04
g +x 394.71
= 15.7884
15.79 g NI3
53
I
Iodine
126.90
Limiting Reactant (Everyday example)
10 pages + 3 staples + 1 cover  booklet
(500 pages)
500/10=50
(240 staples) (25 covers)
240/3=80
??
25/1=25
How many booklets can be made?
25 booklets
Justify your answer:
The covers run out first so you can make only 25.
Limiting Reactant Concept
A reaction stops when you use up one of the
reactants.
The limiting reactant determines the amount
of maximum product (Theoretical Yield).
Limiting Reactant: Reactant that gets “used up”
Excess Reactant: Reactant that is “left over”
Reagent: Another name for “Reactant”.
Limiting Reactant (Chemistry example)
The Haber process is the most widely used
chemical process in the world. It is the
production of ammonia, NH3, from nitrogen and
hydrogen.
Nitrogen is free and hydrogen costs $15 per Liter
Hydrogen gas should be the limiting
reagent because it costs money.
Limiting Reactant (Chemistry example)
Write a balanced chemical equation.
1 N2
3 H2
+
 2 NH3
If you had 12
of
12moles
mol N
30moles
mol Hof
2 N2 and 30
2 H2, how
many grams of NH3 can you make?
Compare moles given in problem to moles from Bal. Eqn.
Moles given in
Problem
H2
N2
=
Ideal situation
(From Bal.Eqn.)
=
2.5
1
H2
N2
=
2.5 is less than 3 so…
30 moles H2 in given problem is the limiting reagent.
Limiting Reactant (Chemistry example)
Write a balanced chemical equation.
1 N2
3 H2
+
 2 NH3
If you had 12
of
12moles
mol N
30moles
mol Hof
2 N2 and 30
2 H2, how
many grams of NH3 can you make?
What if you reversed the order?
Moles given in
Problem
N2
H2
=
Ideal situation
(From Bal.Eqn.)
=
0.4
1
N2
H2
=
=
0.3
1
0.4 is more than 0.3 so…
12 moles N2 in given problem is the excess reagent.
Limiting Reactant (Chemistry example)
Write a balanced chemical equation.
N2
+
3 H2
 2 NH3
If you had 12
of
12moles
mol N
30moles
mol Hof
2 N2 and 30
2 H2, how
many grams of NH3 can you make?
Use limiting reactant to calculate theoretical yield for NH3
x
mol NH3
mol H2
x
g NH3
mol NH3
=
Use Mole Ratio 1 mole = molar mass
g NH3
Limiting Reactant (Chemistry example)
Write a balanced chemical equation.
N2
+
3 H2
 2 NH3
If you had 12
of
12moles
mol N
30moles
mol Hof
2 N2 and 30
2 H2, how
many grams of NH3 can you make?
Use limiting reactant to calculate theoretical yield for NH3
30 mol H2 x
2
mol NH3
3
mol H2
x
17.0 g NH3
1
mol NH3
=
Do the Calculation:
Use Mole Ratio 1 mole = molar mass
NH
30 x 2 ÷ 3 x 17 1= mole
340
3
= 17.0 g NH3
340
g NH3
Theoretical
Yield
Theoretical Yield
30 mol H2 x
2
mol NH3
3
mol H2
17.0 g NH3
x
1 mol NH3
=
340
g NH3
Theoretical
Yield
Suppose you get an actual yield of 285
285g for NH3.
Based on the theoretical yield calculated above,
what is the % yield for this reaction?
% yield = Actual x 100 =
Theo.
x 100 =
83.8 % Yield
Limiting Reactant
Balanced chemical equation.
4 NH3 +
6 NO
 5 N2
+
6 H2O
1. 12 moles of ammonia, NH3 and 15 moles of nitrogen monoxide, NO,
are mixed. Determine the limiting and excess reactants.
Determine limiting reactant
15 mol NO is the
Limiting Reagent
Given
“I have”
Limiting reagent  NO
15
Excess reagent  NH3
12
=
Alternate method
Ideal situation (Bal.Eqn.)
Given
Excess reagent  NH3 12
Limiting reagent  NO
=
15
=
1.25
NO
6
1
NH3
4
“I have”
=
“I need”
0.8
1
=
=
1.5
1
Ideal situation
“I(Bal.Eqn.)
need”
NH3
NO
=
4
6
=
0.67
1