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Chapter 4
Chemical Reactions
Changing Matter
• Chemistry is defined as the study of matter and
changes it undergoes. In chemistry, we
recognize two different types of ways that
matter can change.
– Physical Change
– Chemical Change
Physical Changes
• A physical change is a change in only the appearance
of matter. The matter does not change identity, only
its form.
• Boiling water is a physical change. The water
changes from a liquid to a gas, so it changes its form
or appearance. But it remains water throughout the
entire process.
• Other examples are freezing a liquid, pressing plastic
into a new shape, or breaking an object.
Chemical Changes
• A chemical change is a change in a substance’s
identity. During a chemical change, the atoms in a
compound or molecule rearrange to form a new
compound or molecule.
• Chemical changes are also known as chemical
reactions. They may or may not also involve a
physical change.
• Examples inclued
– An iron nail turning to rust
– Burning propane gas
– Rotting or decomposition of organic materials.
Chemical Equations
• A chemical equation is a recipe for a chemical
reaction. A chemical equation contains all the
same type of information as a recipe.
• What four things are in a recipe?
– Reactants
– Products
– Coefficients
– Arrow
Chemical Reactions
“reactants”
starting material
“products”
produced by rxn.
Chemical Equations
CH4 + 2O2
“coefficients”
CO2 + 2H2O
“subscripts”
coefficients can be changed to achieve mass balance
subscripts are never changed to balance an equation
Chemical Equations
H2O + 3 NO2  2 HNO3 + NO
• Identify:
– Reactants
– Products
– Coefficients
– Arrow
Molar Mass
• The molar mass is the mass of one mole of
anything. For example, the mass of a mole of
neon atoms is known as its molar mass. The
mass of a mole of butane molecules is known
as its molar mass. Also known as
formula weight
• Molecular weight is used only for molecules.
• For elements, the atomic mass (on the periodic
table) is equal to its molar mass in grams.
Molar Mass
• Molar mass can be used as a conversion factor
• The atomic mass of potassium is 39.10. 1
mole of potassium weighs 39.10 grams.
1 mole = 39.10 g
• What is the molar mass of neon?
• What is the molar mass of oxygen?
Molar Mass of Compounds
• For compounds, the molar mass is the sum of the
molar masses of each individual atom in the
compound.
For:
K
Cl
O
KClO3
39.098 amu * 1 =
35.453 amu * 1 =
15.999 amu * 3 =
Formula Weight of KClO3 =
• Problem 4.1
39.098 amu
35.453 amu
47.997 amu
122.548 amu
The Mole
• The mole is exactly like the dozen. It is a way
of counting a large number of items.
• A dozen represents the number 12, 12 eggs can
be called a dozen eggs
• The mole represents the number 6.02 x 1023,
6.02 x 1023 eggs can be called a mole of eggs.
The Mole
• We use the mole to quickly count atoms and
molecules, just like we use the dozen to
quickly count eggs.
• In the same way that eggs are packaged and
sold by the dozen, we often work with atoms
and molecules by the mole.
Avogadro's Number
• 6.02214199 x 1023 is known as Avogadro’s
number.
• Avogadro’s number can be used as a
conversion factor:
1 mole = 6.02 x 1023.
Practice Problems
• How many molecules of aspirin (C9H8O4) are
in 1.75 mol?
• How many atoms of oxygen are in 2.1 mol
aspirin?
• How many atoms of hydrogen are in 1.7 mol
aspirin?
• How many atoms of carbon are in 3.2 mol
aspirin?
Molar Mass as a Conversion
• in 25.0 g of boron?
• in 342.3 g of sucrose?
• in 10.0 g of sucrose?
Molar Mass as a Conversion Factor
•
•
•
•
•
Problem 4.2 p 112
Problem 4.3 p 112
Problem 4.4 p 113
Problem 4.5 p 113
Problem 4.6 p 114
Conservation of Mass in
Chemical reactions
• Atoms cannot be created or destroyed by
ordinary chemical reactions. Therefore, all
atoms which are reacting in a chemical
reaction must also show up as a product of that
reaction.
• When there is an equal number of each type of
atom on both sides of the arrow in a chemical
equation, it is said to be balanced.
Balancing Chemical Equations
• Write formula for each reactant and product on the
correct side of the “reaction arrow”
• Count atoms of each element on both sides of arrow
• Start with the compound which has the most
complex formula
• Add coefficients to chemical formulas to balance
numbers of each atom
• Trial and error begins...
May add subscripts of (g), (s), (l), (aq) or (ppt)
to indicate the physical state of the participants.
Balancing Chemical Equations
Examples:
___Al + ___Cl2 ___AlCl3
___Ca + ___H2O ___Ca(OH)2 + ___H2
___HCl + ___Al(OH)3
___AlCl3 + ___H2O
___Ba(OH)2 + ___H3PO4 ___Ba3(PO4)2 + ___H2O
Combustion Example:
___C5H8O3 + ___O2
___CO2 + ___ H2O
Practice Problems (p 216 & 217)
• Problem 4.7
• Problem 4.8
• Problem 4.9
Types of Reactions
• 4 different types of Reactions:
– Combination: A + B AB
– Decomposition: AB  A + B
– Single Replacement: A + BC  AC + B
– Double Replacement: AB + CD  AD + BC
Mole Relationships in Chemical
Equations
• N2 (g) + 3 H2 (g)  2NH3 (g)
• What are the mole to mole relationships?
• How many moles of hydrogen are needed to
react with 1.3 moles of nitrogen?
• How many moles of nitrogen reacted in 0.60
moles of ammonia is produced?
• How many moles of ammonia are produced
when 1.4 moles of hydrogen react?
Mass Calculations for Reactions
• N2 (g) + 3 H2 (g)  2NH3 (g)
• If you have 1.80 moles of hydrogen, how
many grams of ammonia can be produced?
• How many grams of hydrogen are needed to
react with 2.80 g of nitrogen?
• How many grams of ammonia can be
produced from 12.0 g of hydrogen?
Mass Calculations for
Chemical Reactions
•Problem 4.10 p 119
•Problem 4.11 p 120
•Problem 4.12 p 121
Limiting Reactants and Reagents
• The limiting reactant is the reactant in a
chemical reaction which limits the amounts of
products that can be formed
• The limiting reagent in a chemical reaction is
present in insufficient quantity to consume
the other reactant(s)
Limiting Reagent
Calculating Limiting Reagents
1) calculate moles (or mass) of product
formed by complete reaction of each
reactant.
2) the reactant that yields the least product is
the limiting reagent.
3) the theoretical yield for a reaction is the
maximum amount of product that could be
generated by complete consumption of the
limiting reagent.
Problem 4.13 p 122
Limiting Reactants
• Iron (III) oxide reacts with carbon monoxide to
produce iron and carbon dioxide.
– What is the limiting reactant if 43.2 g iron (III)
oxide reacts with 24.3 g carbon monoxide?
– What is the theoretical yield?
Percent Yield
• Theoretical yield – the maximum amount of product
expected in a chemical reaction.
• In reality, most chemical reactions produce less
product than the theoretical yield predicts. The actual
amount of product obtained from a chemical reaction
is the actual yield. The actual yield is only
determined by doing a reaction and weighing the
amount of product.
• The percent yield is:
(actual yield) / (theoretical yield) x 100.
Percent Yield
• Carbon disulfide and carbon monoxide are
produced by the reaction of carbon and sulfur
dioxide:
– What is the percent yield for the reaction if 40.0 g of
carbon produces 36.0 g of carbon disulfide??
– What is the percent yield for the reaction if 32.0 g of
sulfur dioxide produces 12.0 g of carbon disulfide?
– What is the percent yield for the reaction if 40.0 g of
carbon reacts with 12.0 g of sulfur dioxide to form
37 g of carbon disulfide?
Calculating Percent Yield
When 66.6 g of O2 gas is mixed with 27.8 g of
NH3 gas and 25.1 g of CH4 gas, 36.4 g of HCN
gas is produced by the following reaction:
2CH4 + 2NH3 + 3O2
2HCN + 6H2O
What is the percent yield of HCN in this reaction?
How many grams of NH3 remain?
Calculating Limiting Reagent
66.6 g of O2 = 2.08 mol O2
27.8 g of NH3 = 1.63 mol NH3
25.1 g of CH4 = 1.56 mol CH4
Which reactant is limiting?
2.08 mol O2 can yield 1.39 mol (or 37.5 g) HCN
1.63 mol NH3 can yield 1.63 mol (or 44.1 g) HCN
1.56 mol CH4 can yield 1.56 mol (or 42.2 g) HCN
O2 is the limiting reagent.
O2 is the limiting reagent; thus, the theoretical
yield is based on 100% consumption of O2.
2.08 mol O2 can yield 1.39 mol (or 37.5 g) HCN
% yield =
actual yield
theoretical yield
% yield =
36.4 g HCN
37.5 g HCN
* 100
* 100 = 97.1%
2CH4 + 2NH3 + 3O2
2HCN + 6H2O
How many grams of NH3 remain?
36.4 g (or 1.35 mol) of HCN gas is produced
Since the reaction stoichiometry is 1:1, 1.35 mol of
NH3 is consumed:
(1.63 mol NH3 initially) –(1.35 mol NH3 consumed)
= 0.28 mol NH3 remaining
0.28 mol NH3 * (17.03 NH3 g/mol NH3) = 4.8 g NH3
remain