Mathematical Relationships with
Chemical Equations…Stoichiometry

• You have to make a number of BLTs (bacon, lettuce, tomato sandwiches).
• Each sandwich requires 2 slices of bread, 4 slices of bacon, 2 slices of tomato, 1 leaf of lettuce, and 1 tbsp of mayonnaise.
• How many sandwiches can you make with
24 slices of bacon? 14 pieces of bread? 7 slices of tomato?

Information Contained in a Balanced Equation
Viewed in
Terms of molecules
Reactants
C
3
H
8
(g) + 5O
2
(g)
1 molecule C
3
H
8
+ 5 molecules O
2
Products
3CO
2
(g) + 4H
2
O(g)
3 molecules CO
2
+ 4 molecules H
2
O amount (mol) mass (amu) mass (g) total mass (g)
1 mol C
3
H
8
+ 5 mol O
2
44.09 amu C
3
H
8
+ 160.00 amu O
2
44.09 g C
3
H
8
+ 160.00 g O
2
204.09 g
3 mol CO
2
+ 4 mol H
2
O
132.03 amu CO
2
+ 72.06 amu H
2
O
132.03 g CO
2
+ 72.06 g H
2
O
204.09 g

• The relation between the quantities of substances that take part in a chemical reaction or form a compound
• “"Stoichiometry" is derived from the
Greek words στοιχείον (stoikheion, meaning element) and μέτρον (metron, meaning measure.”
(http://en.wikipedia.org/wiki/Stoichiometry)

2 C
8
H
18
( l
) + 25 O
2
( g
)  16 CO
2
( g
) + 18 H
2
O ( g
)
2 mol C
8
H
18
=
25 mol O
2
=
16 mol CO
2
=
18 mol H
2
O
• So…
25 mol O
2 or
16 mol CO
2 or 2
25 mol O
2

Summary of the mass-mole-number relationships in a chemical reaction.
MASS(g) of compound A
AMOUNT(mol) of compound A
MM (g/mol) of compound
A molar ratio from balanced equation
Avogadro’s number
(molecules/mol)
MOLECULES
(or formula units) of compound A
MASS(g) of compound B
MM (g/mol) of compound
B
AMOUNT(mol) of compound B
Avogadro’s number
(molecules/mol)
MOLECULES
(or formula units) of compound B

• Are grams mentioned?
If yes you will need to use a molar mass conversion.
• Are you comparing two different substances?
If yes you will need a conversion using the coefficients in an equation .
• Are particles (atoms or molecules) involved?
If yes you will need a conversion with Avogadro’s number.
You may need any number of these conversions depending on the problem.

Calculating Amounts of Reactants and Products
PROBLEM:
In a lifetime, the average American uses 1750 lb(794 g) of copper in coins, plumbing, and wiring. Copper is obtained from sulfide ores, such as chalcocite, or copper(I) sulfide, by a multistep process. After an initial grinding, the first step is to “roast” the ore (heat it strongly with oxygen gas) to form powdered copper(I) oxide and gaseous sulfur dioxide.
(a) How many moles of oxygen are required to roast 10.0 mol of copper(I) sulfide?
(b) How many grams of sulfur dioxide are formed when 10.0 mol of copper(I) sulfide is roasted?
(c) How many kilograms of oxygen are required to form 2.86 kg of copper(I) oxide?

Calculating Amounts of Reactants and Products
SOLUTION: 2Cu
2
S( s
) + 3O
2
( g
) 2Cu
2
O( s
) + 2SO
2
( g
)
(a) How many moles of oxygen are required to roast 10.0 mol of copper(I) sulfide?
(a) 10.0mol Cu
2
S
3mol O
2 = 15.0mol O
2
2mol Cu
2
S
(b) How many grams of sulfur dioxide are formed when 10.0 mol of copper(I) sulfide is roasted?
(b) 10.0mol Cu
2
S 2mol SO
2mol Cu
2
S
2
64.07g SO
1 mol SO
2
2
= 641g SO
2
(c) How many kilograms of oxygen are required to form 2.86 kg of copper(I) oxide?
(c) 2.86kg Cu
2
O 10
3 g Cu
2
1kg Cu
2
O
O
1 mol Cu
2
O
143.10g Cu
2
O
3mol O
2
2mol Cu
2
O
32.00g O
1 mol O
2
2
1 kg O
2
10 3 g O
2
= 0.959kg O
2

• You have to make BLTs (bacon, lettuce, tomato sandwiches).
• Each sandwich requires 2 slices of bread, 4 slices of bacon, 2 slices of tomato, 1 leaf of lettuce, and 1 tbsp of mayonnaise.
• How many sandwiches can you make with
24 slices of bacon and 14 pieces of bread?

• In most chemical changes you do not have the exact amounts of all reactants.
• In this case, one reactant will run out first. This reactant will limit the amount of product and so is called the limiting reactant or limiting reagent .
• The reactant(s) left over is called the excess reactant .

Calculating Amounts of Reactant and Product in a
Limiting-Reactant Problem
PROBLEM:
A fuel mixture used in the early days of rocketry is composed of two liquids, hydrazine
(N
2
H
4
) and dinitrogen tetraoxide(N
2
O
4
), which ignite on contact to form nitrogen gas and water vapor. How many grams of nitrogen gas form when 1.00x10
2 g of N
4 and 2.00x10
2 g of N
2
O
4 are mixed?
2
H

Calculating Amounts of Reactant and Product in a
Limiting-Reactant Problem
SOLUTION:
2 N
2
H
4
( l
) + N
1.00x10
2 g N
2
H
4
1 mol N
2
H
4
32.05g N
2
H
4
2
O
4
( l
3
2
( g
3 mol N
2
2mol N
2
H
4
= 4.68mol N
2
4
2
O( l
)
2.00x10
2 g
N
2
O
4
1 mol N
2
92.02g N
O
4
2
O
4
3 mol N
2
1mol N
2
O
4
= 6.51mol N
2
N
2
H
4 is the limiting reactant because it produces less product, N
2 than does N
2
O
4
.
,
4.68mol N
2
28.02g N
2
1 mol N
2
= 131g N
2

•
• How much reactant was left in the previous problem?
6.51 mol N made
2 possible but only 4.68 mol
6.51 - 4.68 = 1.83 mol N
2 not made
2 4
3 mol N
2
x
1 mol N
2 4
= 56.1 g N O

How many grams of aluminum carbonate are formed when 20.0 grams of sodium carbonate react with 25.0 grams of aluminum nitrate?
How much of which reactant is left over?

• Reactions do not always go to completion for many reasons. Percent yield is an expression of how much of the expected product actually formed.
actual amount produced theoretical (expected) amount
x 100 = % yield
• The theoretical amount generally comes from the stoichiometry of the reaction.

Calculating Percent Yield
PROBLEM:
Silicon carbide (SiC) is an important ceramic material that is made by allowing sand (silicon dioxide, SiO
2
) to react with powdered carbon at high temperature. Carbon monoxide is also formed. When 100.0 g of sand is processed, 51.4 g of
SiC is recovered. What is the percent yield of SiC from this process?
SOLUTION:
SiO
2
( s
) + 3C( s
) SiC( s
) + 2CO( g
)
100.0 g SiO
2
1 mol SiO
2
60.09 g SiO
2
= 1.664 mol SiO
2 mol SiO
2
= mol SiC = 1.664
1.664 mol SiC 40.10 g SiC
1 mol SiC
= 66.73 g
51.4 g
66.73 g x 100 =77.0%

Calculating Amounts of Reactants and Products for a
Reaction in Solution
PROBLEM:
Specialized cells in the stomach release HCl to aid digestion. If they release too much, the excess can be neutralized with antacids. A common antacid contains magnesium hydroxide, which reacts with the acid to form water and magnesium chloride solution.
As a government chemist testing commercial antacids, you use 0.10M HCl to simulate the acid concentration in the stomach. How many liters of “stomach acid” react with a tablet containing 0.10g of magnesium hydroxide?

Calculating Amounts of Reactants and Products for a
Reaction in Solution
SOLUTION: Mg(OH)
2
( s
) + 2HCl( aq
) MgCl
2
( aq
) + 2H
2
O( l
)
0.10g Mg(OH)
2
1 mol Mg(OH)
2
58.33g Mg(OH)
2
2 mol HCl
1 mol Mg(OH)
2
= 3.4x10
-3 mol HCl
3.4x10
-3 mol HCl
1L
0.10mol HCl
= 3.4x10
-2 L HCl